Slides from FYS4410 Lectures

Qubic spline interpolation

Morten Hjorth-Jensen

1Department of Physics and Center of Mathematics for ApplicationsUniversity of Oslo N-0316 Oslo Norway

Spring 2008

Computational Physics II FYS4410

Topics for Week 3 January 14-18

Introduction Parallelization MPI and Hartree-Fock theory

Thursday

Introduction to Message Passing Interface (MPI) andparallelization

How to use the local cluster and the Titan cluster

Presentation of topics to be covered in part I (MHJ)

Friday

Hartree-Fock theory and presentation of project 1

Part I 17 January - 14 March

Course overview Part I Computational aspects

Parallelization (MPI) high-performance computing topicsand object orientation Choose between F95 andor C++as programming languages Python also possible asprogramming language

Algorithms for Monte Carlo Simulations (multidimensionalintegrals) Metropolis-Hastings and importance samplingalgorithms Improved Monte Carlo methods

Statistical analysis of data from Monte Carlo calculationsblocking and Jack-Knife methods

Solutions of coupled differential equations for Hartree-Fockand density functional calculations

Parameterizations of statistical data via the Singular-valuedecomposition algorithm

Project 1 deadline April 2Computational Physics II FYS4410

Part I 17 January -14 March

Quantum Mechanical Methods and Systems

1 Hartree-Fock and density functional theories relevant foratomic molecularsolid state materials sciencenanotechnology quantum chemistry and nuclear physics

2 Variational Monte Carlo for rsquoab initiorsquo studies of quantummechanical many-body systems

3 Simulation of atoms like Helium Beryllium and Neon withextensions to solids

4 Aim of project 1 understand how to simulate qauntummechanical systems with many interacting particles Thespecific relevance of project 1 is taylored to the fields ofquantum chemistry materials science nanotechnologysolid state physics and atomic physics The methods wediscuss can however easily be extended to other fields

Lectures and ComputerLab

Lectures Thursday (1415-1600 room FV329) and fridayday (1415-16 room FV311)

Detailed lecture notes all programs presented andprojects can be found at the homepage of the course

Computerlab 16-19 thursday room FV329

Weekly plans and all other information are on the officialwebpage

The thursday lectures will be used to demonstratealgorithms etc while the friday lectures will be more likeclassical blackboard sessions

For the first part (MHJ) chapters 8 9 11 and 16 and 17 ofthe FYS31504150 lecture notes give a good starting pointWe recommend also J M Thijssen text ComputationalPhysics For MPI we recommend Gropp Lusk andSjellumrsquos text and UiO lecture notes on the web

Thijssenrsquos text

J M Thijssenrsquos text

ComputationalPhysics

Chapters 3-6 and 12

see httpwwwtntudelftnltnPeopleStaffThijssencomphybookhtml

MPI text

Gropp Lusk and Sjellum

Using MPI

Chapters 1-5

seehttpmitpressmiteducatalogitemdefaultaspttype=2amptid=10761

Lattice Gas and Cellular Automata text Eirikrsquos part

Rothman and Zaleski

Lattice Gas andCellular Automata

see httpwwwcambridgeorgcataloguecatalogueaspisbn=9780521607605

Selected Texts and lectures on CC++

J J Barton and L R NackmanScientific and Engineering C++ Addison Wesley3rd edition 2000

B Stoustrup The C++ programming language Pearson 1997

H P Langtangen INF-VERK3830httpheimifiuiono˜hplINF-VERK4830

D Yang C++ and Object-oriented Numeric Computing for Scientists andEngineers Springer 2000

More books reviewed at httpwwwaccuorg andhttpwwwcomeaucomputingcombooklist

Quantum Monte Carlo Motivation

Most quantum mechanical problems of interest in eg atomic molecular nuclear andsolid state physics consist of a large number of interacting electrons and ions ornucleons The total number of particles N is usually sufficiently large that an exactsolution cannot be found Typically the expectation value for a chosen hamiltonian for asystem of N particles is

〈H〉 =RdR1dR2 dRNΨlowast(R1R2 RN)H(R1R2 RN)Ψ(R1R2 RN)R

dR1dR2 dRNΨlowast(R1R2 RN)Ψ(R1R2 RN)

an in general intractable problem an in general intractable problem

This integral is actually the starting point in a Variational Monte Carlo calculation

Gaussian quadrature Forget it given 10 particles and 10 mesh points for each

degree of freedom and an ideal 1 Tflops machine (all operations take the same time)

how long will it ta ke to compute the above integral Lifetime of the universe

T asymp 47times 1017s

Quantum Monte Carlo

As an example from the nuclear many-body problem we have Schrodingerrsquos equationas a differential equation

HΨ(r1 rA α1 αA) = EΨ(r1 rA α1 αA)

wherer1 rA

are the coordinates andα1 αA

are sets of relevant quantum numbers such as spin and isospin for a system of A

nucleons (A = N + Z N being the number of neutrons and Z the number of protons)

Quantum Monte Carlo

There are

2A timesbdquo

laquocoupled second-order differential equations in 3A dimensionsFor a nucleus like 10Be this number is 215040 This is a truely challenging many-bodyproblem

Methods like partial differential equations can at most be used for 2-3 particles

Quantum Many-particle(body) Methods

1 Monte-Carlo methods

2 Renormalization group (RG) methods in particular density matrix RG

3 Large-scale diagonalization (Iterative methods Lanczorsquos methoddimensionalities 1010 states)

4 Coupled cluster theory favoured method in quantum chemistry molecular andatomic physics Applications to ab initio calculations in nuclear physics as well forlarge nuclei

5 Perturbative many-body methods

6 Greenrsquos function methods

7 Density functional theoryMean-field theory and Hartree-Fock theory

The physics of the system hints at which many-body methods to use For systems with

strong correlations among the constituents item 5 and 7 are ruled out

Pros and Cons of Monte Carlo

Is physically intuitive

Allows one to study systems with many degrees of freedom Diffusion MonteCarlo (DMC) and Greenrsquos function Monte Carlo (GFMC) yield in principle theexact solution to Schrodingerrsquos equation

Variational Monte Carlo (VMC) is easy to implement but needs a reliable trialwave function can be difficult to obtain This is where we will use Hartree-Focktheory to construct an optimal basis

DMCGFMC for fermions (spin with half-integer values electrons baryonsneutrinos quarks) has a sign problem Nature prefers an anti-symmetric wavefunction PDF in this case given distribution of random walkers (p ge 0)

The solution has a statistical error which can be large

There is a limit for how large systems one can study DMC needs a huge numberof random walkers in order to achieve stable results

Obtain only the lowest-lying states with a given symmetry Difficult to get excitedstates

Where and why do we use Monte Carlo Methods inQuantum Physics

Quantum systems with many particles at finite temperature Path Integral MonteCarlo with applications to dense matter and quantum liquids (phase transitionsfrom normal fluid to superfluid) Strong correlations

Bose-Einstein condensation of dilute gases method transition from non-linearPDE to Diffusion Monte Carlo as density increases

Light atoms molecules solids and nuclei

Lattice Quantum-Chromo Dynamics Impossible to solve without MCcalculations

Simulations of systems in solid state physics from semiconductors to spinsystems Many electrons active and possibly strong correlations

Bose-Einstein Condensation of atoms thousands ofAtoms in one State Project 2 in 2007

Going Parallel with MPI

Task parallelism the work of a global problem can be dividedinto a number of independent tasks which rarely need tosynchronize Monte Carlo simulation or integrations areexamples of this It is almost embarrassingly trivial to parallelizeMonte Carlo codesMPI is a message-passing library where all the routines havecorresponding CC++-binding

MPI_Command_name

and Fortran-binding (routine names are in uppercase but canalso be in lower case)

MPI_COMMAND_NAME

What is Message Passing Interface (MPI) Yetanother library

MPI is a library not a language It specifies the names callingsequences and results of functions or subroutines to be calledfrom C or Fortran programs and the classes and methods thatmake up the MPI C++ library The programs that users write inFortran C or C++ are compiled with ordinary compilers andlinked with the MPI libraryMPI is a specification not a particular implementation MPIprograms should be able to run on all possible machines andrun all MPI implementetations without changeAn MPI computation is a collection of processescommunicating with messagesSee chapter 75 of lecture notes for more details

MPI is a library specification for the message passing interfaceproposed as a standard

independent of hardware

not a language or compiler specification

not a specific implementation or product

A message passing standard for portability and ease-of-useDesigned for high performanceInsert communication and synchronization functions wherenecessary

Demands from the HPC community

In the field of scientific computing there is an ever-lasting wishto do larger simulations using shorter computer timeDevelopment of the capacity for single-processor computerscan hardly keep up with the pace of scientific computing

processor speed

memory sizespeed

Solution parallel computing

The basic ideas of parallel computing

Pursuit of shorter computation time and larger simulationsize gives rise to parallel computing

Multiple processors are involved to solve a global problem

The essence is to divide the entire computation evenlyamong collaborative processors Divide and conquer

A rough classification of hardware models

Conventional single-processor computers can be calledSISD (single-instruction-single-data) machines

SIMD (single-instruction-multiple-data) machinesincorporate the idea of parallel processing which use alarge number of processing units to execute the sameinstruction on different data

Modern parallel computers are so-called MIMD(multiple-instruction- multiple-data) machines and canexecute different instruction streams in parallel on differentdata

Shared memory and distributed memory

One way of categorizing modern parallel computers is tolook at the memory configuration

In shared memory systems the CPUs share the sameaddress space Any CPU can access any data in theglobal memory

In distributed memory systems each CPU has its ownmemory The CPUs are connected by some network andmay exchange messages

A recent trend is ccNUMA(cache-coherent-non-uniform-memory- access) systemswhich are clusters of SMP (symmetric multi-processing)machines and have a virtual shared memory

Different parallel programming paradigms

Task parallelism the work of a global problem can bedivided into a number of independent tasks which rarelyneed to synchronize Monte Carlo simulation is oneexample Integration is another However this paradigm isof limited use

Data parallelism use of multiple threads (eg one threadper processor) to dissect loops over arrays etc Thisparadigm requires a single memory address spaceCommunication and synchronization between processorsare often hidden thus easy to program However the usersurrenders much control to a specialized compilerExamples of data parallelism are compiler-basedparallelization and OpenMP directives

Message-passing all involved processors have anindependent memory address space The user isresponsible for partition- ing the datawork of a globalproblem and distributing the subproblems to theprocessors Collaboration between processors is achievedby explicit message passing which is used for datatransfer plus synchronization

This paradigm is the most general one where the user hasfull control Better parallel efficiency is usually achieved byexplicit message passing However message-passingprogramming is more difficult

Although message-passing programming supports MIMD itsuffices with an SPMD (single-program-multiple-data) modelwhich is flexible enough for practical cases

Same executable for all the processors

Each processor works primarily with its assigned localdata

Progression of code is allowed to differ betweensynchronization points

Possible to have a masterslave model The standardoption in Monte Carlo calculations and numericalintegration

Todayrsquos situation of parallel computing

Distributed memory is the dominant hardwareconfiguration There is a large diversity in these machinesfrom MPP (massively parallel pro cessing) systems toclusters of off-the-shelf PCs which are very cost-effective

Message-passing is a mature programming paradigm andwidely accepted It often provides an efficient match to thehardware It is primarily for the distributed memorysystems but can also be used on shared memory systems

In these lectures we consider only message-passing for writingparallel programs

Some useful concepts

Speed-up

S(P) =T (1)

T (1) is the time for one processor while T (P) is the timefor P processors

Efficiency

η(P) =S(P)

PLatency and bandwidth ndash the cost model of sending amessage of length L between two processors

tC(L) = τ + βL

Overhead present in parallel computing

Uneven load balance not all the processors can performuseful work at all time

Overhead of synchronization

Overhead of communication

Extra computation due to parallelization

Due to the above overhead and that certain part of a sequentialalgorithm cannot be parallelized

T (P) ge T (1)

Prarr S(P) le P

However superlinear speed-up (S(P) gt P) may sometimesoccur due to for example cache effects

Parallelizing a sequential algorithm

Identify the part(s) of a sequential algorithm that can beexecuted in parallel This is the difficult part

Distribute the global work and data among P processors

Process and processor

We refer to process as a logical unit which executes itsown code in an MIMD style

The processor is a physical device on which one or severalprocesses are executed

The MPI standard uses the concept process consistentlythroughout its documentation

However we only consider situations where one processoris responsible for one process and therefore use the twoterms interchangeably

Bindings to MPI routines

MPI is a message-passing library where all the routines havecorresponding CC++-binding

MPI_Command_name

MPI_COMMAND_NAME

The discussion in these slides has its focus on the C++ binding

Communicator

A group of MPI processes with a name (context)

Any process is identified by its rank The rank is onlymeaningful within a particular communicator

By default communicator MPI COMM WORLD contains allthe MPI processes

Mechanism to identify subset of processes

Promotes modular design of parallel libraries

The 6 most important MPI routines

MPI Init - initiate an MPI computation

MPI Finalize - terminate the MPI computation and clean up

MPI Comm size - how many processes participate in agiven MPI communicator

MPI Comm rank - which one am I (A number between 0and size-1)

MPI Send - send a message to a particular pro cess withinan MPI communicator

MPI Recv - receive a message from a particular pro cesswithin an MPI communicator

The first MPI CC++ program

Let every process write rdquoHello worldrdquo on the standard output This is example2cpp

using namespace stdinclude ltmpihgtinclude ltiostreamgtint main (int nargs char args[])int numprocs my_rank MPI initializationsMPI_Init (ampnargs ampargs)MPI_Comm_size (MPI_COMM_WORLD ampnumprocs)MPI_Comm_rank (MPI_COMM_WORLD ampmy_rank)cout ltlt Hello world I have rank ltlt my_rank ltlt out of

ltlt numprocs ltlt endl End MPIMPI_Finalize ()

The Fortran program

PROGRAM helloINCLUDE mpifhINTEGER size my_rank ierr

CALL MPI_INIT(ierr)CALL MPI_COMM_SIZE(MPI_COMM_WORLD size ierr)CALL MPI_COMM_RANK(MPI_COMM_WORLD my_rank ierr)WRITE()Hello world Irsquove rank my_rank out of sizeCALL MPI_FINALIZE(ierr)

END PROGRAM hello

Note 1

The output to screen is not ordered since all processes aretrying to write to screen simultaneously It is then the operatingsystem which opts for an ordering If we wish to have anorganized output starting from the first process we may rewriteour program as in the next example (example3cpp) see againchapter 75 of lecture notes

Ordered output with MPI Barrier

int main (int nargs char args[])

int numprocs my_rank iMPI_Init (ampnargs ampargs)MPI_Comm_size (MPI_COMM_WORLD ampnumprocs)MPI_Comm_rank (MPI_COMM_WORLD ampmy_rank)for (i = 0 i lt numprocs i++) MPI_Barrier (MPI_COMM_WORLD)if (i == my_rank) cout ltlt Hello world I have rank ltlt my_rank ltlt

out of ltlt numprocs ltlt endlMPI_Finalize ()

Note 2

Here we have used the MPI Barrier function to ensure that thatevery process has completed its set of instructions in aparticular order A barrier is a special collective operation thatdoes not allow the processes to continue until all processes inthe communicator (here MPI COMM WORLD have calledMPI Barrier The barriers make sure that all processes havereached the same point in the code Many of the collectiveoperations like MPI ALLREDUCE to be discussed later havethe same property viz no process can exit the operation untilall processes have started However this is slightly moretime-consuming since the processes synchronize betweenthemselves as many times as there are processes In the nextHello world example we use the send and receive functions inorder to a have a synchronized action

Ordered output with MPI Recv and MPI Send

int numprocs my_rank flagMPI_Status statusMPI_Init (ampnargs ampargs)MPI_Comm_size (MPI_COMM_WORLD ampnumprocs)MPI_Comm_rank (MPI_COMM_WORLD ampmy_rank)if (my_rank gt 0)MPI_Recv (ampflag 1 MPI_INT my_rank-1 100

MPI_COMM_WORLD ampstatus)cout ltlt Hello world I have rank ltlt my_rank ltlt out of ltlt numprocs ltlt endlif (my_rank lt numprocs-1)MPI_Send (ampmy_rank 1 MPI_INT my_rank+1

100 MPI_COMM_WORLD)MPI_Finalize ()

Note 3

The basic sending of messages is given by the function MPI SEND which in CC++ isdefined as

int MPI_Send(void buf int count MPI_Datatype datatypeint dest int tag MPI_Comm comm)

This single command allows the passing of any kind of variable even a large array to

any group of tasks The variable buf is the variable we wish to send while count is the

number of variables we are passing If we are passing only a single value this should

be 1 If we transfer an array it is the overall size of the array For example if we want to

send a 10 by 10 array count would be 10times 10 = 100 since we are actually passing

100 values

Note 4

Once you have sent a message you must receive it on another task The functionMPI RECV is similar to the send call

int MPI_Recv( void buf int count MPI_Datatype datatype int sourceint tag MPI_Comm comm MPI_Status status )

The arguments that are different from those in MPI SEND are buf which is the name ofthe variable where you will be storing the received data source which replaces thedestination in the send command This is the return ID of the senderFinally we have used MPI Status status where one can check if the receive wascompletedThe output of this code is the same as the previous example but now process 0 sendsa message to process 1 which forwards it further to process 2 and so forth

Armed with this wisdom performed all hello world greetings we are now ready for

serious work

Integrating π

Examples

Go to the webpageand click on theprograms link

Go to MPI examples

Load downexample5ccp andexample6cpp(Fortran version alsoavailable)

These codescompute π using therectangular andtrapezoidal rules

Integration algos

The trapezoidal rule (example6cpp)

af (x)dx = h (f (a)2 + f (a + h) + f (a + 2h) + middot middot middot+ f (b minus h) + fb2)

Another very simple approach is the so-called midpoint or rectangle method In thiscase the integration area is split in a given number of rectangles with length h andheigth given by the mid-point value of the function This gives the following simple rulefor approximating an integral

af (x)dx asymp h

f (ximinus12)

where f (ximinus12) is the midpoint value of f for a given rectangle This is used in

example5cpp

Message Passing Interface (MPI) and the local cluster

Basic info for CC++ users

The basic instructions for using the local cluster are athttpfolkuionojonknigvdworkdocmpich2-fysmpich2-fyshtml Follow these instructionsin detail The file rsquomachinesrsquo can be pulled down from MHJrsquospart under the webpage of the course see programs and MPIprograms

Use the example code example5cppcompile and load withmpicxx -O2 -o example5x example5cpprun with 4 nodesmpiexec -np 4 example5x

Basic info for Fortran95 usersUse the example code calc_pif90compile and load withmpif90 -O2 -o calcpiexe calc_pif90run with 4 nodesmpirun -np 4 calpiexe

Dissection of example 5

1 Reactangle rule and numerical integration2 using namespace std3 include ltmpihgt4 include ltiostreamgt

5 int main (int nargs char args[])6 7 int numprocs my_rank i n = 10008 double local_sum rectangle_sum x h9 MPI initializations10 MPI_Init (ampnargs ampargs)11 MPI_Comm_size (MPI_COMM_WORLD ampnumprocs)12 MPI_Comm_rank (MPI_COMM_WORLD ampmy_rank)

After the standard initializations with MPI such as MPI Init MPI Comm size andMPI Comm rank MPI COMM WORLD contains now the number of processes definedby using for example

mpirun -np 10 progx

In line 4 we check if we have read in from screen the number of mesh points n Notethat in line 7 we fix n = 1000 however we have the possibility to run the code with adifferent number of mesh points as well If my rank equals zero which correponds tothe master node then we read a new value of n if the number of arguments is largerthan two This can be done as follows when we run the code

mpiexec -np 10 progx 10000

13 Read from screen a possible new vaue of n14 if (my_rank == 0 ampamp nargs gt 1) 15 n = atoi(args[1])16 17 h = 10n18 Broadcast n and h to all processes19 MPI_Bcast (ampn 1 MPI_INT 0 MPI_COMM_WORLD)20 MPI_Bcast (amph 1 MPI_DOUBLE 0 MPI_COMM_WORLD)21 Every process sets up its contribution to the integral22 local_sum = 023 for (i = my_rank i lt n i += numprocs) 24 x = (i+05)h25 local_sum += 40(10+xx)26 27 local_sum = h

In line 17 we define also the step length h In lines 19 and 20 we use the broadcast

function MPI Bcast We use this particular function because we want data on one

processor (our master node) to be shared with all other processors The broadcast

function sends data to a group of processes

The MPI routine MPI Bcast transfers data from one task to a group of others Theformat for the call is in C++ given by the parameters of

MPI_Bcast (ampn 1 MPI_INT 0 MPI_COMM_WORLD)MPI_Bcast (amph 1 MPI_DOUBLE 0 MPI_COMM_WORLD)

in a case of a double The general structure of this function is

MPI_Bcast( void buf int count MPI_Datatype datatype int root MPI_Comm comm)

All processes call this function both the process sending the data (with rank zero) andall the other processes in MPI COMM WORLD Every process has now copies of nand h the number of mesh points and the step length respectively

We transfer the addresses of n and h The second argument represents the number of

data sent In case of a one-dimensional array one needs to transfer the number of

array elements If you have an n timesm matrix you must transfer n timesm We need also to

specify whether the variable type we transfer is a non-numerical such as a logical or

character variable or numerical of the integer real or complex type

28 if (my_rank == 0) 29 MPI_Status status30 rectangle_sum = local_sum31 for (i=1 i lt numprocs i++) 32 MPI_Recv(amplocal_sum1MPI_DOUBLEMPI_ANY_SOURCE500

MPI_COMM_WORLDampstatus)33 rectangle_sum += local_sum34 35 cout ltlt Result ltlt rectangle_sum ltlt endl36 else37 MPI_Send(amplocal_sum1MPI_DOUBLE0500MPI_COMM_WORLD)38 End MPI39 MPI_Finalize ()40 return 041

In lines 23-27 every process sums its own part of the final sum used by the rectanglerule The receive statement collects the sums from all other processes in casemy rank == 0 else an MPI send is performed If we are not the master node wesend the results else they are received and the local results are added to final sumThe above can be rewritten using the MPI allreduce as discussed in the next example

The above function is not very elegant Furthermore the MPI instructions can be

simplified by using the functions MPI Reduce or MPI Allreduce The first function takes

information from all processes and sends the result of the MPI operation to one process

only typically the master node If we use MPI Allreduce the result is sent back to all

processes a feature which is useful when all nodes need the value of a joint operation

We limit ourselves to MPI Reduce since it is only one process which will print out the

final number of our calculation The arguments to MPI Allreduce are the same

MPI reduce

Call as

MPI_reduce( void senddata void resultdata int countMPI_Datatype datatype MPI_Op int root MPI_Comm comm)

The two variables senddata and resultdata are obvious besides the fact that onesends the address of the variable or the first element of an array If they are arrays theyneed to have the same size The variable count represents the total dimensionality 1in case of just one variable while MPI Datatype defines the type of variable which issent and receivedThe new feature is MPI Op It defines the type of operation we want to do In our casesince we are summing the rectangle contributions from every process we defineMPI Op = MPI SUM If we have an array or matrix we can search for the largest ogsmallest element by sending either MPI MAX or MPI MIN If we want the location aswell (which array element) we simply transfer MPI MAXLOC or MPI MINOC If we wantthe product we write MPI PRODMPI Allreduce is defined as

MPI_Alreduce( void senddata void resultdata int countMPI_Datatype datatype MPI_Op MPI_Comm comm)

Trapezoidal rule and numerical integration usign MPI example 6using namespace stdinclude ltmpihgtinclude ltiostreamgt

Here we define various functions called by the main program

double int_function(double )double trapezoidal_rule(double double int double ()(double))

Main function begins hereint main (int nargs char args[])

int n local_n numprocs my_rankdouble a b h local_a local_b total_sum local_sumdouble time_start time_end total_time

MPI initializationsMPI_Init (ampnargs ampargs)MPI_Comm_size (MPI_COMM_WORLD ampnumprocs)MPI_Comm_rank (MPI_COMM_WORLD ampmy_rank)time_start = MPI_Wtime() Fixed values for a b and na = 00 b = 10 n = 1000h = (b-a)n h is the same for all processeslocal_n = nnumprocs make sure n gt numprocs else integer division gives zero Length of each processrsquo interval of integration = local_nhlocal_a = a + my_ranklocal_nhlocal_b = local_a + local_nh

total_sum = 00local_sum = trapezoidal_rule(local_a local_b local_n

ampint_function)MPI_Reduce(amplocal_sum amptotal_sum 1 MPI_DOUBLE

MPI_SUM 0 MPI_COMM_WORLD)time_end = MPI_Wtime()total_time = time_end-time_startif ( my_rank == 0)

cout ltlt Trapezoidal rule = ltlt total_sum ltlt endlcout ltlt Time = ltlt total_time

ltlt on number of processors ltlt numprocs ltlt endl End MPIMPI_Finalize ()return 0

end of main program

We use MPI reduce to collect data from each process Note also the use of the

function MPI Wtime

this function defines the function to integratedouble int_function(double x)

double value = 4(1+xx)return value

end of function to evaluate

Implementation of the trapezoidal rule

this function defines the trapezoidal ruledouble trapezoidal_rule(double a double b int n

double (func)(double))

double trapez_sumdouble fa fb x stepint jstep=(b-a)((double) n)fa=(func)(a)2 fb=(func)(b)2 trapez_sum=0for (j=1 j lt= n-1 j++)

x=jstep+atrapez_sum+=(func)(x)

trapez_sum=(trapez_sum+fb+fa)stepreturn trapez_sum

end trapezoidal_rule

Strategies

Develop codes locally run with some few processes andtest your codes Do benchmarking timing and so forth onlocal nodes for example your laptop You can installMPICH2 on your laptop (most new laptos come with dualcores) or run locally on the clusters here

When you are convinced that your codes run correctly youstart your production runs on available supercomputers inour case titanuiono see Jon Nilsenrsquos introduction

Hartree-Fock theory and computing topics

Thursday

Algorithm for the Hartree-Fock equations

computation of the single-particle equation homogeneousand non-homogeneous part

Numerovrsquos algorithm and Greenrsquos function

More on parallelization in particular how to write out MonteCarlo data in parallel This is very useful for the statisticalanalysis of the data to be performed later (presented byJon Nilsen)

Friday

Hartree-Fock theory derivation of the Hartree-Fockequations New lecture hall store fys aud

Hartree-Fock equations

The single-particle equation

minus1

dr2unl (r) +

bdquol(l + 1)

2r2minus

r+ Φ(r)

laquounl (r) + Fnl = enl unl (r)

is the equation we will discuss today using Numerovrsquos method The function unl is thesolution of the radial part of the Schrodinger equation The total one-particle wavefunction is

ψnlml ms = φnlml(r)ξms (s)

with s the spin (12 for electrons) ms is the spin projection ms = plusmn12 and the spatialpart is

φnlml(r) = Rnl (r)Ylml

with Y the spherical harmonics and unl = rRnl All results are in atomic units meaningthat the energy is given by enl = minusZ 22n2

The boundary values are unl (0) = unl (infin) = 0 This is a tricky boundary value

problem

Lets specialize to the homogeneous case that is we skip Fnl and absorb Φ(r) inV (r) = minusZr + Φ(r)

minus1

dr2unl (r) +

bdquol(l + 1)

laquounl (r) = enl unl (r)

and rewrite as

minus1

dr2unl (r) + k2(r)unl (r) = 0

k2(r) = V (r) +l(l + 1)

2r2minus enl

We can rewrite the equation as and rewrite as the general inhomogeneous equation

dr2unl (r) + k2(r)unl (r) = Snl (r)

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

Note that there are restrictions on k2 If k is real then k2 gt 0 in order to have asolution If k2 lt 0 there is no solution unless k is imaginaryThe term Snl is called the source term and will in general cause problems for us Wewill insert this later (next week) Today we specialize on the homogeneous equationThe codes with a simple matching recipe are under programs for project 1 at thewebpage of the course see numerovcpp or numerovf90

Your task today is to modify these codes in order to prepare for the introduction of the

source term which in our case is the Fock term that is Snl (r) = minusFnl

The other terms in the Hartree-Fock equation (for particle i note this)

V (ri ) = minusZri

Φ(ri ) =Xnprime lprime

2(2l prime + 1)

Z infin

0|unprime lprime (rj )|2

rgtdrj

with rgt = max(ri rj ) This equation with a given guess for unprime lprime can be integrated

numerically with for example gaussian quadrature or Simpsonrsquos method see chapter 7

of lecture notes and the corresponding codes in either libcpp or f90libf90

The tricky part is the Fock term Fnl (r)

Fnl (ri ) =Xnprime lprime

l+lprimeXλ=|lminuslprime|

2l prime + 1

2λ+ 1lt ll prime00|λ0 gt2

raquoZ infin

0ulowastnprime lprime (rj )

(rlt)λ

(rgt)λ+1unl (rj )drj

ndashunprime lprime (ri )

with rgt = max(ri rj ) and with rlt = min(ri rj ) This terms is our source term The

object lt ll prime00|λ0 gt is a vector recoupling coefficient called Clebsch-Gordan You will

find the code for C++ users as clebsch-gordancpp with calling examples For f95 users

it is baked into the f90libf90 library all codes are under the Project 1 directory of the

programs catalogue

How do I solve the Hartree-Fock equations

Here follows the algorithm for doing the Hartree-Fock calculations

Set up the system how many electrons there are in the 1s 2s and 2p orbitalsChoose a charge Z and use hydrogenic wave functions as a guess for unl It canbe convenient to make a class or module which identifies the single-particlequantum numbers and the corresponding single-particle energies

Fix a numerical precision for when you wish to stop the iterative process and amaximum number of iterations For every iteration you compute new energies enland compare these with those from the previous iteration and check with yourchosen numerical precision

At every iteration you have an input value for enl and unl Compute first Φ(ri ) andFnl (ri ) Solve then the differential equation for unl

Compute then new single-particle energies using the expression

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

r+ Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr+Z infin

0unl (r)Fnl (r)dr

Use these energies for the next iteration You will also need to compute the totalenergy and compare this at every iteration

The expression for the total energy E is

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

2Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr + EFock

EFock =1

Z infin

0unl (r)Fnl (r)dr

and Nnl are the number of particles in a given orbit with quantum numbers nl

The easy part is to integrate the above expressions although it may take some timeThe difficulty resides in solving the general inhomogeneous equation Let us first startwith the homogenous equation

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

Fnl is the Fock term term and will in general cause problems for us However we donot need to find the energies They are given by our guess This avoids the searchfor both energies and wave functionsHowever we have still a two-point boudary value problem see chapter 14 of lecturenotes

We will first look at the homogeneous case and introduce Numerovrsquos algorithm

Numerovrsquos algo chap 14 in lecture notes

Let us rewrite the inhomogeneous equation as

dx2y(x) + k2(x)y(x) = F (x)

Our discretised second derivative reads at a step xi = a + ih with i = 0 1

y primeprimei =yi+1 + yiminus1 minus 2yi

h2+ O(h2)

leading to a discretised differential equation

yi+1 + yiminus1 minus 2yi

h2+ O(h2) + k2

i yi = Fi

Recall that the fourth-order Runge-Kutta method has a local error of O(h4)Since we want to integrate our equation from x0 = a to xN = b we rewrite it as

yi+1 asymp minusyiminus1 +ldquo

2minus h2k2i

rdquoyi + h2Fi

Numerovrsquos algo

The local truncation error is O(h2) We can obtain O(h6) without too many additionalfunction evaluationsOur equation is a second order differential equation without any first order derivativesLet us also for the sake of simplicity assume that F (x) = 0 Numerovrsquos method isdesigned to solve such an equation numerically achieving a local truncation errorO(h6)We start with the Taylor expansion of the desired solution

y(x + h) = y(x) + hy (1)(x) +h2

2y (2)(x) +

3y (3)(x) +

4y (4)(x) + middot middot middot

Numerovrsquos algo

We have that y (n)(x) is a shorthand notation for the nth derivative dnydxn Becausethe corresponding Taylor expansion of y(x minus h) has odd powers of h appearing withnegative signs all odd powers cancel when we add y(x + h) and y(x minus h)

y(x + h) + y(x minus h) = 2y(x) + h2y (2)(x) +h4

12y (4)(x) + O(h6)

We obtain

y (2)(x) =y(x + h) + y(x minus h)minus 2y(x)

h2minus

12y (4)(x) + O(h4)

Numerovrsquos algo

To eliminate the fourth-derivative term we apply the operator (1 + h2

dx2 ) on thedifferential equation

y (2)(x) +h2

12y (4)(x) + k2(x)y(x) +

ldquok2(x)y(x)

rdquoasymp 0

Numerovrsquos algo

In this expression the y (4) terms cancel To treat the general x dependence of k2(x)we approximate the second derivative of (k2(x)y(x) by

d2(k2y(x))

dx2asymp`k2(x + h)y(x + h) + k2(x)y(x)

acute+`k2(x minus h)y(x minus h) + k2(x)y(x)

acuteh2

Numerovrsquos algo

We replace then y(x + h) with the shorthand yi+1 etc and obtain a final discretisedalgorithm for obtaining yi+1

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

+ O(h6)

where xi = ih ki = k(xi = ih) and yi = y(xi = ih) etc

Numerovrsquos algo

It is easy to add the term Fi since we need only to take the second derivative The finalalgorithm reads then

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

12(Fi+1 + Fiminus1 minus 2Fi ) + O(h6)

This equation is implemented in the codes numerovcpp and numerovf90 Thisequation carries a local truncation error proportional to h6

Boundary Value Problems

Let us return to Schrodingerrsquos equation In the notation we use the shorthand u(r)instead of unl (r)

minus1

dr2u(r) +

bdquoV (r) +

l(l + 1)

laquou(r) = eu(r)

In our case we are interested in attractive potentials

V (r) = minusV0f (r)

where V0 gt 0 and analyze bound states where e lt 0

The final equation can be written as

dr2u(r) + k2(r)u(r) = 0

k2(r) = γ

bdquof (r)minus

l(l + 1)

r2minus ε

laquoγ = 2V0

ε =|E |V0

We specialize to a box potential but the arguments here apply also to a potential whichgoes like minusZr

f (r) =

forr le ar gt a

and choose α = a Then

k2(r) = γ

(1minus εminus 1

γl(l+1)

minusεminusminus 1γ

l(l+1)

forr le ar gt a

For small r we getd2

dr2u(r)minus

l(l + 1)

r2u(r) = 0

with solutions u(r) = r l+1 or u(r) = rminusl Since the final solution must be finiteeverywhere we get the condition for our numerical solution

u(r) = r l+1 for small r

For large r we getd2

dr2u(r)minus γεu(r) = 0 γ gt 0

with solutions u(r) = exp(plusmnγεr) and the condition for large r means that our numericalsolution must satisfy

u(r) = eminusγεr for large r

In order to find a bound state we start integrating with a trial negative value for the

energy from small values of the variable r usually zero and up to some large value of

r As long as the potential is significantly different from zero the function oscillates

Outside the range of the potential the function will approach an exponential form If we

have chosen a correct eigenvalue the function decreases exponetially as

u(r) = eminusγεr However due to numerical inaccuracy the solution will contain small

admixtures of the undesireable exponential growing function u(r) = e+γεr

The final solution will then become unstable Therefore it is better to generate two

solutions with one starting from small values of r and integrate outwards to some

matching point r = rm We call that function ult(r) The next solution ugt(r) is then

obtained by integrating from some large value r where the potential is of no

importance and inwards to the same matching point rm Due to the quantum

mechanical requirements the logarithmic derivative at the matching point rm should be

well defined

We obtain the following condition

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

We can modify this expression by normalizing the function ultult(rm) = Cugtult(rm)

drult(r) =

drugt(r) at r = rm

We can calculate the first order derivatives by

drult(rm) asymp

ult(rm)minus ult(rm minus h)

drugt(rm) asymp

ugt(rm + h)minus ugt(rm)

Thus the criterium for a proper eigenfunction will be

f = ult(rm minus h)minus ugt(rm + h)

which should be smaller than a fixed number

Numerovcpp main part

quick and dirty set up of parametersr_max = 300 orb_l = 0 n = 1 max_step = 1000 charge = 1 step length energy and orbital momentum factorstep = (r_max - r_min) max_steporb_factor = orb_l (orb_l + 1)energy = -10(20nn)

allocate memory for rthe wave function and k

r = new double[max_step+1]k = new double[max_step+1]wf = new double[max_step+1]

set up r and add the chosen energy to the potential and set up the homogeneous partfor(i = 0 i lt= max_step i++)

r[i] = r_min + i step note stepstep in the definition of k

k[i] = -(orb_factor(r[i]r[i])-2energy-2charger[i])stepstep

calculate the wave function using Numerovrsquos method

wfn(step max_step k wf)

print out wave functionoutput(max_step r wf) delete [] r delete [] wf delete [] kofileclose() close output filereturn 0

Numerovcpp

void wfn(double step int max_step double k double wf)

int loop loop_1matchdouble fac wwf norm

Boundary conditions and value of the wave function at max stepwf[max_step] = 00wf[max_step - 1] = 10E-10 start from right and move towards rminfor(loop = max_step - 2 loop gt 0 loop--)

wf[loop] = (20(10-50k[loop+1]120) wf[loop+1]- (10+k[loop+2]120) wf[loop+2])(10+k[loop]120)

find point were the outer part starts turningin order to determine matching pointif(wf[loop] lt= wf[loop + 1]) break

matching pointmatch = loop + 1wwf = wf[match]

Numerovcpp

start from rmin and move towards the matching point boundary condition and first stepwf[0] = 00wf[1] = 10e-10for(loop = 2 loop lt= match loop++)

wf[loop] = (2(10-50k[loop-1]120) wf[loop-1]- (10+k[loop-2]120) wf[loop-2])(10+k[loop]120)

if(fabs(wf[loop]) gt INFINITY) for(loop_1 = 0 loop_1 lt= loop loop_1++)

wf[loop_1] = INFINITY

Numerovcpp

scale the wave function below the matching pointfac = wwfwf[match]for(loop = 0 loop lt= match loop++) wf[loop] = fac compute the normfor(loop = 0 norm = 00 loop lt= max_step loop++)

norm += wf[loop] wf[loop] stepif(fabs(norm) lt 10e-15)

printf(nnError in norm in function wfn() )exit(1)

norm = 1 sqrt(norm)for(loop = 0 loop lt= max_step loop++)

wf[loop] = norm

Todayrsquos tasks

Get aquainted with numerovcpp try the 2p wave or the 2s wave Try differentnumber of mesh points and length of mesh

