Lecture 15 Solving the time dependent Schrdinger equation Lecture 15 Solving the time dependent Schrödinger equation

Lecture 15Lecture 15

Solving the time dependent SchrSolving the time dependent Schrödinger dinger equationequation

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t

uiVuu

m

22

2

Poisson’s equation

t

uiVuu

m

22

2

2

2

22 1

t

u

cu

Introduction to PDEsIntroduction to PDEs

In many physical situations we encounter quantities which depend on two or more variables, for example the displacement of a string varies with space and time: y(x, t). Handing such functions mathematically involves partial differentiation and partial differential equations (PDEs).

t

u

hu

22 1

02 u

0

2

uAs (4) in regions

containing mass, charge, sources of heat, etc.

Electromagnetism, gravitation,

hydrodynamics, heat flow.

Laplace’s equation

Heat flow, chemical diffusion, etc.

Diffusion equation

Quantum mechanicsSchrödinger’s

equation

Elastic waves, sound waves, electromagnetic

waves, etc.Wave equation

x


A wave equation is an example of a partial differential equation

Think of a photo of waves (i.e. time is fixed)

If you make either x or t constant, then you return to the expected SHM case

2 2

2 2 2

( , ) 1 ( , )y x t y x t

x c t

You all know from last year that a possible solution to the above is

)cos( tkxAy

and that this can be demonstrated by substituting this into the wave equation

Or a movie of the sea in which you focus on a specific spot (i.e. x is fixed)

)cos( constkxAy

)cos( consttAy

y

ty

Step 4: Boundary conditions could then be applied to find A and B

Step 2: The auxiliary is then and so roots are

Step 1: Let the trial solution be So andmtx e mxmedt

dx mt xmemdt

xd mt 222

2

Unstable equilibrium

xxm 22 m

Step 3: General solution for real roots is m

)()( 2

2

2

txdt

txd

tt BeAetx )(


Thing to notice is that x(t) only tends towards x=0 in one direction of t, increasing exponentially in the other

tAetx )(

t

Step 1: Let the trial solution be So and


Harmonic oscillator

mtx e mxmedt

dx mt xmemdt

xd mt 222

2

Step 2: The auxiliary is then and so roots are xxm 22 im

Step 3: General solution for complex is

where = 0 and = so

)cossin( tDtCex t

tDtCx cossin

im

Thing to notice is that x(t) passes through the equilibrium position (x=0) more than once !!!!

Step 4: Boundary conditions could then be applied to find C and D

)()( 2

02

2

txdt

txd

tCx sin


Half-range Fourier sine series

1

sin)(n

n d

xnbxf

d

n dxd

xnxf

db

0sin)(

2

where

3)(x

xf 9

1

9)(

xxf

4

d

12

d

d

d

d

n dxd

xnx

ddx

d

xnx

db

4

4

0sin

99

12sin

3

2

A guitarist plucks a string of length d such that it is displaced from the equilibrium position as shown. This shape can then be represented by the half range sine (or cosine) series.

d

1

sin)(n

n d

xnbxf

2 2

2 2 2

( , ) 1 ( , )y x t y x t

x c t

Find the solution to the wave equation to predict the displacement of the guitar string at any later time t

The One-Dimensional Wave Equation

A guitarist plucks a string of length L such that it is displaced from the equilibrium position as shown at t = 0 and then released.

Let’s go thorugh the steps to solve the PDE for our specific case …..

2 2

2 2 2

( , ) 1 ( , )y x t y x t

x c t

)()(),( tTxXtxy

Ndt

tTd

tTcdx

xXd

xX

2

2

22

2 )(

)(

1)(

)(

1

SUMMARY of the procedure used to solve PDEs

9. The Fourier series can be used to find the particular solution at all times.

1. We have an equation with supplied boundary conditions

2. We look for a solution of the form

3. We find that the variables ‘separate’

4. We use the boundary conditions to deduce the polarity of N. e.g.

5. We use the boundary conditions further to find allowed values of k and hence X(x).

6. We find the corresponding solution of the equation for T(t).

7. We hence write down the special solutions.

8. By the principle of superposition, the general solution is the sum of all special solutions..

2kN

L

xnBxX nn

sin)( kxBkxAxX sincos)( so

kctDkctCtT sincos)(

nn L

ctnEtT cos)(

nnn L

ctn

L

xnBtxY

cossin),(

1

cossin),(n

nn L

ctn

L

xnBtxy

L

ct

L

x

L

ct

L

x

L

ct

L

x

L

ct

L

xdtxy

7cos

7sin

49

15cos

5sin

25

13cos

3sin9

1cossin

8),(

2

www.falstad.com/mathphysics.html

http://www.falstad.com/mathphysics.html

But any superposition such as also satisfies the

TDSE, and thus represents a possible state of the system.

