013 Hamiltonian Formulation Canonical Coordinatescomplex.gmu.edu/ Hamiltonian... · 2017-10-16 · Review of Lagrangian Formulation-Pick n proper (independent) generalized coordinates

PHYS 705: Classical MechanicsHamiltonian Formulation &Canonical Transformation

1

Legendre TransformLet consider the simple case with just a real value function:

2

F x

expresses a relationship between an independent variable x and

its dependent value F

F x

This relation is encoded in the functional form of F x

We will denote this encoding in general by: ,F x

As one has learned in math and physics, it is sometime useful to encode the

information contained in a function in different ways…

e.g., Fourier Transform, Laplace Transform, etc.

The original information of for a given x value is encoded in terms

of the Fourier components and a new set of variables k.

Legendre Transform

3

F x

In our new encoding scheme, we will say that

,F x

As in the Fourier Transform: 1ˆ2

ikxF k e F x dx

F̂ k

1 ˆ2

ikxF x e F k dk

and its inverse transform:

ˆ ,F k encodes the SAME information as

The Legendre Transform is yet another (convenient) encoding scheme

for when the following two conditions are true:

2. It is easier to measure, control, or think about than x itself

1. The function is strictly convex (concave), i.e, 2nd derivative always

positive (negative) and smooth

Legendre Transform

dFdx

4

F x

Since never changes sign (condition 1), there is a one-to-one

correspondence between and x

dFdx dF

dx

Legendre Transform gives us a new encoding scheme with

being a function of instead of x

,G s G sdFsdx

Here is the definition of the Legendre Transform for

Note that is a function of s and we have to express in

terms of s by inverting the function: to get

Legendre Transform

dF xs

dx

5

F x

G s sx s F x s

G s F x s

x s

Geometric Construction of the Legendre Transform:

Legendre Transform

6

G sx F

x

F

sx

slope = s F x

GG

F G sx

y-intercept of slope

- Suppose we have and define

Important Properties of the Legendre Transform:

1. The Legendre Transform is its own inverse transform:

Legendre Transform

dG sy s

ds

7

H y ys y G s y

G s

- Using the inverted relation , the Legendre Transform of is G s s y

- This can be rewritten as and comparing to our original

transform , we can immediate see that

and is F ’s own inverse transform.

G sy H G sx F , ,H y F x

F xs G

Legendre Transform

dG sx s

ds

8

- The two independent variables are two conjugate pair of

variables related to each other through

,x s

or

- Note that there is only ONE independent variable (either x or s) in:

dF xs x

dx

x x xF G s xs

s sF x G xs s

Legendre Transform

9

2. Properties of the Minima [note: is convex, an extrema is a

minimum]

0dFsdx

F x

minF

At the minimum of F xminFwe have the slope

So, we have

min( 0) 0G s x F

min 0F G

F x

Legendre Transform

10

2. Properties of the Minima

minGSimilarly, at a minimum of

We again have correspondingly

min 0G F

G s

0dG slope for G xds

So, we have

F x

minG 0F

y-intercept coincides with tangent pt

Legendre Transform

11

2. Reciprocal Relation of F & G’s curvature

Now take derivatives with respective to their independent variables,

dG sx s

ds

Start with the two reciprocal definitions

and dF xs x

dx

2

2

d G sdxds ds

2

2

d F xdsdx dx

2 2

2 2 1d F x d G s ds dx

dx ds dx ds

So, if is with curvature , then will have curvature F x G s 1

Review of Lagrangian Formulation

- Pick n proper (independent) generalized coordinates to describe the state

of the system this defines a n-dim Configuration Space.

2

1

, , 0j jI L q q t dt

Recall Lagrangian Formulation of Mechanics

0, 1, ,j j

L d L j nq dt q

- Apply the Hamilton’s Principle:

The motion of the system from t1 to t2 in config. space is such that the

Action (I) has a stationary value, i.e.,

- Using Variational Calculus, this implies the Euler-Lagrange Eq:

12

Review of Lagrangian Formulation

0, 1, ,j j

L d L j nq dt q

Notes:

- We have n independent generalized coordinates .

Solving this gives the Equations of Motion

1

n

jq

- The n are time derivatives of the . 1

n

jq

- We have n 2nd-order ODEs for the equations of motion

- We need 2n initial conditions to completely specify the motion.