Use it so solve the differential equation

dr2u(r)minus a2u(r) = minus

2reminusr

with analytic solution

u(r) =

bdquo1

1minus a2

laquo2 bdquoeminusar minus eminusr

raquo1 +

2(1minus a2)r

ndashlaquo

with boundary conditions u(0) = u(infin) = 0 for various values of a Test yourcode when you integrate only outward and only inward

Choose different intermediate matching points and see if your results changedramatically

Before you start solving this equation you should clarify which values of a are possible

Recall that it is only the case where minusa2 gt 0 which yields a solution

Topics for Week 5 January 28 - February 1

Thursday

Continuation from last week Algorithm for theHartree-Fock equations

Numerovrsquos algorithm and Greenrsquos function Finalization ofcode for solving differential equations with boundaryconditions

Friday

Hartree-Fock theory derivation of the Hartree-Fockequations and begin of discussion of density functionaltheory

Reminder on Greenrsquos functions

Consider the differential equation

minusu(x)primeprime = f (x) x isin (0 1) u(0) = u(1) = 0

and using the fundamental theorem of calculus there is a constant c1 such that

u(x) = c1 +

0uprime(y)dy

and a constant c2

uprime(y) = c2 +

0uprimeprime(z)dz

This is true for any twice continuously differentiable function u

If u satisfies the above differential equation we have then

uprime(y) = c2 minusZ y

0f (z)dz

which inserted into the equation for u gives

u(x) = c1 + c2x minusZ x

bdquoZ y

0f (z)dz

laquody

and defining

F (y) =

0f (z)dz

and performing an integration by parts we obtainZ x

0F (y)dy =

bdquoZ y

0f (z)dz

laquody =

0(x minus y)f (y)dy

This gives us

0(x minus y)f (y)dy

The boundary condition u(0) = 0 yields c1 = 0 and u(1) = 0 results in

0(1minus y)f (y)dy

meaning that we can write the solution as

u(x) = xZ 1

0(1minus y)f (y)dy minus

0(x minus y)f (y)dy

If we then define the Greenrsquos function as

G(x y) =

y(1minus x) if 0 le y le xx(1minus y) if x le y le 1

we can write the solution as

u(x) =

0G(x y)f (y)dy

The Greenrsquos function is

1 continuous

2 Is symmetric in the sense that G(x y) = G(y x)

3 G(0 y) = G(1 y) = G(x 0) = G(x 1) = 0

4 Piecewise linear function of x for fixed y and vice versa Gprime is discontinuos aty = x

5 G(x y) ge 0 for all x y isin [0 1]

6 It is the solution of the differential equation

dx2G(x y) = minusδ(x minus y)

For more details see for example Arfken Mathematical methods for Physicists

chapter 165

The Greenrsquos function can now be used to define the solution before and after a specificpoint in the domainThe Greenrsquos function satisfies the homogeneous equation for y 6= x and its derivativeis discontinuous at x = y We can see this if we integrate the differential equation

from x = y minus ε to x = y + ε with ε as an infinitesmally small number We obtain then

dx|x=y+ε minus

dx|x=yminusε = 1

The problem is obvioulsy to find G

We can obtain this by considering two solutions of the homogenous equation Choosea general domain x isin [a b] with a boundary condition on the general solutionu(a) = u(b) = 0One solution is obtained by integrating from a to b (called ult) and one by integratinginward from b to a labelled ugtUsing the continuity requirement on the function and its derivative we can compute theWronskian

W =dugt

dxult minus

and usingdG

dx|x=y+ε minus

dx|x=yminusε = 1

and one can then show that Greenrsquos function is

G(x y) = ult(xlt)ugt(xgt)

where xlt is for x = y minus ε and xgt = y + ε

Simple example

uprimeprime(x) + λu(x) = 0 x isin (0 1) u(0) = u(1) = 0

The solution of the homogenous equation uprimeprime = 0 from x = 0 is

ult(x) = x

and starting from x = 1ugt(x) = 1minus x

The Wronskian isultuprimegt minus ugtuprimelt = minus1

we obtain the Greenrsquos function

G(x y) =

x(1minus y) if 0 le x lt yy(1minus x) if y lt x le 1

u(x) = λ

0G(x y)u(t)dy

The Greenrsquos function

For our particular case discussed last week the Greenrsquos function is also a solution of

dr2G(r r prime)minus a2G(r r prime) = δ(r minus r prime)

and has the property that the solution can be written as

u(r) =

Z infin

0G(r r prime)F (r prime)dr prime

and can be expressed via the matching condition (logarithmic derivative of the wavefunction)

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

asG(r r prime) = ult(rlt)ugt(rgt)

with rlt below the matching point and rgt above

Todayrsquos tasks the Greenrsquos function

The Greenrsquos function satisfies the homogenous equation and can be used to find u(r)using

u(r) =

Z infin

andG(r r prime) = ult(rlt)ugt(rgt)

resulting in

u(r) = ugt(r)Z r

0ult(r prime)F (r prime)dr prime + ult(r)

Z infin

rugt(r prime)F (r prime)dr prime

Your task is to choose a matching point say the midpoint and then compute the

Greensrsquo function after you have used Numerovrsquos algo to find u (inward and outward

integration for all points) Find u integrating with the Greenrsquos function

Algorithm

Compute ugt using Numerovrsquos algorithm inward integration

Boundary conditions and value of the wave function at max stepuin[max_step] = 00uin[max_step - 1] = 10E-10 start from right and move towards rminfor(loop = max_step - 2 loop gt 0 loop--)

uin[loop] = (20(10-50k[loop+1]120) uin[loop+1]- (10+k[loop+2]120) uin[loop+2])(10+k[loop]120)

Algorithm

Compute ult using Numerovrsquos algorithm outward integration

boundary condition and first stepuout[0] = 00uout[1] = 10e-10for(loop = 2 loop lt= max_step loop++)

uout[loop] = (2(10-50k[loop-1]120) uout[loop-1]- (10+k[loop-2]120) uout[loop-2])(10+k[loop]120)

Algorithm

Compute Wronskian at matching point using

dxasymp

u(x + h)minus u(x + h)

for the first derivative and choose matching point as the midpoint (try the stabilityof the solution afterwards)

midpoint = max_step2 first part of Wronskianwronskian = (uin[midpoin+1]-uin[midpoint-1])uout[midpoint](2step) second partwronskian -= (uout[midpoin+1]-uout[midpoint-1])uin[midpoint](2step)

Algorithm

Compute outward integral of Greenrsquos function approach note the source termsource[loop] in code Use the Trapezoidal rule

sum = 00for(loop = 0 loop lt= max_step loop++)

term = uout[loop]source[loop]sum += termwf[loop] = uin[loop]sumstep

Algorithm

Compute inward integral of Greenrsquos function approach Use the Trapezoidal rule

sum = 00for(loop = max_step loop gt= 0 loop--)

term = uin[loop]source[loop]sum += termwf[loop] = (wf[loop]+uout[loop]sumstep)wronskian

Algorithm

Compare with analytic solution

Topics for Week 6 February 4-8

Hartree-Fock density functional theory and computing topics

Thursday

Reminder on variational calculus

Final derivation of Hartree-Fock equations

Continuation from last week seeting up code for theHartree-Fock equations

Friday

Density functional theory and begin of Variational MonteCarlo

Variational Calculus and Lagrangian Multiplier

The calculus of variations involves problems where the quantity to be minimized ormaximized is an integralIn the general case we have an integral of the type

E [Φ] =

af (Φ(x)

partΦ

partx x)dx

where E is the quantity which is sought minimized or maximized The problem is that

although f is a function of the variables Φ partΦpartx and x the exact dependence of Φ

on x is not known This means again that even though the integral has fixed limits a

and b the path of integration is not known In our case the unknown quantities are the

single-particle wave functions and we wish to choose an integration path which makes

the functional E [Φ] stationary This means that we want to find minima or maxima or

saddle points In physics we search normally for minima Our task is therefore to find

the minimum of E [Φ] so that its variation δE is zero subject to specific constraints In

our case the constraints appear as the integral which expresses the orthogonality of

the single-particle wave functions The constraints can be treated via the technique of

Lagrangian multipliers

Euler-Lagrange equations

We assume the existence of an optimum path that is a path for which E [Φ] isstationary There are infinitely many such paths The difference between two paths δΦis called the variation of ΦWe call the variation η(x) and it is scaled by a factor α The function η(x) is arbitraryexcept for

η(a) = η(b) = 0

and we assume that we can model the change in Φ as

Φ(x α) = Φ(x 0) + αη(x)

andδΦ = Φ(x α)minus Φ(x 0) = αη(x)

We choose Φ(x α = 0) as the unkonwn path that will minimize E The valueΦ(x α 6= 0) describes a neighbouring pathWe have

E [Φ(α)] =

af (Φ(x α)

partΦ(x α)

partx x)dx

In the slides I will use the shorthand

Φx (x α) =partΦ(x α)

In our case a = 0 and b = infin and we know the value of the wave function

The condition for an extreme of

E [Φ(α)] =

af (Φ(x α)Φx (x α) x)dx

is raquopartE [Φ(α)]

ndashα=0

The α dependence is contained in Φ(x α) and Φx (x α) meaning thatraquopartE [Φ(α)]

partα

ndash=

bdquopartf

partΦ

partα+

partΦx

partα

laquodx

We have definedpartΦ(x α)

partα= η(x)

and therebypartΦx (x α)

partα=

d(η(x))

UsingpartΦ(x α)

partα= η(x)

andpartΦx (x α)

partα=

d(η(x))

in the integral givesraquopartE [Φ(α)]

partα

ndash=

bdquopartf

partΦη(x) +

partΦx

d(η(x))

laquodx

Integrate the second term by partsZ b

partΦx

d(η(x))

dxdx = η(x)

partΦx|ba minus

aη(x)

partΦxdx

and since the first term dissappears due to η(a) = η(b) = 0 we obtainraquopartE [Φ(α)]

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

can also be written as

raquopartE [Φ(α)]

partα

ndashα=0

bdquopartf

partΦminus

partΦx

laquoδΦ(x)dx = δE = 0

The condition for a stationary value is thus a partial differential equation

partΦminus

partΦx= 0

known as Eulerrsquos equation Can easily be generalized to more variables

Lagrangian Multipliers

Consider a function of three independent variables f (x y z) For the function f to bean extreme we have

df = 0

A necessary and sufficient condition is

partx=partf

party=partf

partz= 0

due to

df =partf

partxdx +

partydy +

partzdz

In physical problems the variables x y z are often subject to constraints (in our case Φ

and the orthogonality constraint) so that they are no longer all independent It is

possible at least in principle to use each constraint to eliminate one variable and to

proceed with a new and smaller set of independent varables

The use of so-called Lagrangian multipliers is an alternative technique when theelimination of of variables is incovenient or undesirable Assume that we have anequation of constraint on the variables x y z

φ(x y z) = 0

resulting in

dφ =partφ

partxdx +

partφ

partydy +

partφ

partzdz = 0

Now we cannot set anymore

partx=partf

party=partf

partz= 0

if df = 0 is wanted because there are now only two independent variables Assume x

and y are the independent variables Then dz is no longer arbitrary

However we can add to

df =partf

partxdx +

partydy +

partzdz

a multiplum of dφ viz λdφ resulting in

df + λdφ = (partf

partz+ λ

partφ

partx)dx + (

party+ λ

partφ

party)dy + (

partz+ λ

partφ

partz)dz = 0

Our multiplier is chosen so that

partz+ λ

partφ

partz= 0

However we took dx and dy as to be arbitrary and thus we must have

partx+ λ

partφ

partx= 0

andpartf

party+ λ

partφ

party= 0

When all these equations are satisfied df = 0 We have four unknowns x y z and λActually we want only x y z λ need not to be determined it is therefore often calledLagrangersquos undetermined multiplier If we have a set of constraints φk we have theequations

partxi+X

λkpartφk

partxi= 0

Variational Calculus and Lagrangian Multipliers

Let us specialize to the expectation value of the energy for one particle inthree-dimensions This expectation value reads

Zdxdydzψlowast(x y z)Hψ(x y z)

with the constraint Zdxdydzψlowast(x y z)ψ(x y z) = 1

and a Hamiltonian

H = minus1

2nabla2 + V (x y z)

I will skip the variables x y z below and write for example V (x y z) = V

The integral involving the kinetic energy can be written as if we assume periodicboundary conditions or that the function ψ vanishes strongly for large values of x y zZ

dxdydzψlowastbdquominus

2nabla2laquoψdxdydz = ψlowastnablaψ|+

Zdxdydz

2nablaψlowastnablaψ

Inserting this expression into the expectation value for the energy and taking thevariational minimum we obtain

δE = δ

Zdxdydz

bdquo1

2nablaψlowastnablaψ + Vψlowastψ

laquoff= 0

The constraint appears in integral form asZdxdydzψlowastψ = constant

and multiplying with a Lagrangian multiplier λ and taking the variational minimum weobtain the final variational equation

Zdxdydz

bdquo1

2nablaψlowastnablaψ + Vψlowastψ minus λψlowastψ

laquoff= 0

Introducing the function f

2nablaψlowastnablaψ + Vψlowastψ minus λψlowastψ =

2(ψlowastxψx + ψlowastyψy + ψlowastzψz) + Vψlowastψ minus λψlowastψ

where we have skipped the dependence on x y z and introduced the shorthand ψx

ψy and ψz for the various derivatives

For ψlowast the Euler equation results in

partψlowastminus

partψlowastxminus

partψlowastyminus

partψlowastz= 0

which yields

minus1

2(ψxx + ψyy + ψzz) + Vψ = λψ

We can then identify the Lagrangian multiplier as the energy of the system Then thelast equation is nothing but the standard Schrodinger equation and the variationalapproach discussed here provides a powerful method for obtaining approximatesolutions of the wave function

Finding the Hartree-Fock functional E [Φ]

We rewrite our Hamiltonian

H = minusNX

2nabla2

i minusNX

NXiltj

H = H1 + H2 =NX

hi +NX

iltj=1

hi = minus1

2nabla2

i minusZ

Let us denote the ground state energy by E0 According to the variational principle wehave

E0 le E [Φ] =

ZΦlowastHΦdτ

where Φ is a trial function which we assume to be normalizedZΦlowastΦdτ = 1

where we have used the shorthand dτ = dr1dr2 drN

In the Hartree-Fock method the trial function is the Slater determinant which can berewritten as

Ψ(r1 r2 rN α β ν) =1

radicN

(minus)PPψα(r1)ψβ(r2) ψν(rN) =radic

where we have introduced the anti-symmetrization operator A defined by thesummation over all possible permutations of two eletrons It is defined as

(minus)PP

with the the Hartree-function given by the simple product of all possible single-particlefunction (two for helium four for beryllium and ten for neon)

ΦH(r1 r2 rN α β ν) = ψα(r1)ψβ(r2) ψν(rN)

Both H1 and H2 are invariant under electron permutations and hence commute with A

[H1A] = [H2A] = 0

Furthermore A satisfiesA2 = A

since every permutation of the Slater determinant reproduces it

The expectation value of H1ZΦlowastH1Φdτ = N

ZΦlowastHAH1AΦHdτ

is readily reduced to ZΦlowastH1Φdτ = N

ZΦlowastH H1AΦHdτ

which can be rewritten as

ZΦlowastH1Φdτ =

(minus)PZ

ΦlowastH hi PΦHdτ

The integral vanishes if two or more electrons are permuted in only one of theHartree-functions ΦH because the individual orbitals are orthogonal We obtain then

ZΦlowastH1Φdτ =

ZΦlowastH hiΦHdτ

Orthogonality allows us to further simplify the integral and we arrive at the followingexpression for the expectation values of the sum of one-body Hamiltonians

ZΦlowastH1Φdτ =

NXmicro=1

Zψlowastmicro(r i )hiψmicro(r i )dr i

or just as ZΦlowastH1Φdτ =

NXmicro=1

〈micro|h|micro〉

The expectation value of the two-body Hamiltonian is obtained in a similar manner Wehave Z

ΦlowastH2Φdτ = N

which reduces to

ZΦlowastH2Φdτ =

NXilej=1

(minus)PZ

ΦlowastH1

rijPΦHdτ

by following the same arguments as for the one-body Hamiltonian Because of thedependence on the inter-electronic distance 1rij permutations of two electrons nolonger vanish and we get

ZΦlowastH2Φdτ =

NXiltj=1

ZΦlowastH

rij(1minus Pij )ΦHdτ

where Pij is the permutation operator that interchanges electrons i and j

We use the assumption that the orbitals are orthogonal and obtain

ZΦlowastH2Φdτ =

NXmicro=1

NXν=1

Zψlowastmicro(r i )ψ

lowastν(r j )

rijψmicro(r i )ψν(r j )dr i r j

minusZψlowastmicro(r i )ψ

lowastν(r j )

rijψν(r i )ψmicro(r i )dx ix j

The first term is the so-called direct term or Hartree term while the second is due tothe Pauli principle and is called exchange term or Fock term The factor 12 isintroduced because we now run over all pairs twiceThe compact notation is

NXmicro=1

NXν=1

〈microν|

rij|microν〉 minus 〈microν|

rij|νmicro〉

Variational Calculus and Lagrangian Multiplier back toHartree-Fock

Our functional is written as

E [Φ] =NX

micro=1

Zψlowastmicro(r i )hiψmicro(r i )dr i +

NXmicro=1

NXν=1

lowastν(r j )

rijψν(r i )ψmicro(r i )dr i r j

The more compact version is

E [Φ] =NX

micro=1

〈micro|h|micro〉+1

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

Variational Strategies

With the given functional we can perform at least two types of variational strategis

Vary the Slater determinat by changing the spatial part of the single-particlewave functions themselves This is what we will do

Expand the single-particle functions in a known basis and vary the coefficientsthat is the new function single-particle wave function |a〉 is written as a linearexpansion in terms of a fixed basis (harmonic oscillator Laguerre polynomialsetc)

ψa =Xλ

Caλψλ

Both cases lead to a new Slater determinant which is related to the previous via aunitary transformation

We look at the first case first The other one willbe presented tomorrow during the

lectures

Exercise

1 Consider a Slater determinant built up of single-particle orbitals ψλ withλ = 1 2 N

The unitary transformationψa =

Caλψλ

brings us into the new basis Show that the new basis is orthonormal

2 Show that the new Slater determinant constructed from the new single-particlewave functions can be written as the determinant based on the previous basisand the determinant of the matrix C

3 Show that the old and the new Slater determinants are equal up to a complexconstant with absolute value unity (Hint C is a unitary matrix)

Hartree-Fock Variational Calculus and LagrangianMultiplier

If we generalize the Euler-Lagrange equations to more variables and introduce N2

Lagrange multipliers which we denote by εmicroν we can write the variational equation forthe functional of E

δE minusNX

micro=1

NXν=1

εmicroνδ

Zψlowastmicroψν = 0

For the orthogonal wave functions ψmicro this reduces to

δE minusNX

micro=1

εmicroδ

Zψlowastmicroψmicro = 0

Variation with respect to the single-particle wave functions ψmicro yields then

NXmicro=1

Zδψlowastmicrohiψmicrodr i +

NXmicro=1

NXν=1

Zδψlowastmicroψ

lowastν

rijψmicroψνdr i dr j minus

Zδψlowastmicroψ

lowastν

rijψνψmicrodr i dr j

micro=1

Zψlowastmicrohiδψmicrodr i +

NXmicro=1

NXν=1

Zψlowastmicroψ

lowastν

rijδψmicroψνdr i dr j minus

Zψlowastmicroψ

lowastν

rijψνδψmicrodr i dr j

minusNX

micro=1

Emicro

Zδψlowastmicroψmicrodr i minus

NXmicro=1

Emicro

Zψlowastmicroδψmicrodr i = 0

Although the variations δψ and δψlowast are not independent they may in fact be treated assuch so that the terms dependent on either δψ and δψlowast individually may be set equalto zero To see this simply replace the arbitrary variation δψ by iδψ so that δψlowast isreplaced by minusiδψlowast and combine the two equations We thus arrive at theHartree-Fock equations24minus1

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

rijψmicro(r j )dr j

35ψν(r i ) = εmicroψmicro(r i )

Notice that the integrationR

dr j implies an integration over the spatial coordinates r j

and a summation over the spin-coordinate of electron j

The two first terms are the one-body kinetic energy and the electron-nucleus potential

The third or direct term is the averaged electronic repulsion of the other electrons This

term is identical to the Coulomb integral introduced in the simple perturbative approach

to the helium atom As written the term includes the rsquoself-interactionrsquo of electrons when

i = j The self-interaction is cancelled in the fourth term or the exchange term The

exchange term results from our inclusion of the Pauli principle and the assumed

determinantal form of the wave-function The effect of exchange is for electrons of

like-spin to avoid each other

A theoretically convenient form of the Hartree-Fock equation is to regard the direct andexchange operator defined through

V dmicro(r i ) =

Zψlowastmicro(r j )

V exmicro (r i )g(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

The function g(r i ) is an arbitrary function and by the substitution g(r i ) = ψν(r i ) we get

V exmicro (r i )ψν(r i ) =

rijψν(r j )dr j

ψmicro(r i )

We may then rewrite the Hartree-Fock equations as

HHFi ψν(r i ) = ενψν(r i )

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

and where hi is the one-body part

Hartree-Fock Explicit expression for Direct Term

We want to show first that

V dmicro(r i ) =

V dmicro(r i ) = V d

nl (ri ) =Xnprime lprime

2(2l prime + 1)

Z infin

rgtdrj

with rgt = max(ri rj )

Hartree-Fock Explicit expression for Direct TermWe need to break down the single-particle wave functions in radial angular and spinparts Note that direct term is diagonal in the spin quantum numbers Thesingle-particle wave functions are

ψα(r) = ψnlml sms (r) = φnlml(r)ξms (s)

withφnlml

(r) = Rnl (r)Ylml(r)

and we defined unl (r) = rRnl (r) The direct term is with a factor two from the spindegrees of freedom gives for a single shell

V dnprime lprime (ri ) = 2

lXmprimel =minuslprime

Z infin

0|φnprime lprimemprimel

(rj )|21

rijdr j

which reads

Z infin

rijdrj

|Ylmprimel(θj φj )|2dΩj

with dΩj the angular part d(cosθj )dφj

The addition theorem ofthe spherical harmonics yields

|Ylmprimel(θj φj )|2 =

2l prime + 1

resulting in

V dnprime lprime (ri ) =

2(2l prime + 1)

Z infin

rijdrj dΩj

The quantity rij depends on the angles of particle i and j

The integral over the angles can be performed by expanding 1rij in terms of sphericalharmonics and using the fact that the functions unprime lprime do not depend on the angles Wehave

infinXl=0

bdquorj

laquol

Pl cos(θ))

if ri gt rj or

infinXl=0

Pl cos(θ))

if rj gt ri In a compact form it reads

infinXl=0

(rlt)l

(rgt)l+1

Pl cos(θ))

with rgt = max(ri rj ) and with rlt = min(ri rj ) Pl is the Legendre polynomial and

cos(θ) = cos(θi )cos(θj ) + sin(θi )sin(θj )cos(φi minus φj )

We can use the expression for spherical harmonics to express the interaction as

infinXl=0

(rlt)l

(rgt)l+1

Ylowastlmprimel

(θi φi )Ylmprimel(θj φj )

and inserting it we obtain the final expression as

2(2l prime + 1)

Z infin

rgtdrj

Hartree-Fock Explicit expression for Exchange Term

Exercise show that the exchange part can be written as

2l prime + 1

raquoZ infin

(rlt)λ

with rgt = max(ri rj ) and with rlt = min(ri rj )

The Clebsch-Gordan coefficient can be written as

〈ll prime00|λ0〉 =1

p2λ+ 1(1 + (minus1)2g)(minus1)g

s(2g minus 2l)(2g minus 2l prime)(2g minus 2l prime)

(2g + 1)

timesg

(g minus l)(g minus l prime)(g minus λ)

with 2g = 2l + λ You can write a small function for this expression or use the enclosed

programs

Hartree-Fock theory DFT and Monte Carlo

Thursday

Discussion of the Hartree-Fock program

Friday

End of DFT discussion and begin of Variational MonteCarlo

Hartree-Fock Explicit expressions for various Atoms

We need to discuss how the Hartree-Fock equations look like for helium beryllium andneon (Exercise from last week)The general equations take the form24minus1

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

Hartree-Fock Equations

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

Hartree-Fock Explicit expressions for various Atomshelium

The Slater determinant for helium reads with two electrons in the 1s state

Φ(r1 r2 α β) =1radic

˛ψα(r1) ψα(r2)ψβ(r1) ψβ(r2)

with α = nlml sms = 1001212 and β = nlml sms = 10012minus 12 or usingms = 12 =uarr and ms = minus12 =darr as α = nlml sms = 10012 uarr andβ = nlml sms = 10012 darr Writing out the Slater determinant we obtain

ˆψα(r1)ψβ(r2)minus ψβ(r1)ψγ(r2)

and we see that the Slater determinant is antisymmetric with respect to thepermutation of two particles that is

Φ(r1 r2 α β) = minusΦ(r2 r1 α β)

In the derivations of the direct and exchange terms we have not discussed the role ofthe electron spinLet us introduce

ψnlml sms = φnlml(r)ξms (s)

with s is the spin (12 for electrons) ms is the spin projection ms = plusmn12 and thespatial part is

with Y the spherical harmonics and unl = rRnl We have for helium

2φ100(r1)φ100(r2)

ˆξuarr(1)ξdarr(2)minus ξuarr(2)ξdarr(1)

The direct term acts on1radic

2φ100(r1)φ100(r2)ξuarr(1)ξdarr(2)

while the exchange term acts on

minus1radic

How do these terms get translated into the Hartree and the Fock terms

The Hartree term

V dmicro(r i ) =

acts on ψλ(r i ) = φnlml(r i )ξms (si ) that is it results in

V dmicro(r i )ψλ(r i ) =

ψλ(r i )

and accounting for spins we have

V dnlmluarr(r i )ψλ(r i ) =

Zψlowastnlmluarr(r j )

rijψnlmluarr(r j )dr j

ψλ(r i )

V dnlmldarr(r i )ψλ(r i ) =

Zψlowastnlmldarr(r j )

rijψnlmldarr(r j )dr j

ψλ(r i )

If the state we act on has spin up we obtain two terms from the Hartree part

NXmicro=1

V dmicro(r i )

and since the interaction does not depend on spin we end up with a total contributionfor helium

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψλ(r i )

one from spin up and one from spin down Since the energy for spin up or spin down isthe same we can then write the action of the Hartree term as

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψ100uarr(r i )

(the spin in ψ100uarr is irrelevant)

What we need to code for helium is then

Φ(ri )u10 = 2V d10(ri )u10(ri ) = 2

Z infin

0|u10(rj )|2

rgtdrj )u10(ri )

with rgt = max(ri rj ) What about the exchange or Fock term

V exmicro (r i )ψλ(r i ) =

rijψλ(r j )dr j

ψmicro(r i )

We must be careful here with

rijψλ(r j )dr j

ψmicro(r i )

because the spins of micro and λ have to be the same due to the constraint

〈smicrommicros |sλmλ

s 〉 = δmmicros mλ

This means that if mmicros =uarr then mλ

s =uarr and if mmicros =darr then mλ

s =darr That is

V exmicro (r i )ψλ(r i ) = δmmicro

s mλs

rijψλ(r j )dr j

ψmicro(r i )

The consequence is that for the 1s uarr (and the same for 1s darr) state we get only onecontribution from the Fock term

NXmicro=1

V exmicro (r i )ψλ(r i )

NXmicro=1

V exmicro (r i )ψ100uarr(r i ) = δmmicro

s uarr

rijψ100uarr(r j )dr j

ψmicro(r i )

resulting in

NXmicro=1

V exmicro (r i )ψ100uarr(r i ) =

Zψlowast100uarr(r j )

ψ100uarr(r i )

The final Fock term for helium is then

NXmicro=1

ψ100uarr(r i )

which is exactly the same as the Hartree term except for a factor of 2 Else the integralis the same We can then write the differential equationbdquo

minus1

l(l + 1)

2r2minus

r+ Φnl (r)minus Fnl (r)

as bdquominus

l(l + 1)

2r2minus

r+ 2V d

10(r)laquo

u10(r)minus V ex10 (r) = e10u10(r)

or bdquominus

dr2minus

r+ V d

10(r)laquo

u10(r) = e10u10(r)

since l = 0 The shorthand V ex10 (r) contains the 1s wave function and can be

dangerous later

Hartree-Fock Explicit expressions for various Atomsberyllium

The expression we have obtained are independent of the spin projections and we haveskipped them in the equations Last weekrsquos exercise was to derive the correspondingequations for beryllium with two electrons in 1s as in helium but now also twoelectrons in 2sThe Slater determinant takes the form

Φ(r1 r2 r3 r4 α β γ δ) =1radic

˛˛ ψ100uarr(r1) ψ100uarr(r2) ψ100uarr(r3) ψ100uarr(r4)ψ100darr(r1) ψ100darr(r2) ψ100darr(r3) ψ100darr(r4)ψ200uarr(r1) ψ200uarr(r2) ψ200uarr(r3) ψ200uarr(r4)ψ200darr(r1) ψ200darr(r2) ψ200darr(r3) ψ200darr(r4)

When we now spell out the Hartree-Fock equations we get two coupled differentialequations one for u10 and one for u20The 1s wave function has the same Hartree-Fock contribution as in helium for the 1sstate but the 2s state gives two times the Hartree term and one time the Fock termThat is we get

NXmicro=1

V dmicro(r i )ψ100uarr(r i ) = 2

Z infin

φlowast100(r j )

rijφ100(r j ) + φlowast200(r j )

rijφ200(r j )

ψ100uarr(r i )

= (2V d10(r i ) + 2V d

20(r i ))ψ100uarr(r i )

for the Hartree part

For the Fock term we get (we fix the spin)

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

rijφ100(r j )ψ100uarr(r i )+

Z infin

0dr jφ

lowast200(r j )

rijφ100(r j )ψ200uarr(r i ) = V ex

10 (r i ) + V ex20 (r i )

The first term is the same as we have for the Hartree term with 1s except the factor oftwo The final differential equation isbdquo

minus1

dr2minus

r+ V d

10(r) + 2V d20(r)

laquou10(r)minus V ex

20 (r) = e10u10(r)

Note again that the V ex20 (r) contains the 1s function in the integral that is

V ex20 (r) =

Z infin

0dr jφ

lowast200(r j )

r minus rjφ100(r j )ψ200uarr(r)

The 2s wave function obtains the following Hartree term (recall that the interaction hasno spin dependence)

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ200uarr(r i ) =

(2V d10(r i ) + 2V d

20(r i ))ψ200uarr(r i )

For the Fock term we get

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

The second term is the same as we have for the Hartree term with 2s The finaldifferential equation isbdquo

minus1

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

V ex10 (r) =

Z infin

0dr jφ

lowast100(r j )

Hartree-Fock Final expressions for beryllium

We have two coupled differential equationsbdquominus

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

Recall again that the interaction does not depend on spin This means that thesingle-particle energies and single-particle function u do not depend on spin Also

Exercise derive the equivalent expressions for neon

Hartree-Fock Final expressions for neon

Exercise fill in the missing partsbdquominus

dr2minus

r+ V d

10(r) + 2V d20(r)+

20 (r)+ = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)+

10 (r)+ = e20u20(r)

andbdquominus

2r2minus

r+ 2V d

10(r) + 2V d20(r)+

laquou21(r)minusV ex

10 (r)minusV ex20 (r)+ = e21u21(r)

Useful Tests

The 1s hydrogen like wave function

R10(r) = 2bdquo

laquo32

exp (minusZra0) = u10r

The total energy for helium (not the Hartree or Fock terms) from the direct and theexchange term should give 5Z8The single-particle energy with no interactions should give minusZ 22n2The 2s hydrogen-like wave function is

R20(r) = 2bdquo

laquo32 bdquo1minus

laquoexp (minusZr2a0) = u20r

and the 2p hydrogen -like wave function is

R21(r) =1radic

bdquoZ

laquo32 Zr

a0exp (minusZr2a0) = u21r

We use a0 = 1

Effective Charge and first Iteration

If we compute the total energy of the helium atom with the function

R10(r) = 2bdquo

laquo32

as a trial single-particle wave fuction we obtain a total energy (one-body and two-body)

E [Z ] = Z 2 minus 4Z +5

The minimum is not at Z = 2 Take the derivative wrt Z and we find that the minimumis at

Z = 2minus5

16= 16875

and represents an optimal effective charge When we do the Hartree-Fock calculations

and use the optimal single-particle wave function in a variational Monte Carlo

calculation we should have the wave function calculated at the optimal value How do

we do this

Before the first iteration we compute the total energy of the system using thehydrogen wave function

Then we use the virial theorem for the Coulomb potential which states that

minus〈V 〉 = 2〈T 〉

where the expectation values are the total potential (one-body and two-body)and total kinetic energy

The effective charge is then

Zeff = minusZ 〈V 〉2〈T 〉

We use this effective charge in the remaining Hartree-Fock equations

Exercise Repeat the virial theorem(FYS3110 lectures)

Density Functional Theory (DFT)

Hohenberg and Kohn proved that the total energy of a system including that of themany-body effects of electrons (exchange and correlation) in the presence of staticexternal potential (for example the atomic nuclei) is a unique functional of the chargedensity The minimum value of the total energy functional is the ground state energy ofthe system The electronic charge density which yields this minimum is then the exactsingle particle ground state energyIn Hartree-Fock theory one works with large basis sets This poses a problem for largesystems An alternative to the HF methods is DFT DFT takes into account electroncorrelations but is less demanding computationally than full scale diagonalization orMonte Carlo methods

Density Functional Theory

The electronic energy E is said to be a functional of the electronic density E [ρ] in thesense that for a given function ρ(r) there is a single corresponding energy TheHohenberg-Kohn theorem confirms that such a functional exists but does not tell usthe form of the functional As shown by Kohn and Sham the exact ground-state energyE of an N-electron system can be written as

E [ρ] = minus1

ZΨlowasti (r1)nabla2

1Ψi (r1)dr1minusZ

r1ρ(r1)dr1+

Zρ(r1)ρ(r2)

r12dr1dr2+EEXC [ρ]

with Ψi the Kohn-Sham (KS) orbitals

The ground-state charge density is given by

ρ(r) =NX

|Ψi (r)|2

where the sum is over the occupied Kohn-Sham orbitals The last term EEXC [ρ] is the

exchange-correlation energy which in theory takes into account all non-classical

electron-electron interaction However we do not know how to obtain this term exactly

and are forced to approximate it The KS orbitals are found by solving the Kohn-Sham

equations which can be found by applying a variational principle to the electronic

energy E [ρ] This approach is similar to the one used for obtaining the HF equation

The KS equations readsminus

2nabla2

1 minusZ

Zρ(r2)

r12dr2 + VEXC(r1)

ffΨi (r1) = εiΨi (r1)

where εi are the KS orbital energies and where the exchange-correlation potential isgiven by

VEXC [ρ] =δEEXC [ρ]

The KS equations are solved in a self-consistent fashion A variety of basis setfunctions can be used and the experience gained in HF calculations are often usefulThe computational time needed for a DFT calculation formally scales as the thirdpower of the number of basis functionsThe main source of error in DFT usually arises from the approximate nature of EEXC Inthe local density approximation (LDA) it is approximated as

EEXC =

Zρ(r)εEXC [ρ(r)]dr

where εEXC [ρ(r)] is the exchange-correlation energy per electron in a homogeneous

electron gas of constant density The LDA approach is clearly an approximation as the

charge is not continuously distributed To account for the inhomogeneity of the electron

density a nonlocal correction involving the gradient of ρ is often added to the

exchange-correlation energy

Thursday

Variational Monte Carlo and the Metropolis algorithm

Discussion of the Blocking method and repeat onparallelization

Discussion of variational Monte Carlo and the codesvmc paracpp and vmc blockingcpp

Friday

Variational Monte Carlo statistical data analysis andimportance sampling

Background chapters 8-11 in Lecture notes Emphasis onchapter 11

Brownian Motion and Markov Processes

A Markov process is a random walk with a selected probability for making a move Thenew move is independent of the previous history of the system The Markov process isused repeatedly in Monte Carlo simulations in order to generate new random statesThe reason for choosing a Markov process is that when it is run for a long enough timestarting with a random state we will eventually reach the most likely state of thesystem In thermodynamics this means that after a certain number of Markovprocesses we reach an equilibrium distribution This mimicks the way a real systemreaches its most likely state at a given temperature of the surroundingsTo reach this distribution the Markov process needs to obey two important conditionsthat of ergodicity and detailed balance These conditions impose then constraints onour algorithms for accepting or rejecting new random states The Metropolis algorithmdiscussed here abides to both these constraints The Metropolis algorithm is widelyused in Monte Carlo simulations and the understanding of it rests within theinterpretation of random walks and Markov processes

In a random walk one defines a mathematical entity called a walker whose attributescompletely define the state of the system in question The state of the system can referto any physical quantities from the vibrational state of a molecule specified by a set ofquantum numbers to the brands of coffee in your favourite supermarketThe walker moves in an appropriate state space by a combination of deterministic andrandom displacements from its previous position