Solving the time dependent SchrSolving the time dependent Schrödinger equationdinger equation

L

x

Lx

sin

2)0,(1

L

x

Lx

2sin

2)0,(2

)0,()0,()0,( 2211 xCxCx

1),(2

dxtx

12

2

2

1 CC

The TDSE is a linear equation, so any superposition of solutions is also a solution. For example, consider two different energy eigenstates, with energies E1 and E2. Their complete normalised

wavefunctions at t = 0 are:

When we superpose, the resulting wavefunction is no longer normalised.

Recall that all wavefunctions must obey the normalization condition:

However it can be shown that the normalisation condition is fulfilled so long as:


t

txitxtxV

x

tx

m

),(

),(),(),(

2 2

22

t

txi

x

tx

m

),(),(

2 2

22

Consider the time dependent Schrödinger equation in 1 dimensional space:

Within a quantum well in a region of zero potential, V(x,t) = 0, this simplifies to:

Let’s solve the TDSE subject to boundary conditions (0, t) = (L, t) = 0 (as for the infinite potential well) For all real values of time tand for the condition that the particle exists in a superposition of eigenstates given below at t = 0 .

Question

L

x

L

x

L

x

Lx

3sin3

12sin3

1sin3

12)0,(


What does the wavefunction look like?

L

x

L

x

L

x

Lx

3sin3

12sin3

1sin3

12)0,(

-1.5

-1

-0.5

0

0.5

1

1.5

displacement from x = 0 to x = L

tota

l am

plit

ud

e

-1.5

-1

-0.5

0

0.5

1

1.5


tota

l am

plit

ud

e

-1.5

-1

-0.5

0

0.5

1

1.5


tota

l am

plit

ud

e

-1.5

-1

-0.5

0

0.5

1

1.5


tota

l am

plit

ud

e

n = 3

n = 2

n = 1

Superposition at t = 0

If we measure the energy of the state Ψ(x,t) described above we will measure either E1 or E2 or E3 each with the probability of 1/3.

These curves arent normalised – figs intended just to show shape

Step 1: Separation of the Variables

Our boundary conditions are true at special values of x, for all values of time, so we look for solutions of the form (x, t) = X(x)T(t). Substitute this into the Schrödinger equation:


t

txi

x

tx

m

),(),(

2 2

22

In a region of zero potential, V(x,t) = 0, so :

dt

tdTxXitT

dx

xXd

m

)()()(

)(

2 2

22

dt

dT

T

i

dx

Xd

Xm

2

22 1

2

Separating variables:

Step 2: Rearrange the equation


Step 3: Equate to a constant

Now we have separated the variables. The above equation can only be true for all x, t if both sides are equal to a constant. It is conventional (see PHY202!) to call the constant E.

Edx

Xd

Xm

2

22 1

2

X

mE

dx

Xd22

2 2

Edt

dT

T

i

T

iE

dt

dT

So we have:

which rearranges to (i)

which rearranges to (ii)

dt

dT

T

i

dx

Xd

Xm

2

22 1

2

where giving

For X(x)Our boundary conditions are (0, t) =(L, t) = 0, which means X(0) = X(L) = 0. So clearly we need E > 0, so that equation (i) has the form of the harmonic oscillator equation. It is simpler to write (i) as:


Step 4: Decide based on situation if E is positive or negative

We have ordinary differential equations for X(x) and T(t) which we can solve but the polarity of E affects the solution …..

Xkdx

Xd 22

2

22 2

mE

k kxBkxAxX sincos)(

XmE

dx

Xd22

2 2

If where then applying boundary conditions

gives

For X(x)

Our boundary conditions are (0, t) =(L, t) = 0, which means X(0) = X(L) = 0.


22 2

mE

k kxBkxAxX sincos)(

Step 5: Solve for the boundary conditions for X(x)

X(0) = 0 gives A = 0 ; we must have B ≠ 0 so X(L) = 0 requires ,

i.e. so L

xnBxX nn

sin)( for n = 1, 2, 3, ….

0sin kL

L

nkn

Step 6: Solve for the boundary conditions for T(t)


iEteTT 0Equation (ii) has solution as it’s only a 1st order ODE T

iE

dt

dT

A couple of slides back we decided that in order to have LHO style solutions for X(x) we must have E > 0. So here we must also take E > 0.

dtiE

T

dT

dt

iE

T

dT

ciEt

T

ln tiE

ceeT

Proof of statement above

tiE

eTT

0Replace the constant with T0

Step 7: Write down the special solution for (x, t)


tiEnnnn

neL

xnBtTxXtx

sin)()(),( 2

22222

22 mL

n

m

kE nn

where

(These are the energy eigenvalues of the system.)

