Mr. A. Square Unbound

Continuum States in 1-D Quantum Mechanics

With Apologies to Shelley

In the previous section, we assumedThat a particle exists in a 1-d spaceThat it experiences a real potential, V(x)That its wavefunction is a solution of the

TISE or TDSEThat at infinity, its wavefunction is zero.

In this section, those are removed

The consequences

If the boundary condition at infinity is removed,Then a quantum system is not limited to a

discrete set of states butA continuum of energies is allowed.

Normalizing Infinity

One problem if (x)∞, how do you normalize it?

Well, Postulate 7 (wherein we discuss normalization) is based on the proviso that it mainly applies to bound states.

Mathematically, if we have to find a matrix element, we perform the following operation:

a x ax

a a

The Free Particle

2 2

2 2

2

2 2

22

2

22

22

2

If V(x)=0 then the TDSE reduces to

( , ) ( , )-

2m i

( , ) ( )

Now the TISE:

2

2where

0 's solution is sinusoidal so

( ) ( )

iEt

ikx ikx

x t x t

x t

x t x e

mE

x

kx

mEk

kx

A k e B k e

( , ) ( )

( , ) ( ) ( )

where

iEt

i kx t i kx t

x t x e

x t A k e B k e

E

Assume k>0 & real, and B(k)=0, then describes a wave moving from –x to +x

*

*

*

*

* *

* *

What about p ?

p dxa p a

pa a

dx

dxi x

p

dx

k dx dx

p k

dx dx

p k

Obviously, <p2>=2k2

So p=<p2>-<p>2 =0

There is no variance in momentum, thus the free particle has mixed momentum

This is in agreement with Newton’s 1st Law

Assume k<0 & real, and A(k)=0, then describes a wave moving from +x to -x

*

*

*

*

* *

* *

What about p ?

p dxa p a

pa a

dx

dxi x

p

dx

k dx dx

p k

dx dx

p k

Obviously, <p2>=2k2

So p=<p2>-<p>2 =0

There is no variance in momentum, thus the free particle has mixed momentum

This is in agreement with Newton’s 1st Law

Obviously

eikx represents a particle moving from right to left

e-ikx represents a particle moving from left to right

The Wave Packet as a solution

Another solution to the TDSE is a “wave packet”

As an example, let B(k)=0 and the solution is in the form of the integral:

Note that this is the inverse Fourier transform

A complication arises in that is not really independent of k

( )( , ) ( ) i kx tx t A k e dk

The Wave Packet cont’d

Typically, the form of A(k) is chosen to be a Gaussian

We also assume that (k) can be expanded in a Taylor series about a specific value of k

0 0

22

0 0 0 2

1( ) ( ) ( ) ( )

2k k

k k k k k kk k

The Wave Packet cont’d

The packet consists of “ripples” contained within an “envelope”

“the phase velocity” is the velocity of the ripples “the group velocity” is the velocity of the envelope In the earlier expansion, the group velocity is d/dk

The phase velocity

2

2 2 2 2 222

22 2 2 2 2

2

2

22

2

2

2 2

2 2

1Classically, E=

22

4 42

2

phase

phase

c

c phase

c phase

d x x tvdt t k k

x

EE

vmE m

mv

E Ev v

m mv v

So the ripple travels at ½ the speed of the particle

Also, note if <p2>=2k2

then I can find a “quantum velocity”= <p2> /m2

2k2/m2= E/2m=vq

So vq is the phase velocity or the quantum mechanical wave function travels at the phase speed

The Group Velocity

22

2

2 22 2

2

2 2

2

2

2

group

group c

mE mk

k

mk

d dkm

d kv

dk m

k Ev v

m m

The group velocity (the velocity of the envelope) is velocity of the particle and is twice the ripple velocity.

BTW the formula for in terms of k is called the dispersion relation

The Step Potential

Region 2

x=0

V(x)=V0

Region 1

0 x>0( )

0 x<0

VV x

Region 1

1 1

221

1 1 12 2

1

2

Soik x ik x

mEk

x

Ae Be

“A” is the amplitude of the incident wave “B” is the amplitude of the reflected wave

Region 2

2

20 22

2 2 22 2

1

2

Soik x

m E Vk

x

Ce

“C” is the amplitude of the transmitted wave

Matching Boundary Conditions

1 2

1 2 1 1 2

Condition 1: (0) (0)

Condition 2: '(0) '(0)

A B C

ik A ik B ik C

The problem is that we have 2 equations and 3 unknowns.

“A” is controlled by the experimenter so we will always solve ALL equations in terms of the amplitude of the incident wave

Applying some algebra

1 1 2 1 1 2

1 1 2

1 2 1 2

1 2

1 2

1 2 1 2 1

1 2 1 2 1 2

1

1

21

B CA B C

A AB C

ik A ik B ik C k k kA A

B Bk k k

A A

Bk k k k

Ak kB

A k k

k k k k kC B

A A k k k k k k

If E>V0 then E-V0>0 or “+” Then k2 is real and 2 is

an oscillator propagation

If E<V0 Classically, the particle

is repelled In QM, k2 is imaginary

and 2 describes an attenuating wave

Graphically

If E>V0 then E-V0>0 or “+” Then k2 is real and 2 is

an oscillator propagation

If E<V0 Classically, the particle

is repelled In QM, k2 is imaginary

and 2 describes an attenuating wave

Region 2

x=0

V(x)=V0

Region 1

Region 2

x=0

V(x)=V0

Region 1

Reflection and Transmission Coefficients

2

**

21

21

2 2Im( )22

Recall

2

Define 3 currents, , ,

Re( )