This sequence of steps forms a chain

A simple Example

The obvious case is that of a random walker on a one- or two- or three-dimensionallattice (dubbed coordinate space hereafter)Consider a system whose energy is defined by the orientation of single spins Considerthe state i with given energy Ei represented by the following N spins

uarr uarr uarr uarr darr uarr uarr darr1 2 3 k minus 1 k k + 1 N minus 1 N

We may be interested in the transition with one single spinflip to a new state j withenergy Ej

uarr uarr uarr uarr uarr uarr uarr darr1 2 3 k minus 1 k k + 1 N minus 1 N

This change from one microstate i (or spin configuration) to another microstate j is the

configuration space analogue to a random walk on a lattice Instead of jumping from

one place to another in space we rsquojumprsquo from one microstate to another

We wish to study the time-development of a PDF after a given number of time stepsWe define our PDF by the function w(t) In addition we define a transition probabilityW The time development of our PDF w(t) after one time-step from t = 0 is given by

wi (t = ε) = W (j rarr i)wj (t = 0)

This equation represents the discretized time-development of an original PDF We canrewrite this as a

wi (t = ε) = Wij wj (t = 0)

with the transition matrix W for a random walk left or right (cannot stay in the sameposition) given by

Wij (ε) = W (il minus jl ε) =

12 |i minus j| = 10 else

We call Wij for the transition probability and we represent it as a matrix

Both W and w represent probabilities and they have to be normalized meaning thatthat at each time step we have X

wi (t) = 1

and Xj

W (j rarr i) = 1

Further constraints are 0 le Wij le 1 and 0 le wj le 1 We can thus write the action of Was

wi (t + 1) =X

Wij wj (t)

or as vector-matrix relationw(t + 1) = Ww(t)

and if we have that ||w(t + 1)minus w(t)|| rarr 0 we say that we have reached the mostlikely state of the system the so-called steady state or equilibrium state Another wayof phrasing this is

w(t = infin) = Ww(t = infin)

Brownian Motion and Markov Processes a simpleExample

Consider the simple 3times 3 matrix W

0 14 18 2334 58 00 14 13

and we choose our initial state as

w(t = 0) =

The first iteration iswi (t = ε) = W (j rarr i)wj (t = 0)

resulting in

w(t = ε) =

0 14340

The next iteration results in

wi (t = 2ε) = W (j rarr i)wj (t = ε)

resulting in

w(t = 2ε) =

0 5232132632

Note that the vector w is always normalized to 1 We find the steady state of thesystem by solving the linear set of equations

This linear set of equations reads

W11w1(t = infin) + W12w2(t = infin) + W13w3(t = infin) = w1(t = infin)

with the constraint that Xi

wi (t = infin) = 1

yielding as solution

w(t = infin) =

0 415815315

Convergence of the simple example

Iteration w1 w2 w30 100000 000000 0000001 025000 075000 0000002 015625 062625 0187503 024609 052734 0226564 027848 051416 0207365 027213 053021 0197666 026608 053548 0198447 026575 053424 0200028 026656 053321 0200239 026678 053318 020005

10 026671 053332 01999811 026666 053335 02000012 026666 053334 02000013 026667 053333 020000

w(t = infin) 026667 053333 020000

Exercise make a small program where you perform these iterationsbut change the

initial vector and study the convergence

Brownian Motion and Markov Processes what ishappening

We have after t-stepsw(t) = Wtw(0)

with w(0) the distribution at t = 0 and W representing the transition probability matrixWe can always expand w(0) in terms of the right eigenvectors v of W as

w(0) =X

αi v i

resulting inw(t) = Wt w(0) = Wt

αi v i =X

λtiαi v i

with λi the i th eigenvalue corresponding to the eigenvector v i

If we assume that λ0 is the largest eigenvector we see that in the limit t rarrinfin w(t)becomes proportional to the corresponding eigenvector v0 This is our steady state orfinal distributionIn our discussion below in connection with the entropy of a system and tomorrowrsquoslecture on physics applications we will relate these properties to correlation functionssuch as the time-correlation function

That will allow us to define the so-called equilibration timeviz the time needed for the

system to reach its most likely state Form that state and on we can can compute

contributions to various statistical variables

We can relate this property to an observable like the mean magnetization of say amagnetic material With the probabilty w(t) we can write the mean magnetization as

〈M(t)〉 =Xmicro

w(t)microMmicro

or as the scalar of a vector product

〈M(t)〉 = w(t)m

with m being the vector whose elements are the values of Mmicro in its variousmicrostates microRecall our definition of an expectation value with a discrete PDF p(xi )

E [xk ] = 〈xk 〉 =1

xki p(xi )

provided that the sums (or integrals)PN

i=1 p(xi ) converge absolutely (viz PN

i=1 |p(xi )|converges)

We rewrite the last relation as

〈M(t)〉 = w(t)m =X

λtiαi v i m i

If we define mi = v i m i as the expectation value of M in the i th eigenstate we canrewrite the last equation as

〈M(t)〉 =X

λtiαi mi

Since we have that in the limit t rarrinfin the mean magnetization is dominated by thelargest eigenvalue λ0 we can rewrite the last equation as

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

λtiαi mi

We define the quantity

τi = minus1

logλi

and rewrite the last expectation value as

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

αi mi eminustτi

The quantities τi are the correlation times for the system They control also thetime-correlation functions to be discussed later

The longest correlation time is obviously given by the second largest eigenvalue τ1

which normally defines the correlation time discussed above For large times this is

the only correlation time that survives If higher eigenvalues of the transition matrix are

well separated from λ1 and we simulate long enough τ1 may well define the

correlation time In other cases we may not be able to extract a reliable result for τ1

Detailed BalanceAn important condition we require is that our Markov chain should satisfy is that ofdetailed balance In statistical physics this condition ensures that it is eg theBoltzmann distribution which is generated when equilibrium is reached The definitionfor being in equilibrium is that the rates at which a system makes a transition to or froma given state i have to be equal that isX

W (j rarr i)wj =X

W (i rarr j)wi

However the condition that the rates should equal each other is in general not sufficientto guarantee that we after many simulations generate the correct distribution Wetherefore introduce an additional condition namely that of detailed balance

W (j rarr i)wj = W (i rarr j)wi

At equilibrium detailed balance gives thus

W (j rarr i)

W (i rarr j)=

Proof Xi

W (i rarr j)wi =X

W (j rarr i)wj = wj

W (j rarr i) = wj

In a Markov process w is known while W is the unknown

Ergodicity

It should be possible for any Markov process to reach everypossible state of the system from any starting point if thesimulations is carried out for a long enough time If any state ina distribution which has a probability different from zero andthis state cannot be reached from any given starting point if wesimulate long enough then the system is not ergodic

The Boltzmann Distribution as Example

We introduce the Boltzmann distribution

wi =exp (minusβ(Ei ))

which states that probability of finding the system in a state i with energy Ei at aninverse temperature β = 1kBT is wi prop exp (minusβ(Ei )) The denominator Z is anormalization constant which ensures that the sum of all probabilities is normalized toone It is defined as the sum of probabilities over all microstates j of the system

exp (minusβ(Ei ))

From the partition function we can in principle generate all interesting quantities for a

given system in equilibrium with its surroundings at a temperature T

Boltzmann Distribution as Example

With the probability distribution given by the Boltzmann distribution we are now in theposition where we can generate expectation values for a given variable A through thedefinition

〈A〉 =X

Aj wj =

Pj Aj exp (minusβ(Ej )

In general most systems have an infinity of microstates making thereby thecomputation of Z practically impossible and a brute force Monte Carlo calculation overa given number of randomly selected microstates may therefore not yield thosemicrostates which are important at equilibrium To select the most importantcontributions we need to use the condition for detailed balance Since this is just givenby the ratios of probabilities we never need to evaluate the partition function Z For theBoltzmann distribution detailed balance results in

wj= exp (minusβ(Ei minus Ej ))

Let us now specialize to a system whose energy is defined by the orientation of singlespins Consider the state i with given energy Ei represented by the following N spins

We are interested in the transition with one single spinflip to a new state j with energyEj

We saw previously that this change from one microstate i (or spin configuration) to

another microstate j is the configuration space analogue to a random walk on a lattice

However the selection of states has to generate a final distribution which is theBoltzmann distribution This is again the same we saw for a random walker for thediscrete case we had always a binomial distribution whereas for the continuous casewe had a normal distribution The way we sample configurations should result in whenequilibrium is established in the Boltzmann distribution Else our algorithm forselecting microstates has to be wrongSince we do not know the analytic form of the transition rate we are free to model it as

W (i rarr j) = g(i rarr j)A(i rarr j)

where g is a selection probability while A is the probability for accepting a move It is

also called the acceptance ratio

The selection probability should be same for all possible spin orientations namely

g(i rarr j) =1

With detailed balance this gives

g(j rarr i)A(j rarr i)

g(i rarr j)A(i rarr j)= exp (minusβ(Ei minus Ej ))

but since the selection ratio is the same for both transitions we have

A(j rarr i)

A(i rarr j)= exp (minusβ(Ei minus Ej ))

In general we are looking for those spin orientations which correspond to the average

energy at equilibrium

We are in this case interested in a new state Ej whose energy is lower than Ei viz∆E = Ej minus Ei le 0 A simple test would then be to accept only those microstates whichlower the energy Suppose we have ten microstates with energyE0 le E1 le E2 le E3 le middot middot middot le E9 Our desired energy is E0 At a given temperature Twe start our simulation by randomly choosing state E9 Flipping spins we may then finda path from E9 rarr E8 rarr E7 middot middot middot rarr E1 rarr E0 This would however lead to biasedstatistical averages since it would violate the ergodic hypothesis discussed above Thisprinciple states that it should be possible for any Markov process to reach everypossible state of the system from any starting point if the simulations is carried out for along enough time

Any state in a Boltzmann distribution has a probability different from zero and if such astate cannot be reached from a given starting point then the system is not ergodicThis means that another possible path to E0 could beE9 rarr E7 rarr E8 middot middot middot rarr E9 rarr E5 rarr E0 and so forth Even though such a path couldhave a negligible probability it is still a possibility and if we simulate long enough itshould be included in our computation of an expectation value

Thus we require that our algorithm should satisfy the principle of detailed balance and

be ergodic One possible way is the Metropolis algorithm

Metropolis Algorithm

The equation for detailed balance

A(microrarr ν)

A(ν rarr micro)= exp (minusβ(Eν minus Emicro))

is general and we could replace the Boltzmann distribution with other distributions aswell It specifies the ratio of pairs of acceptance probabilities which leaves us withquite some room to manouvre

We give the largest of the two acceptance ratios the value 1 and adjust the otherto satisfy the constraint

If Emicro lt Eν then the larger of the two acceptance ratios is A(ν rarr micro) and we setto 1

Then we must have

A(microrarr ν) = exp (minusβ(Eν minus Emicro))

if Eν minus Emicro gt 0 and 1 otherwise And that is the Metropolis algorithm

More general Definition

A Markov chain Monte Carlo method for the simulation of a distribution p is any methodproducing an ergodic Markov chain of events x whose stationary distribution is p

Generate an initial value x (i)

Generate a trial value yt with probability f (yt |x (i))

Take a new value

x (i+1) =

yt with probability= ρ(x (i) yt )

x (i) with probability= 1minus ρ(x (i) yt )

We have defined

ρ(x y) = min

p(y)f (x |y)

p(x)f (y |x) 1ff

The distribution f is often called the instrumental (we will relate it to the jumpingof a walker) or proposal distribution while ρ is the Metropolis-Hastingsacceptance probability When f (y |x) is symmetric it is just called the Metropolisalgo

Implementation

Establish an initial state with some selected features to test

Do a random change of this initial state

Compute the Metropolis-Hastings acceptance probability ρ

Compare ρ with a random number r If r le ρ accept else keepthe old configuration

Compute the terms needed to obtain expectations values

Repeat the above steps in order to have as good statistics aspossible

For a given number of MC cycles compute then the finalexpectation values

Quantum Monte Carlo

Given a hamiltonian H and a trial wave function ΨT the variational principle states thatthe expectation value of 〈H〉 defined through

E [H] = 〈H〉 =

RdRΨlowastT (R)H(R)ΨT (R)R

dRΨlowastT (R)ΨT (R)

is an upper bound to the ground state energy E0 of the hamiltonian H that is

E0 le 〈H〉

In general the integrals involved in the calculation of various expectation values are

multi-dimensional ones Traditional integration methods such as the Gauss-Legendre

will not be adequate for say the computation of the energy of a many-body system

Quantum Monte Carlo

The trial wave function can be expanded in the eigenstates of the hamiltonian sincethey form a complete set viz

ΨT (R) =X

aiΨi (R)

and assuming the set of eigenfunctions to be normalized one obtainsPnm alowastman

RdRΨlowastm(R)H(R)Ψn(R)P

nm alowastmanR

dRΨlowastm(R)Ψn(R)=

nEnPn a2

nge E0

where we used that H(R)Ψn(R) = EnΨn(R) In general the integrals involved in the

calculation of various expectation values are multi-dimensional ones The variational

principle yields the lowest state of a given symmetry

Quantum Monte Carlo

In most cases a wave function has only small values in large parts of configurationspace and a straightforward procedure which uses homogenously distributed randompoints in configuration space will most likely lead to poor results This may suggest thatsome kind of importance sampling combined with eg the Metropolis algorithm maybe a more efficient way of obtaining the ground state energy The hope is then thatthose regions of configurations space where the wave function assumes appreciablevalues are sampled more efficiently

The tedious part in a VMC calculation is the search for the variational minimum A

good knowledge of the system is required in order to carry out reasonable VMC

calculations This is not always the case and often VMC calculations serve rather as

the starting point for so-called diffusion Monte Carlo calculations (DMC) DMC is a way

of solving exactly the many-body Schrodinger equation by means of a stochastic

procedure A good guess on the binding energy and its wave function is however

necessary A carefully performed VMC calculation can aid in this context

Quantum Monte Carlo

Construct first a trial wave function ψαT (R) for a many-body system consisting of

N particles located at positions R = (R1 RN) The trial wave functiondepends on α variational parameters α = (α1 αN)

Then we evaluate the expectation value of the hamiltonian H

E [H] = 〈H〉 =

RdRΨlowastTα

(R)H(R)ΨTα(R)R

dRΨlowastTα(R)ΨTα

Thereafter we vary α according to some minimization algorithm and return to thefirst step

Quantum Monte Carlo

Choose a trial wave function ψT (R)

P(R) =|ψT (R)|2R|ψT (R)|2 dR

This is our new probability distribution function (PDF) The approximation to theexpectation value of the Hamiltonian is now

E [H] asympR

dRΨlowastT (R)H(R)ΨT (R)RdRΨlowastT (R)ΨT (R)

Define a new quantity

EL(R) =1

ψT (R)HψT (R)

called the local energy which together with our trial PDF yields

E [H] = 〈H〉 asympZ

P(R)EL(R)dR asymp1

P(Ri)EL(Ri)

with N being the number of Monte Carlo samples

Quantum Monte Carlo

Initialisation Fix the number of Monte Carlo steps Choose an initial R andvariational parameters α and calculate

˛ψα

T (R)˛2

Initialise the energy and the variance and start the Monte Carlo calculation(thermalize)

1 Calculate a trial position Rp = R + r lowast step where r is a random variabler isin [0 1]

2 Metropolis algorithm to accept or reject this move

w = P(Rp)P(R)

3 If the step is accepted then we set R = Rp Update averages

Finish and compute final averages

Observe that the jumping in space is governed by the variable step Called brute-force

sampling Need importance sampling

Quantum Monte Carlo

The radial Schrodinger equation for the hydrogen atom can be written as

minus~2

part2u(r)

partr2minusbdquo

rminus

~2l(l + 1)

laquou(r) = Eu(r)

or with dimensionless variables

minus1

part2u(ρ)

partρ2minus

l(l + 1)

2ρ2u(ρ)minus λu(ρ) = 0

with the hamiltonian

H = minus1

partρ2minus

l(l + 1)

Use variational parameter α in the trial wave function

uαT (ρ) = αρeminusαρ

Quantum Monte Carlo

Inserting this wave function into the expression for the local energy EL gives

EL(ρ) = minus1

ρminusα

bdquoαminus

α 〈H〉 σ2 σradic

N700000E-01 -457759E-01 451201E-02 671715E-04800000E-01 -481461E-01 305736E-02 552934E-04900000E-01 -495899E-01 820497E-03 286443E-04100000E-00 -500000E-01 000000E+00 000000E+00110000E+00 -493738E-01 116989E-02 342036E-04120000E+00 -475563E-01 885899E-02 941222E-04130000E+00 -454341E-01 145171E-01 120487E-03

Quantum Monte Carlo

We note that at α = 1 we obtain the exact result and the variance is zero as it shouldThe reason is that we then have the exact wave function and the action of thehamiltionan on the wave function

Hψ = constanttimes ψ

yields just a constant The integral which defines various expectation values involvingmoments of the hamiltonian becomes then

〈Hn〉 =

RdRΨlowastT (R)Hn(R)ΨT (R)R

dRΨlowastT (R)ΨT (R)= constanttimes

RdRΨlowastT (R)ΨT (R)RdRΨlowastT (R)ΨT (R)

= constant

This gives an important information the exact wave function leads to zero

variance Variation is then performed by minimizing both the energy and the variance

Quantum Monte Carlo

The helium atom consists of two electrons and a nucleus with charge Z = 2 Thecontribution to the potential energy due to the attraction from the nucleus is

minus2ke2

r1minus

and if we add the repulsion arising from the two interacting electrons we obtain thepotential energy

V (r1 r2) = minus2ke2

r1minus

with the electrons separated at a distance r12 = |r1 minus r2|

Quantum Monte Carlo

The hamiltonian becomes then

bH = minus~2nabla2

2mminus

~2nabla22

2mminus

r1minus

and Schrodingers equation reads bHψ = Eψ

All observables are evaluated with respect to the probability distribution

P(R) =|ψT (R)|2R|ψT (R)|2 dR

generated by the trial wave function The trial wave function must approximate an exact

eigenstate in order that accurate results are to be obtained Improved trial wave

functions also improve the importance sampling reducing the cost of obtaining a

certain statistical accuracy

Quantum Monte Carlo

Choice of trial wave function for Helium Assume r1 rarr 0

EL(R) =1

ψT (R)HψT (R) =

ψT (R)

bdquominus

2nabla2

1 minusZ

laquoψT (R) + finite terms

EL(R) =1

RT (r1)

minus1

dr1minus

RT (r1) + finite terms

For small values of r1 the terms which dominate are

limr1rarr0

EL(R) =1

RT (r1)

bdquominus

dr1minus

laquoRT (r1)

since the second derivative does not diverge due to the finiteness of Ψ at the origin

Quantum Monte Carlo

This results in1

RT (r1)

dRT (r1)

dr1= minusZ

andRT (r1) prop eminusZr1

A similar condition applies to electron 2 as well For orbital momenta l gt 0 we have

RT (r)

dRT (r)

dr= minus

Similalry studying the case r12 rarr 0 we can write a possible trial wave function as

ψT (R) = eminusα(r1+r2)eβr12

The last equation can be generalized to

ψT (R) = φ(r1)φ(r2) φ(rN)Yiltj

f (rij )

for a system with N electrons or particles

Quantum Monte Carlo and Hartree-Fock

What do we want to do and where do we stand

In Hartree-Fock for helium we have reduced the six dimensional integral to aone-dimensional one We solve an integro-differential equation

The variational HF procedure finds a minimum for the energy by varying thewave function (in our case the parameter α in eminusαri of the single-particle wavefunction) This parameter has the physical interpretation of an effective charge

In variational Monte Carlo with no correlation part we should get the sameresults as the Hartree-Fock code if we start with the same guess for the Slaterdeterminant

But in VMC we compute by stochastic methods the full multi-dimensionalintegral

Our first step is to perform a VMC calculation with the same Slater determinantas we did for helium beryllium and neon for the Hartree-Fock calculations Thatis the wave function is

ψT (R) = φ(r1)φ(r2) φ(rN)

for a system with N electrons where the single-particle functions are thehydrogen-like functions we start the Hartree-Fock equations with

We discuss the parallel VMC code vmc paracpp which implements this

In addition we discuss the statistical analysis of the Monte Carlo dato by theso-called Blocking technique and the code vmc blockingcpp

We do a brute force Metropolis sampling meaning that the points we select forthe integral are given by the uniform distribution

We will then replace the hydrogen-like wave functions with the optimalHartree-Fock equations You should get the same result as the Hartree-Fockcalculation (test of having implemented the new wave functions correctly)

We will need to interpolate in an array (next thursday) by using splines seechapter 6 of lecture notes The spline code is available in the c++ or F95 libraries

We will add the correlation part to the trial wave function

f (rij )

and perform the variational Monte Carlo calculation on the correlation part onlyWe will use only one variational parameter The results with its standarddeviation willbe parametrized using the singular value decomposition method

We will need to improve the brute force technique with importance sampling

Finally we need to code a Slater determinant for the Monte Carlo calculation

Status of Hartree-Fock

Many of you have working Hartree-Fock codes

If your code works jump to the VMC part

Else use today to finish the Hartree-Fock part

VMC code for helium vmc paracpp

Here we define global variables used in various functions These can be changed by reading from file the different parametersint dimension = 3 three-dimensional systemint charge = 2 we fix the charge to be that of the helium atomint my_rank numprocs these are the parameters used by MPI to

define which node and how manydouble step_length = 10 we fix the brute force jump to 1 Bohr radiusint number_particles = 2 we fix also the number of electrons to be 2

VMC code for helium vmc paracpp main part

MPI initializationsMPI_Init (ampargc ampargv)MPI_Comm_size (MPI_COMM_WORLD ampnumprocs)MPI_Comm_rank (MPI_COMM_WORLD ampmy_rank)time_start = MPI_Wtime()

if (my_rank == 0 ampamp argc lt= 2) cout ltlt Bad Usage ltlt argv[0] ltlt

read also output file on same line ltlt endlexit(1)

if (my_rank == 0 ampamp argc gt 2)

outfilename=argv[1]ofileopen(outfilename)

Setting output file name for this rankostringstream ostost ltlt blocks_rank ltlt my_rank ltlt dat Open file for writingblockofileopen(oststr()c_str() iosout | iosbinary)

total_cumulative_e = new double[max_variations+1]total_cumulative_e2 = new double[max_variations+1]cumulative_e = new double[max_variations+1]cumulative_e2 = new double[max_variations+1]

initialize the arrays by zeroing themfor( i=1 i lt= max_variations i++)

cumulative_e[i] = cumulative_e2[i] = 00total_cumulative_e[i] = total_cumulative_e2[i] = 00

broadcast the total number of variationsMPI_Bcast (ampmax_variations 1 MPI_INT 0 MPI_COMM_WORLD)MPI_Bcast (ampnumber_cycles 1 MPI_INT 0 MPI_COMM_WORLD)total_number_cycles = number_cyclesnumprocs array to store all energies for last variation of alphaall_energies = new double[number_cycles+1] Do the mc sampling and accumulate data with MPI_Reducemc_sampling(max_variations number_cycles cumulative_e

cumulative_e2 all_energies) Collect data in total averagesfor( i=1 i lt= max_variations i++)

MPI_Reduce(ampcumulative_e[i] amptotal_cumulative_e[i] 1 MPI_DOUBLEMPI_SUM 0 MPI_COMM_WORLD)

MPI_Reduce(ampcumulative_e2[i] amptotal_cumulative_e2[i] 1 MPI_DOUBLEMPI_SUM 0 MPI_COMM_WORLD)

blockofilewrite((char)(all_energies+1)number_cyclessizeof(double))

blockofileclose()delete [] total_cumulative_e delete [] total_cumulative_e2delete [] cumulative_e delete [] cumulative_e2 delete [] all_energies End MPIMPI_Finalize ()return 0 end of main function

VMC code for helium vmc paracpp sampling

alpha = 05charge every node has its own seed for the random numbersidum = -1-my_rank allocate matrices which contain the position of the particlesr_old =(double )matrix(number_particlesdimensionsizeof(double))r_new =(double )matrix(number_particlesdimensionsizeof(double))for (i = 0 i lt number_particles i++)

for ( j=0 j lt dimension j++) r_old[i][j] = r_new[i][j] = 0

loop over variational parameters

for (variate=1 variate lt= max_variations variate++) initialisations of variational parameters and energiesalpha += 01energy = energy2 = 0 accept =0 delta_e=0 initial trial position note calling with alphafor (i = 0 i lt number_particles i++)

for ( j=0 j lt dimension j++) r_old[i][j] = step_length(ran2(ampidum)-05)

wfold = wave_function(r_old alpha)

loop over monte carlo cyclesfor (cycles = 1 cycles lt= number_cycles cycles++)

new positionfor (i = 0 i lt number_particles i++)

for ( j=0 j lt dimension j++) r_new[i][j] = r_old[i][j]+step_length(ran2(ampidum)-05)

for the other particles we need to set the position to the old position since we move only one particle at the time

for (k = 0 k lt number_particles k++) if ( k = i)

for ( j=0 j lt dimension j++) r_new[k][j] = r_old[k][j]

wfnew = wave_function(r_new alpha)

The Metropolis test is performed by moving one particle at the timeif(ran2(ampidum) lt= wfnewwfnewwfoldwfold )

for ( j=0 j lt dimension j++) r_old[i][j]=r_new[i][j]

wfold = wfnew

end of loop over particles

compute local energydelta_e = local_energy(r_old alpha wfold) save all energies on last variateif(variate==max_variations)

all_energies[cycles] = delta_e update energiesenergy += delta_eenergy2 += delta_edelta_e

end of loop over MC trials update the energy average and its squaredcumulative_e[variate] = energycumulative_e2[variate] = energy2

end of loop over variational steps

VMC code for helium vmc paracpp wave function

Function to compute the squared wave function simplest form

double wave_function(double r double alpha)

int i j kdouble wf argument r_single_particle r_12

argument = wf = 0for (i = 0 i lt number_particles i++)

r_single_particle = 0for (j = 0 j lt dimension j++)

r_single_particle += r[i][j]r[i][j]argument += sqrt(r_single_particle)

wf = exp(-argumentalpha) return wf

VMC code for helium vmc paracpp local energy

Function to calculate the local energy with num derivative

double local_energy(double r double alpha double wfold)

int i j kdouble e_local wfminus wfplus e_kinetic e_potential r_12

r_single_particledouble r_plus r_minus

allocate matrices which contain the position of the particles the function matrix is defined in the progam libraryr_plus =(double )matrix(number_particlesdimensionsizeof(double))r_minus =(double )matrix(number_particlesdimensionsizeof(double))for (i = 0 i lt number_particles i++)

for ( j=0 j lt dimension j++) r_plus[i][j] = r_minus[i][j] = r[i][j]

compute the kinetic energye_kinetic = 0for (i = 0 i lt number_particles i++)

for (j = 0 j lt dimension j++) r_plus[i][j] = r[i][j]+hr_minus[i][j] = r[i][j]-hwfminus = wave_function(r_minus alpha)wfplus = wave_function(r_plus alpha)e_kinetic -= (wfminus+wfplus-2wfold)r_plus[i][j] = r[i][j]r_minus[i][j] = r[i][j]

include electron mass and hbar squared and divide by wave functione_kinetic = 05h2e_kineticwfold

compute the potential energye_potential = 0 contribution from electron-proton potentialfor (i = 0 i lt number_particles i++)

r_single_particle += r[i][j]r[i][j]e_potential -= chargesqrt(r_single_particle)

contribution from electron-electron potentialfor (i = 0 i lt number_particles-1 i++)

for (j = i+1 j lt number_particles j++) r_12 = 0for (k = 0 k lt dimension k++)

r_12 += (r[i][k]-r[j][k])(r[i][k]-r[j][k])e_potential += 1sqrt(r_12)

Monte Carlo Blocking and Importance Sampling

Thursday

Discussion of the spline method to interpolate data

Importance sampling

Lab Repeat of parallelization

Lab Discussion of Jon Nilsenrsquos blocking code seeadditional slides

Lab Discussion of variational Monte Carlo and the codesvmc paracpp and vmc blockingcpp

Friday

Variational Monte Carlo theoretical background forimportance sampling

Merging Hartree-Fock with the Monte Carlo code

Things to do

Helium first write out the final HF wave functions to filewith mesh points and values of the wave function

Read this in from file in the VMC code (or merge the twocodes Every node must have access to the full wf)

Implement the spline method to interpolate in your dataset the spline functions are declared in libh and defined inlibcpp

Run just a brute force Monte Carlo with the Slaterdeterminant as your trial function See that you get thesame energy as the HF code gives

Things to do

Now change the trial wave function to include thecorrelation part as discussed in the project Run thecalculation for various values of β and try to find avariational minimum Use some few millions of cycles

Implement then blocking and perform a statistical analysisof the data

Implement importance sampling

If everything ok submit your production run

Repeat for Beryllium (next week HF= -1452 au)Remember to orthogonalize 1s and 2s in HF

Slater det and VMC (Neon HF= -12828 au) and finalsmoothing of data with data fitting procedures Beware ofcomplex spherical harmonics

Orthogonalize 1s and 2s States (why not 2p)

Simple way to orthogonalize (pseudo code)

if ( state == 2s) thensum = 0 overlap between 1s and 2sfor i = 1 meshpoints

sum = sum + psi1s(i)psi2s(i)end forsum = sumstep project out unwanted solutionfor i = 1 meshpoints

psi2s(i) = psi2s(i) - sumpsi1s(i)end for

Qubic Splines Chapter 6

Qubic spline interpolation is among one of the mostly used methods for interpolatingbetween data points where the arguments are organized as ascending series In thelibrary program we supply such a function based on the so-called qubic spline methodto be described belowA spline function consists of polynomial pieces defined on subintervals The differentsubintervals are connected via various continuity relationsAssume we have at our disposal n + 1 points x0 x1 xn arranged so thatx0 lt x1 lt x2 lt xnminus1 lt xn (such points are called knots) A spline function s ofdegree k with n + 1 knots is defined as follows

On every subinterval [ximinus1 xi ) s is a polynomial of degree le k

s has k minus 1 continuous derivatives in the whole interval [x0 xn]

Splines

As an example consider a spline function of degree k = 1 defined as follows

s(x) =

8gtgtltgtgts0(x) = a0x + b0 x isin [x0 x1)s1(x) = a1x + b1 x isin [x1 x2)

snminus1(x) = anminus1x + bnminus1 x isin [xnminus1 xn]

In this case the polynomial consists of series of straight lines connected to each other

at every endpoint The number of continuous derivatives is then k minus 1 = 0 as

expected when we deal with straight lines Such a polynomial is quite easy to construct

given n + 1 points x0 x1 xn and their corresponding function values

Splines

The most commonly used spline function is the one with k = 3 the so-called qubicspline function Assume that we have in adddition to the n + 1 knots a series offunctions values y0 = f (x0) y1 = f (x1) yn = f (xn) By definition the polynomialssiminus1 and si are thence supposed to interpolate the same point i ie

siminus1(xi ) = yi = si (xi )

with 1 le i le n minus 1 In total we have n polynomials of the type

si (x) = ai0 + ai1x + ai2x2 + ai2x3

yielding 4n coefficients to determine

Splines

Every subinterval provides in addition the 2n conditions

yi = s(xi )

ands(xi+1) = yi+1

to be fulfilled If we also assume that sprime and sprimeprime are continuous then

sprimeiminus1(xi ) = sprimei (xi )

yields n minus 1 conditions Similarly

sprimeprimeiminus1(xi ) = sprimeprimei (xi )

results in additional n minus 1 conditions In total we have 4n coefficients and 4n minus 2

equations to determine them leaving us with 2 degrees of freedom to be determined

Splines

Using the last equation we define two values for the second derivative namely

sprimeprimei (xi ) = fi

andsprimeprimei (xi+1) = fi+1

and setting up a straight line between fi and fi+1 we have

sprimeprimei (x) =fi

xi+1 minus xi(xi+1 minus x) +

xi+1 minus xi(x minus xi )

and integrating twice one obtains

si (x) =fi

6(xi+1 minus xi )(xi+1 minus x)3 +

6(xi+1 minus xi )(x minus xi )

3 + c(x minus xi ) + d(xi+1 minus x)

Splines

Using the conditions si (xi ) = yi and si (xi+1) = yi+1 we can in turn determine theconstants c and d resulting in

si (x) =fi

6(xi+1minusxi )(xi+1 minus x)3 +

fi+16(xi+1minusxi )

(x minus xi )3

+ (yi+1

xi+1minusximinus fi+1(xi+1minusxi )

6 )(x minus xi ) + (yi

xi+1minusximinus fi (xi+1minusxi )

6 )(xi+1 minus x) (2)

Splines

How to determine the values of the second derivatives fi and fi+1 We use thecontinuity assumption of the first derivatives

and set x = xi Defining hi = xi+1 minus xi we obtain finally the following expression

himinus1fiminus1 + 2(hi + himinus1)fi + hi fi+1 =6

hi(yi+1 minus yi )minus

himinus1(yi minus yiminus1)

and introducing the shorthands ui = 2(hi + himinus1) vi = 6hi

(yi+1 minus yi )minus 6himinus1

(yi minus yiminus1)

we can reformulate the problem as a set of linear equations to be solved through eg

Gaussian elemination

Splines

Gaussian elimination2666664u1 h1 0 h1 u2 h2 0 0 h2 u3 h3 0

0 hnminus3 unminus2 hnminus20 hnminus2 unminus1

3777775

2666664f1f2f3

fnminus2fnminus1

3777775 =

2666664v1v2v3

vnminus2vnminus1

3777775

Note that this is a set of tridiagonal equations and can be solved through only O(n)

operations

Splines

The functions supplied in the program library are spline and splint In order to usequbic spline interpolation you need first to call

spline(double x[] double y[] int n double yp1double yp2 double y2[])

This function takes as input x [0 n minus 1] and y [0 n minus 1] containing a tabulation

yi = f (xi ) with x0 lt x1 lt lt xnminus1 together with the first derivatives of f (x) at x0 and

xnminus1 respectively Then the function returns y2[0 n minus 1] which contains the second

derivatives of f (xi ) at each point xi n is the number of points This function provides

the qubic spline interpolation for all subintervals and is called only once

Splines

Thereafter if you wish to make various interpolations you need to call the function

splint(double x[] double y[] double y2a[] int ndouble x double y)

which takes as input the tabulated values x [0 n minus 1] and y [0 n minus 1] and the

output y2a[0n - 1] from spline It returns the value y corresponding to the point x

Importance SamplingWe need to replace the brute force Metropolis algorithm with a walk in coordinatespace biased by the trial wave function This approach is based on the Fokker-Planckequation and the Langevin equation for generating a trajectory in coordinate spaceFor a diffusion process characterized by a time-dependent probability density P(x t) inone dimension the Fokker-Planck equation reads (for one particlewalker)

partt= D

bdquopartP

partxminus F

laquoP(x t)

where F is a drift term and D is the diffusion coefficient The drift term is

F = 21

ΨTnablaΨT

where ΨT is our trial wave function The new positions in coordinate space are givenas the solutions of the Langevin equation using Eulerrsquos method namely we go from theLangevin equation

partx(t)

partt= DF (x(t)) + η

with η a random variable yielding a new position

y = x + DF (x)∆t + ξ

where ξ is gaussian random variable and ∆t is a chosen time step

Importance Sampling

The Fokker-Planck equation yields a transition probability given by the Greenrsquos function

G(y x ∆t) =1

(4πD∆t)3N2exp

ldquominus(y minus x minus D∆tF (x))24D∆t

rdquowhich in turn means that our brute force Metropolis algorithm

A(y x) = min(1 q(y x)))

with q(y x) = |ΨT (y)|2|ΨT (x)|2 is now replaced by

q(y x) =G(x y ∆t)|ΨT (y)|2

G(y x ∆t)|ΨT (x)|2

Topics for Week 10 March 3-7

Thursday

Lab only today brief reminder of tasks from last week

Lab Continue with implementing the spline interpolationand importance sampling for the helium atom

Lab Implement blocking as well When this is done youcan jump on beryllium

Lab Spherical harmonics for neon and the need to changeto solid harmonics

Friday

Theoretical background for importance sampling anddiffusion Monte Carlo

Begin of discussion on how to implement the Slaterdeterminant (optional in codes)

Problems with neon and HF solution for VMC

In the Hartree-Fock code we have used spherical coordinates

x = rsinθcosφ

y = rsinθsinφ

andz = rcosθ

The reason we introduce spherical coordinates is the spherical symmetry of theCoulomb potential

4πε0r=

4πε0p

x2 + y2 + z2

where we have used r =p

x2 + y2 + z2 It is not possible to find a separable solutionof the type

ψ(x y z) = ψ(x)ψ(y)ψ(z)

However with spherical coordinates we can find a solution of the form

ψ(r θ φ) = R(r)P(θ)F (φ)

The angle-dependent differential equations result in the spherical harmonic functionsas solutions with quantum numbers l and ml These functions are given by

Ylml(θ φ) = P(θ)F (φ) =

s(2l + 1)(l minusml )

4π(l + ml )Pml

l (cos(θ)) exp (imlφ)

with Pmll being the associated Legendre polynomials They can be rewritten as

Ylml(θ φ) = sin|ml |(θ)times (polynom(cosθ)) exp (imlφ)

We have the following selected examples

for l = ml = 0

4πcos(θ)

for l = 1 og ml = 0

Y1plusmn1 =

8πsin(θ)exp(plusmniφ)

for l = 1 og ml = plusmn1

A problem with the spherical harmonics is that they are complex The introduction ofsolid harmonics allows the use of real orbital wave-functions for a wide range ofapplications The complex solid harmonics Ylml

(r) are related to the sphericalharmonics Ylml

(r) through

Ylml(r) = r l Ylml

By factoring out the leading r -dependency of the radial-function

Rnl (r) = rminusl Rnl (r)

we obtainΨnlml

(r θ φ) = Rnl (r) middot Ylml(r)

Remember that you get unl (r) = rRnl (r) from the solution of the Hartree-Fock

equation

For the theoretical development of the real solid harmonics we first express thecomplex solid harmonics Clml

by (complex) Cartesian coordinates and arrive at thereal solid harmonics Slml

through the unitary transformation Slml

Slminusml

1radic

(minus1)m

minus(minus1)ml i i

Clminusml

This transformation will not alter any physical quantities that are degenerate in the

subspace consisting of opposite magnetic quantum numbers (the angular momentum l

is equal for both these cases) This means for example that the above transformation

does not alter the energies unless an external magnetic field is applied to the system