1

n

jq

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Hamiltonian Formulation

where jj

Lpq

- Instead of n 2nd-order ODEs as EOM with n independent generalized

coordinates in configuration space.

1

n

jq

(Note: In Hamiltonian Formulation, we need to start with a set of

independent generalized coordinates. If they are not proper, a reduced

set of m < n proper coordinates must be chosen first.)

1

n

jp

- The most natural choice for this set of 2n variables in phase space is:

- Seek 2n 1nd-order ODEs as EOM with 2n independent generalized

coordinates in phase space.

generalized coordinates generalized conjugate momenta

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Hamiltonian Formulation

Notes:

- As we will see… this is NOT the only choice

- Coordinates in this 2n-dim phase space are call “canonical variables”

- Later, we will investigate invariant ways to transform the system to other

2n canonical variables using “canonical transformations”

(In Landau & Lifshitz: they are called “canonical” since the EOM

resulted from them has the simplicity and symmetry in form.)

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Configuration Space vs. Phase Space

A given point in configuration space prescribes fully the

“configuration” of the system at a given time t.

Different paths crossing P will have

the same set of but diff

However, the specification of a point in this space does NOT specify the

time evolution of the system completely !

1 1, , nq q

(a unique soln for a n-dim 2nd – order ODE needs 2n ICs)

Many different paths can go thru a given point in config space

1

n

jq 1

n

jq

1q

jq

P

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Configuration Space vs. Phase Space

To specify the state AND time evolution of a system uniquely at a given

time, one needs to specify BOTH AND or equivalently,

GOAL: to find the EOM that applies to

points in phase space.

2

,j jq p

The 2n-dim space where both and are independent variables

is called phase space.

jq jq

jq

jp

jq jp

Thru any given point in phase space, there can only be ONE unique path !

17

Hamiltonian Formulation- Instead of using the Lagrangian, , we will introduce a new

function that depends on , , and t:

One can think of this as a coordinate transformation from to

To be more specific, is the Legendre Transform of .

H is defined “similarly” to the Jacobi (energy) function h BUT h is a

function of and H is a function of .

, ,j jL L q q t

- This new function is call the Hamiltonian and it is defined by:

jq , ,j jH H q p tjp

- Plugging in the definition for the generalized momenta:

jj

LH q Lq

jj

Lpq

j jH p q L

, jjq q , jjq p

,j jq q ,j jq p

(Einstein’ s Convention:

Repeated indices are summed)

( )sum

,H q p ,L q q

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L and H are a Legendre Transform Pair

19

Consider a simple Lagrangian for a single particle m under the influence of

a conservative potential

Inverting the above relation and arriving at

Then, we haveLp mq

dq

Then, the Legendre Transform of is: ,L q q

, ,H q p pq p L q q p

2

,2

mqL q q V q

q p p m

V q

2 2

2 2p m p pp V q V qm m m

, ,L q H p

(q is an irrelevant variable in

this L-Trans)

x sF G

q L q pL H

G sx H q LF p

Hamiltonian Formulation- Taking the differential of our definition for , we have

- Now, we require that so that we should have

- To be consistent, let try to resolve this by expanding dL:

jj

Lpq

j j j jdH p dq q dp dL

, ,j jH H q p t

j jj j

H H HdH dq dp dtq p t

j jj j

L L LdL dq dq dtq q t

( )sum

- Plug in :

jj

jj pL LdL dq dq dtq t

( )sum

( )sum

( )sum

j jH p q L

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Hamiltonian Formulation- Now, since the system satisfies the Euler-Lagrange Equation, we have

- Substituting this into our previous equation, we have:

( )sum

- Now, plug this into our equation for dH:

j jj jLdL dq p dq dtt

p

jj j

L d L pq dt q

j jdH p dq jj jj j jp dd pq q qp d L dtt

j j j jLdH q dp p dq dtt

( )sum

(** this is an important step if we want our formalism to describe the same dynamic.)

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Hamiltonian Formulation- If H describes the same dynamics as is given by the EL equation, the two

expressions must equal to each other:

- Comparing gives the condition:

j j j jLdH p dq q dp dtt

j j

j j

H H HdH dq dp dtq p t

jj

jj

H pq

H qp

and H L

t t

These are called the Hamilton’s Equations of Motion and they are the desired

set of equations describing the EOM in phase space.