Define

A B C

A

B

k xC

CB

A A

Jmi x x

J J J

kJ A

mk

J Bmk

J C k em

JJR T

J J

If k2 is

imaginary, T=0 If k2 is real, then

2

2

1

k CT

k A

2

22

1 2

1 2

22Im( )2

1

Re( )

B

A

k xC

A

J k kBR

J A k k

J k CT e

J k A

In terms of Energy,

If E>V0 then

12

0 01 22 21

21 2

0 0

4 14

1

E EV Vk k

Tk k

E EV V

If E<V0 then R=1 and T=0

The Step Potential

Region 2

x=0

V(x)=V0

Region 1

0

0 x<0

( ) 0<x<a

0 x>a

V x V

x=a

Region 3

The Wave Function

0 0

0

20 2

022

2

2

ik x ik x

ikx ikx

ik x

mEe Re kAe Be

m E VkTe

Boundary Conditions

1 2 2 3

1 2 2 3

) (0) (0) ) ( ) ( )

) '(0) '(0) ) '( ) '( )

a b a a

c d a a

Apply Boundary Conditions

0

0

0

0

) 1

)

) 1

)

ik a ika ika

ik a ika ika

a R A B

b Te Ae Be

c ik R ik A B

d ik Te ik Ae Be

Solving

0

0

2

2 2

2 2

Let 1

1 sin 2

1 sin 2 cos 1 sin 2 cos

1 1

1 sin 2 cos 1 sin 2 cos

ika

ika ika

Vk

k E

ka iR T e

ka i ka ka i ka

i iA e B e

ka i ka ka i ka

Reflection and Transmission Coefficients

22 22

2 22 2 2

22

2 22 2 2

1 sin

1 sin 2 cos

2

1 sin 2 cos

kaR

ka ka

Tka ka

Some Consequences

22 22

2 22 2 2

22

2 22 2 2

1 sin

1 sin 2 cos

2

1 sin 2 cos

kaR

ka ka

Tka ka

When ka=n*, n=integer, implies T=1 and R=0

This happens because there are 2 edges where reflection occur and these components can add destructively

Called “Ramsauer-Townsend” effect

For E<V0

22

2 22 2 2

2

1 sinh 2 coshT

ka ka

Classically, the particle must always be reflected

QM says that there is a nonvanishing T

In region 2, k is imaginarySince

cos(iz)=cosh(z) sin(iz)=isinh(z)

Since cosh2z-sinh2z=1

T cannot be unity so there is no Ramsauer-Townsend effect

What happens if the barrier height is high and the length is long?

Consequence: T is very small; barrier is nearly opaque.

What if V0<0? Then the problem reduces to the finite boxPoles (or infinities) in T correspond to

discrete states

An Alternate Method

We could have skipped over the Mr. A Square Bound and gone straight to Mr. A Square Unbound. We would identify poles in the scattering amplitude as bound states.

This approach is difficult to carry out in practice

The Dirac Delta Potential

The delta barrier can either be treated as a bound state problem or considered as a scattering problem.

The potential is given by V(x)=-(x-x0)

x=x0

Region 1 Region 2

Wavefunctions and Boundary Conditions

1

2

1 2

1 2

(0) (0)

'(0) '(0)

ikx ikx

ikx

Ae Be

Ce

From the previous lecture, the discontinuity at the singularity is given by:

02

2' ( )

mx

Applying the boundary conditions

0 0 0

0 0 0 0

0

2

22 2 4

2 2 4 2 2

2 2 22

2 2 4 2 2

2( )

Elimination of is straight forward and

C C

A A

B B

A A

ikx ikx ikx

ikx ikx ikx ikx

ikx

Ae Be Ce

mikCe ikAe ikBe Ce

B

ik kT

ik m k m

m me R

ik m k m

R cannot vanish or only vanishes if k is very large so there is always some reflection

Solving for k and E

0

22

2 2

22

2 2 22

4 2 2

C B

A ABoth of these quantities become infinite if

the divisor goes to zero

0

2

2

ikxik me

ik m ik m

mik m k

i

m mE mk E

This is in agreement with the result of the previous section. If is negative, then the spike is repulsive and there are

no bound states

A Matrix Approach to Scattering

Consider a general, localized scattering problem

Region 1

V(x)

Region 2 Region 3

Wavefunctions

22

22

2Region 1 ( ) ( )

2Region 3 ( ) ( )

Region 2 ( ) ( ) ( )

where f(x) and g(x) are two linearly independent

functions which depend on the potential.

ikx ikx

ikx ikx

mEx Ae Be k

mEx Fe Ge k

x C f x D g x

Boundary Conditions

There are four boundary conditions in this problem and we can use them to solve for “B” and “F” in terms of “A” and “G”.

B=S11A+S12G F=S21A+S22GSij are the various coefficients which

depend on k. They seem to form a 2 x 2 matrix

Called the scattering matrix (s-matrix for short)

11 12

21 22

S S

S S

S

Consequences

The case of scattering from the left, G=0 so RL=|S11|2 and TL=|S21|2

The case of scattering from the right, F=0 so RR=|S22|2 and TR=|S12|2

The S-matrix tells you everything that you need to know about scattering from a localized potential.

It also contains information about the bound states If you have the S-matrix and you want to locate

bound states, let ki and look for the energies where the S-matrix blows up.