Henceforth we will use the solid harmonics and note that changing the spherical

potential beyond the Coulomb potential will not alter the solid harmonics

We have definedYlml

(r) = r l Ylml(r)

The real-valued spherical harmonics are defined as

2l + 1Yl0(r)

Slml= (minus1)ml

2l + 1ReYl0(r)

Slml= (minus1)ml

2l + 1ImYl0(r)

for ml gt 0

The lowest-order real solid harmonics are listed in here

Real Solid Harmonics

mll 0 1 2 3

q52 (x2 minus 3y2)x

radic3(x2 minus y2) 1

radic15(x2 minus y2)z

+1 xradic

3xz 12

q32 (5z2 minus r2)x

0 1 y 12 (3z2 minus r2) 1

2 (5z2 minus 3r2)x

-1 zradic

3yz 12

q32 (5z2 minus r2)y

-2radic

3xyradic

q52 (3x2 minus y2)y

Things to do repeat from last week

Change the trial wave function to include the correlationpart as discussed in the project Run the calculation forvarious values of β and try to find a variational minimumUse some few millions of cycles

If everything ok submit your production run for helium

Repeat for Beryllium (HF= -1452 au) Remember toorthogonalize 1s and 2s in HF

Slater det and VMC (Neon HF= -12828 au) and finalsmoothing of data with data fitting procedures(singular-value decomposition Beware of complexspherical harmonics If we run into time problems withneon we make the VMC part of neon optional We awaitthis decision to next week

Last week of project 1

Thursday

Lab brief reminder of tasks from last week

Lab Continue with implementing the spline interpolationand importance sampling for the helium atom and start ofberyllium calculations

Construction of the Slater determinant last part neededfor beryllium and neon

Summary of things to do for the final report

Discussion of the report see also revised report text

If time Final note on the Langevin and Fokker-Planckequations

How to parameterize Monte Carlo with singular valuedecomposition will not be included in the project

Slater determinant Explicit expressions for variousAtoms beryllium

The Slater determinant takes the form

Note that the Slater determinant as written is zero since the spatial wave functions for

the spin up and spin down states are equal But we can rewrite it the product of two

Slater determinants one for spin up and one for spin down

We can rewrite it as

Φ(r1 r2 r3 r4 α β γ δ) = Det uarr (1 2)Det darr (3 4)minus Det uarr (1 3)Det darr (2 4)

minusDet uarr (1 4)Det darr (3 2) + Det uarr (2 3)Det darr (1 4)minus Det uarr (2 4)Det darr (1 3)

+Det uarr (3 4)Det darr (1 2)

where we have defined

Det uarr (1 2) =

˛1radic

ψ100uarr(r1) ψ100uarr(r2)ψ200uarr(r1) ψ200uarr(r2)

Det darr (3 4) =

˛1radic

ψ100darr(r3) ψ100darr(r4)ψ200darr(r3) ψ200darr(r4)

The total determinant is still zero

We want to avoid to sum over spin variables in particular when the interaction does notdepend on spinIt can be shown see for example Moskowitz and Kalos Int J Quantum Chem 20(1981) 1107 that for the variational energy we can approximate the Slater determinantas

Φ(r1 r2 r3 r4 α β γ δ) prop Det uarr (1 2)Det darr (3 4)

or more generally asΦ(r1 r2 rN) prop Det uarr Det darr

where we have the Slater determinant as the product of a spin up part involving thenumber of electrons with spin up only (2 in Beryllium and 5 in neon) and a spin downpart involving the electrons with spin downThis ansatz is not antisymmetric under the exchange of electrons with opposite spinsbut it can be shown that it gives the same expectation value for the energy as the fullSlater determinant

As long as the Hamiltonian is spin independent the above is correct

Code vmc becpp at the webpage

In this code I have included the Slater determinant for the beryllium atom Note that theimplementation is rather brute force like

for (i = 0 i lt number_particles i++) argument[i] = 00r_single_particle = 0for (j = 0 j lt dimension j++)

r_single_particle += r[i][j]r[i][j]argument[i] = sqrt(r_single_particle)

Slater determinant no factors as they vanish in Metropolis ratiowf = (psi1s(argument[0])psi2s(argument[1])

-psi1s(argument[1])psi2s(argument[0]))(psi1s(argument[2])psi2s(argument[3])

-psi1s(argument[3])psi2s(argument[2]))

For beryllium we can easily implement the explicit evaluation of the Slater determinant

For neon we need to think

Slater Determinant and VMC

Determining a determinant of an N times N matrix by standard Gaussian elimination is ofthe order of O(N3) calculations As there are N middot d independent coordinates we needto evaluate N middot d Slater determinants for the gradient (quantum force) and N middot d for theLaplacian (kinetic energy)We therefore need 2d middot N + 1 evaluations of the Slater determinant These evaluationsmay be reduced by an efficient evaluation of the Slater determinant Instead of workingwith the 2d middot N + 1 we perform most of the manipulations on the inverse of the Slatermatrix

By changing the position of only one particle at the time we may limit the number of

calculations to update the inverse matrix to approximately N3 per electron or

approximately N4 calculations for all electrons

The trial wave function takes then the form

Ψα = DuarrαDdarrαGβ

This allow us to treat the Slater determinant part and the part concering thecorrelations separately since

Ψα(Xnew )

Ψα(Xold )=

Duarrα(Xnew )

Duarrα(Xold )

Ddarrα(Xnew )

Ddarrα(Xold )

Gβ(Xnew )

Gβ(Xold )

with variational parameters α and β For the quantum force

nablaΨα

Ψα=nablaDuarrα

Duarrα+nablaDdarrα

Ddarrα+nablaGβ

and finally for the kinetic energy

nabla2Ψα

Ψα=nabla2Duarrα

Duarrα+nabla2Ddarrα

Ddarrα+nabla2Gβ

Gβ+ (

nablaGβ

Gβ)(nablaDuarrα

Ddarrα)

First define the Slater matrix to be

Dij (x) equiv φj (xi )

The inverse of the Slater matrix is the adjoint of the cofactors Cji divided by thedeterminant of the Slater matrix Dα ie

Dminus1ij =

By definition the Slater matrix and its inverse must satisfy the relation

Dik (x)Dminus1kj (x) = δij

The inverse matrix may therefore be used to evaluate the determinant in the usualcofactor expansion Consider the fraction of two such expansions

Dα(y)

Dα(x)=

Dij (y)Cij (y)

Dij (x)Cij (x)

with y and x two electron configurations that are equal except for the coordinates ofelectron i

The cofactors Cij (y) and Cij (x) are independent of the coordinates of electron i andtherefore equal This observation yields

Dα(y)

Dα(x)=

Dij (y)Cij (x)

Dij (x)Cij (x)

Dij (y)Dminus1ji (x)

Dij (x)Dminus1ji (x)

By applying the inverse of the Slater matrix we arrive at

Dα(y)

Dα(x)=

for the ratio In a similar manner we compute the derivatives

nablai Dα(x)

Dα(x)=

nablaφj (xi )Dminus1ji (x)

nabla2i Dα(x)

Dα(x)=

nabla2φj (xi )Dminus1ji (x)

For the new position we get the following expressions

nablai Dα(y)

Dα(y)=

nablaφj (yi )Dminus1ji (x)

andnabla2

i Dα(y)

Dα(y)=

nabla2φj (yi )Dminus1ji (x)

All the determinant derived quantities required at each step of the walk can becomputed using the single inverse matrix Dminus1

The new inverse matrix is evaluated only if the move from x to y is accepted It ispossible to update the inverse matrix one column at a time rather than recomputingthe entire matrixThe ratio of the determinant at y to the determinant at x is

R equivNX

If this move is rejected by the Monte Carlo algorithm then no further computations areneeded If the move is accepted the new inverse matrix is related to the old inversematrix by

Dminus1kj (y) =

8gtgtgtltgtgtgtDminus1

kj (x)minus1

RDminus1

ki (x)NX

Dil (x)Dminus1lj (x) for j 6= i

RDminus1

ki (x) for j = i

To compute the initial inverse matrix start with the identity matrix and replace eachcolumn one at a time

The two spin-degrees of freedom allow us to split the Slater matrix in two separatematrices

D = Duarr middot Ddarr

reducing the number of calculations by up to a factor of eight for systems with equalnumber of spin-up and spin-down electrons we now have two matrices that each need(N2)3 = N38 calculations to update their inverses

When moving one electron only one of these need to be updated thereof the factor18 Each matrix is treated independently but have the same structure The relation

R =D(y)

Duarr(y) middot Ddarr(y)

Duarr(x) middot Ddarr(x)

follows from the orthogonality of the spin-up and spin-down states We see that when

we move only one particle the ratio R is either Duarr(y)Duarr(x) or Ddarr(y)Ddarr(x)

How to compute the Slater Determinant

If you choose to implement the above recipe for the computation of the Slaterdeterminant you need to LU decompose the Slater matrix This is described in chapter4 of the lecture notes

You need to call the function ludcmp in libcpp You need to transfer the Slater matrix

and its dimension You get back an LU decomposed matrix

LU Decomposition

The LU decomposition method means that we can rewrite this matrix as the product oftwo matrices B and C where0BB

a11 a12 a13 a14a21 a22 a23 a24a31 a32 a33 a34a41 a42 a43 a44

1CCA =

0BB1 0 0 0

b21 1 0 0b31 b32 1 0b41 b42 b43 1

1CCA0BB

c11 c12 c13 c140 c22 c23 c240 0 c33 c340 0 0 c44

The matrix A isin Rntimesn has an LU factorization if the determinant is different from zero Ifthe LU factorization exists and A is non-singular then the LU factorization is uniqueand the determinant is given by

detA = c11c22 cnn

More improvements optional but useful

We can speed up the computation of the correlation part by noting that we can write itas

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

9=which means that the gradient needed for the quantum force and local energy is

nablaGβ

Gβ=Xiltj

(1 + βrij )2

nabla2Gβ

Gβ=Xiltj

a(2rminus1ij (1 + βrij ) + a)

(1 + βrij )4

This will speed up your code since the computation of the correlation part and the

Slater determinant are the most time consuming parts in your code You can choose

whether you wish to implement this or not

Important feature

For the correlation part

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

9=we need to take into account whether electrons have equal or opposite spins since wehave to obey the electron-electron cusp condition as well For Beryllium you can fixelectrons 1 and 2 to have spin up while electrons 3 and 4 have spin down When theelectrons have equal spins

a = 14

while for opposite spins (as we did for helium)

a = 12

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

9=you must therefore pay attention in the sum

for (j = i+1 j lt number_particles j++) r_ij = 0for (k = 0 k lt dimension k++)

r_ij += (r[i][k]-r[j][k])(r[i][k]-r[j][k])r_ij=sqrt(r_ij)wf = exp(r_ij(2(1+betar_ij)))

What to include in the report

Things to do

Description of the algorithms and how you implementedthem both HF and VMC

Donrsquot write too much keep the report short and answer thequestions in brief

Remember to label figures and tables properly (labels onaxes)

You must have importance sampling and blocking andyour VMC code should run in parallel

Discuss the correlation energy in helium and beryllium wrtto the Hartree-Fock results Neon is optional All VMCenergies should be accompanied by the standarddeviation

Things to do

Remember to use solid harmonics for the 2p orbit if you doVMC for neon

Test that your code has a good parallel scaling Try also tofind time consuming bottlenecks by profiling your codeCompile the code with the minuspg option for profiling butremove it in production runs

Use the minusO3 option for faster executable

Improvements and criticism in the discussion

A small Note on the Langevin and the Fokker-PlanckEquations

Consider a particle suspended in a liquid On its path through the liquid it willcontinuously collide with the liquid molecules Because on average the particle willcollide more often on the front side than on the back side it will experience asystematic force proportional with its velocity and directed opposite to its velocityBesides this systematic force the particle will experience a stochastic force ~F (t) Theequations of motion then read

dt= ~v

dt= minusξ~v + ~F

From hydrodynamics we know that the friction constant ξ is given by

ξ = 6πηam

where η is the viscosity of the solvent and a is the radius of the particle You will seequantities like these in Eirikrsquospart of the course

Solving the second equation in the previous slide we get

~v(t) = ~v0eminusξt +

0dτeminusξ(tminusτ)~F (τ)

If we want to get some useful information out of this we have to average over allpossible realizations of ~F (t) with the initial velocity as a condition A useful quantity forexample is

〈~v(t) middot ~v(t)〉~v0= vminusξ2t

0 + 2Z t

0dτeminusξ(2tminusτ)~v0 middot 〈~F (τ)〉~v0

0dτ prime

0dτeminusξ(2tminusτminusτ prime)〈~F (τ) middot ~F (τ prime)〉~v0

In order to continue we have to make some assumptions about the conditionalaverages of the stochastic forces In view of the chaotic character of the stochasticforces the following assumptions seem to be appropriate

〈~F (t)〉 = 0

〈~F (t) middot ~F (t prime)〉~v0= C~v0

δ(t minus t prime)

We omit the subscript ~v0 when the quantity of interest turns out to be independent of~v0 Using the last three equations we get

〈~v(t) middot ~v(t)〉~v0= v2

0 eminus2ξt +C~v0

2ξ(1minus eminus2ξt )

For large t this should be equal to 3kTm from which it follows that

〈~F (t) middot ~F (t prime)〉 = 6kT

mξδ(t minus t prime)

This result is called the fluctuation-dissipation theorem Integrating

we get

~r(t) = ~r0 + ~v01

ξ(1minus eminusξt ) +

0dτZ τ

0τ primeeminusξ(τminusτ prime)~F (τ prime)

from which we calculate the mean square displacement

〈(~r(t)minus~r0)2〉~v0

ξ(1minus eminusξt )2 +

mξ2(2ξt minus 3 + 4eminusξt minus eminus2ξt )

For very large t this becomes

〈(~r(t)minus~r0)2〉 =

from which we get the Einstein relation

where we have used 〈(~r(t)minus~r0)2〉 = 6Dt

Introduction Parallelization MPI and Hartree-Fock theory

Thursday

Introduction to Message Passing Interface (MPI) andparallelization

How to use the local cluster and the Titan cluster

Presentation of topics to be covered in part I (MHJ)

Friday

Hartree-Fock theory and presentation of project 1

Thijssenrsquos text

Chapters 3-6 and 12

MPI text

Using MPI

Chapters 1-5

Rothman and Zaleski

Quantum Monte Carlo

wherer1 rA

Quantum Monte Carlo

There are

2A timesbdquo

MPI_Command_name

MPI_COMMAND_NAME

processor speed

memory sizespeed

Speed-up

S(P) =T (1)

Efficiency

η(P) =S(P)

tC(L) = τ + βL

T (P) ge T (1)

Prarr S(P) le P

MPI_Command_name

MPI_COMMAND_NAME

Communicator

The Fortran program

END PROGRAM hello

Note 1

Note 2

Note 3

100 values

Note 4

serious work

Integrating π

Examples

Go to MPI examples

Integration algos

af (x)dx asymp h

f (ximinus12)

example5cpp

mpirun -np 10 progx

MPI reduce

Call as

end of main program

function MPI Wtime

Strategies

Thursday

Friday

minus1

dr2unl (r) +

bdquol(l + 1)

2r2minus

r+ Φ(r)

problem

minus1

dr2unl (r) +

bdquol(l + 1)

and rewrite as

minus1

k2(r) = V (r) +l(l + 1)

2r2minus enl

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

V (ri ) = minusZri

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

programs catalogue

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

r+ Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr+Z infin

0unl (r)Fnl (r)dr

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

2Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr + EFock

EFock =1

Z infin

0unl (r)Fnl (r)dr

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

dx2y(x) + k2(x)y(x) = F (x)

h2+ O(h2)

h2+ O(h2) + k2

i yi = Fi

2minus h2k2i

rdquoyi + h2Fi

Numerovrsquos algo

y(x + h) = y(x) + hy (1)(x) +h2

2y (2)(x) +

3y (3)(x) +

Numerovrsquos algo

12y (4)(x) + O(h6)

We obtain

h2minus

12y (4)(x) + O(h4)

Numerovrsquos algo

y (2)(x) +h2

12y (4)(x) + k2(x)y(x) +

ldquok2(x)y(x)

rdquoasymp 0

Numerovrsquos algo

d2(k2y(x))

acuteh2

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

+ O(h6)

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

minus1

dr2u(r) +

bdquoV (r) +

l(l + 1)

laquou(r) = eu(r)

dr2u(r) + k2(r)u(r) = 0

k2(r) = γ

bdquof (r)minus

l(l + 1)

r2minus ε

laquoγ = 2V0

ε =|E |V0

f (r) =

forr le ar gt a

k2(r) = γ

(1minus εminus 1

γl(l+1)

l(l+1)

forr le ar gt a

dr2u(r)minus

l(l + 1)

r2u(r) = 0

well defined

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

drult(r) =

drugt(r) at r = rm

drult(rm) asymp

drugt(rm) asymp

Numerovcpp

wf[loop] = norm

Todayrsquos tasks

2reminusr

u(r) =

bdquo1

1minus a2

raquo1 +

2(1minus a2)r

ndashlaquo

Thursday

Friday

u(x) = c1 +

0uprime(y)dy

and a constant c2

uprime(y) = c2 +

0uprimeprime(z)dz

0f (z)dz

bdquoZ y

0f (z)dz

laquody

and defining

F (y) =

0f (z)dz

0F (y)dy =

bdquoZ y

0f (z)dz

laquody =

0(x minus y)f (y)dy

This gives us

0(x minus y)f (y)dy

0(1minus y)f (y)dy

u(x) = xZ 1

0(x minus y)f (y)dy

G(x y) =

u(x) =

0G(x y)f (y)dy

1 continuous

3 G(0 y) = G(1 y) = G(x 0) = G(x 1) = 0

chapter 165

dx|x=y+ε minus

dx|x=yminusε = 1

W =dugt

dxult minus

and usingdG

dx|x=y+ε minus

dx|x=yminusε = 1

Simple example

ult(x) = x

G(x y) =

u(x) = λ

0G(x y)u(t)dy

u(r) =

Z infin

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

u(r) =

Z infin

resulting in

u(r) = ugt(r)Z r

Z infin

Algorithm

dxasymp

Algorithm

Thursday

Friday

E [Φ] =

af (Φ(x)

partΦ

partx x)dx

η(a) = η(b) = 0

Φ(x α) = Φ(x 0) + αη(x)

E [Φ(α)] =

af (Φ(x α)

partΦ(x α)

partx x)dx

E [Φ(α)] =

ndashα=0

partα

ndash=

bdquopartf

partΦ

partα+

partΦx

partα

laquodx

partα= η(x)

partα=

d(η(x))

UsingpartΦ(x α)

partα= η(x)

andpartΦx (x α)

partα=

d(η(x))

partα

ndash=

bdquopartf

partΦη(x) +

partΦx

d(η(x))

laquodx

partΦx

d(η(x))

dxdx = η(x)

partΦx|ba minus

aη(x)

partΦxdx

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndashα=0

bdquopartf

partΦminus

partΦx

partΦminus

partΦx= 0

df = 0

partx=partf

party=partf

partz= 0

due to

df =partf

partxdx +

partydy +

partzdz

φ(x y z) = 0

resulting in

dφ =partφ

partxdx +

partφ

partydy +

partφ

partzdz = 0

partx=partf

party=partf

partz= 0

df =partf

partxdx +

partydy +

partzdz

df + λdφ = (partf

partz+ λ

partφ

partx)dx + (

party+ λ

partφ

party)dy + (

partz+ λ

partφ

partz)dz = 0

partz+ λ

partφ

partz= 0

partx+ λ

partφ

partx= 0

andpartf

party+ λ

partφ

party= 0

partxi+X

λkpartφk

partxi= 0

and a Hamiltonian

H = minus1

2nabla2 + V (x y z)

Zdxdydz

δE = δ

Zdxdydz

bdquo1

laquoff= 0

Zdxdydz

bdquo1

laquoff= 0

partψlowastminus

partψlowastxminus

partψlowastyminus

partψlowastz= 0

which yields

minus1

H = minusNX

2nabla2

i minusNX

NXiltj

H = H1 + H2 =NX

hi +NX

iltj=1

hi = minus1

2nabla2

i minusZ

E0 le E [Φ] =

ZΦlowastHΦdτ

radicN

(minus)PP

[H1A] = [H2A] = 0

ZΦlowastH1Φdτ =

(minus)PZ

ZΦlowastH1Φdτ =

ZΦlowastH hiΦHdτ

ZΦlowastH1Φdτ =

NXmicro=1

〈micro|h|micro〉

ΦlowastH2Φdτ = N

which reduces to

ZΦlowastH2Φdτ =

NXilej=1

(minus)PZ

ΦlowastH1

rijPΦHdτ

ZΦlowastH2Φdτ =

NXiltj=1

ZΦlowastH

ZΦlowastH2Φdτ =

NXmicro=1

NXν=1

lowastν(r j )

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

lowastν(r j )

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

ψa =Xλ

Caλψλ

lectures

Exercise

Caλψλ

δE minusNX

micro=1

NXν=1

εmicroνδ

δE minusNX

micro=1

εmicroδ

NXmicro=1

NXν=1

Zδψlowastmicroψ

lowastν

Zδψlowastmicroψ

lowastν

micro=1

NXmicro=1

NXν=1

Zψlowastmicroψ

lowastν

Zψlowastmicroψ

lowastν

minusNX

micro=1

Emicro

NXmicro=1

Emicro

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

rijψν(r j )dr j

ψmicro(r i )

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

V dmicro(r i ) =

2(2l prime + 1)

Z infin

rgtdrj

withφnlml

Z infin

(rj )|21

rijdr j

which reads

Z infin

rijdrj

2l prime + 1

resulting in

2(2l prime + 1)

Z infin

rijdrj dΩj

infinXl=0

bdquorj

laquol

Pl cos(θ))

if ri gt rj or

infinXl=0

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Ylowastlmprimel

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

(2g + 1)

timesg

programs

Thursday

Friday

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

2φ100(r1)φ100(r2)

minus1radic

The Hartree term

V dmicro(r i ) =

ψλ(r i )

NXmicro=1

V dmicro(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψλ(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψ100uarr(r i )

Φ(ri )u10 = 2V d10(ri )u10(ri ) = 2

Z infin

0|u10(rj )|2

rgtdrj )u10(ri )

rijψλ(r j )dr j

ψmicro(r i )

rijψλ(r j )dr j

ψmicro(r i )

s =darr That is

s mλs

rijψλ(r j )dr j

ψmicro(r i )

NXmicro=1

s uarr

ψmicro(r i )

resulting in

NXmicro=1

ψ100uarr(r i )

NXmicro=1

ψ100uarr(r i )

minus1

l(l + 1)

2r2minus

as bdquominus

l(l + 1)

2r2minus

r+ 2V d

10(r)laquo

or bdquominus

dr2minus

r+ V d

10(r)laquo

u10(r) = e10u10(r)

dangerous later

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ100uarr(r i )

= (2V d10(r i ) + 2V d

20(r i ))ψ100uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

V ex20 (r) =

Z infin

0dr jφ

lowast200(r j )

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ200uarr(r i ) =

(2V d10(r i ) + 2V d

20(r i ))ψ200uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

V ex10 (r) =

Z infin

0dr jφ

lowast100(r j )

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

dr2minus

r+ V d

10(r) + 2V d20(r)+

20 (r)+ = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)+

10 (r)+ = e20u20(r)

andbdquominus

2r2minus

r+ 2V d

10(r) + 2V d20(r)+

10 (r)minusV ex20 (r)+ = e21u21(r)

Useful Tests

R10(r) = 2bdquo

laquo32

R20(r) = 2bdquo

laquo32 bdquo1minus

R21(r) =1radic

bdquoZ

laquo32 Zr

We use a0 = 1

R10(r) = 2bdquo

laquo32

E [Z ] = Z 2 minus 4Z +5

Z = 2minus5

16= 16875

we do this

E [ρ] = minus1

1Ψi (r1)dr1minusZ

r1ρ(r1)dr1+

Zρ(r1)ρ(r2)

r12dr1dr2+EEXC [ρ]

ρ(r) =NX

|Ψi (r)|2

2nabla2

1 minusZ

Zρ(r2)

r12dr2 + VEXC(r1)

EEXC =

Thursday

Friday

A simple Example

wi (t = ε) = Wij wj (t = 0)

wi (t) = 1

and Xj

W (j rarr i) = 1

wi (t + 1) =X

Wij wj (t)

0 14 18 2334 58 00 14 13

w(t = 0) =

resulting in

w(t = ε) =

0 14340

resulting in

w(t = 2ε) =

0 5232132632

wi (t = infin) = 1

w(t = infin) =

0 415815315

Iteration w1 w2 w30 100000 000000 0000001 025000 075000 0000002 015625 062625 0187503 024609 052734 0226564 027848 051416 0207365 027213 053021 0197666 026608 053548 0198447 026575 053424 0200028 026656 053321 0200239 026678 053318 020005

10 026671 053332 01999811 026666 053335 02000012 026666 053334 02000013 026667 053333 020000

w(t = infin) 026667 053333 020000

w(0) =X

αi v i

αi v i =X

λtiαi v i

〈M(t)〉 =Xmicro

w(t)microMmicro

〈M(t)〉 = w(t)m

E [xk ] = 〈xk 〉 =1

xki p(xi )

〈M(t)〉 = w(t)m =X

λtiαi v i m i

〈M(t)〉 =X

λtiαi mi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

λtiαi mi

τi = minus1

logλi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

αi mi eminustτi

W (j rarr i)wj =X

W (i rarr j)wi

W (j rarr i)

W (i rarr j)=

Proof Xi

W (i rarr j)wi =X

W (j rarr i)wj = wj

W (j rarr i) = wj

Ergodicity

exp (minusβ(Ei ))

〈A〉 =X

Aj wj =

g(i rarr j) =1

A(j rarr i)

A(microrarr ν)

Then we must have

Take a new value

x (i+1) =

We have defined

ρ(x y) = min

p(y)f (x |y)

p(x)f (y |x) 1ff

Implementation

Quantum Monte Carlo

E [H] = 〈H〉 =

E0 le 〈H〉

Quantum Monte Carlo

ΨT (R) =X

aiΨi (R)

nm alowastmanR

nEnPn a2

nge E0

Quantum Monte Carlo

E [H] = 〈H〉 =

RdRΨlowastTα

(R)H(R)ΨTα(R)R

Quantum Monte Carlo

P(R) =|ψT (R)|2R|ψT (R)|2 dR

E [H] asympR

EL(R) =1

ψT (R)HψT (R)

P(R)EL(R)dR asymp1

P(Ri)EL(Ri)

Quantum Monte Carlo

˛ψα

T (R)˛2

w = P(Rp)P(R)

Quantum Monte Carlo

minus~2

part2u(r)

partr2minusbdquo

rminus

~2l(l + 1)

laquou(r) = Eu(r)

minus1

part2u(ρ)

partρ2minus

l(l + 1)

H = minus1

partρ2minus

l(l + 1)

Quantum Monte Carlo

EL(ρ) = minus1

ρminusα

bdquoαminus

Quantum Monte Carlo

〈Hn〉 =

= constant

Quantum Monte Carlo

minus2ke2

r1minus

Quantum Monte Carlo

bH = minus~2nabla2

2mminus

~2nabla22

2mminus

r1minus

P(R) =|ψT (R)|2R|ψT (R)|2 dR

Quantum Monte Carlo

EL(R) =1

ψT (R)HψT (R) =

ψT (R)

bdquominus

2nabla2

1 minusZ

EL(R) =1

RT (r1)

minus1

dr1minus

limr1rarr0

EL(R) =1

RT (r1)

bdquominus

dr1minus

laquoRT (r1)

Quantum Monte Carlo

This results in1

RT (r1)

dRT (r1)

dr1= minusZ

RT (r)

dRT (r)

dr= minus

f (rij )

wfold = wfnew

Thursday

Importance sampling

Friday

Things to do

Splines

s(x) =

Splines

yi = s(xi )

ands(xi+1) = yi+1

Splines

si (x) =fi

Splines

si (x) =fi

fi+16(xi+1minusxi )

(x minus xi )3

+ (yi+1

6 )(xi+1 minus x) (2)

Splines

(yi minus yiminus1)

Splines

3777775

2666664f1f2f3

fnminus2fnminus1

3777775 =

2666664v1v2v3

vnminus2vnminus1

3777775

operations

Splines

partt= D

bdquopartP

partxminus F

laquoP(x t)

F = 21

ΨTnablaΨT

partx(t)

y = x + DF (x)∆t + ξ

Importance Sampling

G(y x ∆t) =1

(4πD∆t)3N2exp

A(y x) = min(1 q(y x)))

q(y x) =G(x y ∆t)|ΨT (y)|2

G(y x ∆t)|ΨT (x)|2

Thursday

Friday

x = rsinθcosφ

y = rsinθsinφ

andz = rcosθ

4πε0r=

4πε0p

x2 + y2 + z2

4π(l + ml )Pml

for l = ml = 0

4πcos(θ)

for l = 1 og ml = 0

Y1plusmn1 =

(r) through

Ylml(r) = r l Ylml

we obtainΨnlml

equation

Slminusml

1radic

(minus1)m

minus(minus1)ml i i

Clminusml

We have definedYlml

(r) = r l Ylml(r)

2l + 1Yl0(r)

Slml= (minus1)ml

2l + 1ReYl0(r)

Slml= (minus1)ml

2l + 1ImYl0(r)

for ml gt 0

mll 0 1 2 3

q52 (x2 minus 3y2)x

+1 xradic

3xz 12

q32 (5z2 minus r2)x

0 1 y 12 (3z2 minus r2) 1

2 (5z2 minus 3r2)x

-1 zradic

3yz 12

q32 (5z2 minus r2)y

-2radic

3xyradic

q52 (3x2 minus y2)y

Thursday

Det uarr (1 2) =

˛1radic

Det darr (3 4) =

˛1radic

Ψα(Xnew )

Ψα(Xold )=

Duarrα(Xnew )

Duarrα(Xold )

Ddarrα(Xnew )

Ddarrα(Xold )

Gβ(Xnew )

Gβ(Xold )

nablaΨα

Ψα=nablaDuarrα

Ddarrα+nablaGβ

nabla2Ψα

Ψα=nabla2Duarrα

Ddarrα+nabla2Gβ

Gβ+ (

nablaGβ

Gβ)(nablaDuarrα

Ddarrα)

Dminus1ij =

Dα(y)

Dα(x)=

Dij (y)Cij (y)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

Dij (y)Cij (x)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

nablai Dα(x)

Dα(x)=

nabla2i Dα(x)

Dα(x)=

nablai Dα(y)

Dα(y)=

andnabla2

i Dα(y)

Dα(y)=

R equivNX

Dminus1kj (y) =

kj (x)minus1

RDminus1

ki (x)NX

RDminus1

ki (x) for j = i

R =D(y)

LU Decomposition

1CCA =

0BB1 0 0 0

b21 1 0 0b31 b32 1 0b41 b42 b43 1

1CCA0BB

c11 c12 c13 c140 c22 c23 c240 0 c33 c340 0 0 c44

detA = c11c22 cnn

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

nablaGβ

Gβ=Xiltj

(1 + βrij )2

nabla2Gβ

Gβ=Xiltj

(1 + βrij )4

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

a = 14

a = 12

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

Things to do

dt= ~v

dt= minusξ~v + ~F

ξ = 6πηam

0 + 2Z t

0dτ prime

〈~F (t)〉 = 0

δ(t minus t prime)

0 eminus2ξt +C~v0

we get

~r(t) = ~r0 + ~v01

0dτZ τ

Thijssenrsquos text

Chapters 3-6 and 12

MPI text

Using MPI

Chapters 1-5

Rothman and Zaleski

Quantum Monte Carlo

wherer1 rA

Quantum Monte Carlo

There are

2A timesbdquo

MPI_Command_name

MPI_COMMAND_NAME

processor speed

memory sizespeed

Speed-up

S(P) =T (1)

Efficiency

η(P) =S(P)

tC(L) = τ + βL

T (P) ge T (1)

Prarr S(P) le P

MPI_Command_name

MPI_COMMAND_NAME

Communicator

The Fortran program

END PROGRAM hello

Note 1

Note 2

Note 3

100 values

Note 4

serious work

Integrating π

Examples

Go to MPI examples

Integration algos

af (x)dx asymp h

f (ximinus12)

example5cpp

mpirun -np 10 progx

MPI reduce

Call as

end of main program

function MPI Wtime

Strategies

Thursday

Friday

minus1

dr2unl (r) +

bdquol(l + 1)

2r2minus

r+ Φ(r)

problem

minus1

dr2unl (r) +

bdquol(l + 1)

and rewrite as

minus1

k2(r) = V (r) +l(l + 1)

2r2minus enl

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

V (ri ) = minusZri

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

programs catalogue

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

r+ Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr+Z infin

0unl (r)Fnl (r)dr

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

2Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr + EFock

EFock =1

Z infin

0unl (r)Fnl (r)dr

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

dx2y(x) + k2(x)y(x) = F (x)

h2+ O(h2)

h2+ O(h2) + k2

i yi = Fi

2minus h2k2i

rdquoyi + h2Fi

Numerovrsquos algo

y(x + h) = y(x) + hy (1)(x) +h2

2y (2)(x) +

3y (3)(x) +

Numerovrsquos algo

12y (4)(x) + O(h6)

We obtain

h2minus

12y (4)(x) + O(h4)

Numerovrsquos algo

y (2)(x) +h2

12y (4)(x) + k2(x)y(x) +

ldquok2(x)y(x)

rdquoasymp 0

Numerovrsquos algo

d2(k2y(x))

acuteh2

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

+ O(h6)

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

minus1

dr2u(r) +

bdquoV (r) +

l(l + 1)

laquou(r) = eu(r)

dr2u(r) + k2(r)u(r) = 0

k2(r) = γ

bdquof (r)minus

l(l + 1)

r2minus ε

laquoγ = 2V0

ε =|E |V0

f (r) =

forr le ar gt a

k2(r) = γ

(1minus εminus 1

γl(l+1)

l(l+1)

forr le ar gt a

dr2u(r)minus

l(l + 1)

r2u(r) = 0

well defined

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

drult(r) =

drugt(r) at r = rm

drult(rm) asymp

drugt(rm) asymp

Numerovcpp

wf[loop] = norm

Todayrsquos tasks

2reminusr

u(r) =

bdquo1

1minus a2

raquo1 +

2(1minus a2)r

ndashlaquo

Thursday

Friday

u(x) = c1 +

0uprime(y)dy

and a constant c2

uprime(y) = c2 +

0uprimeprime(z)dz

0f (z)dz

bdquoZ y

0f (z)dz

laquody

and defining

F (y) =

0f (z)dz

0F (y)dy =

bdquoZ y

0f (z)dz

laquody =

0(x minus y)f (y)dy

This gives us

0(x minus y)f (y)dy

0(1minus y)f (y)dy

u(x) = xZ 1

0(x minus y)f (y)dy

G(x y) =

u(x) =

0G(x y)f (y)dy

1 continuous

3 G(0 y) = G(1 y) = G(x 0) = G(x 1) = 0

chapter 165

dx|x=y+ε minus

dx|x=yminusε = 1

W =dugt

dxult minus

and usingdG

dx|x=y+ε minus

dx|x=yminusε = 1

Simple example

ult(x) = x

G(x y) =

u(x) = λ

0G(x y)u(t)dy

u(r) =

Z infin

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

u(r) =

Z infin

resulting in

u(r) = ugt(r)Z r

Z infin

Algorithm

dxasymp

Algorithm

Thursday

Friday

E [Φ] =

af (Φ(x)

partΦ

partx x)dx

η(a) = η(b) = 0

Φ(x α) = Φ(x 0) + αη(x)

E [Φ(α)] =

af (Φ(x α)

partΦ(x α)

partx x)dx

E [Φ(α)] =

ndashα=0

partα

ndash=

bdquopartf

partΦ

partα+

partΦx

partα

laquodx

partα= η(x)

partα=

d(η(x))

UsingpartΦ(x α)

partα= η(x)

andpartΦx (x α)

partα=

d(η(x))

partα

ndash=

bdquopartf

partΦη(x) +

partΦx

d(η(x))

laquodx

partΦx

d(η(x))

dxdx = η(x)

partΦx|ba minus

aη(x)

partΦxdx

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndashα=0

bdquopartf

partΦminus

partΦx

partΦminus

partΦx= 0

df = 0

partx=partf

party=partf

partz= 0

due to

df =partf

partxdx +

partydy +

partzdz

φ(x y z) = 0

resulting in

dφ =partφ

partxdx +

partφ

partydy +

partφ

partzdz = 0

partx=partf

party=partf

partz= 0

df =partf

partxdx +

partydy +

partzdz

df + λdφ = (partf

partz+ λ

partφ

partx)dx + (

party+ λ

partφ

party)dy + (

partz+ λ

partφ

partz)dz = 0

partz+ λ

partφ

partz= 0

partx+ λ

partφ

partx= 0

andpartf

party+ λ

partφ

party= 0

partxi+X

λkpartφk

partxi= 0

and a Hamiltonian

H = minus1

2nabla2 + V (x y z)

Zdxdydz

δE = δ

Zdxdydz

bdquo1

laquoff= 0

Zdxdydz

bdquo1

laquoff= 0

partψlowastminus

partψlowastxminus

partψlowastyminus

partψlowastz= 0

which yields

minus1

H = minusNX

2nabla2

i minusNX

NXiltj

H = H1 + H2 =NX

hi +NX

iltj=1

hi = minus1

2nabla2

i minusZ

E0 le E [Φ] =

ZΦlowastHΦdτ

radicN

(minus)PP

[H1A] = [H2A] = 0

ZΦlowastH1Φdτ =

(minus)PZ

ZΦlowastH1Φdτ =

ZΦlowastH hiΦHdτ

ZΦlowastH1Φdτ =

NXmicro=1

〈micro|h|micro〉

ΦlowastH2Φdτ = N

which reduces to

ZΦlowastH2Φdτ =

NXilej=1

(minus)PZ

ΦlowastH1

rijPΦHdτ

ZΦlowastH2Φdτ =

NXiltj=1

ZΦlowastH

ZΦlowastH2Φdτ =

NXmicro=1

NXν=1

lowastν(r j )