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Hamiltonian FormulationSummary of Steps:

1. Pick a proper set of and form the Lagrangian L

2. Obtain the conjugate momenta by calculating

3. Form

4. Eliminate from H using the inverse of so as to have

5. Apply the Hamilton’s Equation of Motion:

jj

Lpq

As you will see, this formulation does not necessary simplify practical

calculations but it forms a theoretical bridge to QM and SM.

jq

j jH p q L

jq jj

Lpq

( , , )jj pH H q t

j jj j

H Hp and qq p

23

Hamilton Equations in Matrix (Symplectic) Notation

The pair of Hamilton equations look almost symmetric (except the “-” sign).

, ; 1, ,j j j n jq p j n

The following is an elegant way to write the Hamilton equations into a single

matrix equation:

, ; 1, ,j jj j n

H H H H j nq p

η η

For a system with n dofs, we group all of our qi’s and pi’s into a 2n-dim

vector :η

Similarly, we will define another 2n-dimensional column vector :H η

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Now, we define a anti-symmetric matrix ,

Note that the transpose of J is its own inverse (orthogonal):

In summary, we have the following properties for J:

J2 2n n

0 IJ

I 0where I is a identity matrixn n

0 is a zero matrixn n

T

0 IJ

I 0T T

I 0J J JJ

0 Iand

1,T J J J 2 , J I det 1 J

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The Hamilton equation can then be written in a compact form:

1 1 1

2 2 2

1 1 1

2 2 2

0 0 1 0 0 0 1 00 0 0 1 0 0 0 11 0 0 0 1 0 0 0

0 1 0 0 0 1 0 0

q H q pq H q pp H p qp H p q

As an example, with only 2 dofs, this matrix equation expands into:

H

η J

η

This is typically referred to as the matrix or symplectic notation for the

Hamilton equations.

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Example: Spherical Pendulum

2 2 2 2 2 21 1 1 sin2 2 2

T mv mb mb

E-L equations gives

two 2nd – order ODEs:

Pick Generalized Coordinates: , 0U

cosU mgb

2 2 2 2 21 1 sin cos2 2

L T U mb mb mgb

m

b ˆ ˆsind b d b d s θ ds

ˆ ˆsind b bdt

sv θ

Thus,

2

2

sin cos sin 0

sin 2 sin cos 0

gb

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Practically, we can do one integration immediately

through the two constants of motion:

0U

2 2 2 2 21 1 sin cos2 2

E mb mb mgb

m

bds

These two 1st – order ODEs can be solved easily by substitute from the equation into the E equation.

2 2sinLp mb const

Conservation of E:

is cyclic, i.e :0d Ldt

p

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Now, let consider the Hamiltonian Formalism.0U

m

bds

2 22 2sinsinpLp mb

mb

The boxed equations are the desired transformation from the config. space to

the canonical variables in phase space.

We first need to calculate the conjugate momenta:

22

pLp mbmb

2 2 2 2 21 1 sin cos2 2

L mb mb mgb

29


Construct the Hamiltonian:

0U

m

bds

(H is properly expressed in terms of the canonical variables .)

j jH q p L sum

22

2 22 2 2 2 2

2 cossi

1 1 sin2n 2

mb

p p L

pp mgbmb mb

mb

22

2 2

22

2 2 22 cossi in 2 2 s n

ppm

pp mgbmb m mbb b

22

2 2 2 cos2 2 sin

ppH mgbmb mb

, , , ,p p t

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Apply Hamilton’s Equations to get four 1st – order ODEs:

0U

m

bds

2

32 2 sin cos sin

2pH p mgb pmb

2

2 3

cossin

sinp

p mgbmb

0H p p

2

pHp mb

2 2sinpH

p mb

Just repeated what we had

Since is cyclic, P is a constant and can be integrated to get (t). We will exploit this later.

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2 2 2 cos2 2 sin

ppH mgbmb mb

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Symmetry and Conservation Theorem Again

- Consider the full-time derivative of H,

j jj j

dH H H Hq pdt q p t

Using the Hamilton’s Equations:

j jdH p qdt

j jq p Ht

dH Hdt t

(so if time does not explicitly appears in H

(time is cyclic), H is conserved!)

