Download Mr. A. Square Unbound

Mr. A. Square Unbound
Continuum States in 1-D Quantum
Mechanics
With Apologies to Shelley

In the previous section, we assumed
That a particle exists in a 1-d space
 That it experiences a real potential, V(x)
 That its wavefunction is a solution of the
TISE or TDSE
 That 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 but
 A continuum of energies is allowed.

Normalizing Infinity
One problem if y(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:

x 
a x a
a a
The Free Particle
If V(x)=0 then the TDSE reduces to
 2  ( x, t )
 ( x, t )


2m x 2
i t 2
2
 ( x, t )  y ( x )  e
Now the TISE:

iEt
 2y
2mE


y
2
2
x
 2y
2


k
y
2
x
2mE
where k 2  2
 2y
 k 2y  0 's solution is sinusoidal so
2
x
y  A(k )eikx  B (k )e  ikx
 ( x, t )  y ( x )  e

iEt
 ( x, t )  A(k )ei kx t   B(k )e i  kx t 
E
where  
Assume k>0 & real, and B(k)=0, then 
describes a wave moving from –x to +x
What about p ?


p 
a p a
a a

*
y
 py dx




*
y
 y dx



p 
 
*
y
  i x y dx

 y y dx
*


y 
*
p 


*
y
 y dx

p  k

ky  dx
 y y dx
*
 k


*
y
 y dx

Obviously, <p2>=2k2
So
Dp=<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 
a p a

a a
y
*
py dx




*
 y y dx



p 
  
y dx
i

x


*
y
 


*
y
 y dx


 y 
*
p 


 y y dx
*

p  k

ky  dx
 y y dx
*
 k
Obviously, <p2>=2k2


 y y dx
*

So
Dp=<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
 ( x, t )  A(k )ei ( kx t ) dk
integral:

Note that this is the
inverse Fourier
transform
A complication arises in
that  is not really
independent of k



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

 ( k )   ( k0 )  ( k  k0 )
k
k0
2
1
2  
 ( k  k0 )
2
k 2

k0
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
v 2phase
v 2phase
 2y
2
2
2
d 2 x x 2  2y


y

 2  2
 2t 
 2
2
2
y
dt
 y t
k y
k
x 2
E2
2
E


2mE 2m
2
1
Classically, E= mvc2
2
2E
E
vc2 
 4
 4v 2phase
m
2m
vc  2v phase




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
k 
2
2mE
2

2m
k2

2m
k
d 
dk
2m
d
k

 vgroup
dk
m
2 2
k
2E
2
vgroup  2 
 vc2
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
V(x)=V0
Region 1
Region 2
x=0
V0 x>0
V ( x)  
 0 x<0
Region 1
 y 1 2mE
2

y

k
1
1y1
2
2
x
So
2
y 1  Aeik x  Beik x
1
1
“A” is the amplitude of the incident wave
 “B” is the amplitude of the reflected wave

Region 2
 2y 2 2m  E  V0 
2

y

k
y2
2
2
2
2
x
So
y 1  Ceik x
2

“C” is the amplitude of the transmitted wave
Matching Boundary Conditions
Condition 1: y 1 (0)  y 2 (0)  A  B  C
Condition 2: y 1 '(0)  y 2 '(0)  ik1 A  ik1 B  ik2C

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
A  B  C  1
B C

A A
ik1 A  ik1 B  ik2C  k1  k1

B
C
 k2
A
A
B
B

 k2  1  
A
A

B
k1  k2   k1  k2 
A
B k1  k2

A k1  k2
If E>V0 then E-V0>0 or
“+”

k1  k1
2k1
C
B k k k k
 1  1 2  1 2 
A
A k1  k2 k1  k2 k1  k2

Then k2 is real and y2
is an oscillator
propagation
If E<V0


Classically, the particle
is repelled
In QM, k2 is imaginary
and y2 describes an
attenuating wave
Graphically

V(x)=V0
Region 1

Region 2
x=0
If E>V0 then E-V0>0 or
“+”

If E<V0

V(x)=V0

Region 1
Region 2
x=0
Then k2 is real and y2
is an oscillator
propagation
Classically, the particle
is repelled
In QM, k2 is imaginary
and y2 describes an
attenuating wave
Reflection and Transmission
Coefficients
Recall
 * y
y * 
J
y
y

2mi 
x
x 
Define 3 currents, J A , J B , J C
k1 2
A
m
k
2
JB   1 B
m
k
2
J C  2 C Re(k2 )e 2Im( k2 x )
m
Define
JA 
R
JB
JA
T
JC
JA
JB
k1  k2
B
R