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

lowastν(r j )

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

ψa =Xλ

Caλψλ

lectures

Exercise

Caλψλ

δE minusNX

micro=1

NXν=1

εmicroνδ

δE minusNX

micro=1

εmicroδ

NXmicro=1

NXν=1

Zδψlowastmicroψ

lowastν

Zδψlowastmicroψ

lowastν

micro=1

NXmicro=1

NXν=1

Zψlowastmicroψ

lowastν

Zψlowastmicroψ

lowastν

minusNX

micro=1

Emicro

NXmicro=1

Emicro

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

rijψν(r j )dr j

ψmicro(r i )

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

V dmicro(r i ) =

2(2l prime + 1)

Z infin

rgtdrj

withφnlml

Z infin

(rj )|21

rijdr j

which reads

Z infin

rijdrj

2l prime + 1

resulting in

2(2l prime + 1)

Z infin

rijdrj dΩj

infinXl=0

bdquorj

laquol

Pl cos(θ))

if ri gt rj or

infinXl=0

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Ylowastlmprimel

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

(2g + 1)

timesg

programs

Thursday

Friday

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

2φ100(r1)φ100(r2)

minus1radic

The Hartree term

V dmicro(r i ) =

ψλ(r i )

NXmicro=1

V dmicro(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψλ(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψ100uarr(r i )

Φ(ri )u10 = 2V d10(ri )u10(ri ) = 2

Z infin

0|u10(rj )|2

rgtdrj )u10(ri )

rijψλ(r j )dr j

ψmicro(r i )

rijψλ(r j )dr j

ψmicro(r i )

s =darr That is

s mλs

rijψλ(r j )dr j

ψmicro(r i )

NXmicro=1

s uarr

ψmicro(r i )

resulting in

NXmicro=1

ψ100uarr(r i )

NXmicro=1

ψ100uarr(r i )

minus1

l(l + 1)

2r2minus

as bdquominus

l(l + 1)

2r2minus

r+ 2V d

10(r)laquo

or bdquominus

dr2minus

r+ V d

10(r)laquo

u10(r) = e10u10(r)

dangerous later

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ100uarr(r i )

= (2V d10(r i ) + 2V d

20(r i ))ψ100uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

V ex20 (r) =

Z infin

0dr jφ

lowast200(r j )

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ200uarr(r i ) =

(2V d10(r i ) + 2V d

20(r i ))ψ200uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

V ex10 (r) =

Z infin

0dr jφ

lowast100(r j )

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

dr2minus

r+ V d

10(r) + 2V d20(r)+

20 (r)+ = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)+

10 (r)+ = e20u20(r)

andbdquominus

2r2minus

r+ 2V d

10(r) + 2V d20(r)+

10 (r)minusV ex20 (r)+ = e21u21(r)

Useful Tests

R10(r) = 2bdquo

laquo32

R20(r) = 2bdquo

laquo32 bdquo1minus

R21(r) =1radic

bdquoZ

laquo32 Zr

We use a0 = 1

R10(r) = 2bdquo

laquo32

E [Z ] = Z 2 minus 4Z +5

Z = 2minus5

16= 16875

we do this

E [ρ] = minus1

1Ψi (r1)dr1minusZ

r1ρ(r1)dr1+

Zρ(r1)ρ(r2)

r12dr1dr2+EEXC [ρ]

ρ(r) =NX

|Ψi (r)|2

2nabla2

1 minusZ

Zρ(r2)

r12dr2 + VEXC(r1)

EEXC =

Thursday

Friday

A simple Example

wi (t = ε) = Wij wj (t = 0)

wi (t) = 1

and Xj

W (j rarr i) = 1

wi (t + 1) =X

Wij wj (t)

0 14 18 2334 58 00 14 13

w(t = 0) =

resulting in

w(t = ε) =

0 14340

resulting in

w(t = 2ε) =

0 5232132632

wi (t = infin) = 1

w(t = infin) =

0 415815315

Iteration w1 w2 w30 100000 000000 0000001 025000 075000 0000002 015625 062625 0187503 024609 052734 0226564 027848 051416 0207365 027213 053021 0197666 026608 053548 0198447 026575 053424 0200028 026656 053321 0200239 026678 053318 020005

10 026671 053332 01999811 026666 053335 02000012 026666 053334 02000013 026667 053333 020000

w(t = infin) 026667 053333 020000

w(0) =X

αi v i

αi v i =X

λtiαi v i

〈M(t)〉 =Xmicro

w(t)microMmicro

〈M(t)〉 = w(t)m

E [xk ] = 〈xk 〉 =1

xki p(xi )

〈M(t)〉 = w(t)m =X

λtiαi v i m i

〈M(t)〉 =X

λtiαi mi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

λtiαi mi

τi = minus1

logλi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

αi mi eminustτi

W (j rarr i)wj =X

W (i rarr j)wi

W (j rarr i)

W (i rarr j)=

Proof Xi

W (i rarr j)wi =X

W (j rarr i)wj = wj

W (j rarr i) = wj

Ergodicity

exp (minusβ(Ei ))

〈A〉 =X

Aj wj =

g(i rarr j) =1

A(j rarr i)

A(microrarr ν)

Then we must have

Take a new value

x (i+1) =

We have defined

ρ(x y) = min

p(y)f (x |y)

p(x)f (y |x) 1ff

Implementation

Quantum Monte Carlo

E [H] = 〈H〉 =

E0 le 〈H〉

Quantum Monte Carlo

ΨT (R) =X

aiΨi (R)

nm alowastmanR

nEnPn a2

nge E0

Quantum Monte Carlo

E [H] = 〈H〉 =

RdRΨlowastTα

(R)H(R)ΨTα(R)R

Quantum Monte Carlo

P(R) =|ψT (R)|2R|ψT (R)|2 dR

E [H] asympR

EL(R) =1

ψT (R)HψT (R)

P(R)EL(R)dR asymp1

P(Ri)EL(Ri)

Quantum Monte Carlo

˛ψα

T (R)˛2

w = P(Rp)P(R)

Quantum Monte Carlo

minus~2

part2u(r)

partr2minusbdquo

rminus

~2l(l + 1)

laquou(r) = Eu(r)

minus1

part2u(ρ)

partρ2minus

l(l + 1)

H = minus1

partρ2minus

l(l + 1)

Quantum Monte Carlo

EL(ρ) = minus1

ρminusα

bdquoαminus

Quantum Monte Carlo

〈Hn〉 =

= constant

Quantum Monte Carlo

minus2ke2

r1minus

Quantum Monte Carlo

bH = minus~2nabla2

2mminus

~2nabla22

2mminus

r1minus

P(R) =|ψT (R)|2R|ψT (R)|2 dR

Quantum Monte Carlo

EL(R) =1

ψT (R)HψT (R) =

ψT (R)

bdquominus

2nabla2

1 minusZ

EL(R) =1

RT (r1)

minus1

dr1minus

limr1rarr0

EL(R) =1

RT (r1)

bdquominus

dr1minus

laquoRT (r1)

Quantum Monte Carlo

This results in1

RT (r1)

dRT (r1)

dr1= minusZ

RT (r)

dRT (r)

dr= minus

f (rij )

wfold = wfnew

Thursday

Importance sampling

Friday

Things to do

Splines

s(x) =

Splines

yi = s(xi )

ands(xi+1) = yi+1

Splines

si (x) =fi

Splines

si (x) =fi

fi+16(xi+1minusxi )

(x minus xi )3

+ (yi+1

6 )(xi+1 minus x) (2)

Splines

(yi minus yiminus1)

Splines

3777775

2666664f1f2f3

fnminus2fnminus1

3777775 =

2666664v1v2v3

vnminus2vnminus1

3777775

operations

Splines

partt= D

bdquopartP

partxminus F

laquoP(x t)

F = 21

ΨTnablaΨT

partx(t)

y = x + DF (x)∆t + ξ

Importance Sampling

G(y x ∆t) =1

(4πD∆t)3N2exp

A(y x) = min(1 q(y x)))

q(y x) =G(x y ∆t)|ΨT (y)|2

G(y x ∆t)|ΨT (x)|2

Thursday

Friday

x = rsinθcosφ

y = rsinθsinφ

andz = rcosθ

4πε0r=

4πε0p

x2 + y2 + z2

4π(l + ml )Pml

for l = ml = 0

4πcos(θ)

for l = 1 og ml = 0

Y1plusmn1 =

(r) through

Ylml(r) = r l Ylml

we obtainΨnlml

equation

Slminusml

1radic

(minus1)m

minus(minus1)ml i i

Clminusml

We have definedYlml

(r) = r l Ylml(r)

2l + 1Yl0(r)

Slml= (minus1)ml

2l + 1ReYl0(r)

Slml= (minus1)ml

2l + 1ImYl0(r)

for ml gt 0

mll 0 1 2 3

q52 (x2 minus 3y2)x

+1 xradic

3xz 12

q32 (5z2 minus r2)x

0 1 y 12 (3z2 minus r2) 1

2 (5z2 minus 3r2)x

-1 zradic

3yz 12

q32 (5z2 minus r2)y

-2radic

3xyradic

q52 (3x2 minus y2)y

Thursday

Det uarr (1 2) =

˛1radic

Det darr (3 4) =

˛1radic

Ψα(Xnew )

Ψα(Xold )=

Duarrα(Xnew )

Duarrα(Xold )

Ddarrα(Xnew )

Ddarrα(Xold )

Gβ(Xnew )

Gβ(Xold )

nablaΨα

Ψα=nablaDuarrα

Ddarrα+nablaGβ

nabla2Ψα

Ψα=nabla2Duarrα

Ddarrα+nabla2Gβ

Gβ+ (

nablaGβ

Gβ)(nablaDuarrα

Ddarrα)

Dminus1ij =

Dα(y)

Dα(x)=

Dij (y)Cij (y)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

Dij (y)Cij (x)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

nablai Dα(x)

Dα(x)=

nabla2i Dα(x)

Dα(x)=

nablai Dα(y)

Dα(y)=

andnabla2

i Dα(y)

Dα(y)=

R equivNX

Dminus1kj (y) =

kj (x)minus1

RDminus1

ki (x)NX

RDminus1

ki (x) for j = i

R =D(y)

LU Decomposition

1CCA =

0BB1 0 0 0

b21 1 0 0b31 b32 1 0b41 b42 b43 1

1CCA0BB

c11 c12 c13 c140 c22 c23 c240 0 c33 c340 0 0 c44

detA = c11c22 cnn

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

nablaGβ

Gβ=Xiltj

(1 + βrij )2

nabla2Gβ

Gβ=Xiltj

(1 + βrij )4

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

a = 14

a = 12

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

Things to do

dt= ~v

dt= minusξ~v + ~F

ξ = 6πηam

0 + 2Z t

0dτ prime

〈~F (t)〉 = 0

δ(t minus t prime)

0 eminus2ξt +C~v0

we get

~r(t) = ~r0 + ~v01

0dτZ τ

Thijssenrsquos text

Chapters 3-6 and 12

MPI text

Using MPI

Chapters 1-5

Rothman and Zaleski

Quantum Monte Carlo

wherer1 rA

Quantum Monte Carlo

There are

2A timesbdquo

MPI_Command_name

MPI_COMMAND_NAME

processor speed

memory sizespeed

Speed-up

S(P) =T (1)

Efficiency

η(P) =S(P)

tC(L) = τ + βL

T (P) ge T (1)

Prarr S(P) le P

MPI_Command_name

MPI_COMMAND_NAME

Communicator

The Fortran program

END PROGRAM hello

Note 1

Note 2

Note 3

100 values

Note 4

serious work

Integrating π

Examples

Go to MPI examples

Integration algos

af (x)dx asymp h

f (ximinus12)

example5cpp

mpirun -np 10 progx

MPI reduce

Call as

end of main program

function MPI Wtime

Strategies

Thursday

Friday

minus1

dr2unl (r) +

bdquol(l + 1)

2r2minus

r+ Φ(r)

problem

minus1

dr2unl (r) +

bdquol(l + 1)

and rewrite as

minus1

k2(r) = V (r) +l(l + 1)

2r2minus enl

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

V (ri ) = minusZri

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

programs catalogue

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

r+ Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr+Z infin

0unl (r)Fnl (r)dr

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

2Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr + EFock

EFock =1

Z infin

0unl (r)Fnl (r)dr

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

dx2y(x) + k2(x)y(x) = F (x)

h2+ O(h2)

h2+ O(h2) + k2

i yi = Fi

2minus h2k2i

rdquoyi + h2Fi

Numerovrsquos algo

y(x + h) = y(x) + hy (1)(x) +h2

2y (2)(x) +

3y (3)(x) +

Numerovrsquos algo

12y (4)(x) + O(h6)

We obtain

h2minus

12y (4)(x) + O(h4)

Numerovrsquos algo

y (2)(x) +h2

12y (4)(x) + k2(x)y(x) +

ldquok2(x)y(x)

rdquoasymp 0

Numerovrsquos algo

d2(k2y(x))

acuteh2

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

+ O(h6)

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

minus1

dr2u(r) +

bdquoV (r) +

l(l + 1)

laquou(r) = eu(r)

dr2u(r) + k2(r)u(r) = 0

k2(r) = γ

bdquof (r)minus

l(l + 1)

r2minus ε

laquoγ = 2V0

ε =|E |V0

f (r) =

forr le ar gt a

k2(r) = γ

(1minus εminus 1

γl(l+1)

l(l+1)

forr le ar gt a

dr2u(r)minus

l(l + 1)

r2u(r) = 0

well defined

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

drult(r) =

drugt(r) at r = rm

drult(rm) asymp

drugt(rm) asymp

Numerovcpp

wf[loop] = norm

Todayrsquos tasks

2reminusr

u(r) =

bdquo1

1minus a2

raquo1 +

2(1minus a2)r

ndashlaquo

Thursday

Friday

u(x) = c1 +

0uprime(y)dy

and a constant c2

uprime(y) = c2 +

0uprimeprime(z)dz

0f (z)dz

bdquoZ y

0f (z)dz

laquody

and defining

F (y) =

0f (z)dz

0F (y)dy =

bdquoZ y

0f (z)dz

laquody =

0(x minus y)f (y)dy

This gives us

0(x minus y)f (y)dy

0(1minus y)f (y)dy

u(x) = xZ 1

0(x minus y)f (y)dy

G(x y) =

u(x) =

0G(x y)f (y)dy

1 continuous

3 G(0 y) = G(1 y) = G(x 0) = G(x 1) = 0

chapter 165

dx|x=y+ε minus

dx|x=yminusε = 1

W =dugt

dxult minus

and usingdG

dx|x=y+ε minus

dx|x=yminusε = 1

Simple example

ult(x) = x

G(x y) =

u(x) = λ

0G(x y)u(t)dy

u(r) =

Z infin

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

u(r) =

Z infin

resulting in

u(r) = ugt(r)Z r

Z infin

Algorithm

dxasymp

Algorithm

Thursday

Friday

E [Φ] =

af (Φ(x)

partΦ

partx x)dx

η(a) = η(b) = 0

Φ(x α) = Φ(x 0) + αη(x)

E [Φ(α)] =

af (Φ(x α)

partΦ(x α)

partx x)dx

E [Φ(α)] =

ndashα=0

partα

ndash=

bdquopartf

partΦ

partα+

partΦx

partα

laquodx

partα= η(x)

partα=

d(η(x))

UsingpartΦ(x α)

partα= η(x)

andpartΦx (x α)

partα=

d(η(x))

partα

ndash=

bdquopartf

partΦη(x) +

partΦx

d(η(x))

laquodx

partΦx

d(η(x))

dxdx = η(x)

partΦx|ba minus

aη(x)

partΦxdx

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndashα=0

bdquopartf

partΦminus

partΦx

partΦminus

partΦx= 0

df = 0

partx=partf

party=partf

partz= 0

due to

df =partf

partxdx +

partydy +

partzdz

φ(x y z) = 0

resulting in

dφ =partφ

partxdx +

partφ

partydy +

partφ

partzdz = 0

partx=partf

party=partf

partz= 0

df =partf

partxdx +

partydy +

partzdz

df + λdφ = (partf

partz+ λ

partφ

partx)dx + (

party+ λ

partφ

party)dy + (

partz+ λ

partφ

partz)dz = 0

partz+ λ

partφ

partz= 0

partx+ λ

partφ

partx= 0

andpartf

party+ λ

partφ

party= 0

partxi+X

λkpartφk

partxi= 0

and a Hamiltonian

H = minus1

2nabla2 + V (x y z)

Zdxdydz

δE = δ

Zdxdydz

bdquo1

laquoff= 0

Zdxdydz

bdquo1

laquoff= 0

partψlowastminus

partψlowastxminus

partψlowastyminus

partψlowastz= 0

which yields

minus1

H = minusNX

2nabla2

i minusNX

NXiltj

H = H1 + H2 =NX

hi +NX

iltj=1

hi = minus1

2nabla2

i minusZ

E0 le E [Φ] =

ZΦlowastHΦdτ

radicN

(minus)PP

[H1A] = [H2A] = 0

ZΦlowastH1Φdτ =

(minus)PZ

ZΦlowastH1Φdτ =

ZΦlowastH hiΦHdτ

ZΦlowastH1Φdτ =

NXmicro=1

〈micro|h|micro〉

ΦlowastH2Φdτ = N

which reduces to

ZΦlowastH2Φdτ =

NXilej=1

(minus)PZ

ΦlowastH1

rijPΦHdτ

ZΦlowastH2Φdτ =

NXiltj=1

ZΦlowastH

ZΦlowastH2Φdτ =

NXmicro=1

NXν=1

lowastν(r j )

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

lowastν(r j )

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

ψa =Xλ

Caλψλ

lectures

Exercise

Caλψλ

δE minusNX

micro=1

NXν=1

εmicroνδ

δE minusNX

micro=1

εmicroδ

NXmicro=1

NXν=1

Zδψlowastmicroψ

lowastν

Zδψlowastmicroψ

lowastν

micro=1

NXmicro=1

NXν=1

Zψlowastmicroψ

lowastν

Zψlowastmicroψ

lowastν

minusNX

micro=1

Emicro

NXmicro=1

Emicro

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

rijψν(r j )dr j

ψmicro(r i )

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

V dmicro(r i ) =

2(2l prime + 1)

Z infin

rgtdrj

withφnlml

Z infin

(rj )|21

rijdr j

which reads

Z infin

rijdrj

2l prime + 1

resulting in

2(2l prime + 1)

Z infin

rijdrj dΩj

infinXl=0

bdquorj

laquol

Pl cos(θ))

if ri gt rj or

infinXl=0

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Ylowastlmprimel

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

(2g + 1)

timesg

programs

Thursday

Friday

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

2φ100(r1)φ100(r2)

minus1radic

The Hartree term

V dmicro(r i ) =

ψλ(r i )

NXmicro=1

V dmicro(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψλ(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψ100uarr(r i )

Φ(ri )u10 = 2V d10(ri )u10(ri ) = 2

Z infin

0|u10(rj )|2

rgtdrj )u10(ri )

rijψλ(r j )dr j

ψmicro(r i )

rijψλ(r j )dr j

ψmicro(r i )

s =darr That is

s mλs

rijψλ(r j )dr j

ψmicro(r i )

NXmicro=1

s uarr

ψmicro(r i )

resulting in

NXmicro=1

ψ100uarr(r i )

NXmicro=1

ψ100uarr(r i )

minus1

l(l + 1)

2r2minus

as bdquominus

l(l + 1)

2r2minus

r+ 2V d

10(r)laquo

or bdquominus

dr2minus

r+ V d

10(r)laquo

u10(r) = e10u10(r)

dangerous later

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ100uarr(r i )

= (2V d10(r i ) + 2V d

20(r i ))ψ100uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

V ex20 (r) =

Z infin

0dr jφ

lowast200(r j )

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ200uarr(r i ) =

(2V d10(r i ) + 2V d

20(r i ))ψ200uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

V ex10 (r) =

Z infin

0dr jφ

lowast100(r j )

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

dr2minus

r+ V d

10(r) + 2V d20(r)+

20 (r)+ = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)+

10 (r)+ = e20u20(r)

andbdquominus

2r2minus

r+ 2V d

10(r) + 2V d20(r)+

10 (r)minusV ex20 (r)+ = e21u21(r)

Useful Tests

R10(r) = 2bdquo

laquo32

R20(r) = 2bdquo

laquo32 bdquo1minus

R21(r) =1radic

bdquoZ

laquo32 Zr

We use a0 = 1

R10(r) = 2bdquo

laquo32

E [Z ] = Z 2 minus 4Z +5

Z = 2minus5

16= 16875

we do this

E [ρ] = minus1

1Ψi (r1)dr1minusZ

r1ρ(r1)dr1+

Zρ(r1)ρ(r2)

r12dr1dr2+EEXC [ρ]

ρ(r) =NX

|Ψi (r)|2

2nabla2

1 minusZ

Zρ(r2)

r12dr2 + VEXC(r1)

EEXC =

Thursday

Friday

A simple Example

wi (t = ε) = Wij wj (t = 0)

wi (t) = 1

and Xj

W (j rarr i) = 1

wi (t + 1) =X

Wij wj (t)

0 14 18 2334 58 00 14 13

w(t = 0) =

resulting in

w(t = ε) =

0 14340

resulting in

w(t = 2ε) =

0 5232132632

wi (t = infin) = 1

w(t = infin) =

0 415815315

Iteration w1 w2 w30 100000 000000 0000001 025000 075000 0000002 015625 062625 0187503 024609 052734 0226564 027848 051416 0207365 027213 053021 0197666 026608 053548 0198447 026575 053424 0200028 026656 053321 0200239 026678 053318 020005

10 026671 053332 01999811 026666 053335 02000012 026666 053334 02000013 026667 053333 020000

w(t = infin) 026667 053333 020000

w(0) =X

αi v i

αi v i =X

λtiαi v i

〈M(t)〉 =Xmicro

w(t)microMmicro

〈M(t)〉 = w(t)m

E [xk ] = 〈xk 〉 =1

xki p(xi )

〈M(t)〉 = w(t)m =X

λtiαi v i m i

〈M(t)〉 =X

λtiαi mi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

λtiαi mi

τi = minus1

logλi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

αi mi eminustτi

W (j rarr i)wj =X

W (i rarr j)wi

W (j rarr i)

W (i rarr j)=

Proof Xi

W (i rarr j)wi =X

W (j rarr i)wj = wj

W (j rarr i) = wj

Ergodicity

exp (minusβ(Ei ))

〈A〉 =X

Aj wj =

g(i rarr j) =1

A(j rarr i)

A(microrarr ν)

Then we must have

Take a new value

x (i+1) =

We have defined

ρ(x y) = min

p(y)f (x |y)

p(x)f (y |x) 1ff

Implementation

Quantum Monte Carlo

E [H] = 〈H〉 =

E0 le 〈H〉

Quantum Monte Carlo

ΨT (R) =X

aiΨi (R)

nm alowastmanR

nEnPn a2

nge E0

Quantum Monte Carlo

E [H] = 〈H〉 =

RdRΨlowastTα

(R)H(R)ΨTα(R)R

Quantum Monte Carlo

P(R) =|ψT (R)|2R|ψT (R)|2 dR

E [H] asympR

EL(R) =1

ψT (R)HψT (R)

P(R)EL(R)dR asymp1

P(Ri)EL(Ri)

Quantum Monte Carlo

˛ψα

T (R)˛2

w = P(Rp)P(R)

Quantum Monte Carlo

minus~2

part2u(r)

partr2minusbdquo

rminus

~2l(l + 1)

laquou(r) = Eu(r)

minus1

part2u(ρ)

partρ2minus

l(l + 1)

H = minus1

partρ2minus

l(l + 1)

Quantum Monte Carlo

EL(ρ) = minus1

ρminusα

bdquoαminus

Quantum Monte Carlo

〈Hn〉 =

= constant

Quantum Monte Carlo

minus2ke2

r1minus

Quantum Monte Carlo

bH = minus~2nabla2

2mminus

~2nabla22

2mminus

r1minus

P(R) =|ψT (R)|2R|ψT (R)|2 dR

Quantum Monte Carlo

EL(R) =1

ψT (R)HψT (R) =

ψT (R)

bdquominus

2nabla2

1 minusZ

EL(R) =1

RT (r1)

minus1

dr1minus

limr1rarr0

EL(R) =1

RT (r1)

bdquominus

dr1minus

laquoRT (r1)

Quantum Monte Carlo

This results in1

RT (r1)

dRT (r1)

dr1= minusZ

RT (r)

dRT (r)

dr= minus

f (rij )

wfold = wfnew

Thursday

Importance sampling

Friday

Things to do

Splines

s(x) =

Splines

yi = s(xi )

ands(xi+1) = yi+1

Splines

si (x) =fi

Splines

si (x) =fi

fi+16(xi+1minusxi )

(x minus xi )3

+ (yi+1

6 )(xi+1 minus x) (2)

Splines

(yi minus yiminus1)

Splines

3777775

2666664f1f2f3

fnminus2fnminus1

3777775 =

2666664v1v2v3

vnminus2vnminus1

3777775

operations

Splines

partt= D

bdquopartP

partxminus F

laquoP(x t)

F = 21

ΨTnablaΨT

partx(t)

y = x + DF (x)∆t + ξ

Importance Sampling

G(y x ∆t) =1

(4πD∆t)3N2exp

A(y x) = min(1 q(y x)))

q(y x) =G(x y ∆t)|ΨT (y)|2

G(y x ∆t)|ΨT (x)|2

Thursday

Friday

x = rsinθcosφ

y = rsinθsinφ

andz = rcosθ

4πε0r=

4πε0p

x2 + y2 + z2

4π(l + ml )Pml

for l = ml = 0

4πcos(θ)

for l = 1 og ml = 0

Y1plusmn1 =

(r) through

Ylml(r) = r l Ylml

we obtainΨnlml

equation

Slminusml

1radic

(minus1)m

minus(minus1)ml i i

Clminusml

We have definedYlml

(r) = r l Ylml(r)

2l + 1Yl0(r)

Slml= (minus1)ml

2l + 1ReYl0(r)

Slml= (minus1)ml

2l + 1ImYl0(r)

for ml gt 0

mll 0 1 2 3

q52 (x2 minus 3y2)x

+1 xradic

3xz 12

q32 (5z2 minus r2)x

0 1 y 12 (3z2 minus r2) 1

2 (5z2 minus 3r2)x

-1 zradic

3yz 12

q32 (5z2 minus r2)y

-2radic

3xyradic

q52 (3x2 minus y2)y

Thursday

Det uarr (1 2) =

˛1radic

Det darr (3 4) =

˛1radic

Ψα(Xnew )

Ψα(Xold )=

Duarrα(Xnew )

Duarrα(Xold )

Ddarrα(Xnew )

Ddarrα(Xold )

Gβ(Xnew )

Gβ(Xold )

nablaΨα

Ψα=nablaDuarrα

Ddarrα+nablaGβ

nabla2Ψα

Ψα=nabla2Duarrα

Ddarrα+nabla2Gβ

Gβ+ (

nablaGβ

Gβ)(nablaDuarrα

Ddarrα)

Dminus1ij =

Dα(y)

Dα(x)=

Dij (y)Cij (y)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

Dij (y)Cij (x)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

nablai Dα(x)

Dα(x)=

nabla2i Dα(x)

Dα(x)=

nablai Dα(y)

Dα(y)=

andnabla2

i Dα(y)

Dα(y)=

R equivNX

Dminus1kj (y) =

kj (x)minus1

RDminus1

ki (x)NX

RDminus1

ki (x) for j = i

R =D(y)

LU Decomposition

1CCA =

0BB1 0 0 0

b21 1 0 0b31 b32 1 0b41 b42 b43 1

1CCA0BB

c11 c12 c13 c140 c22 c23 c240 0 c33 c340 0 0 c44

detA = c11c22 cnn

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

nablaGβ

Gβ=Xiltj

(1 + βrij )2

nabla2Gβ

Gβ=Xiltj

(1 + βrij )4

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

a = 14

a = 12

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

Things to do

dt= ~v

dt= minusξ~v + ~F

ξ = 6πηam

0 + 2Z t

0dτ prime

〈~F (t)〉 = 0

δ(t minus t prime)

0 eminus2ξt +C~v0

we get

~r(t) = ~r0 + ~v01

0dτZ τ

Thijssenrsquos text

Chapters 3-6 and 12

MPI text

Using MPI

Chapters 1-5

Rothman and Zaleski

Quantum Monte Carlo

wherer1 rA

Quantum Monte Carlo

There are

2A timesbdquo

MPI_Command_name

MPI_COMMAND_NAME

processor speed

memory sizespeed

Speed-up

S(P) =T (1)

Efficiency

η(P) =S(P)

tC(L) = τ + βL

T (P) ge T (1)

Prarr S(P) le P

MPI_Command_name

MPI_COMMAND_NAME

Communicator

The Fortran program

END PROGRAM hello

Note 1

Note 2

Note 3

100 values

Note 4

serious work

Integrating π

Examples

Go to MPI examples

Integration algos

af (x)dx asymp h

f (ximinus12)

example5cpp

mpirun -np 10 progx

MPI reduce

Call as

end of main program

function MPI Wtime

Strategies

Thursday

Friday

minus1

dr2unl (r) +

bdquol(l + 1)

2r2minus

r+ Φ(r)

problem

minus1

dr2unl (r) +

bdquol(l + 1)

and rewrite as

minus1

k2(r) = V (r) +l(l + 1)

2r2minus enl

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

V (ri ) = minusZri

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

programs catalogue

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

r+ Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr+Z infin

0unl (r)Fnl (r)dr

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

2Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr + EFock

EFock =1

Z infin

0unl (r)Fnl (r)dr

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

dx2y(x) + k2(x)y(x) = F (x)

h2+ O(h2)

h2+ O(h2) + k2

i yi = Fi

2minus h2k2i

rdquoyi + h2Fi

Numerovrsquos algo

y(x + h) = y(x) + hy (1)(x) +h2

2y (2)(x) +

3y (3)(x) +

Numerovrsquos algo

12y (4)(x) + O(h6)

We obtain

h2minus

12y (4)(x) + O(h4)

Numerovrsquos algo

y (2)(x) +h2

12y (4)(x) + k2(x)y(x) +

ldquok2(x)y(x)

rdquoasymp 0

Numerovrsquos algo

d2(k2y(x))

acuteh2

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

+ O(h6)

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

minus1

dr2u(r) +

bdquoV (r) +

l(l + 1)

laquou(r) = eu(r)

dr2u(r) + k2(r)u(r) = 0

k2(r) = γ

bdquof (r)minus

l(l + 1)

r2minus ε

laquoγ = 2V0

ε =|E |V0

f (r) =

forr le ar gt a

k2(r) = γ

(1minus εminus 1

γl(l+1)

l(l+1)

forr le ar gt a

dr2u(r)minus

l(l + 1)

r2u(r) = 0

well defined

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

drult(r) =

drugt(r) at r = rm

drult(rm) asymp

drugt(rm) asymp

Numerovcpp

wf[loop] = norm

Todayrsquos tasks

2reminusr

u(r) =

bdquo1

1minus a2

raquo1 +

2(1minus a2)r

ndashlaquo

Thursday

Friday

u(x) = c1 +

0uprime(y)dy

and a constant c2

uprime(y) = c2 +

0uprimeprime(z)dz

0f (z)dz

bdquoZ y

0f (z)dz

laquody

and defining

F (y) =

0f (z)dz

0F (y)dy =

bdquoZ y

0f (z)dz

laquody =

0(x minus y)f (y)dy

This gives us

0(x minus y)f (y)dy

0(1minus y)f (y)dy

u(x) = xZ 1

0(x minus y)f (y)dy

G(x y) =

u(x) =

0G(x y)f (y)dy

1 continuous

3 G(0 y) = G(1 y) = G(x 0) = G(x 1) = 0

chapter 165

dx|x=y+ε minus

dx|x=yminusε = 1

W =dugt

dxult minus

and usingdG

dx|x=y+ε minus

dx|x=yminusε = 1

Simple example

ult(x) = x

G(x y) =

u(x) = λ

0G(x y)u(t)dy

u(r) =

Z infin

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

u(r) =

Z infin

resulting in

u(r) = ugt(r)Z r

Z infin

Algorithm

dxasymp

Algorithm

Thursday

Friday

E [Φ] =

af (Φ(x)

partΦ

partx x)dx

η(a) = η(b) = 0

Φ(x α) = Φ(x 0) + αη(x)

E [Φ(α)] =

af (Φ(x α)

partΦ(x α)

partx x)dx

E [Φ(α)] =

ndashα=0

partα

ndash=

bdquopartf

partΦ

partα+

partΦx

partα

laquodx

partα= η(x)

partα=

d(η(x))

UsingpartΦ(x α)

partα= η(x)

andpartΦx (x α)

partα=

d(η(x))

partα

ndash=

bdquopartf

partΦη(x) +

partΦx

d(η(x))

laquodx

partΦx

d(η(x))

dxdx = η(x)

partΦx|ba minus

aη(x)

partΦxdx

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndashα=0

bdquopartf

partΦminus

partΦx

partΦminus

partΦx= 0

df = 0

partx=partf

party=partf

partz= 0

due to

df =partf

partxdx +

partydy +

partzdz

φ(x y z) = 0

resulting in

dφ =partφ

partxdx +

partφ

partydy +

partφ

partzdz = 0

partx=partf

party=partf

partz= 0

df =partf

partxdx +

partydy +

partzdz

df + λdφ = (partf

partz+ λ

partφ

partx)dx + (

party+ λ

partφ

party)dy + (

partz+ λ

partφ

partz)dz = 0

partz+ λ

partφ

partz= 0

partx+ λ

partφ

partx= 0

andpartf

party+ λ

partφ

party= 0

partxi+X

λkpartφk

partxi= 0

and a Hamiltonian

H = minus1

2nabla2 + V (x y z)

Zdxdydz

δE = δ

Zdxdydz

bdquo1

laquoff= 0

Zdxdydz

bdquo1

laquoff= 0

partψlowastminus

partψlowastxminus

partψlowastyminus

partψlowastz= 0

which yields

minus1

H = minusNX

2nabla2

i minusNX

NXiltj

H = H1 + H2 =NX

hi +NX

iltj=1

hi = minus1

2nabla2

i minusZ

E0 le E [Φ] =

ZΦlowastHΦdτ

radicN

(minus)PP

[H1A] = [H2A] = 0

ZΦlowastH1Φdτ =

(minus)PZ

ZΦlowastH1Φdτ =

ZΦlowastH hiΦHdτ

ZΦlowastH1Φdτ =

NXmicro=1

〈micro|h|micro〉

ΦlowastH2Φdτ = N

which reduces to

ZΦlowastH2Φdτ =

NXilej=1

(minus)PZ

ΦlowastH1

rijPΦHdτ

ZΦlowastH2Φdτ =

NXiltj=1

ZΦlowastH

ZΦlowastH2Φdτ =

NXmicro=1

NXν=1

lowastν(r j )

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

lowastν(r j )

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

ψa =Xλ

Caλψλ

lectures

Exercise

Caλψλ

δE minusNX

micro=1

NXν=1

εmicroνδ

δE minusNX

micro=1

εmicroδ

NXmicro=1

NXν=1

Zδψlowastmicroψ

lowastν

Zδψlowastmicroψ

lowastν

micro=1

NXmicro=1

NXν=1

Zψlowastmicroψ

lowastν

Zψlowastmicroψ

lowastν

minusNX

micro=1

Emicro

NXmicro=1

Emicro

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

rijψν(r j )dr j

ψmicro(r i )

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

V dmicro(r i ) =

2(2l prime + 1)

Z infin

rgtdrj

withφnlml

Z infin

(rj )|21

rijdr j

which reads

Z infin

rijdrj

2l prime + 1

resulting in

2(2l prime + 1)

Z infin

rijdrj dΩj

infinXl=0

bdquorj

laquol

Pl cos(θ))

if ri gt rj or

infinXl=0

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Ylowastlmprimel

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

(2g + 1)

timesg

programs

Thursday

Friday

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

2φ100(r1)φ100(r2)

minus1radic

The Hartree term

V dmicro(r i ) =

ψλ(r i )

NXmicro=1

V dmicro(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψλ(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψ100uarr(r i )

Φ(ri )u10 = 2V d10(ri )u10(ri ) = 2

Z infin

0|u10(rj )|2

rgtdrj )u10(ri )

rijψλ(r j )dr j

ψmicro(r i )

rijψλ(r j )dr j

ψmicro(r i )

s =darr That is

s mλs

rijψλ(r j )dr j

ψmicro(r i )

NXmicro=1

s uarr

ψmicro(r i )

resulting in

NXmicro=1

ψ100uarr(r i )

NXmicro=1

ψ100uarr(r i )

minus1

l(l + 1)

2r2minus

as bdquominus

l(l + 1)

2r2minus

r+ 2V d

10(r)laquo

or bdquominus

dr2minus

r+ V d

10(r)laquo

u10(r) = e10u10(r)

dangerous later

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ100uarr(r i )

= (2V d10(r i ) + 2V d

20(r i ))ψ100uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

V ex20 (r) =

Z infin

0dr jφ

lowast200(r j )

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ200uarr(r i ) =

(2V d10(r i ) + 2V d

20(r i ))ψ200uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

V ex10 (r) =

Z infin

0dr jφ

lowast100(r j )

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

dr2minus

r+ V d

10(r) + 2V d20(r)+

20 (r)+ = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)+

10 (r)+ = e20u20(r)

andbdquominus

2r2minus

r+ 2V d

10(r) + 2V d20(r)+

10 (r)minusV ex20 (r)+ = e21u21(r)