Note: H = E if … 1. U does not depend on the generalized velocities

2. Transformation defining does not depend on t

explicitly. jq

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Subtle Difference between h and H

Recall that the Jacobi integral h is a function of instead of

, ,j j jj j

Lh q q t q Lq

j

j j

h L Lqt t q t

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, ,j jq p t , ,j jq q t

Thus, we can rewrite the above equation as,

0ht

0dh

dt

Also, from the E-L Eq, we know that L dht dt

jj j

h dd

q hq tttL

Thus, unless the red term is explicitly zero,


Example:

Lh q Lq

L mq tq

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212

L mq qt V q

So, 212

h q mq t mq qt V q

2mq tq 212

mq qt V q

212

h mq V q


0L qt

35

212

L mq qt V q

So,

212

h mq V q

but 0ht

So, the LHS is zero and we can also explicitly check that the RHS=0 too

but with .

From our previous calculations, we have the following relationship,

h L dhqt t q dt

0

0dhdt


36

L dhqt

hq dtt

Lqt q

q t mqt

212

dh d m Vdt

q qdt

Vmqq qq

q

This has to be zero since and indeed, it does. From EL Eq, we have

Thus, we have,L dh Vq q mq

t q dt q

0ht

0d L Ldt q q

0d Vt mq

dt q

0Vmqq

0


0Lt

37

212

L mq qt V q

So, for this specific example, we have

212

h mq V q

but 0ht

and 0dhdt

0Vmqq qq

0q

One can also calculate using , ,H q p tLp t mqq

2

2p t

H V qm

0H dHt dt

Symmetry and Conservation Theorem Again

- Other cyclic coordinates:

1 1 1( , , , , , , , )n nL L q q q q t If is cyclic, the Lagrangian is

n

L constq

(The LHS is usually a complicated function

of and much efforts still needed to

solve for .)

In contrast, if is cyclic in the Hamiltonian Formalism,

nq

missingnq

is still herenqThe n-th EOM is then given by:

& j jq q

( )nq t

Lagr

angi

an F

orm

alis

m

nq

0 nn

H pq

(conserved)np const

(This equation is immediately “solved” from IC is gone from the prob.)npHam

ilton

ian

(The Hamiltonian Formalism has a much nicer structure for cyclic coords)

38

Routh’s Procedure

Goal: Transform only the cyclic coordinates to take advantage of simplicity in

phase space for the cyclic variables

1 1( , , , , , , , )s nL L q q q q t

Define the Routhian,1

n

j jj s

R p q L

where R depends on

Let say we have: 1where , , are cyclics nq q

0 for 1, ,j jj j

R Rp and q j s nq p

jp const

similar to Hamiltonian

11 1, , , , , , , , ,s s s nq q q q p p t Then, apply the EL equation to non-cyclic variables , 1, , sq q

0 for 1, ,j j

d R R j sdt q q

And, apply the Hamilton’s equation to cyclic variables , 1, ,s nq q

(missing)

39

Routh’s Procedure: example

A single particle in a central potential: ( ) nV r k r

is cyclic and r is not, so we define

R p q L

From previous chapter, we have:

The Routhian is:

2 2 2

2 2 n

m m kL r rr

22

pLp mrmr

22 22 2 2

2 2 22 2 2 2n n

p p pm m k m kR r r rmr mr r mr r

Depends on only as advertised

, ,r r p

40

Routh’s Procedure: example

Applying the EL equation to the non-cyclic variable r :

0R p

p const l

And, the Hamilton’s Equation for :

2

3 1

10 2 02 n

pd R R d kmr ndt r r dt m r r

22

22 2 n

p m kR rmr r

2

3 1 0n

p nkmrmr r

2

pRp mr

2

lmr

41

Cyclic Variables: Action-Angle Coordinates

- Note that a given system can be described by several different sets of

generalized coordinates

Generalized coordinates are not unique !

- Recall also that the # of cyclic variables can depend on the choice of the

generalized coordinates

e.g., in the previous central force problem:

Rect coord (x, y): both x, y are not cyclic

polar coord (r, ): is cyclic

42


- Although one might not be able to pick a set of generalized coordinates

(from a given physical system) to have ALL qj being cyclic,

Canonical Transformation (next Chapter)

one can imagine transforming them to an ideal set such that they are all

cyclic.