JA
A
k1  k2
2
2
2
J C Re(k2 ) C 2Im( k2 x )
T

e
JA
k1
A


If k2 is
imaginary, T=0
If k2 is real, then
k2 C
T
k1 A
2
In terms of Energy,

If E>V0 then

If E<V0 then R=1
and T=0
T
4k1k2
 k1  k2 
2

E
4 
 V0 
1
2
E
1
V0
   12

E
E
  
1 
  V0 

V0


2
The Step Potential
V(x)=V0
Region 1
Region 3
Region 2
x=0
 0 x<0

V ( x)  V0 0<x<a
 0 x>a

x=a
The Wave Function
 e  Re
 ikx
 ikx
y   Ae  Be
 Teik0 x

ik0 x
 ik0 x
k 
2
0

k 
2
2mE
2
2m  E  V0 
2
Boundary Conditions
a) y 1 (0)  y 2 (0)
b) y 2 (a)  y 3 (a)
c) y 1 '(0)  y 2 '(0)
d ) y 2 '(a)  y 3 '(a)
Apply Boundary Conditions
a) 1  R  A  B
b) Te
ik0 a
 Ae
ika
 Be
 ika
c) ik0 1  R   ik  A  B 
d ) ik0Te
ik0 a

 ik Ae
ika
 Be
 ika

Solving
Let  
V
k
 1 0
k0
E
1    sin ka

R
1    sin ka  2i cos ka
2
2
Ae
 ika
i 1   
1    sin ka  2i cos ka
2
T  e ika
Be
2i 
1   2 sin ka  2i  cos ka

ika

i 1   
1    sin ka  2i cos ka
2
Reflection and Transmission
Coefficients
R 
2
T 

1 

1 
2

2

1 


2
sin 2 ka
sin ka   2   cos 2 ka
2
2
2
 2 
2
2
2
2
sin ka   2   cos 2 ka
2
2
Some Consequences

R 
2
T 

1  2
1   
2
2


sin 2 ka
2

sin 2 ka   2   cos 2 ka
2
 2 
2
1  2

2
2
sin 2 ka   2   cos 2 ka
2

When ka=n*p, 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 “RamsauerTownsend” effect
For E<V0

Classically, the
particle must always
be reflected
 QM says that there
is a nonvanishing T
 In region 2, k is
imaginary


Since
cos(iz)=cosh(z)
sin(iz)=isinh(z)
T 
 2 
2


1  2

2
2
sinh 2 ka   2  cosh 2 ka
2
Since
cosh2z-sinh2z=1
 T cannot be unity so
there is no
RamsauerTownsend 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 box


Poles (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)=-ad(x-x0)

x=x0
Region 1
Region 2
Wavefunctions and Boundary
Conditions
y 1  Ae  Be
ikx
y 2  Ce
y 1 (0)  y 2 (0)
y 1 '(0)  y 2 '(0)
ikx
 ikx
From the previous lecture, the discontinuity at
the singularity is given by:
Dy '  
2ma
2
y (x )
0
Applying the boundary conditions
Aeikx0  Be  ikx0  Ceikx0
ikCeikx0  (ikAeikx0  ikBe  ikx0 )  
2ma
2
Ceikx0
Elimination of B is straight forward and
C
ik 2

A ik 2  ma
B
ma
2 ikx0

e
A ik 2  ma

2
C
k2 4
 T
 2 4
A
k
 m 2a 2
B
m 2a 2
 R
 2 4
A
k
 m 2a 2
2
R cannot vanish or only vanishes if k is very
large so there is always some reflection
Solving for k and E
C
ik 2
B
ma
2 ikx0


e
A ik 2  ma A ik 2  ma
Both of these quantities become infinite if
the divisor goes to zero
ik
2
 ma  0  k 
k2  


m 2a 2
4

2mE
2
ma
i 2
ma 2
 E 2
2
This is in agreement with the result of the previous section.
If a is negative, then the spike is repulsive and there are no
bound states
A Matrix Approach to Scattering
Consider a general, localized scattering problem
V(x)
Region 1
Region 2
Region 3
Wavefunctions
Region 1 y ( x)  Ae  Be
ikx
 ikx
Region 3 y ( x)  Fe  Ge
ikx
 ikx
(k 
2mE
(k 
2mE
2
2
2
2
)
)
Region 2 y ( x)  C f ( x)  D g ( x)
where f(x) and g(x) are two linearly independent
functions which depend on the potential.
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+S22G

Sij are the various coefficients which
depend on k. They seem to form a 2 x 2
matrix
 S11
S
 S21

S12 

S22 
Called the scattering matrix (s-matrix for short)
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 kik and look for the energies
where the S-matrix blows up.