Useful Tests

R10(r) = 2bdquo

laquo32

R20(r) = 2bdquo

laquo32 bdquo1minus

R21(r) =1radic

bdquoZ

laquo32 Zr

We use a0 = 1

R10(r) = 2bdquo

laquo32

E [Z ] = Z 2 minus 4Z +5

Z = 2minus5

16= 16875

we do this

E [ρ] = minus1

1Ψi (r1)dr1minusZ

r1ρ(r1)dr1+

Zρ(r1)ρ(r2)

r12dr1dr2+EEXC [ρ]

ρ(r) =NX

|Ψi (r)|2

2nabla2

1 minusZ

Zρ(r2)

r12dr2 + VEXC(r1)

EEXC =

Thursday

Friday

A simple Example

wi (t = ε) = Wij wj (t = 0)

wi (t) = 1

and Xj

W (j rarr i) = 1

wi (t + 1) =X

Wij wj (t)

0 14 18 2334 58 00 14 13

w(t = 0) =

resulting in

w(t = ε) =

0 14340

resulting in

w(t = 2ε) =

0 5232132632

wi (t = infin) = 1

w(t = infin) =

0 415815315

Iteration w1 w2 w30 100000 000000 0000001 025000 075000 0000002 015625 062625 0187503 024609 052734 0226564 027848 051416 0207365 027213 053021 0197666 026608 053548 0198447 026575 053424 0200028 026656 053321 0200239 026678 053318 020005

10 026671 053332 01999811 026666 053335 02000012 026666 053334 02000013 026667 053333 020000

w(t = infin) 026667 053333 020000

w(0) =X

αi v i

αi v i =X

λtiαi v i

〈M(t)〉 =Xmicro

w(t)microMmicro

〈M(t)〉 = w(t)m

E [xk ] = 〈xk 〉 =1

xki p(xi )

〈M(t)〉 = w(t)m =X

λtiαi v i m i

〈M(t)〉 =X

λtiαi mi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

λtiαi mi

τi = minus1

logλi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

αi mi eminustτi

W (j rarr i)wj =X

W (i rarr j)wi

W (j rarr i)

W (i rarr j)=

Proof Xi

W (i rarr j)wi =X

W (j rarr i)wj = wj

W (j rarr i) = wj

Ergodicity

exp (minusβ(Ei ))

〈A〉 =X

Aj wj =

g(i rarr j) =1

A(j rarr i)

A(microrarr ν)

Then we must have

Take a new value

x (i+1) =

We have defined

ρ(x y) = min

p(y)f (x |y)

p(x)f (y |x) 1ff

Implementation

Quantum Monte Carlo

E [H] = 〈H〉 =

E0 le 〈H〉

Quantum Monte Carlo

ΨT (R) =X

aiΨi (R)

nm alowastmanR

nEnPn a2

nge E0

Quantum Monte Carlo

E [H] = 〈H〉 =

RdRΨlowastTα

(R)H(R)ΨTα(R)R

Quantum Monte Carlo

P(R) =|ψT (R)|2R|ψT (R)|2 dR

E [H] asympR

EL(R) =1

ψT (R)HψT (R)

P(R)EL(R)dR asymp1

P(Ri)EL(Ri)

Quantum Monte Carlo

˛ψα

T (R)˛2

w = P(Rp)P(R)

Quantum Monte Carlo

minus~2

part2u(r)

partr2minusbdquo

rminus

~2l(l + 1)

laquou(r) = Eu(r)

minus1

part2u(ρ)

partρ2minus

l(l + 1)

H = minus1

partρ2minus

l(l + 1)

Quantum Monte Carlo

EL(ρ) = minus1

ρminusα

bdquoαminus

Quantum Monte Carlo

〈Hn〉 =

= constant

Quantum Monte Carlo

minus2ke2

r1minus

Quantum Monte Carlo

bH = minus~2nabla2

2mminus

~2nabla22

2mminus

r1minus

P(R) =|ψT (R)|2R|ψT (R)|2 dR

Quantum Monte Carlo

EL(R) =1

ψT (R)HψT (R) =

ψT (R)

bdquominus

2nabla2

1 minusZ

EL(R) =1

RT (r1)

minus1

dr1minus

limr1rarr0

EL(R) =1

RT (r1)

bdquominus

dr1minus

laquoRT (r1)

Quantum Monte Carlo

This results in1

RT (r1)

dRT (r1)

dr1= minusZ

RT (r)

dRT (r)

dr= minus

f (rij )

wfold = wfnew

Thursday

Importance sampling

Friday

Things to do

Splines

s(x) =

Splines

yi = s(xi )

ands(xi+1) = yi+1

Splines

si (x) =fi

Splines

si (x) =fi

fi+16(xi+1minusxi )

(x minus xi )3

+ (yi+1

6 )(xi+1 minus x) (2)

Splines

(yi minus yiminus1)

Splines

3777775

2666664f1f2f3

fnminus2fnminus1

3777775 =

2666664v1v2v3

vnminus2vnminus1

3777775

operations

Splines

partt= D

bdquopartP

partxminus F

laquoP(x t)

F = 21

ΨTnablaΨT

partx(t)

y = x + DF (x)∆t + ξ

Importance Sampling

G(y x ∆t) =1

(4πD∆t)3N2exp

A(y x) = min(1 q(y x)))

q(y x) =G(x y ∆t)|ΨT (y)|2

G(y x ∆t)|ΨT (x)|2

Thursday

Friday

x = rsinθcosφ

y = rsinθsinφ

andz = rcosθ

4πε0r=

4πε0p

x2 + y2 + z2

4π(l + ml )Pml

for l = ml = 0

4πcos(θ)

for l = 1 og ml = 0

Y1plusmn1 =

(r) through

Ylml(r) = r l Ylml

we obtainΨnlml

equation

Slminusml

1radic

(minus1)m

minus(minus1)ml i i

Clminusml

We have definedYlml

(r) = r l Ylml(r)

2l + 1Yl0(r)

Slml= (minus1)ml

2l + 1ReYl0(r)

Slml= (minus1)ml

2l + 1ImYl0(r)

for ml gt 0

mll 0 1 2 3

q52 (x2 minus 3y2)x

+1 xradic

3xz 12

q32 (5z2 minus r2)x

0 1 y 12 (3z2 minus r2) 1

2 (5z2 minus 3r2)x

-1 zradic

3yz 12

q32 (5z2 minus r2)y

-2radic

3xyradic

q52 (3x2 minus y2)y

Thursday

Det uarr (1 2) =

˛1radic

Det darr (3 4) =

˛1radic

Ψα(Xnew )

Ψα(Xold )=

Duarrα(Xnew )

Duarrα(Xold )

Ddarrα(Xnew )

Ddarrα(Xold )

Gβ(Xnew )

Gβ(Xold )

nablaΨα

Ψα=nablaDuarrα

Ddarrα+nablaGβ

nabla2Ψα

Ψα=nabla2Duarrα

Ddarrα+nabla2Gβ

Gβ+ (

nablaGβ

Gβ)(nablaDuarrα

Ddarrα)

Dminus1ij =

Dα(y)

Dα(x)=

Dij (y)Cij (y)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

Dij (y)Cij (x)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

nablai Dα(x)

Dα(x)=

nabla2i Dα(x)

Dα(x)=

nablai Dα(y)

Dα(y)=

andnabla2

i Dα(y)

Dα(y)=

R equivNX

Dminus1kj (y) =

kj (x)minus1

RDminus1

ki (x)NX

RDminus1

ki (x) for j = i

R =D(y)

LU Decomposition

1CCA =

0BB1 0 0 0

b21 1 0 0b31 b32 1 0b41 b42 b43 1

1CCA0BB

c11 c12 c13 c140 c22 c23 c240 0 c33 c340 0 0 c44

detA = c11c22 cnn

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

nablaGβ

Gβ=Xiltj

(1 + βrij )2

nabla2Gβ

Gβ=Xiltj

(1 + βrij )4

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

a = 14

a = 12

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

Things to do

dt= ~v

dt= minusξ~v + ~F

ξ = 6πηam

0 + 2Z t

0dτ prime

〈~F (t)〉 = 0

δ(t minus t prime)

0 eminus2ξt +C~v0

we get

~r(t) = ~r0 + ~v01

0dτZ τ

Thijssenrsquos text

Chapters 3-6 and 12

MPI text

Using MPI

Chapters 1-5

Rothman and Zaleski

Quantum Monte Carlo

wherer1 rA

Quantum Monte Carlo

There are

2A timesbdquo

MPI_Command_name

MPI_COMMAND_NAME

processor speed

memory sizespeed

Speed-up

S(P) =T (1)

Efficiency

η(P) =S(P)

tC(L) = τ + βL

T (P) ge T (1)

Prarr S(P) le P

MPI_Command_name

MPI_COMMAND_NAME

Communicator

The Fortran program

END PROGRAM hello

Note 1

Note 2

Note 3

100 values

Note 4

serious work

Integrating π

Examples

Go to MPI examples

Integration algos

af (x)dx asymp h

f (ximinus12)

example5cpp

mpirun -np 10 progx

MPI reduce

Call as

end of main program

function MPI Wtime

Strategies

Thursday

Friday

minus1

dr2unl (r) +

bdquol(l + 1)

2r2minus

r+ Φ(r)

problem

minus1

dr2unl (r) +

bdquol(l + 1)

and rewrite as

minus1

k2(r) = V (r) +l(l + 1)

2r2minus enl

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

V (ri ) = minusZri

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

programs catalogue

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

r+ Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr+Z infin

0unl (r)Fnl (r)dr

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

2Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr + EFock

EFock =1

Z infin

0unl (r)Fnl (r)dr

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

dx2y(x) + k2(x)y(x) = F (x)

h2+ O(h2)

h2+ O(h2) + k2

i yi = Fi

2minus h2k2i

rdquoyi + h2Fi

Numerovrsquos algo

y(x + h) = y(x) + hy (1)(x) +h2

2y (2)(x) +

3y (3)(x) +

Numerovrsquos algo

12y (4)(x) + O(h6)

We obtain

h2minus

12y (4)(x) + O(h4)

Numerovrsquos algo

y (2)(x) +h2

12y (4)(x) + k2(x)y(x) +

ldquok2(x)y(x)

rdquoasymp 0

Numerovrsquos algo

d2(k2y(x))

acuteh2

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

+ O(h6)

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

minus1

dr2u(r) +

bdquoV (r) +

l(l + 1)

laquou(r) = eu(r)

dr2u(r) + k2(r)u(r) = 0

k2(r) = γ

bdquof (r)minus

l(l + 1)

r2minus ε

laquoγ = 2V0

ε =|E |V0

f (r) =

forr le ar gt a

k2(r) = γ

(1minus εminus 1

γl(l+1)

l(l+1)

forr le ar gt a

dr2u(r)minus

l(l + 1)

r2u(r) = 0

well defined

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

drult(r) =

drugt(r) at r = rm

drult(rm) asymp

drugt(rm) asymp

Numerovcpp

wf[loop] = norm

Todayrsquos tasks

2reminusr

u(r) =

bdquo1

1minus a2

raquo1 +

2(1minus a2)r

ndashlaquo

Thursday

Friday

u(x) = c1 +

0uprime(y)dy

and a constant c2

uprime(y) = c2 +

0uprimeprime(z)dz

0f (z)dz

bdquoZ y

0f (z)dz

laquody

and defining

F (y) =

0f (z)dz

0F (y)dy =

bdquoZ y

0f (z)dz

laquody =

0(x minus y)f (y)dy

This gives us

0(x minus y)f (y)dy

0(1minus y)f (y)dy

u(x) = xZ 1

0(x minus y)f (y)dy

G(x y) =

u(x) =

0G(x y)f (y)dy

1 continuous

3 G(0 y) = G(1 y) = G(x 0) = G(x 1) = 0

chapter 165

dx|x=y+ε minus

dx|x=yminusε = 1

W =dugt

dxult minus

and usingdG

dx|x=y+ε minus

dx|x=yminusε = 1

Simple example

ult(x) = x

G(x y) =

u(x) = λ

0G(x y)u(t)dy

u(r) =

Z infin

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

u(r) =

Z infin

resulting in

u(r) = ugt(r)Z r

Z infin

Algorithm

dxasymp

Algorithm

Thursday

Friday

E [Φ] =

af (Φ(x)

partΦ

partx x)dx

η(a) = η(b) = 0

Φ(x α) = Φ(x 0) + αη(x)

E [Φ(α)] =

af (Φ(x α)

partΦ(x α)

partx x)dx

E [Φ(α)] =

ndashα=0

partα

ndash=

bdquopartf

partΦ

partα+

partΦx

partα

laquodx

partα= η(x)

partα=

d(η(x))

UsingpartΦ(x α)

partα= η(x)

andpartΦx (x α)

partα=

d(η(x))

partα

ndash=

bdquopartf

partΦη(x) +

partΦx

d(η(x))

laquodx

partΦx

d(η(x))

dxdx = η(x)

partΦx|ba minus

aη(x)

partΦxdx

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndashα=0

bdquopartf

partΦminus

partΦx

partΦminus

partΦx= 0

df = 0

partx=partf

party=partf

partz= 0

due to

df =partf

partxdx +

partydy +

partzdz

φ(x y z) = 0

resulting in

dφ =partφ

partxdx +

partφ

partydy +

partφ

partzdz = 0

partx=partf

party=partf

partz= 0

df =partf

partxdx +

partydy +

partzdz

df + λdφ = (partf

partz+ λ

partφ

partx)dx + (

party+ λ

partφ

party)dy + (

partz+ λ

partφ

partz)dz = 0

partz+ λ

partφ

partz= 0

partx+ λ

partφ

partx= 0

andpartf

party+ λ

partφ

party= 0

partxi+X

λkpartφk

partxi= 0

and a Hamiltonian

H = minus1

2nabla2 + V (x y z)

Zdxdydz

δE = δ

Zdxdydz

bdquo1

laquoff= 0

Zdxdydz

bdquo1

laquoff= 0

partψlowastminus

partψlowastxminus

partψlowastyminus

partψlowastz= 0

which yields

minus1

H = minusNX

2nabla2

i minusNX

NXiltj

H = H1 + H2 =NX

hi +NX

iltj=1

hi = minus1

2nabla2

i minusZ

E0 le E [Φ] =

ZΦlowastHΦdτ

radicN

(minus)PP

[H1A] = [H2A] = 0

ZΦlowastH1Φdτ =

(minus)PZ

ZΦlowastH1Φdτ =

ZΦlowastH hiΦHdτ

ZΦlowastH1Φdτ =

NXmicro=1

〈micro|h|micro〉

ΦlowastH2Φdτ = N

which reduces to

ZΦlowastH2Φdτ =

NXilej=1

(minus)PZ

ΦlowastH1

rijPΦHdτ

ZΦlowastH2Φdτ =

NXiltj=1

ZΦlowastH

ZΦlowastH2Φdτ =

NXmicro=1

NXν=1

lowastν(r j )

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

lowastν(r j )

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

ψa =Xλ

Caλψλ

lectures

Exercise

Caλψλ

δE minusNX

micro=1

NXν=1

εmicroνδ

δE minusNX

micro=1

εmicroδ

NXmicro=1

NXν=1

Zδψlowastmicroψ

lowastν

Zδψlowastmicroψ

lowastν

micro=1

NXmicro=1

NXν=1

Zψlowastmicroψ

lowastν

Zψlowastmicroψ

lowastν

minusNX

micro=1

Emicro

NXmicro=1

Emicro

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

rijψν(r j )dr j

ψmicro(r i )

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

V dmicro(r i ) =

2(2l prime + 1)

Z infin

rgtdrj

withφnlml

Z infin

(rj )|21

rijdr j

which reads

Z infin

rijdrj

2l prime + 1

resulting in

2(2l prime + 1)

Z infin

rijdrj dΩj

infinXl=0

bdquorj

laquol

Pl cos(θ))

if ri gt rj or

infinXl=0

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Ylowastlmprimel

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

(2g + 1)

timesg

programs

Thursday

Friday

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

2φ100(r1)φ100(r2)

minus1radic

The Hartree term

V dmicro(r i ) =

ψλ(r i )

NXmicro=1

V dmicro(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψλ(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψ100uarr(r i )

Φ(ri )u10 = 2V d10(ri )u10(ri ) = 2

Z infin

0|u10(rj )|2

rgtdrj )u10(ri )

rijψλ(r j )dr j

ψmicro(r i )

rijψλ(r j )dr j

ψmicro(r i )

s =darr That is

s mλs

rijψλ(r j )dr j

ψmicro(r i )

NXmicro=1

s uarr

ψmicro(r i )

resulting in

NXmicro=1

ψ100uarr(r i )

NXmicro=1

ψ100uarr(r i )

minus1

l(l + 1)

2r2minus

as bdquominus

l(l + 1)

2r2minus

r+ 2V d

10(r)laquo

or bdquominus

dr2minus

r+ V d

10(r)laquo

u10(r) = e10u10(r)

dangerous later

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ100uarr(r i )

= (2V d10(r i ) + 2V d

20(r i ))ψ100uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

V ex20 (r) =

Z infin

0dr jφ

lowast200(r j )

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ200uarr(r i ) =

(2V d10(r i ) + 2V d

20(r i ))ψ200uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

V ex10 (r) =

Z infin

0dr jφ

lowast100(r j )

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

dr2minus

r+ V d

10(r) + 2V d20(r)+

20 (r)+ = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)+

10 (r)+ = e20u20(r)

andbdquominus

2r2minus

r+ 2V d

10(r) + 2V d20(r)+

10 (r)minusV ex20 (r)+ = e21u21(r)

Useful Tests

R10(r) = 2bdquo

laquo32

R20(r) = 2bdquo

laquo32 bdquo1minus

R21(r) =1radic

bdquoZ

laquo32 Zr

We use a0 = 1

R10(r) = 2bdquo

laquo32

E [Z ] = Z 2 minus 4Z +5

Z = 2minus5

16= 16875

we do this

E [ρ] = minus1

1Ψi (r1)dr1minusZ

r1ρ(r1)dr1+

Zρ(r1)ρ(r2)

r12dr1dr2+EEXC [ρ]

ρ(r) =NX

|Ψi (r)|2

2nabla2

1 minusZ

Zρ(r2)

r12dr2 + VEXC(r1)

EEXC =

Thursday

Friday

A simple Example

wi (t = ε) = Wij wj (t = 0)

wi (t) = 1

and Xj

W (j rarr i) = 1

wi (t + 1) =X

Wij wj (t)

0 14 18 2334 58 00 14 13

w(t = 0) =

resulting in

w(t = ε) =

0 14340

resulting in

w(t = 2ε) =

0 5232132632

wi (t = infin) = 1

w(t = infin) =

0 415815315

Iteration w1 w2 w30 100000 000000 0000001 025000 075000 0000002 015625 062625 0187503 024609 052734 0226564 027848 051416 0207365 027213 053021 0197666 026608 053548 0198447 026575 053424 0200028 026656 053321 0200239 026678 053318 020005

10 026671 053332 01999811 026666 053335 02000012 026666 053334 02000013 026667 053333 020000

w(t = infin) 026667 053333 020000

w(0) =X

αi v i

αi v i =X

λtiαi v i

〈M(t)〉 =Xmicro

w(t)microMmicro

〈M(t)〉 = w(t)m

E [xk ] = 〈xk 〉 =1

xki p(xi )

〈M(t)〉 = w(t)m =X

λtiαi v i m i

〈M(t)〉 =X

λtiαi mi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

λtiαi mi

τi = minus1

logλi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

αi mi eminustτi

W (j rarr i)wj =X

W (i rarr j)wi

W (j rarr i)

W (i rarr j)=

Proof Xi

W (i rarr j)wi =X

W (j rarr i)wj = wj

W (j rarr i) = wj

Ergodicity

exp (minusβ(Ei ))

〈A〉 =X

Aj wj =

g(i rarr j) =1

A(j rarr i)

A(microrarr ν)

Then we must have

Take a new value

x (i+1) =

We have defined

ρ(x y) = min

p(y)f (x |y)

p(x)f (y |x) 1ff

Implementation

Quantum Monte Carlo

E [H] = 〈H〉 =

E0 le 〈H〉

Quantum Monte Carlo

ΨT (R) =X

aiΨi (R)

nm alowastmanR

nEnPn a2

nge E0

Quantum Monte Carlo

E [H] = 〈H〉 =

RdRΨlowastTα

(R)H(R)ΨTα(R)R

Quantum Monte Carlo

P(R) =|ψT (R)|2R|ψT (R)|2 dR

E [H] asympR

EL(R) =1

ψT (R)HψT (R)

P(R)EL(R)dR asymp1

P(Ri)EL(Ri)

Quantum Monte Carlo

˛ψα

T (R)˛2

w = P(Rp)P(R)

Quantum Monte Carlo

minus~2

part2u(r)

partr2minusbdquo

rminus

~2l(l + 1)

laquou(r) = Eu(r)

minus1

part2u(ρ)

partρ2minus

l(l + 1)

H = minus1

partρ2minus

l(l + 1)

Quantum Monte Carlo

EL(ρ) = minus1

ρminusα

bdquoαminus

Quantum Monte Carlo

〈Hn〉 =

= constant

Quantum Monte Carlo

minus2ke2

r1minus

Quantum Monte Carlo

bH = minus~2nabla2

2mminus

~2nabla22

2mminus

r1minus

P(R) =|ψT (R)|2R|ψT (R)|2 dR

Quantum Monte Carlo

EL(R) =1

ψT (R)HψT (R) =

ψT (R)

bdquominus

2nabla2

1 minusZ

EL(R) =1

RT (r1)

minus1

dr1minus

limr1rarr0

EL(R) =1

RT (r1)

bdquominus

dr1minus

laquoRT (r1)

Quantum Monte Carlo

This results in1

RT (r1)

dRT (r1)

dr1= minusZ

RT (r)

dRT (r)

dr= minus

f (rij )

wfold = wfnew

Thursday

Importance sampling

Friday

Things to do

Splines

s(x) =

Splines

yi = s(xi )

ands(xi+1) = yi+1

Splines

si (x) =fi

Splines

si (x) =fi

fi+16(xi+1minusxi )

(x minus xi )3

+ (yi+1

6 )(xi+1 minus x) (2)

Splines

(yi minus yiminus1)

Splines

3777775

2666664f1f2f3

fnminus2fnminus1

3777775 =

2666664v1v2v3

vnminus2vnminus1

3777775

operations

Splines

partt= D

bdquopartP

partxminus F

laquoP(x t)

F = 21

ΨTnablaΨT

partx(t)

y = x + DF (x)∆t + ξ

Importance Sampling

G(y x ∆t) =1

(4πD∆t)3N2exp

A(y x) = min(1 q(y x)))

q(y x) =G(x y ∆t)|ΨT (y)|2

G(y x ∆t)|ΨT (x)|2

Thursday

Friday

x = rsinθcosφ

y = rsinθsinφ

andz = rcosθ

4πε0r=

4πε0p

x2 + y2 + z2

4π(l + ml )Pml

for l = ml = 0

4πcos(θ)

for l = 1 og ml = 0

Y1plusmn1 =

(r) through

Ylml(r) = r l Ylml

we obtainΨnlml

equation

Slminusml

1radic

(minus1)m

minus(minus1)ml i i

Clminusml

We have definedYlml

(r) = r l Ylml(r)

2l + 1Yl0(r)

Slml= (minus1)ml

2l + 1ReYl0(r)

Slml= (minus1)ml

2l + 1ImYl0(r)

for ml gt 0

mll 0 1 2 3

q52 (x2 minus 3y2)x

+1 xradic

3xz 12

q32 (5z2 minus r2)x

0 1 y 12 (3z2 minus r2) 1

2 (5z2 minus 3r2)x

-1 zradic

3yz 12

q32 (5z2 minus r2)y

-2radic

3xyradic

q52 (3x2 minus y2)y

Thursday

Det uarr (1 2) =

˛1radic

Det darr (3 4) =

˛1radic

Ψα(Xnew )

Ψα(Xold )=

Duarrα(Xnew )

Duarrα(Xold )

Ddarrα(Xnew )

Ddarrα(Xold )

Gβ(Xnew )

Gβ(Xold )

nablaΨα

Ψα=nablaDuarrα

Ddarrα+nablaGβ

nabla2Ψα

Ψα=nabla2Duarrα

Ddarrα+nabla2Gβ

Gβ+ (

nablaGβ

Gβ)(nablaDuarrα

Ddarrα)

Dminus1ij =

Dα(y)

Dα(x)=

Dij (y)Cij (y)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

Dij (y)Cij (x)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

nablai Dα(x)

Dα(x)=

nabla2i Dα(x)

Dα(x)=

nablai Dα(y)

Dα(y)=

andnabla2

i Dα(y)

Dα(y)=

R equivNX

Dminus1kj (y) =

kj (x)minus1

RDminus1

ki (x)NX

RDminus1

ki (x) for j = i

R =D(y)

LU Decomposition

1CCA =

0BB1 0 0 0

b21 1 0 0b31 b32 1 0b41 b42 b43 1

1CCA0BB

c11 c12 c13 c140 c22 c23 c240 0 c33 c340 0 0 c44

detA = c11c22 cnn

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

nablaGβ

Gβ=Xiltj

(1 + βrij )2

nabla2Gβ

Gβ=Xiltj

(1 + βrij )4

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

a = 14

a = 12

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

Things to do

dt= ~v

dt= minusξ~v + ~F

ξ = 6πηam

0 + 2Z t

0dτ prime

〈~F (t)〉 = 0

δ(t minus t prime)

0 eminus2ξt +C~v0

we get

~r(t) = ~r0 + ~v01

0dτZ τ

MPI text

Using MPI

Chapters 1-5

Rothman and Zaleski

Quantum Monte Carlo

wherer1 rA

Quantum Monte Carlo

There are

2A timesbdquo

MPI_Command_name

MPI_COMMAND_NAME

processor speed

memory sizespeed

Speed-up

S(P) =T (1)

Efficiency

η(P) =S(P)

tC(L) = τ + βL

T (P) ge T (1)

Prarr S(P) le P

MPI_Command_name

MPI_COMMAND_NAME

Communicator

The Fortran program

END PROGRAM hello

Note 1

Note 2

Note 3

100 values

Note 4

serious work

Integrating π

Examples

Go to MPI examples

Integration algos

af (x)dx asymp h

f (ximinus12)

example5cpp

mpirun -np 10 progx

MPI reduce

Call as

end of main program

function MPI Wtime

Strategies

Thursday

Friday

minus1

dr2unl (r) +

bdquol(l + 1)

2r2minus

r+ Φ(r)

problem

minus1

dr2unl (r) +

bdquol(l + 1)

and rewrite as

minus1

k2(r) = V (r) +l(l + 1)

2r2minus enl

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

V (ri ) = minusZri

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

programs catalogue

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

r+ Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr+Z infin

0unl (r)Fnl (r)dr

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

2Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr + EFock

EFock =1

Z infin

0unl (r)Fnl (r)dr

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

dx2y(x) + k2(x)y(x) = F (x)

h2+ O(h2)

h2+ O(h2) + k2

i yi = Fi

2minus h2k2i

rdquoyi + h2Fi

Numerovrsquos algo

y(x + h) = y(x) + hy (1)(x) +h2

2y (2)(x) +

3y (3)(x) +

Numerovrsquos algo

12y (4)(x) + O(h6)

We obtain

h2minus

12y (4)(x) + O(h4)

Numerovrsquos algo

y (2)(x) +h2

12y (4)(x) + k2(x)y(x) +

ldquok2(x)y(x)

rdquoasymp 0

Numerovrsquos algo

d2(k2y(x))

acuteh2

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

+ O(h6)

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

minus1

dr2u(r) +

bdquoV (r) +

l(l + 1)

laquou(r) = eu(r)

dr2u(r) + k2(r)u(r) = 0

k2(r) = γ

bdquof (r)minus

l(l + 1)

r2minus ε

laquoγ = 2V0

ε =|E |V0

f (r) =

forr le ar gt a

k2(r) = γ

(1minus εminus 1

γl(l+1)

l(l+1)

forr le ar gt a

dr2u(r)minus

l(l + 1)

r2u(r) = 0

well defined

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

drult(r) =

drugt(r) at r = rm

drult(rm) asymp

drugt(rm) asymp

Numerovcpp

wf[loop] = norm

Todayrsquos tasks

2reminusr

u(r) =

bdquo1

1minus a2

raquo1 +

2(1minus a2)r

ndashlaquo

Thursday

Friday

u(x) = c1 +

0uprime(y)dy

and a constant c2

uprime(y) = c2 +

0uprimeprime(z)dz

0f (z)dz

bdquoZ y

0f (z)dz

laquody

and defining

F (y) =

0f (z)dz

0F (y)dy =

bdquoZ y

0f (z)dz

laquody =

0(x minus y)f (y)dy

This gives us

0(x minus y)f (y)dy

0(1minus y)f (y)dy

u(x) = xZ 1

0(x minus y)f (y)dy

G(x y) =

u(x) =

0G(x y)f (y)dy

1 continuous

3 G(0 y) = G(1 y) = G(x 0) = G(x 1) = 0

chapter 165

dx|x=y+ε minus

dx|x=yminusε = 1

W =dugt

dxult minus

and usingdG

dx|x=y+ε minus

dx|x=yminusε = 1

Simple example

ult(x) = x

G(x y) =

u(x) = λ

0G(x y)u(t)dy

u(r) =

Z infin

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

u(r) =

Z infin

resulting in

u(r) = ugt(r)Z r

Z infin

Algorithm

dxasymp

Algorithm

Thursday

Friday

E [Φ] =

af (Φ(x)

partΦ

partx x)dx

η(a) = η(b) = 0

Φ(x α) = Φ(x 0) + αη(x)

E [Φ(α)] =

af (Φ(x α)

partΦ(x α)

partx x)dx

E [Φ(α)] =

ndashα=0

partα

ndash=

bdquopartf

partΦ

partα+

partΦx

partα

laquodx

partα= η(x)

partα=

d(η(x))

UsingpartΦ(x α)

partα= η(x)

andpartΦx (x α)

partα=

d(η(x))

partα

ndash=

bdquopartf

partΦη(x) +

partΦx

d(η(x))

laquodx

partΦx

d(η(x))

dxdx = η(x)

partΦx|ba minus

aη(x)

partΦxdx

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndashα=0

bdquopartf

partΦminus

partΦx

partΦminus

partΦx= 0

df = 0

partx=partf

party=partf

partz= 0

due to

df =partf

partxdx +

partydy +

partzdz

φ(x y z) = 0

resulting in

dφ =partφ

partxdx +

partφ

partydy +

partφ

partzdz = 0

partx=partf

party=partf

partz= 0

df =partf

partxdx +

partydy +

partzdz

df + λdφ = (partf

partz+ λ

partφ

partx)dx + (

party+ λ

partφ

party)dy + (

partz+ λ

partφ

partz)dz = 0

partz+ λ

partφ

partz= 0

partx+ λ

partφ

partx= 0

andpartf

party+ λ

partφ

party= 0

partxi+X

λkpartφk

partxi= 0

and a Hamiltonian

H = minus1

2nabla2 + V (x y z)

Zdxdydz

δE = δ

Zdxdydz

bdquo1

laquoff= 0

Zdxdydz

bdquo1

laquoff= 0

partψlowastminus

partψlowastxminus

partψlowastyminus

partψlowastz= 0

which yields

minus1

H = minusNX

2nabla2

i minusNX

NXiltj

H = H1 + H2 =NX

hi +NX

iltj=1

hi = minus1

2nabla2

i minusZ

E0 le E [Φ] =

ZΦlowastHΦdτ

radicN

(minus)PP

[H1A] = [H2A] = 0

ZΦlowastH1Φdτ =

(minus)PZ

ZΦlowastH1Φdτ =

ZΦlowastH hiΦHdτ

ZΦlowastH1Φdτ =

NXmicro=1

〈micro|h|micro〉

ΦlowastH2Φdτ = N

which reduces to

ZΦlowastH2Φdτ =

NXilej=1

(minus)PZ

ΦlowastH1

rijPΦHdτ

ZΦlowastH2Φdτ =

NXiltj=1

ZΦlowastH

ZΦlowastH2Φdτ =

NXmicro=1

NXν=1

lowastν(r j )

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

lowastν(r j )

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

ψa =Xλ

Caλψλ

lectures

Exercise

Caλψλ

δE minusNX

micro=1

NXν=1

εmicroνδ

δE minusNX

micro=1

εmicroδ

NXmicro=1

NXν=1

Zδψlowastmicroψ

lowastν

Zδψlowastmicroψ

lowastν

micro=1

NXmicro=1

NXν=1

Zψlowastmicroψ

lowastν

Zψlowastmicroψ

lowastν

minusNX

micro=1

Emicro

NXmicro=1

Emicro

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

rijψν(r j )dr j

ψmicro(r i )

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

V dmicro(r i ) =

2(2l prime + 1)

Z infin

rgtdrj

withφnlml

Z infin

(rj )|21

rijdr j

which reads

Z infin

rijdrj

2l prime + 1

resulting in

2(2l prime + 1)

Z infin

rijdrj dΩj

infinXl=0

bdquorj

laquol

Pl cos(θ))

if ri gt rj or

infinXl=0

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Ylowastlmprimel

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

(2g + 1)

timesg

programs

Thursday

Friday

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

2φ100(r1)φ100(r2)

minus1radic

The Hartree term

V dmicro(r i ) =

ψλ(r i )

NXmicro=1

V dmicro(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψλ(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψ100uarr(r i )

Φ(ri )u10 = 2V d10(ri )u10(ri ) = 2

Z infin

0|u10(rj )|2

rgtdrj )u10(ri )

rijψλ(r j )dr j

ψmicro(r i )

rijψλ(r j )dr j

ψmicro(r i )

s =darr That is

s mλs

rijψλ(r j )dr j

ψmicro(r i )

NXmicro=1

s uarr

ψmicro(r i )

resulting in

NXmicro=1

ψ100uarr(r i )

NXmicro=1

ψ100uarr(r i )

minus1

l(l + 1)

2r2minus

as bdquominus

l(l + 1)

2r2minus

r+ 2V d

10(r)laquo

or bdquominus

dr2minus

r+ V d

10(r)laquo

u10(r) = e10u10(r)

dangerous later

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ100uarr(r i )

= (2V d10(r i ) + 2V d

20(r i ))ψ100uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

V ex20 (r) =

Z infin

0dr jφ

lowast200(r j )

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ200uarr(r i ) =

(2V d10(r i ) + 2V d

20(r i ))ψ200uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

V ex10 (r) =

Z infin

0dr jφ

lowast100(r j )

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

dr2minus

r+ V d

10(r) + 2V d20(r)+

20 (r)+ = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)+

10 (r)+ = e20u20(r)

andbdquominus

2r2minus

r+ 2V d

10(r) + 2V d20(r)+

10 (r)minusV ex20 (r)+ = e21u21(r)

Useful Tests

R10(r) = 2bdquo

laquo32

R20(r) = 2bdquo

laquo32 bdquo1minus

R21(r) =1radic

bdquoZ

laquo32 Zr

We use a0 = 1

R10(r) = 2bdquo

laquo32

E [Z ] = Z 2 minus 4Z +5

Z = 2minus5

16= 16875

we do this

E [ρ] = minus1

1Ψi (r1)dr1minusZ

r1ρ(r1)dr1+

Zρ(r1)ρ(r2)

r12dr1dr2+EEXC [ρ]

ρ(r) =NX

|Ψi (r)|2

2nabla2

1 minusZ

Zρ(r2)

r12dr2 + VEXC(r1)

EEXC =

Thursday

Friday

A simple Example

wi (t = ε) = Wij wj (t = 0)

wi (t) = 1

and Xj

W (j rarr i) = 1

wi (t + 1) =X

Wij wj (t)

0 14 18 2334 58 00 14 13

w(t = 0) =

resulting in

w(t = ε) =

0 14340

resulting in

w(t = 2ε) =

0 5232132632

wi (t = infin) = 1

w(t = infin) =

0 415815315

Iteration w1 w2 w30 100000 000000 0000001 025000 075000 0000002 015625 062625 0187503 024609 052734 0226564 027848 051416 0207365 027213 053021 0197666 026608 053548 0198447 026575 053424 0200028 026656 053321 0200239 026678 053318 020005

10 026671 053332 01999811 026666 053335 02000012 026666 053334 02000013 026667 053333 020000

w(t = infin) 026667 053333 020000

w(0) =X

αi v i

αi v i =X

λtiαi v i

〈M(t)〉 =Xmicro

w(t)microMmicro

〈M(t)〉 = w(t)m

E [xk ] = 〈xk 〉 =1

xki p(xi )

〈M(t)〉 = w(t)m =X

λtiαi v i m i

〈M(t)〉 =X

λtiαi mi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

λtiαi mi

τi = minus1

logλi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

αi mi eminustτi

W (j rarr i)wj =X

W (i rarr j)wi

W (j rarr i)

W (i rarr j)=

Proof Xi

W (i rarr j)wi =X

W (j rarr i)wj = wj

W (j rarr i) = wj

Ergodicity

exp (minusβ(Ei ))

〈A〉 =X

Aj wj =

g(i rarr j) =1

A(j rarr i)

A(microrarr ν)

Then we must have

Take a new value

x (i+1) =

We have defined

ρ(x y) = min

p(y)f (x |y)

p(x)f (y |x) 1ff

Implementation

Quantum Monte Carlo

E [H] = 〈H〉 =

E0 le 〈H〉

Quantum Monte Carlo

ΨT (R) =X

aiΨi (R)

nm alowastmanR

nEnPn a2

nge E0

Quantum Monte Carlo

E [H] = 〈H〉 =

RdRΨlowastTα

(R)H(R)ΨTα(R)R

Quantum Monte Carlo

P(R) =|ψT (R)|2R|ψT (R)|2 dR

E [H] asympR

EL(R) =1

ψT (R)HψT (R)

P(R)EL(R)dR asymp1

P(Ri)EL(Ri)

Quantum Monte Carlo

˛ψα

T (R)˛2

w = P(Rp)P(R)

Quantum Monte Carlo

minus~2

part2u(r)

partr2minusbdquo

rminus

~2l(l + 1)

laquou(r) = Eu(r)

minus1

part2u(ρ)

partρ2minus

l(l + 1)