- If possible, then,

all the conjugate momenta are constant: j jp const

additionally, if H is a constant of motion, then

1, , nH H cannot depend on t and qj explicitly !

constant in time

cyclic

43


consequently, the EOM for the are simple:

- Recall that the “natural” choice of the 2n-dim phase space variables is with

being the regular generalized coordinates and being their conjugate momenta.

The Hamiltonian Formalism can be extended to other possibilities:

, , , ,j j j jQ Q q p t and P P q p t

BUT, this is NOT the only choice !

1,j n jj j

H Hq func of constsp

are integration constants depending on IC

jq

( )j j jq t t j

jq

jp

(indices for are suppressed here),j jq p

44


are the “canonically” transformed variables from the original ones

The canonical transformation treats both equally.

, , , ,j j j jQ Q q p t and P P q p t

are on the same theoretical footing in the

Hamiltonian Formalism !

,j jQ P

needs to satisfy the corresponding Hamilton’s Equations in the

transformed coordinates.

,j jQ P

There are no preference between the transformation above for

&j jQ P

&j jQ P

&j jQ P

The Hamiltonian Formalism is the starting point in analyzing QM systems.

45

Connection to Statistical Mechanics

- The Hamilton’s Equations describe motion in phase space

- a point in phase space uniquely determines the state of the system

AND its future evolution.

,j jq p

- nearby points represent system states with similar but slightly different initial

conditions.- One can imagine a cloud of points bounded by a closed surface S with nearly

identical initial conditions moving in time.

0 0,q p 0 0' , 'q p

,q p ', 'q p

S0t

t t

46


- Let say at time t, a small subset of these points crosses a differential area da on

S in a given outward normal direction such that n̂

- Let be the mean “velocity” for points in

- Then, after a given time dt, this small subset of points will trace out a

differential volume in phase space given by

t t

ˆd daa n

n̂daS

da ,j jq pv

davdtv dt dv a

47


- Summing up all points in the cloud bounded by S, we have,

or,

rate of change of “phase-space volume” due to the motion of the points

S

dV d dt

v a

S

dV ddt

v asurface integral of the velocity field v for the points in the cloud

- Then, by Gauss’s Law, we have

S V

dV d dVdt

v a v

V enclosed by S

48

Liouville’s Theorem

- In phase space, we have

S V

dV d dVdt

v a v 1 1 1 1ˆ ˆ ˆ ˆn n n nq q p p v q q p p

- So that,

(*)k k

k k kVV

q pdV dVq p

v

- Using the Hamilton’s Equations, we have:

2

2

k k k k

k k

k

k

k k k

kq Hp

p Hq

Hq q q p

Hp p p q

49


- Substituting into Eq (*), we have,

2

k k

dV Hdt q p

2

k k

Hp q

k

V

dV

(H is a smooth function)

This is the Liouville’s Theorem: collection of phase-space points move

as an incompressible fluid.

0 !dVdt

Phase space volume occupied by a set of points in phase

space is constant in time.

50


This is the starting point for statistical mechanics

- Imagine many (N) identical mechanical systems but with

different initial conditions (ensemble of systems).

- Each is a different point in phase space with ,j jq p

- Statistical properties can be specified by a “density of states” function

per unit volume in phase space

, ,j jq p t dV # system points in phase space volume dV

located at at time t. ,j jq p

51

Then, since N is fixed, V stays constant, = N/V = constant as well !

Thus, Liouville’s Thm implies: density in a neighborhood of any

system state = const as the system evolutes in phase space.


At statistical equilibrium,

The # of points in the ensemble (N) does not change

Thus, equilibrium is uniform along the flow lines of

the system points. (SM’s Master Eq. or Fokker-Planck Eq.)

This condition is typically used to solve for the equilibrium

distribution eq for a statistical mechanical problem in phase space

with which various statistical averages, P, T, U, S,… can be calculated.

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Documents

013 Hamiltonian Formulation Canonical Coordinatescomplex.gmu.edu/ Hamiltonian... · 2017-10-16 · Review of Lagrangian Formulation-Pick n proper (independent) generalized coordinates