Question asks for the solutions of the TDSE for real values of time

sincos ie i

tEnwhere

Real values are thereforetE

L

xnBtTxXtx nnnnn cossin)()(),(


Step 8: Constructing the general solution for (x, t)

We have special solutions:

The general solution of our equation is the sum of all special solutions:

11

cossin),(),(n

nn

nn

tE

L

xnBtxtx

(In general therefore a particle will be in a superposition of eigenstates.)

tE

L

xnBtTxXtx nnnnn cossin)()(),(

-1.5

-1

-0.5

0

0.5

1

1.5


tota

l a

mp

litu

de


tE

L

xtE

L

xtE

L

x

Ltx 321 cos

3sin3

1cos

2sin3

1cossin

3

12),(

2

22

2 2

4

mLE

2

22

3 2

9

mLE

If we know the state of the system at t = 0, we can find the state at any later time.

Since we said that

Then we can say:

and

Step 10: Finding the full solution for all times

L

x

L

x

L

x

Lx

3sin3

12sin3

1sin3

12)0,(

Superposition at t = 0

11

cossin),(),(n

nn

nn

tE

L

xnBtxtx

where2

22

1 2mLE

The general solution is

11

sin)0,()0,(n

nn

n L

xnBxx

At t = 0 the general solution is

Particular solution to the time dependent Particular solution to the time dependent SchrSchrödinger equationdinger equation

2

22

1 2mLE

2

22

3 2

9

mLE

1st Eigenvalue 3rd Eigenvalue

1st Eigenfunction

3rd Eigenfunction

2

22

2 2

4

mLE

2nd Eigenvalue

tE

L

xtE

L

xtE

L

x

Ltx 321 cos

3sin3

1cos

2sin3

1cossin

3

12),(

-1.5

-1

-0.5

0

0.5

1

1.5


tota

l a

mp

litu

de


tE

L

xtE

L

xtE

L

x

Ltx 321 cos

3sin3

1cos

2sin3

1cossin

3

12),(

2

22

2 2

4

mLE

2

22

3 2

9

mLE

and Superposition at t = 0

where2

22

1 2mLE

In this particular example Ψ(x,t) is composed of eigenstates with different parity (even and odd). Therefore Ψ(x,t) does not have a definite parity and P(x,t) oscillates from side to side.

www.falstad.com/mathphysics.html

2. Mixed states (namely, superpositions of energy eigenstates) do not have a definite energy but have a probability of being in any one of the energy states when measured.

3. The probability densities of mixed states vary with time as do therefore the < x >.

1. Energy eigenstates (namely, states with definite energy) are stationary states: they have constant probability densities and definite energies.


The expectation value is interpreted as the average value of x that we would expect to obtain from a large number of measurements.

NB. In order to have time dependence in any observable such as position, it is necessary for the wavefunction to contain a superposition of states with different energies.

For example if we have the single eigenfunction within

an infinite potential well then

and . Notice how there is no time dependence.

tiEeL

xBtx 2

2sin),( 22

tiEeL

xBtx 2

2sin),( 2

*2

L

xxBtxxtx

2sin),(),( 22

22*2

Just for Quantum Mechanics course

This is because the probability density for a mixed state varies with time, whereas for a pure state it is constant in time. Pure states are known as stationary states.

This means < x > is invariant with timewww.falstad.com/mathphysics.html


But if we add another eigenfunction for example:

The complex conjugate is written as:

A superposition of eigenstates having different energies is required in order to have a time dependence in the probability density and therefore in < x >.

Just for Quantum Mechanics course

),(),(),( 21 txtxtx

),(),(),( *2

*1

* txtxtx

),(),(),(),(),(),( 21*2

*1

* txtxtxtxtxtx

tiEtiE eL

xBe

L

xBtxtxtx 21

2sinsin),(),(),( 2121

tiEtiE eL

xBe

L

xBtxtxtx 21

2sinsin),(),(),( 21

*2

*1

*

tiEtiEtiEtiE e

L

xBe

L

xBe

L

xBe

L

xBtxtx 2121

2sinsin

2sinsin),(),( 2121

*

tiEtiEtiEtiE eeee

L

xB

L

xB

L

xB

L

xBtxtx 2121

2sinsin

2sinsin),(),( 21

222

221

*

Therefore:

Let eigenfunctions be:

So:

tEEitEEi ee

L

xB

L

xB

L

xB

L

xBtxtx )()(

2122

222

1* 2121

2sinsin

2sinsin),(),(

Non zero so long as E1≠ E2

Documents

Lecture 15 Solving the time dependent Schrdinger equation Lecture 15 Solving the time dependent Schrödinger equation