H = minus1

partρ2minus

l(l + 1)

Quantum Monte Carlo

EL(ρ) = minus1

ρminusα

bdquoαminus

Quantum Monte Carlo

〈Hn〉 =

= constant

Quantum Monte Carlo

minus2ke2

r1minus

Quantum Monte Carlo

bH = minus~2nabla2

2mminus

~2nabla22

2mminus

r1minus

P(R) =|ψT (R)|2R|ψT (R)|2 dR

Quantum Monte Carlo

EL(R) =1

ψT (R)HψT (R) =

ψT (R)

bdquominus

2nabla2

1 minusZ

EL(R) =1

RT (r1)

minus1

dr1minus

limr1rarr0

EL(R) =1

RT (r1)

bdquominus

dr1minus

laquoRT (r1)

Quantum Monte Carlo

This results in1

RT (r1)

dRT (r1)

dr1= minusZ

RT (r)

dRT (r)

dr= minus

f (rij )

wfold = wfnew

Thursday

Importance sampling

Friday

Things to do

Splines

s(x) =

Splines

yi = s(xi )

ands(xi+1) = yi+1

Splines

si (x) =fi

Splines

si (x) =fi

fi+16(xi+1minusxi )

(x minus xi )3

+ (yi+1

6 )(xi+1 minus x) (2)

Splines

(yi minus yiminus1)

Splines

3777775

2666664f1f2f3

fnminus2fnminus1

3777775 =

2666664v1v2v3

vnminus2vnminus1

3777775

operations

Splines

partt= D

bdquopartP

partxminus F

laquoP(x t)

F = 21

ΨTnablaΨT

partx(t)

y = x + DF (x)∆t + ξ

Importance Sampling

G(y x ∆t) =1

(4πD∆t)3N2exp

A(y x) = min(1 q(y x)))

q(y x) =G(x y ∆t)|ΨT (y)|2

G(y x ∆t)|ΨT (x)|2

Thursday

Friday

x = rsinθcosφ

y = rsinθsinφ

andz = rcosθ

4πε0r=

4πε0p

x2 + y2 + z2

4π(l + ml )Pml

for l = ml = 0

4πcos(θ)

for l = 1 og ml = 0

Y1plusmn1 =

(r) through

Ylml(r) = r l Ylml

we obtainΨnlml

equation

Slminusml

1radic

(minus1)m

minus(minus1)ml i i

Clminusml

We have definedYlml

(r) = r l Ylml(r)

2l + 1Yl0(r)

Slml= (minus1)ml

2l + 1ReYl0(r)

Slml= (minus1)ml

2l + 1ImYl0(r)

for ml gt 0

mll 0 1 2 3

q52 (x2 minus 3y2)x

+1 xradic

3xz 12

q32 (5z2 minus r2)x

0 1 y 12 (3z2 minus r2) 1

2 (5z2 minus 3r2)x

-1 zradic

3yz 12

q32 (5z2 minus r2)y

-2radic

3xyradic

q52 (3x2 minus y2)y

Thursday

Det uarr (1 2) =

˛1radic

Det darr (3 4) =

˛1radic

Ψα(Xnew )

Ψα(Xold )=

Duarrα(Xnew )

Duarrα(Xold )

Ddarrα(Xnew )

Ddarrα(Xold )

Gβ(Xnew )

Gβ(Xold )

nablaΨα

Ψα=nablaDuarrα

Ddarrα+nablaGβ

nabla2Ψα

Ψα=nabla2Duarrα

Ddarrα+nabla2Gβ

Gβ+ (

nablaGβ

Gβ)(nablaDuarrα

Ddarrα)

Dminus1ij =

Dα(y)

Dα(x)=

Dij (y)Cij (y)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

Dij (y)Cij (x)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

nablai Dα(x)

Dα(x)=

nabla2i Dα(x)

Dα(x)=

nablai Dα(y)

Dα(y)=

andnabla2

i Dα(y)

Dα(y)=

R equivNX

Dminus1kj (y) =

kj (x)minus1

RDminus1

ki (x)NX

RDminus1

ki (x) for j = i

R =D(y)

LU Decomposition

1CCA =

0BB1 0 0 0

b21 1 0 0b31 b32 1 0b41 b42 b43 1

1CCA0BB

c11 c12 c13 c140 c22 c23 c240 0 c33 c340 0 0 c44

detA = c11c22 cnn

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

nablaGβ

Gβ=Xiltj

(1 + βrij )2

nabla2Gβ

Gβ=Xiltj

(1 + βrij )4

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

a = 14

a = 12

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

Things to do

dt= ~v

dt= minusξ~v + ~F

ξ = 6πηam

0 + 2Z t

0dτ prime

〈~F (t)〉 = 0

δ(t minus t prime)

0 eminus2ξt +C~v0

we get

~r(t) = ~r0 + ~v01

0dτZ τ

Rothman and Zaleski

Quantum Monte Carlo

wherer1 rA

Quantum Monte Carlo

There are

2A timesbdquo

MPI_Command_name

MPI_COMMAND_NAME

processor speed

memory sizespeed

Speed-up

S(P) =T (1)

Efficiency

η(P) =S(P)

tC(L) = τ + βL

T (P) ge T (1)

Prarr S(P) le P

MPI_Command_name

MPI_COMMAND_NAME

Communicator

The Fortran program

END PROGRAM hello

Note 1

Note 2

Note 3

100 values

Note 4

serious work

Integrating π

Examples

Go to MPI examples

Integration algos

af (x)dx asymp h

f (ximinus12)

example5cpp

mpirun -np 10 progx

MPI reduce

Call as

end of main program

function MPI Wtime

Strategies

Thursday

Friday

minus1

dr2unl (r) +

bdquol(l + 1)

2r2minus

r+ Φ(r)

problem

minus1

dr2unl (r) +

bdquol(l + 1)

and rewrite as

minus1

k2(r) = V (r) +l(l + 1)

2r2minus enl

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

V (ri ) = minusZri

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

programs catalogue

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

r+ Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr+Z infin

0unl (r)Fnl (r)dr

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

2Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr + EFock

EFock =1

Z infin

0unl (r)Fnl (r)dr

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

dx2y(x) + k2(x)y(x) = F (x)

h2+ O(h2)

h2+ O(h2) + k2

i yi = Fi

2minus h2k2i

rdquoyi + h2Fi

Numerovrsquos algo

y(x + h) = y(x) + hy (1)(x) +h2

2y (2)(x) +

3y (3)(x) +

Numerovrsquos algo

12y (4)(x) + O(h6)

We obtain

h2minus

12y (4)(x) + O(h4)

Numerovrsquos algo

y (2)(x) +h2

12y (4)(x) + k2(x)y(x) +

ldquok2(x)y(x)

rdquoasymp 0

Numerovrsquos algo

d2(k2y(x))

acuteh2

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

+ O(h6)

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

minus1

dr2u(r) +

bdquoV (r) +

l(l + 1)

laquou(r) = eu(r)

dr2u(r) + k2(r)u(r) = 0

k2(r) = γ

bdquof (r)minus

l(l + 1)

r2minus ε

laquoγ = 2V0

ε =|E |V0

f (r) =

forr le ar gt a

k2(r) = γ

(1minus εminus 1

γl(l+1)

l(l+1)

forr le ar gt a

dr2u(r)minus

l(l + 1)

r2u(r) = 0

well defined

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

drult(r) =

drugt(r) at r = rm

drult(rm) asymp

drugt(rm) asymp

Numerovcpp

wf[loop] = norm

Todayrsquos tasks

2reminusr

u(r) =

bdquo1

1minus a2

raquo1 +

2(1minus a2)r

ndashlaquo

Thursday

Friday

u(x) = c1 +

0uprime(y)dy

and a constant c2

uprime(y) = c2 +

0uprimeprime(z)dz

0f (z)dz

bdquoZ y

0f (z)dz

laquody

and defining

F (y) =

0f (z)dz

0F (y)dy =

bdquoZ y

0f (z)dz

laquody =

0(x minus y)f (y)dy

This gives us

0(x minus y)f (y)dy

0(1minus y)f (y)dy

u(x) = xZ 1

0(x minus y)f (y)dy

G(x y) =

u(x) =

0G(x y)f (y)dy

1 continuous

3 G(0 y) = G(1 y) = G(x 0) = G(x 1) = 0

chapter 165

dx|x=y+ε minus

dx|x=yminusε = 1

W =dugt

dxult minus

and usingdG

dx|x=y+ε minus

dx|x=yminusε = 1

Simple example

ult(x) = x

G(x y) =

u(x) = λ

0G(x y)u(t)dy

u(r) =

Z infin

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

u(r) =

Z infin

resulting in

u(r) = ugt(r)Z r

Z infin

Algorithm

dxasymp

Algorithm

Thursday

Friday

E [Φ] =

af (Φ(x)

partΦ

partx x)dx

η(a) = η(b) = 0

Φ(x α) = Φ(x 0) + αη(x)

E [Φ(α)] =

af (Φ(x α)

partΦ(x α)

partx x)dx

E [Φ(α)] =

ndashα=0

partα

ndash=

bdquopartf

partΦ

partα+

partΦx

partα

laquodx

partα= η(x)

partα=

d(η(x))

UsingpartΦ(x α)

partα= η(x)

andpartΦx (x α)

partα=

d(η(x))

partα

ndash=

bdquopartf

partΦη(x) +

partΦx

d(η(x))

laquodx

partΦx

d(η(x))

dxdx = η(x)

partΦx|ba minus

aη(x)

partΦxdx

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndashα=0

bdquopartf

partΦminus

partΦx

partΦminus

partΦx= 0

df = 0

partx=partf

party=partf

partz= 0

due to

df =partf

partxdx +

partydy +

partzdz

φ(x y z) = 0

resulting in

dφ =partφ

partxdx +

partφ

partydy +

partφ

partzdz = 0

partx=partf

party=partf

partz= 0

df =partf

partxdx +

partydy +

partzdz

df + λdφ = (partf

partz+ λ

partφ

partx)dx + (

party+ λ

partφ

party)dy + (

partz+ λ

partφ

partz)dz = 0

partz+ λ

partφ

partz= 0

partx+ λ

partφ

partx= 0

andpartf

party+ λ

partφ

party= 0

partxi+X

λkpartφk

partxi= 0

and a Hamiltonian

H = minus1

2nabla2 + V (x y z)

Zdxdydz

δE = δ

Zdxdydz

bdquo1

laquoff= 0

Zdxdydz

bdquo1

laquoff= 0

partψlowastminus

partψlowastxminus

partψlowastyminus

partψlowastz= 0

which yields

minus1

H = minusNX

2nabla2

i minusNX

NXiltj

H = H1 + H2 =NX

hi +NX

iltj=1

hi = minus1

2nabla2

i minusZ

E0 le E [Φ] =

ZΦlowastHΦdτ

radicN

(minus)PP

[H1A] = [H2A] = 0

ZΦlowastH1Φdτ =

(minus)PZ

ZΦlowastH1Φdτ =

ZΦlowastH hiΦHdτ

ZΦlowastH1Φdτ =

NXmicro=1

〈micro|h|micro〉

ΦlowastH2Φdτ = N

which reduces to

ZΦlowastH2Φdτ =

NXilej=1

(minus)PZ

ΦlowastH1

rijPΦHdτ

ZΦlowastH2Φdτ =

NXiltj=1

ZΦlowastH

ZΦlowastH2Φdτ =

NXmicro=1

NXν=1

lowastν(r j )

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

lowastν(r j )

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

ψa =Xλ

Caλψλ

lectures

Exercise

Caλψλ

δE minusNX

micro=1

NXν=1

εmicroνδ

δE minusNX

micro=1

εmicroδ

NXmicro=1

NXν=1

Zδψlowastmicroψ

lowastν

Zδψlowastmicroψ

lowastν

micro=1

NXmicro=1

NXν=1

Zψlowastmicroψ

lowastν

Zψlowastmicroψ

lowastν

minusNX

micro=1

Emicro

NXmicro=1

Emicro

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

rijψν(r j )dr j

ψmicro(r i )

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

V dmicro(r i ) =

2(2l prime + 1)

Z infin

rgtdrj

withφnlml

Z infin

(rj )|21

rijdr j

which reads

Z infin

rijdrj

2l prime + 1

resulting in

2(2l prime + 1)

Z infin

rijdrj dΩj

infinXl=0

bdquorj

laquol

Pl cos(θ))

if ri gt rj or

infinXl=0

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Ylowastlmprimel

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

(2g + 1)

timesg

programs

Thursday

Friday

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

2φ100(r1)φ100(r2)

minus1radic

The Hartree term

V dmicro(r i ) =

ψλ(r i )

NXmicro=1

V dmicro(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψλ(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψ100uarr(r i )

Φ(ri )u10 = 2V d10(ri )u10(ri ) = 2

Z infin

0|u10(rj )|2

rgtdrj )u10(ri )

rijψλ(r j )dr j

ψmicro(r i )

rijψλ(r j )dr j

ψmicro(r i )

s =darr That is

s mλs

rijψλ(r j )dr j

ψmicro(r i )

NXmicro=1

s uarr

ψmicro(r i )

resulting in

NXmicro=1

ψ100uarr(r i )

NXmicro=1

ψ100uarr(r i )

minus1

l(l + 1)

2r2minus

as bdquominus

l(l + 1)

2r2minus

r+ 2V d

10(r)laquo

or bdquominus

dr2minus

r+ V d

10(r)laquo

u10(r) = e10u10(r)

dangerous later

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ100uarr(r i )

= (2V d10(r i ) + 2V d

20(r i ))ψ100uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

V ex20 (r) =

Z infin

0dr jφ

lowast200(r j )

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ200uarr(r i ) =

(2V d10(r i ) + 2V d

20(r i ))ψ200uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

V ex10 (r) =

Z infin

0dr jφ

lowast100(r j )

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

dr2minus

r+ V d

10(r) + 2V d20(r)+

20 (r)+ = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)+

10 (r)+ = e20u20(r)

andbdquominus

2r2minus

r+ 2V d

10(r) + 2V d20(r)+

10 (r)minusV ex20 (r)+ = e21u21(r)

Useful Tests

R10(r) = 2bdquo

laquo32

R20(r) = 2bdquo

laquo32 bdquo1minus

R21(r) =1radic

bdquoZ

laquo32 Zr

We use a0 = 1

R10(r) = 2bdquo

laquo32

E [Z ] = Z 2 minus 4Z +5

Z = 2minus5

16= 16875

we do this

E [ρ] = minus1

1Ψi (r1)dr1minusZ

r1ρ(r1)dr1+

Zρ(r1)ρ(r2)

r12dr1dr2+EEXC [ρ]

ρ(r) =NX

|Ψi (r)|2

2nabla2

1 minusZ

Zρ(r2)

r12dr2 + VEXC(r1)

EEXC =

Thursday

Friday

A simple Example

wi (t = ε) = Wij wj (t = 0)

wi (t) = 1

and Xj

W (j rarr i) = 1

wi (t + 1) =X

Wij wj (t)

0 14 18 2334 58 00 14 13

w(t = 0) =

resulting in

w(t = ε) =

0 14340

resulting in

w(t = 2ε) =

0 5232132632

wi (t = infin) = 1

w(t = infin) =

0 415815315

Iteration w1 w2 w30 100000 000000 0000001 025000 075000 0000002 015625 062625 0187503 024609 052734 0226564 027848 051416 0207365 027213 053021 0197666 026608 053548 0198447 026575 053424 0200028 026656 053321 0200239 026678 053318 020005

10 026671 053332 01999811 026666 053335 02000012 026666 053334 02000013 026667 053333 020000

w(t = infin) 026667 053333 020000

w(0) =X

αi v i

αi v i =X

λtiαi v i

〈M(t)〉 =Xmicro

w(t)microMmicro

〈M(t)〉 = w(t)m

E [xk ] = 〈xk 〉 =1

xki p(xi )

〈M(t)〉 = w(t)m =X

λtiαi v i m i

〈M(t)〉 =X

λtiαi mi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

λtiαi mi

τi = minus1

logλi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

αi mi eminustτi

W (j rarr i)wj =X

W (i rarr j)wi

W (j rarr i)

W (i rarr j)=

Proof Xi

W (i rarr j)wi =X

W (j rarr i)wj = wj

W (j rarr i) = wj

Ergodicity

exp (minusβ(Ei ))

〈A〉 =X

Aj wj =

g(i rarr j) =1

A(j rarr i)

A(microrarr ν)

Then we must have

Take a new value

x (i+1) =

We have defined

ρ(x y) = min

p(y)f (x |y)

p(x)f (y |x) 1ff

Implementation

Quantum Monte Carlo

E [H] = 〈H〉 =

E0 le 〈H〉

Quantum Monte Carlo

ΨT (R) =X

aiΨi (R)

nm alowastmanR

nEnPn a2

nge E0

Quantum Monte Carlo

E [H] = 〈H〉 =

RdRΨlowastTα

(R)H(R)ΨTα(R)R

Quantum Monte Carlo

P(R) =|ψT (R)|2R|ψT (R)|2 dR

E [H] asympR

EL(R) =1

ψT (R)HψT (R)

P(R)EL(R)dR asymp1

P(Ri)EL(Ri)

Quantum Monte Carlo

˛ψα

T (R)˛2

w = P(Rp)P(R)

Quantum Monte Carlo

minus~2

part2u(r)

partr2minusbdquo

rminus

~2l(l + 1)

laquou(r) = Eu(r)

minus1

part2u(ρ)

partρ2minus

l(l + 1)

H = minus1

partρ2minus

l(l + 1)

Quantum Monte Carlo

EL(ρ) = minus1

ρminusα

bdquoαminus

Quantum Monte Carlo

〈Hn〉 =

= constant

Quantum Monte Carlo

minus2ke2

r1minus

Quantum Monte Carlo

bH = minus~2nabla2

2mminus

~2nabla22

2mminus

r1minus

P(R) =|ψT (R)|2R|ψT (R)|2 dR

Quantum Monte Carlo

EL(R) =1

ψT (R)HψT (R) =

ψT (R)

bdquominus

2nabla2

1 minusZ

EL(R) =1

RT (r1)

minus1

dr1minus

limr1rarr0

EL(R) =1

RT (r1)

bdquominus

dr1minus

laquoRT (r1)

Quantum Monte Carlo

This results in1

RT (r1)

dRT (r1)

dr1= minusZ

RT (r)

dRT (r)

dr= minus

f (rij )

wfold = wfnew

Thursday

Importance sampling

Friday

Things to do

Splines

s(x) =

Splines

yi = s(xi )

ands(xi+1) = yi+1

Splines

si (x) =fi

Splines

si (x) =fi

fi+16(xi+1minusxi )

(x minus xi )3

+ (yi+1

6 )(xi+1 minus x) (2)

Splines

(yi minus yiminus1)

Splines

3777775

2666664f1f2f3

fnminus2fnminus1

3777775 =

2666664v1v2v3

vnminus2vnminus1

3777775

operations

Splines

partt= D

bdquopartP

partxminus F

laquoP(x t)

F = 21

ΨTnablaΨT

partx(t)

y = x + DF (x)∆t + ξ

Importance Sampling

G(y x ∆t) =1

(4πD∆t)3N2exp

A(y x) = min(1 q(y x)))

q(y x) =G(x y ∆t)|ΨT (y)|2

G(y x ∆t)|ΨT (x)|2

Thursday

Friday

x = rsinθcosφ

y = rsinθsinφ

andz = rcosθ

4πε0r=

4πε0p

x2 + y2 + z2

4π(l + ml )Pml

for l = ml = 0

4πcos(θ)

for l = 1 og ml = 0

Y1plusmn1 =

(r) through

Ylml(r) = r l Ylml

we obtainΨnlml

equation

Slminusml

1radic

(minus1)m

minus(minus1)ml i i

Clminusml

We have definedYlml

(r) = r l Ylml(r)

2l + 1Yl0(r)

Slml= (minus1)ml

2l + 1ReYl0(r)

Slml= (minus1)ml

2l + 1ImYl0(r)

for ml gt 0

mll 0 1 2 3

q52 (x2 minus 3y2)x

+1 xradic

3xz 12

q32 (5z2 minus r2)x

0 1 y 12 (3z2 minus r2) 1

2 (5z2 minus 3r2)x

-1 zradic

3yz 12

q32 (5z2 minus r2)y

-2radic

3xyradic

q52 (3x2 minus y2)y

Thursday

Det uarr (1 2) =

˛1radic

Det darr (3 4) =

˛1radic

Ψα(Xnew )

Ψα(Xold )=

Duarrα(Xnew )

Duarrα(Xold )

Ddarrα(Xnew )

Ddarrα(Xold )

Gβ(Xnew )

Gβ(Xold )

nablaΨα

Ψα=nablaDuarrα

Ddarrα+nablaGβ

nabla2Ψα

Ψα=nabla2Duarrα

Ddarrα+nabla2Gβ

Gβ+ (

nablaGβ

Gβ)(nablaDuarrα

Ddarrα)

Dminus1ij =

Dα(y)

Dα(x)=

Dij (y)Cij (y)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

Dij (y)Cij (x)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

nablai Dα(x)

Dα(x)=

nabla2i Dα(x)

Dα(x)=

nablai Dα(y)

Dα(y)=

andnabla2

i Dα(y)

Dα(y)=

R equivNX

Dminus1kj (y) =

kj (x)minus1

RDminus1

ki (x)NX

RDminus1

ki (x) for j = i

R =D(y)

LU Decomposition

1CCA =

0BB1 0 0 0

b21 1 0 0b31 b32 1 0b41 b42 b43 1

1CCA0BB

c11 c12 c13 c140 c22 c23 c240 0 c33 c340 0 0 c44

detA = c11c22 cnn

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

nablaGβ

Gβ=Xiltj

(1 + βrij )2

nabla2Gβ

Gβ=Xiltj

(1 + βrij )4

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

a = 14

a = 12

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

Things to do

dt= ~v

dt= minusξ~v + ~F

ξ = 6πηam

0 + 2Z t

0dτ prime

〈~F (t)〉 = 0

δ(t minus t prime)

0 eminus2ξt +C~v0

we get

~r(t) = ~r0 + ~v01

0dτZ τ

Quantum Monte Carlo

wherer1 rA

Quantum Monte Carlo

There are

2A timesbdquo

MPI_Command_name

MPI_COMMAND_NAME

processor speed

memory sizespeed

Speed-up

S(P) =T (1)

Efficiency

η(P) =S(P)

tC(L) = τ + βL

T (P) ge T (1)

Prarr S(P) le P

MPI_Command_name

MPI_COMMAND_NAME

Communicator

The Fortran program

END PROGRAM hello

Note 1

Note 2

Note 3

100 values

Note 4

serious work

Integrating π

Examples

Go to MPI examples

Integration algos

af (x)dx asymp h

f (ximinus12)

example5cpp

mpirun -np 10 progx

MPI reduce

Call as

end of main program

function MPI Wtime

Strategies

Thursday

Friday

minus1

dr2unl (r) +

bdquol(l + 1)

2r2minus

r+ Φ(r)

problem

minus1

dr2unl (r) +

bdquol(l + 1)

and rewrite as

minus1

k2(r) = V (r) +l(l + 1)

2r2minus enl

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

V (ri ) = minusZri

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

programs catalogue

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

r+ Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr+Z infin

0unl (r)Fnl (r)dr

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

2Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr + EFock

EFock =1

Z infin

0unl (r)Fnl (r)dr

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

dx2y(x) + k2(x)y(x) = F (x)

h2+ O(h2)

h2+ O(h2) + k2

i yi = Fi

2minus h2k2i

rdquoyi + h2Fi

Numerovrsquos algo

y(x + h) = y(x) + hy (1)(x) +h2

2y (2)(x) +

3y (3)(x) +

Numerovrsquos algo

12y (4)(x) + O(h6)

We obtain

h2minus

12y (4)(x) + O(h4)

Numerovrsquos algo

y (2)(x) +h2

12y (4)(x) + k2(x)y(x) +

ldquok2(x)y(x)

rdquoasymp 0

Numerovrsquos algo

d2(k2y(x))

acuteh2

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

+ O(h6)

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

minus1

dr2u(r) +

bdquoV (r) +

l(l + 1)

laquou(r) = eu(r)

dr2u(r) + k2(r)u(r) = 0

k2(r) = γ

bdquof (r)minus

l(l + 1)

r2minus ε

laquoγ = 2V0

ε =|E |V0

f (r) =

forr le ar gt a

k2(r) = γ

(1minus εminus 1

γl(l+1)

l(l+1)

forr le ar gt a

dr2u(r)minus

l(l + 1)

r2u(r) = 0

well defined

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

drult(r) =

drugt(r) at r = rm

drult(rm) asymp

drugt(rm) asymp

Numerovcpp

wf[loop] = norm

Todayrsquos tasks

2reminusr

u(r) =

bdquo1

1minus a2

raquo1 +

2(1minus a2)r

ndashlaquo

Thursday

Friday

u(x) = c1 +

0uprime(y)dy

and a constant c2

uprime(y) = c2 +

0uprimeprime(z)dz

0f (z)dz

bdquoZ y

0f (z)dz

laquody

and defining

F (y) =

0f (z)dz

0F (y)dy =

bdquoZ y

0f (z)dz

laquody =

0(x minus y)f (y)dy

This gives us

0(x minus y)f (y)dy

0(1minus y)f (y)dy

u(x) = xZ 1

0(x minus y)f (y)dy

G(x y) =

u(x) =

0G(x y)f (y)dy

1 continuous

3 G(0 y) = G(1 y) = G(x 0) = G(x 1) = 0

chapter 165

dx|x=y+ε minus

dx|x=yminusε = 1

W =dugt

dxult minus

and usingdG

dx|x=y+ε minus

dx|x=yminusε = 1

Simple example

ult(x) = x

G(x y) =

u(x) = λ

0G(x y)u(t)dy

u(r) =

Z infin

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

u(r) =

Z infin

resulting in

u(r) = ugt(r)Z r

Z infin

Algorithm

dxasymp

Algorithm

Thursday

Friday

E [Φ] =

af (Φ(x)

partΦ

partx x)dx

η(a) = η(b) = 0

Φ(x α) = Φ(x 0) + αη(x)

E [Φ(α)] =

af (Φ(x α)

partΦ(x α)

partx x)dx

E [Φ(α)] =

ndashα=0

partα

ndash=

bdquopartf

partΦ

partα+

partΦx

partα

laquodx

partα= η(x)

partα=

d(η(x))

UsingpartΦ(x α)

partα= η(x)

andpartΦx (x α)

partα=

d(η(x))

partα

ndash=

bdquopartf

partΦη(x) +

partΦx

d(η(x))

laquodx

partΦx

d(η(x))

dxdx = η(x)

partΦx|ba minus

aη(x)

partΦxdx

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndashα=0

bdquopartf

partΦminus

partΦx

partΦminus

partΦx= 0

df = 0

partx=partf

party=partf

partz= 0

due to

df =partf

partxdx +

partydy +

partzdz

φ(x y z) = 0

resulting in

dφ =partφ

partxdx +

partφ

partydy +

partφ

partzdz = 0

partx=partf

party=partf

partz= 0

df =partf

partxdx +

partydy +

partzdz

df + λdφ = (partf

partz+ λ

partφ

partx)dx + (

party+ λ

partφ

party)dy + (

partz+ λ

partφ

partz)dz = 0

partz+ λ

partφ

partz= 0

partx+ λ

partφ

partx= 0

andpartf

party+ λ

partφ

party= 0

partxi+X

λkpartφk

partxi= 0

and a Hamiltonian

H = minus1

2nabla2 + V (x y z)

Zdxdydz

δE = δ

Zdxdydz

bdquo1

laquoff= 0

Zdxdydz

bdquo1

laquoff= 0

partψlowastminus

partψlowastxminus

partψlowastyminus

partψlowastz= 0

which yields

minus1

H = minusNX

2nabla2

i minusNX

NXiltj

H = H1 + H2 =NX

hi +NX

iltj=1

hi = minus1

2nabla2

i minusZ

E0 le E [Φ] =

ZΦlowastHΦdτ

radicN

(minus)PP

[H1A] = [H2A] = 0

ZΦlowastH1Φdτ =

(minus)PZ

ZΦlowastH1Φdτ =

ZΦlowastH hiΦHdτ

ZΦlowastH1Φdτ =

NXmicro=1

〈micro|h|micro〉

ΦlowastH2Φdτ = N

which reduces to

ZΦlowastH2Φdτ =

NXilej=1

(minus)PZ

ΦlowastH1

rijPΦHdτ

ZΦlowastH2Φdτ =

NXiltj=1

ZΦlowastH

ZΦlowastH2Φdτ =

NXmicro=1

NXν=1

lowastν(r j )

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

lowastν(r j )

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

ψa =Xλ

Caλψλ

lectures

Exercise

Caλψλ

δE minusNX

micro=1

NXν=1

εmicroνδ

δE minusNX

micro=1

εmicroδ

NXmicro=1

NXν=1

Zδψlowastmicroψ

lowastν

Zδψlowastmicroψ

lowastν

micro=1

NXmicro=1

NXν=1

Zψlowastmicroψ

lowastν

Zψlowastmicroψ

lowastν

minusNX

micro=1

Emicro

NXmicro=1

Emicro

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

rijψν(r j )dr j

ψmicro(r i )

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

V dmicro(r i ) =

2(2l prime + 1)

Z infin

rgtdrj

withφnlml

Z infin

(rj )|21

rijdr j

which reads

Z infin

rijdrj

2l prime + 1

resulting in

2(2l prime + 1)

Z infin

rijdrj dΩj

infinXl=0

bdquorj

laquol

Pl cos(θ))

if ri gt rj or

infinXl=0

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Ylowastlmprimel

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

(2g + 1)

timesg

programs

Thursday

Friday

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

2φ100(r1)φ100(r2)

minus1radic

The Hartree term

V dmicro(r i ) =

ψλ(r i )

NXmicro=1

V dmicro(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψλ(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψ100uarr(r i )

Φ(ri )u10 = 2V d10(ri )u10(ri ) = 2

Z infin

0|u10(rj )|2

rgtdrj )u10(ri )

rijψλ(r j )dr j

ψmicro(r i )

rijψλ(r j )dr j

ψmicro(r i )

s =darr That is

s mλs

rijψλ(r j )dr j

ψmicro(r i )

NXmicro=1

s uarr

ψmicro(r i )

resulting in

NXmicro=1

ψ100uarr(r i )

NXmicro=1

ψ100uarr(r i )

minus1

l(l + 1)

2r2minus

as bdquominus

l(l + 1)

2r2minus

r+ 2V d

10(r)laquo

or bdquominus

dr2minus

r+ V d

10(r)laquo

u10(r) = e10u10(r)

dangerous later

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ100uarr(r i )

= (2V d10(r i ) + 2V d

20(r i ))ψ100uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

V ex20 (r) =

Z infin

0dr jφ

lowast200(r j )

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ200uarr(r i ) =

(2V d10(r i ) + 2V d

20(r i ))ψ200uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

V ex10 (r) =

Z infin

0dr jφ

lowast100(r j )

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

dr2minus

r+ V d

10(r) + 2V d20(r)+

20 (r)+ = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)+

10 (r)+ = e20u20(r)

andbdquominus

2r2minus

r+ 2V d

10(r) + 2V d20(r)+

10 (r)minusV ex20 (r)+ = e21u21(r)

Useful Tests

R10(r) = 2bdquo

laquo32

R20(r) = 2bdquo

laquo32 bdquo1minus

R21(r) =1radic

bdquoZ

laquo32 Zr

We use a0 = 1

R10(r) = 2bdquo

laquo32

E [Z ] = Z 2 minus 4Z +5

Z = 2minus5

16= 16875

we do this

E [ρ] = minus1

1Ψi (r1)dr1minusZ

r1ρ(r1)dr1+

Zρ(r1)ρ(r2)

r12dr1dr2+EEXC [ρ]

ρ(r) =NX

|Ψi (r)|2

2nabla2

1 minusZ

Zρ(r2)

r12dr2 + VEXC(r1)

EEXC =

Thursday

Friday

A simple Example

wi (t = ε) = Wij wj (t = 0)

wi (t) = 1

and Xj

W (j rarr i) = 1

wi (t + 1) =X

Wij wj (t)

0 14 18 2334 58 00 14 13

w(t = 0) =

resulting in

w(t = ε) =

0 14340

resulting in

w(t = 2ε) =

0 5232132632

wi (t = infin) = 1

w(t = infin) =

0 415815315

Iteration w1 w2 w30 100000 000000 0000001 025000 075000 0000002 015625 062625 0187503 024609 052734 0226564 027848 051416 0207365 027213 053021 0197666 026608 053548 0198447 026575 053424 0200028 026656 053321 0200239 026678 053318 020005

10 026671 053332 01999811 026666 053335 02000012 026666 053334 02000013 026667 053333 020000

w(t = infin) 026667 053333 020000

w(0) =X

αi v i

αi v i =X

λtiαi v i

〈M(t)〉 =Xmicro

w(t)microMmicro

〈M(t)〉 = w(t)m

E [xk ] = 〈xk 〉 =1

xki p(xi )

〈M(t)〉 = w(t)m =X

λtiαi v i m i

〈M(t)〉 =X

λtiαi mi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

λtiαi mi

τi = minus1

logλi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

αi mi eminustτi

W (j rarr i)wj =X

W (i rarr j)wi

W (j rarr i)

W (i rarr j)=

Proof Xi

W (i rarr j)wi =X

W (j rarr i)wj = wj

W (j rarr i) = wj

Ergodicity

exp (minusβ(Ei ))

〈A〉 =X

Aj wj =

g(i rarr j) =1

A(j rarr i)

A(microrarr ν)

Then we must have

Take a new value

x (i+1) =

We have defined

ρ(x y) = min

p(y)f (x |y)

p(x)f (y |x) 1ff

Implementation

Quantum Monte Carlo

E [H] = 〈H〉 =

E0 le 〈H〉

Quantum Monte Carlo

ΨT (R) =X

aiΨi (R)

nm alowastmanR

nEnPn a2

nge E0

Quantum Monte Carlo

E [H] = 〈H〉 =

RdRΨlowastTα

(R)H(R)ΨTα(R)R

Quantum Monte Carlo

P(R) =|ψT (R)|2R|ψT (R)|2 dR

E [H] asympR

EL(R) =1

ψT (R)HψT (R)

P(R)EL(R)dR asymp1

P(Ri)EL(Ri)

Quantum Monte Carlo

˛ψα

T (R)˛2

w = P(Rp)P(R)

Quantum Monte Carlo

minus~2

part2u(r)

partr2minusbdquo

rminus

~2l(l + 1)

laquou(r) = Eu(r)

minus1

part2u(ρ)

partρ2minus

l(l + 1)

H = minus1

partρ2minus

l(l + 1)

Quantum Monte Carlo

EL(ρ) = minus1

ρminusα

bdquoαminus

Quantum Monte Carlo

〈Hn〉 =

= constant

Quantum Monte Carlo

minus2ke2

r1minus

Quantum Monte Carlo

bH = minus~2nabla2

2mminus

~2nabla22

2mminus

r1minus

P(R) =|ψT (R)|2R|ψT (R)|2 dR

Quantum Monte Carlo

EL(R) =1

ψT (R)HψT (R) =

ψT (R)

bdquominus

2nabla2

1 minusZ

EL(R) =1

RT (r1)

minus1

dr1minus

limr1rarr0

EL(R) =1

RT (r1)

bdquominus

dr1minus

laquoRT (r1)

Quantum Monte Carlo

This results in1

RT (r1)

dRT (r1)

dr1= minusZ

RT (r)

dRT (r)

dr= minus

f (rij )

wfold = wfnew

Thursday

Importance sampling

Friday

Things to do

Splines

s(x) =

Splines

yi = s(xi )

ands(xi+1) = yi+1

Splines

si (x) =fi

Splines

si (x) =fi

fi+16(xi+1minusxi )

(x minus xi )3

+ (yi+1

6 )(xi+1 minus x) (2)

Splines

(yi minus yiminus1)

Splines

3777775

2666664f1f2f3

fnminus2fnminus1

3777775 =

2666664v1v2v3

vnminus2vnminus1

3777775

operations

Splines

partt= D

bdquopartP

partxminus F

laquoP(x t)

F = 21

ΨTnablaΨT

partx(t)

y = x + DF (x)∆t + ξ

Importance Sampling

G(y x ∆t) =1

(4πD∆t)3N2exp

A(y x) = min(1 q(y x)))

q(y x) =G(x y ∆t)|ΨT (y)|2

G(y x ∆t)|ΨT (x)|2

Thursday

Friday

x = rsinθcosφ

y = rsinθsinφ

andz = rcosθ

4πε0r=

4πε0p

x2 + y2 + z2

4π(l + ml )Pml

for l = ml = 0

4πcos(θ)

for l = 1 og ml = 0

Y1plusmn1 =

(r) through

Ylml(r) = r l Ylml

we obtainΨnlml

equation

Slminusml

1radic

(minus1)m

minus(minus1)ml i i

Clminusml

We have definedYlml

(r) = r l Ylml(r)

2l + 1Yl0(r)

Slml= (minus1)ml

2l + 1ReYl0(r)

Slml= (minus1)ml

2l + 1ImYl0(r)

for ml gt 0

mll 0 1 2 3

q52 (x2 minus 3y2)x

+1 xradic

3xz 12

q32 (5z2 minus r2)x

0 1 y 12 (3z2 minus r2) 1

2 (5z2 minus 3r2)x

-1 zradic

3yz 12

q32 (5z2 minus r2)y

-2radic

3xyradic

q52 (3x2 minus y2)y

Thursday

Det uarr (1 2) =

˛1radic

Det darr (3 4) =

˛1radic

Ψα(Xnew )

Ψα(Xold )=

Duarrα(Xnew )

Duarrα(Xold )

Ddarrα(Xnew )

Ddarrα(Xold )

Gβ(Xnew )

Gβ(Xold )

nablaΨα

Ψα=nablaDuarrα

Ddarrα+nablaGβ

nabla2Ψα

Ψα=nabla2Duarrα

Ddarrα+nabla2Gβ

Gβ+ (

nablaGβ

Gβ)(nablaDuarrα

Ddarrα)

Dminus1ij =

Dα(y)

Dα(x)=

Dij (y)Cij (y)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

Dij (y)Cij (x)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

nablai Dα(x)

Dα(x)=

nabla2i Dα(x)

Dα(x)=

nablai Dα(y)

Dα(y)=

andnabla2

i Dα(y)

Dα(y)=

R equivNX

Dminus1kj (y) =

kj (x)minus1

RDminus1

ki (x)NX

RDminus1

ki (x) for j = i

R =D(y)

LU Decomposition

1CCA =

0BB1 0 0 0

b21 1 0 0b31 b32 1 0b41 b42 b43 1

1CCA0BB

c11 c12 c13 c140 c22 c23 c240 0 c33 c340 0 0 c44

detA = c11c22 cnn

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

nablaGβ

Gβ=Xiltj

(1 + βrij )2

nabla2Gβ

Gβ=Xiltj

(1 + βrij )4

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

a = 14

a = 12

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

Things to do

dt= ~v

dt= minusξ~v + ~F

ξ = 6πηam

0 + 2Z t

0dτ prime

〈~F (t)〉 = 0

δ(t minus t prime)

0 eminus2ξt +C~v0

we get

~r(t) = ~r0 + ~v01

0dτZ τ

Quantum Monte Carlo

wherer1 rA

Quantum Monte Carlo

There are

2A timesbdquo

MPI_Command_name

MPI_COMMAND_NAME

processor speed

memory sizespeed

Speed-up

S(P) =T (1)

Efficiency

η(P) =S(P)

tC(L) = τ + βL

T (P) ge T (1)

Prarr S(P) le P

MPI_Command_name

MPI_COMMAND_NAME

Communicator

The Fortran program

END PROGRAM hello

Note 1

Note 2

Note 3

100 values

Note 4

serious work

Integrating π

Examples

Go to MPI examples

Integration algos

af (x)dx asymp h

f (ximinus12)

example5cpp

mpirun -np 10 progx

MPI reduce

Call as

end of main program

function MPI Wtime

Strategies

Thursday

Friday

minus1

dr2unl (r) +

bdquol(l + 1)

2r2minus

r+ Φ(r)

problem

minus1

dr2unl (r) +

bdquol(l + 1)

and rewrite as

minus1

k2(r) = V (r) +l(l + 1)

2r2minus enl

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

V (ri ) = minusZri

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

programs catalogue

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

r+ Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr+Z infin

0unl (r)Fnl (r)dr

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

2Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr + EFock

EFock =1

Z infin

0unl (r)Fnl (r)dr

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

dx2y(x) + k2(x)y(x) = F (x)

h2+ O(h2)

h2+ O(h2) + k2

i yi = Fi

2minus h2k2i

rdquoyi + h2Fi

Numerovrsquos algo

y(x + h) = y(x) + hy (1)(x) +h2

2y (2)(x) +

3y (3)(x) +

Numerovrsquos algo

12y (4)(x) + O(h6)

We obtain

h2minus

12y (4)(x) + O(h4)

Numerovrsquos algo

y (2)(x) +h2

12y (4)(x) + k2(x)y(x) +

ldquok2(x)y(x)

rdquoasymp 0

Numerovrsquos algo

d2(k2y(x))

acuteh2

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

+ O(h6)

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

minus1

dr2u(r) +

bdquoV (r) +

l(l + 1)

laquou(r) = eu(r)

dr2u(r) + k2(r)u(r) = 0

k2(r) = γ

bdquof (r)minus

l(l + 1)

r2minus ε

laquoγ = 2V0

ε =|E |V0

f (r) =

forr le ar gt a

k2(r) = γ

(1minus εminus 1

γl(l+1)

l(l+1)

forr le ar gt a

dr2u(r)minus

l(l + 1)

r2u(r) = 0

well defined

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

drult(r) =

drugt(r) at r = rm

drult(rm) asymp

drugt(rm) asymp

Numerovcpp

wf[loop] = norm

Todayrsquos tasks

2reminusr

u(r) =

bdquo1

1minus a2

raquo1 +

2(1minus a2)r

ndashlaquo

Thursday

Friday

u(x) = c1 +

0uprime(y)dy

and a constant c2

uprime(y) = c2 +

0uprimeprime(z)dz

0f (z)dz

bdquoZ y

0f (z)dz

laquody

and defining

F (y) =

0f (z)dz

0F (y)dy =

bdquoZ y

0f (z)dz

laquody =

0(x minus y)f (y)dy

This gives us

0(x minus y)f (y)dy

0(1minus y)f (y)dy

u(x) = xZ 1

0(x minus y)f (y)dy

G(x y) =

u(x) =

0G(x y)f (y)dy

1 continuous

3 G(0 y) = G(1 y) = G(x 0) = G(x 1) = 0

chapter 165

dx|x=y+ε minus

dx|x=yminusε = 1

W =dugt

dxult minus

and usingdG

dx|x=y+ε minus

dx|x=yminusε = 1

Simple example

ult(x) = x

G(x y) =

u(x) = λ

0G(x y)u(t)dy

u(r) =

Z infin

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

u(r) =

Z infin

resulting in

u(r) = ugt(r)Z r

Z infin

Algorithm

dxasymp

Algorithm

Thursday

Friday

E [Φ] =

af (Φ(x)

partΦ

partx x)dx

η(a) = η(b) = 0

Φ(x α) = Φ(x 0) + αη(x)

E [Φ(α)] =

af (Φ(x α)

partΦ(x α)

partx x)dx

E [Φ(α)] =

ndashα=0

partα

ndash=

bdquopartf

partΦ

partα+

partΦx

partα

laquodx

partα= η(x)

partα=

d(η(x))

UsingpartΦ(x α)

partα= η(x)

andpartΦx (x α)

partα=

d(η(x))

partα

ndash=

bdquopartf

partΦη(x) +

partΦx

d(η(x))

laquodx

partΦx

d(η(x))

dxdx = η(x)

partΦx|ba minus

aη(x)

partΦxdx

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndashα=0

bdquopartf

partΦminus

partΦx

partΦminus

partΦx= 0

df = 0

partx=partf

party=partf

partz= 0

due to

df =partf

partxdx +

partydy +

partzdz

φ(x y z) = 0

resulting in

dφ =partφ

partxdx +

partφ

partydy +

partφ

partzdz = 0

partx=partf

party=partf

partz= 0

df =partf

partxdx +

partydy +

partzdz

df + λdφ = (partf

partz+ λ

partφ

partx)dx + (

party+ λ

partφ

party)dy + (

partz+ λ

partφ

partz)dz = 0

partz+ λ

partφ

partz= 0

partx+ λ

partφ

partx= 0

andpartf

party+ λ

partφ

party= 0

partxi+X

λkpartφk

partxi= 0

and a Hamiltonian

H = minus1

2nabla2 + V (x y z)

Zdxdydz

δE = δ

Zdxdydz

bdquo1

laquoff= 0

Zdxdydz

bdquo1

laquoff= 0

partψlowastminus

partψlowastxminus

partψlowastyminus

partψlowastz= 0

which yields

minus1

H = minusNX

2nabla2

i minusNX

NXiltj

H = H1 + H2 =NX

hi +NX

iltj=1

hi = minus1

2nabla2

i minusZ

E0 le E [Φ] =

ZΦlowastHΦdτ

radicN

(minus)PP

[H1A] = [H2A] = 0

ZΦlowastH1Φdτ =

(minus)PZ

ZΦlowastH1Φdτ =

ZΦlowastH hiΦHdτ

ZΦlowastH1Φdτ =

NXmicro=1

〈micro|h|micro〉

ΦlowastH2Φdτ = N

which reduces to

ZΦlowastH2Φdτ =

NXilej=1

(minus)PZ

ΦlowastH1

rijPΦHdτ

ZΦlowastH2Φdτ =

NXiltj=1

ZΦlowastH

ZΦlowastH2Φdτ =

NXmicro=1

NXν=1

lowastν(r j )

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

lowastν(r j )

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

ψa =Xλ

Caλψλ

lectures

Exercise

Caλψλ

δE minusNX

micro=1

NXν=1

εmicroνδ

δE minusNX

micro=1

εmicroδ

NXmicro=1

NXν=1

Zδψlowastmicroψ

lowastν

Zδψlowastmicroψ

lowastν

micro=1

NXmicro=1

NXν=1

Zψlowastmicroψ

lowastν

Zψlowastmicroψ

lowastν

minusNX

micro=1

Emicro

NXmicro=1

Emicro

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

rijψν(r j )dr j

ψmicro(r i )

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

V dmicro(r i ) =

2(2l prime + 1)

Z infin

rgtdrj

withφnlml

Z infin

(rj )|21

rijdr j

which reads

Z infin

rijdrj

2l prime + 1

resulting in

2(2l prime + 1)

Z infin

rijdrj dΩj

infinXl=0

bdquorj

laquol

Pl cos(θ))

if ri gt rj or

infinXl=0

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Ylowastlmprimel

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

(2g + 1)

timesg

programs

Thursday

Friday

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

2φ100(r1)φ100(r2)

minus1radic

The Hartree term

V dmicro(r i ) =

ψλ(r i )

NXmicro=1

V dmicro(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψλ(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψ100uarr(r i )

Φ(ri )u10 = 2V d10(ri )u10(ri ) = 2

Z infin

0|u10(rj )|2

rgtdrj )u10(ri )

rijψλ(r j )dr j

ψmicro(r i )

rijψλ(r j )dr j

ψmicro(r i )

s =darr That is

s mλs

rijψλ(r j )dr j

ψmicro(r i )

NXmicro=1

s uarr

ψmicro(r i )

resulting in

NXmicro=1

ψ100uarr(r i )

NXmicro=1

ψ100uarr(r i )

minus1

l(l + 1)

2r2minus

as bdquominus

l(l + 1)

2r2minus

r+ 2V d

10(r)laquo

or bdquominus

dr2minus

r+ V d

10(r)laquo

u10(r) = e10u10(r)

dangerous later

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ100uarr(r i )

= (2V d10(r i ) + 2V d

20(r i ))ψ100uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

V ex20 (r) =

Z infin

0dr jφ

lowast200(r j )

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ200uarr(r i ) =

(2V d10(r i ) + 2V d

20(r i ))ψ200uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

V ex10 (r) =

Z infin

0dr jφ

lowast100(r j )

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

dr2minus

r+ V d

10(r) + 2V d20(r)+

20 (r)+ = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)+

10 (r)+ = e20u20(r)

andbdquominus

2r2minus

r+ 2V d

10(r) + 2V d20(r)+

10 (r)minusV ex20 (r)+ = e21u21(r)

Useful Tests

R10(r) = 2bdquo

laquo32

R20(r) = 2bdquo

laquo32 bdquo1minus

R21(r) =1radic

bdquoZ

laquo32 Zr

We use a0 = 1

R10(r) = 2bdquo

laquo32

E [Z ] = Z 2 minus 4Z +5

Z = 2minus5

16= 16875

we do this

E [ρ] = minus1

1Ψi (r1)dr1minusZ

r1ρ(r1)dr1+

Zρ(r1)ρ(r2)

r12dr1dr2+EEXC [ρ]

ρ(r) =NX

|Ψi (r)|2

2nabla2

1 minusZ

Zρ(r2)

r12dr2 + VEXC(r1)

EEXC =

Thursday

Friday

A simple Example

wi (t = ε) = Wij wj (t = 0)

wi (t) = 1

and Xj

W (j rarr i) = 1

wi (t + 1) =X

Wij wj (t)

0 14 18 2334 58 00 14 13

w(t = 0) =

resulting in

w(t = ε) =

0 14340

resulting in

w(t = 2ε) =

0 5232132632

wi (t = infin) = 1

w(t = infin) =

0 415815315

Iteration w1 w2 w30 100000 000000 0000001 025000 075000 0000002 015625 062625 0187503 024609 052734 0226564 027848 051416 0207365 027213 053021 0197666 026608 053548 0198447 026575 053424 0200028 026656 053321 0200239 026678 053318 020005

10 026671 053332 01999811 026666 053335 02000012 026666 053334 02000013 026667 053333 020000

w(t = infin) 026667 053333 020000

w(0) =X

αi v i

αi v i =X

λtiαi v i

〈M(t)〉 =Xmicro

w(t)microMmicro

〈M(t)〉 = w(t)m

E [xk ] = 〈xk 〉 =1

xki p(xi )

〈M(t)〉 = w(t)m =X

λtiαi v i m i

〈M(t)〉 =X

λtiαi mi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

λtiαi mi

τi = minus1

logλi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

αi mi eminustτi

W (j rarr i)wj =X

W (i rarr j)wi

W (j rarr i)

W (i rarr j)=

Proof Xi

W (i rarr j)wi =X

W (j rarr i)wj = wj

W (j rarr i) = wj

Ergodicity

exp (minusβ(Ei ))

〈A〉 =X

Aj wj =

g(i rarr j) =1

A(j rarr i)

A(microrarr ν)

Then we must have

Take a new value

x (i+1) =

We have defined

ρ(x y) = min

p(y)f (x |y)

p(x)f (y |x) 1ff

Implementation

Quantum Monte Carlo

E [H] = 〈H〉 =

E0 le 〈H〉

Quantum Monte Carlo

ΨT (R) =X

aiΨi (R)

nm alowastmanR

nEnPn a2

nge E0

Quantum Monte Carlo

E [H] = 〈H〉 =

RdRΨlowastTα

(R)H(R)ΨTα(R)R

Quantum Monte Carlo

P(R) =|ψT (R)|2R|ψT (R)|2 dR

E [H] asympR

EL(R) =1

ψT (R)HψT (R)

P(R)EL(R)dR asymp1

P(Ri)EL(Ri)

Quantum Monte Carlo

˛ψα

T (R)˛2

w = P(Rp)P(R)

Quantum Monte Carlo

minus~2

part2u(r)

partr2minusbdquo

rminus

~2l(l + 1)

laquou(r) = Eu(r)

minus1

part2u(ρ)

partρ2minus

l(l + 1)

H = minus1

partρ2minus

l(l + 1)

Quantum Monte Carlo

EL(ρ) = minus1

ρminusα

bdquoαminus

Quantum Monte Carlo

〈Hn〉 =

= constant

Quantum Monte Carlo

minus2ke2

r1minus

Quantum Monte Carlo

bH = minus~2nabla2

2mminus

~2nabla22

2mminus

r1minus

P(R) =|ψT (R)|2R|ψT (R)|2 dR

Quantum Monte Carlo

EL(R) =1

ψT (R)HψT (R) =

ψT (R)

bdquominus

2nabla2

1 minusZ

EL(R) =1

RT (r1)

minus1

dr1minus

limr1rarr0

EL(R) =1

RT (r1)

bdquominus

dr1minus

laquoRT (r1)

Quantum Monte Carlo

This results in1

RT (r1)

dRT (r1)

dr1= minusZ

RT (r)

dRT (r)

dr= minus

f (rij )

wfold = wfnew

Thursday

Importance sampling

Friday

Things to do

Splines

s(x) =

Splines

yi = s(xi )

ands(xi+1) = yi+1

Splines

si (x) =fi

Splines

si (x) =fi

fi+16(xi+1minusxi )

(x minus xi )3

+ (yi+1

6 )(xi+1 minus x) (2)

Splines

(yi minus yiminus1)

Splines

3777775

2666664f1f2f3

fnminus2fnminus1

3777775 =

2666664v1v2v3

vnminus2vnminus1

3777775

operations

Splines

partt= D

bdquopartP

partxminus F

laquoP(x t)

F = 21

ΨTnablaΨT

partx(t)

y = x + DF (x)∆t + ξ

Importance Sampling

G(y x ∆t) =1

(4πD∆t)3N2exp

A(y x) = min(1 q(y x)))

q(y x) =G(x y ∆t)|ΨT (y)|2

G(y x ∆t)|ΨT (x)|2

Thursday

Friday

x = rsinθcosφ

y = rsinθsinφ

andz = rcosθ

4πε0r=

4πε0p

x2 + y2 + z2

4π(l + ml )Pml

for l = ml = 0

4πcos(θ)

for l = 1 og ml = 0

Y1plusmn1 =

(r) through

Ylml(r) = r l Ylml

we obtainΨnlml

equation

Slminusml

1radic

(minus1)m

minus(minus1)ml i i

Clminusml

We have definedYlml

(r) = r l Ylml(r)

2l + 1Yl0(r)

Slml= (minus1)ml

2l + 1ReYl0(r)

Slml= (minus1)ml

2l + 1ImYl0(r)

for ml gt 0

mll 0 1 2 3

q52 (x2 minus 3y2)x

+1 xradic

3xz 12

q32 (5z2 minus r2)x

0 1 y 12 (3z2 minus r2) 1

2 (5z2 minus 3r2)x

-1 zradic

3yz 12

q32 (5z2 minus r2)y

-2radic

3xyradic

q52 (3x2 minus y2)y

Thursday

Det uarr (1 2) =

˛1radic

Det darr (3 4) =

˛1radic

Ψα(Xnew )

Ψα(Xold )=

Duarrα(Xnew )

Duarrα(Xold )

Ddarrα(Xnew )

Ddarrα(Xold )

Gβ(Xnew )

Gβ(Xold )

nablaΨα

Ψα=nablaDuarrα

Ddarrα+nablaGβ

nabla2Ψα

Ψα=nabla2Duarrα

Ddarrα+nabla2Gβ

Gβ+ (

nablaGβ

Gβ)(nablaDuarrα

Ddarrα)

Dminus1ij =

Dα(y)

Dα(x)=

Dij (y)Cij (y)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

Dij (y)Cij (x)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

nablai Dα(x)

Dα(x)=

nabla2i Dα(x)

Dα(x)=

nablai Dα(y)

Dα(y)=

andnabla2

i Dα(y)

Dα(y)=

R equivNX

Dminus1kj (y) =

kj (x)minus1

RDminus1

ki (x)NX

RDminus1

ki (x) for j = i

R =D(y)

LU Decomposition

1CCA =

0BB1 0 0 0

b21 1 0 0b31 b32 1 0b41 b42 b43 1

1CCA0BB

c11 c12 c13 c140 c22 c23 c240 0 c33 c340 0 0 c44

detA = c11c22 cnn

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

nablaGβ

Gβ=Xiltj

(1 + βrij )2

nabla2Gβ

Gβ=Xiltj

(1 + βrij )4

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

a = 14

a = 12

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

Things to do

dt= ~v

dt= minusξ~v + ~F

ξ = 6πηam

0 + 2Z t

0dτ prime

〈~F (t)〉 = 0

δ(t minus t prime)

0 eminus2ξt +C~v0

we get

~r(t) = ~r0 + ~v01

0dτZ τ

Quantum Monte Carlo

wherer1 rA

Quantum Monte Carlo

There are

2A timesbdquo

MPI_Command_name

MPI_COMMAND_NAME

processor speed

memory sizespeed

Speed-up

S(P) =T (1)

Efficiency

η(P) =S(P)

tC(L) = τ + βL

T (P) ge T (1)

Prarr S(P) le P

MPI_Command_name

MPI_COMMAND_NAME

Communicator

The Fortran program

END PROGRAM hello

Note 1

Note 2

Note 3

100 values

Note 4

serious work

Integrating π

Examples

Go to MPI examples

Integration algos

af (x)dx asymp h

f (ximinus12)

example5cpp

mpirun -np 10 progx

MPI reduce

Call as

end of main program

function MPI Wtime

Strategies

Thursday

Friday

minus1

dr2unl (r) +

bdquol(l + 1)

2r2minus

r+ Φ(r)

problem

minus1

dr2unl (r) +

bdquol(l + 1)

and rewrite as

minus1

k2(r) = V (r) +l(l + 1)

2r2minus enl

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

V (ri ) = minusZri

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

programs catalogue

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

r+ Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr+Z infin

0unl (r)Fnl (r)dr

Z infin

raquo1

dunl (r)

dr)2 +

bdquominus

2Φ(r) +

l(l + 1)

laquou2

nl (r)ndash

dr + EFock

EFock =1

Z infin

0unl (r)Fnl (r)dr

k2(r) = minus2(V (r) +l(l + 1)

2r2minus enl )

dx2y(x) + k2(x)y(x) = F (x)

h2+ O(h2)

h2+ O(h2) + k2

i yi = Fi

2minus h2k2i

rdquoyi + h2Fi

Numerovrsquos algo

y(x + h) = y(x) + hy (1)(x) +h2

2y (2)(x) +

3y (3)(x) +

Numerovrsquos algo

12y (4)(x) + O(h6)

We obtain

h2minus

12y (4)(x) + O(h4)

Numerovrsquos algo

y (2)(x) +h2

12y (4)(x) + k2(x)y(x) +

ldquok2(x)y(x)

rdquoasymp 0

Numerovrsquos algo

d2(k2y(x))

acuteh2

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

+ O(h6)

Numerovrsquos algo

yi+1 =2ldquo

1minus 512 h2k2

rdquoyi minus

ldquo1 + 1

12 h2k2iminus1

rdquoyiminus1

1 + h2

12 k2i+1

minus1

dr2u(r) +

bdquoV (r) +

l(l + 1)

laquou(r) = eu(r)

dr2u(r) + k2(r)u(r) = 0

k2(r) = γ

bdquof (r)minus

l(l + 1)

r2minus ε

laquoγ = 2V0

ε =|E |V0

f (r) =

forr le ar gt a

k2(r) = γ

(1minus εminus 1

γl(l+1)

l(l+1)

forr le ar gt a

dr2u(r)minus

l(l + 1)

r2u(r) = 0

well defined

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

drult(r) =

drugt(r) at r = rm

drult(rm) asymp

drugt(rm) asymp

Numerovcpp

wf[loop] = norm

Todayrsquos tasks

2reminusr

u(r) =

bdquo1

1minus a2

raquo1 +

2(1minus a2)r

ndashlaquo

Thursday

Friday

u(x) = c1 +

0uprime(y)dy

and a constant c2

uprime(y) = c2 +

0uprimeprime(z)dz

0f (z)dz

bdquoZ y

0f (z)dz

laquody

and defining

F (y) =

0f (z)dz

0F (y)dy =

bdquoZ y

0f (z)dz

laquody =

0(x minus y)f (y)dy

This gives us

0(x minus y)f (y)dy

0(1minus y)f (y)dy

u(x) = xZ 1

0(x minus y)f (y)dy

G(x y) =

u(x) =

0G(x y)f (y)dy

1 continuous

3 G(0 y) = G(1 y) = G(x 0) = G(x 1) = 0

chapter 165

dx|x=y+ε minus

dx|x=yminusε = 1

W =dugt

dxult minus

and usingdG

dx|x=y+ε minus

dx|x=yminusε = 1

Simple example

ult(x) = x

G(x y) =

u(x) = λ

0G(x y)u(t)dy

u(r) =

Z infin

ddr ult(r)

ult(r)=

ddr ugt(r)

ugt(r)at r = rm

u(r) =

Z infin

resulting in

u(r) = ugt(r)Z r

Z infin

Algorithm

dxasymp

Algorithm

Thursday

Friday

E [Φ] =

af (Φ(x)

partΦ

partx x)dx

η(a) = η(b) = 0

Φ(x α) = Φ(x 0) + αη(x)

E [Φ(α)] =

af (Φ(x α)

partΦ(x α)

partx x)dx

E [Φ(α)] =

ndashα=0

partα

ndash=

bdquopartf

partΦ

partα+

partΦx

partα

laquodx

partα= η(x)

partα=

d(η(x))

UsingpartΦ(x α)

partα= η(x)

andpartΦx (x α)

partα=

d(η(x))

partα

ndash=

bdquopartf

partΦη(x) +

partΦx

d(η(x))

laquodx

partΦx

d(η(x))

dxdx = η(x)

partΦx|ba minus

aη(x)

partΦxdx

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndash=

bdquopartf

partΦminus

partΦx

laquoη(x)dx = 0

raquopartE [Φ(α)]

partα

ndashα=0

bdquopartf

partΦminus

partΦx

partΦminus

partΦx= 0

df = 0

partx=partf

party=partf

partz= 0

due to

df =partf

partxdx +

partydy +

partzdz

φ(x y z) = 0

resulting in

dφ =partφ

partxdx +

partφ

partydy +

partφ

partzdz = 0

partx=partf

party=partf

partz= 0

df =partf

partxdx +

partydy +

partzdz

df + λdφ = (partf

partz+ λ

partφ

partx)dx + (

party+ λ

partφ

party)dy + (

partz+ λ

partφ

partz)dz = 0

partz+ λ

partφ

partz= 0

partx+ λ

partφ

partx= 0

andpartf

party+ λ

partφ

party= 0

partxi+X

λkpartφk

partxi= 0

and a Hamiltonian

H = minus1

2nabla2 + V (x y z)

Zdxdydz

δE = δ

Zdxdydz

bdquo1

laquoff= 0

Zdxdydz

bdquo1

laquoff= 0

partψlowastminus

partψlowastxminus

partψlowastyminus

partψlowastz= 0

which yields

minus1

H = minusNX

2nabla2

i minusNX

NXiltj

H = H1 + H2 =NX

hi +NX

iltj=1

hi = minus1

2nabla2

i minusZ

E0 le E [Φ] =

ZΦlowastHΦdτ

radicN

(minus)PP

[H1A] = [H2A] = 0

ZΦlowastH1Φdτ =

(minus)PZ

ZΦlowastH1Φdτ =

ZΦlowastH hiΦHdτ

ZΦlowastH1Φdτ =

NXmicro=1

〈micro|h|micro〉

ΦlowastH2Φdτ = N

which reduces to

ZΦlowastH2Φdτ =

NXilej=1

(minus)PZ

ΦlowastH1

rijPΦHdτ

ZΦlowastH2Φdτ =

NXiltj=1

ZΦlowastH

ZΦlowastH2Φdτ =

NXmicro=1

NXν=1

lowastν(r j )

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

lowastν(r j )

E [Φ] =NX

micro=1

NXmicro=1

NXν=1

〈microν|

rij|νmicro〉

ψa =Xλ

Caλψλ

lectures

Exercise

Caλψλ

δE minusNX

micro=1

NXν=1

εmicroνδ

δE minusNX

micro=1

εmicroδ

NXmicro=1

NXν=1

Zδψlowastmicroψ

lowastν

Zδψlowastmicroψ

lowastν

micro=1

NXmicro=1

NXν=1

Zψlowastmicroψ

lowastν

Zψlowastmicroψ

lowastν

minusNX

micro=1

Emicro

NXmicro=1

Emicro

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

rijψν(r j )dr j

ψmicro(r i )

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

V dmicro(r i ) =

2(2l prime + 1)

Z infin

rgtdrj

withφnlml

Z infin

(rj )|21

rijdr j

which reads

Z infin

rijdrj

2l prime + 1

resulting in

2(2l prime + 1)

Z infin

rijdrj dΩj

infinXl=0

bdquorj

laquol

Pl cos(θ))

if ri gt rj or

infinXl=0

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Pl cos(θ))

infinXl=0

(rlt)l

(rgt)l+1

Ylowastlmprimel

2(2l prime + 1)

Z infin

rgtdrj

2l prime + 1

raquoZ infin

(rlt)λ

(2g + 1)

timesg

programs

Thursday

Friday

2nabla2

i minusZ

NXν=1

Zψlowastν(r j )

rijψν(r j )dr j

35ψmicro(r i )

24 NXν=1

Zψlowastν(r j )

V dmicro(r i ) =

rijg(r j )dr j

ψmicro(r i )

respectively

HHFi = hi +

NXmicro=1

V dmicro(r i )minus

NXmicro=1

V exmicro (r i )

2φ100(r1)φ100(r2)

minus1radic

The Hartree term

V dmicro(r i ) =

ψλ(r i )

NXmicro=1

V dmicro(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψλ(r i )

NXmicro=1

2Zφlowast100(r j )

rijφ100(r j )dr j

ψ100uarr(r i )

Φ(ri )u10 = 2V d10(ri )u10(ri ) = 2

Z infin

0|u10(rj )|2

rgtdrj )u10(ri )

rijψλ(r j )dr j

ψmicro(r i )

rijψλ(r j )dr j

ψmicro(r i )

s =darr That is

s mλs

rijψλ(r j )dr j

ψmicro(r i )

NXmicro=1

s uarr

ψmicro(r i )

resulting in

NXmicro=1

ψ100uarr(r i )

NXmicro=1

ψ100uarr(r i )

minus1

l(l + 1)

2r2minus

as bdquominus

l(l + 1)

2r2minus

r+ 2V d

10(r)laquo

or bdquominus

dr2minus

r+ V d

10(r)laquo

u10(r) = e10u10(r)

dangerous later

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ100uarr(r i )

= (2V d10(r i ) + 2V d

20(r i ))ψ100uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

V ex20 (r) =

Z infin

0dr jφ

lowast200(r j )

NXmicro=1

Z infin

φlowast100(r j )

rijφ200(r j )

ψ200uarr(r i ) =

(2V d10(r i ) + 2V d

20(r i ))ψ200uarr(r i )

NXmicro=1

Z infin

0dr jφ

lowast100(r j )

Z infin

0dr jφ

lowast200(r j )

10 (r i ) + V ex20 (r i )

minus1

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

V ex10 (r) =

Z infin

0dr jφ

lowast100(r j )

dr2minus

r+ V d

10(r) + 2V d20(r)

20 (r) = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)

10 (r) = e20u20(r)

dr2minus

r+ V d

10(r) + 2V d20(r)+

20 (r)+ = e10u10(r)

and bdquominus

dr2minus

r+ 2V d

10(r) + V d20(r)+

10 (r)+ = e20u20(r)

andbdquominus

2r2minus

r+ 2V d

10(r) + 2V d20(r)+

10 (r)minusV ex20 (r)+ = e21u21(r)

Useful Tests

R10(r) = 2bdquo

laquo32

R20(r) = 2bdquo

laquo32 bdquo1minus

R21(r) =1radic

bdquoZ

laquo32 Zr

We use a0 = 1

R10(r) = 2bdquo

laquo32

E [Z ] = Z 2 minus 4Z +5

Z = 2minus5

16= 16875

we do this

E [ρ] = minus1

1Ψi (r1)dr1minusZ

r1ρ(r1)dr1+

Zρ(r1)ρ(r2)

r12dr1dr2+EEXC [ρ]

ρ(r) =NX

|Ψi (r)|2

2nabla2

1 minusZ

Zρ(r2)

r12dr2 + VEXC(r1)

EEXC =

Thursday

Friday

A simple Example

wi (t = ε) = Wij wj (t = 0)

wi (t) = 1

and Xj

W (j rarr i) = 1

wi (t + 1) =X

Wij wj (t)

0 14 18 2334 58 00 14 13

w(t = 0) =

resulting in

w(t = ε) =

0 14340

resulting in

w(t = 2ε) =

0 5232132632

wi (t = infin) = 1

w(t = infin) =

0 415815315

Iteration w1 w2 w30 100000 000000 0000001 025000 075000 0000002 015625 062625 0187503 024609 052734 0226564 027848 051416 0207365 027213 053021 0197666 026608 053548 0198447 026575 053424 0200028 026656 053321 0200239 026678 053318 020005

10 026671 053332 01999811 026666 053335 02000012 026666 053334 02000013 026667 053333 020000

w(t = infin) 026667 053333 020000

w(0) =X

αi v i

αi v i =X

λtiαi v i

〈M(t)〉 =Xmicro

w(t)microMmicro

〈M(t)〉 = w(t)m

E [xk ] = 〈xk 〉 =1

xki p(xi )

〈M(t)〉 = w(t)m =X

λtiαi v i m i

〈M(t)〉 =X

λtiαi mi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

λtiαi mi

τi = minus1

logλi

〈M(t)〉 = 〈M(infin)〉+Xi 6=0

αi mi eminustτi

W (j rarr i)wj =X

W (i rarr j)wi

W (j rarr i)

W (i rarr j)=

Proof Xi

W (i rarr j)wi =X

W (j rarr i)wj = wj

W (j rarr i) = wj

Ergodicity

exp (minusβ(Ei ))

〈A〉 =X

Aj wj =

g(i rarr j) =1

A(j rarr i)

A(microrarr ν)

Then we must have

Take a new value

x (i+1) =

We have defined

ρ(x y) = min

p(y)f (x |y)

p(x)f (y |x) 1ff

Implementation

Quantum Monte Carlo

E [H] = 〈H〉 =

E0 le 〈H〉

Quantum Monte Carlo

ΨT (R) =X

aiΨi (R)

nm alowastmanR

nEnPn a2

nge E0

Quantum Monte Carlo

E [H] = 〈H〉 =

RdRΨlowastTα

(R)H(R)ΨTα(R)R

Quantum Monte Carlo

P(R) =|ψT (R)|2R|ψT (R)|2 dR

E [H] asympR

EL(R) =1

ψT (R)HψT (R)

P(R)EL(R)dR asymp1

P(Ri)EL(Ri)

Quantum Monte Carlo

˛ψα

T (R)˛2

w = P(Rp)P(R)

Quantum Monte Carlo

minus~2

part2u(r)

partr2minusbdquo

rminus

~2l(l + 1)

laquou(r) = Eu(r)

minus1

part2u(ρ)

partρ2minus

l(l + 1)

H = minus1

partρ2minus

l(l + 1)

Quantum Monte Carlo

EL(ρ) = minus1

ρminusα

bdquoαminus

Quantum Monte Carlo

〈Hn〉 =

= constant

Quantum Monte Carlo

minus2ke2

r1minus

Quantum Monte Carlo

bH = minus~2nabla2

2mminus

~2nabla22

2mminus

r1minus

P(R) =|ψT (R)|2R|ψT (R)|2 dR

Quantum Monte Carlo

EL(R) =1

ψT (R)HψT (R) =

ψT (R)

bdquominus

2nabla2

1 minusZ

EL(R) =1

RT (r1)

minus1

dr1minus

limr1rarr0

EL(R) =1

RT (r1)

bdquominus

dr1minus

laquoRT (r1)

Quantum Monte Carlo

This results in1

RT (r1)

dRT (r1)

dr1= minusZ

RT (r)

dRT (r)

dr= minus

f (rij )

wfold = wfnew

Thursday

Importance sampling

Friday

Things to do

Splines

s(x) =

Splines

yi = s(xi )

ands(xi+1) = yi+1

Splines

si (x) =fi

Splines

si (x) =fi

fi+16(xi+1minusxi )

(x minus xi )3

+ (yi+1

6 )(xi+1 minus x) (2)

Splines

(yi minus yiminus1)

Splines

3777775

2666664f1f2f3

fnminus2fnminus1

3777775 =

2666664v1v2v3

vnminus2vnminus1

3777775

operations

Splines

partt= D

bdquopartP

partxminus F

laquoP(x t)

F = 21

ΨTnablaΨT

partx(t)

y = x + DF (x)∆t + ξ

Importance Sampling

G(y x ∆t) =1

(4πD∆t)3N2exp

A(y x) = min(1 q(y x)))

q(y x) =G(x y ∆t)|ΨT (y)|2

G(y x ∆t)|ΨT (x)|2

Thursday

Friday

x = rsinθcosφ

y = rsinθsinφ

andz = rcosθ

4πε0r=

4πε0p

x2 + y2 + z2

4π(l + ml )Pml

for l = ml = 0

4πcos(θ)

for l = 1 og ml = 0

Y1plusmn1 =

(r) through

Ylml(r) = r l Ylml

we obtainΨnlml

equation

Slminusml

1radic

(minus1)m

minus(minus1)ml i i

Clminusml

We have definedYlml

(r) = r l Ylml(r)

2l + 1Yl0(r)

Slml= (minus1)ml

2l + 1ReYl0(r)

Slml= (minus1)ml

2l + 1ImYl0(r)

for ml gt 0

mll 0 1 2 3

q52 (x2 minus 3y2)x

+1 xradic

3xz 12

q32 (5z2 minus r2)x

0 1 y 12 (3z2 minus r2) 1

2 (5z2 minus 3r2)x

-1 zradic

3yz 12

q32 (5z2 minus r2)y

-2radic

3xyradic

q52 (3x2 minus y2)y

Thursday

Det uarr (1 2) =

˛1radic

Det darr (3 4) =

˛1radic

Ψα(Xnew )

Ψα(Xold )=

Duarrα(Xnew )

Duarrα(Xold )

Ddarrα(Xnew )

Ddarrα(Xold )

Gβ(Xnew )

Gβ(Xold )

nablaΨα

Ψα=nablaDuarrα

Ddarrα+nablaGβ

nabla2Ψα

Ψα=nabla2Duarrα

Ddarrα+nabla2Gβ

Gβ+ (

nablaGβ

Gβ)(nablaDuarrα

Ddarrα)

Dminus1ij =

Dα(y)

Dα(x)=

Dij (y)Cij (y)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

Dij (y)Cij (x)

Dij (x)Cij (x)

Dα(y)

Dα(x)=

nablai Dα(x)

Dα(x)=

nabla2i Dα(x)

Dα(x)=

nablai Dα(y)

Dα(y)=

andnabla2

i Dα(y)

Dα(y)=

R equivNX

Dminus1kj (y) =

kj (x)minus1

RDminus1

ki (x)NX

RDminus1

ki (x) for j = i

R =D(y)

LU Decomposition

1CCA =

0BB1 0 0 0

b21 1 0 0b31 b32 1 0b41 b42 b43 1

1CCA0BB

c11 c12 c13 c140 c22 c23 c240 0 c33 c340 0 0 c44

detA = c11c22 cnn

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

nablaGβ

Gβ=Xiltj

(1 + βrij )2

nabla2Gβ

Gβ=Xiltj

(1 + βrij )4

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

a = 14

a = 12

Important feature

Gβ =Yiltj

f (rij ) = exp

8ltXiltj

1 + βrij

Things to do

dt= ~v

dt= minusξ~v + ~F

ξ = 6πηam

0 + 2Z t

0dτ prime

〈~F (t)〉 = 0

δ(t minus t prime)

0 eminus2ξt +C~v0

we get

~r(t) = ~r0 + ~v01

0dτZ τ

Documents

Slides from FYS4410 Lectures