2. Solving Schrödinger’s Equation
Superposition
2

i
  r, t   
 2   r, t   V  r, t    r, t 
t
2m
•Given a few solutions of Schrödinger’s equation, we can make more of them
•Let 1 and 2 be two solutions of Schrödinger’s equation
•Let c1 and c2 be any pair of complex numbers
•Then the following is also a solution of Schrödinger’s equation
  r, t   c11  r, t   c2  r, t 
•We can generalize this to an arbitrary number of solutions:
  r, t    cn  n  r, t 
n
•We can even generalize to a continuum of solutions:
  r , t    d  c      r , t 
2A. The Free Schrödinger’s Equation
We know a lot of solutions
2
•Consider the free Schrödinger equation:

i
  r, t   
 2   r, t 
•We know a lot of solutions of this equation:
t
2m
2 2
k
ik r it
  r, t   exp  ik  r  i k 2t 2m 
 
  r, t   e
2m
•By combining these, we can make a lot more solutions*
d 3k
2
  r, t   
c
k
exp
i
k

r

i
k
t 2m 



3/2
 2 
•The function c(k) can be (almost) any function of k
•This actually includes all possible solutions
*The factor of (2)3/2 is arbitrary and inserted here for convenience. In 1D, it would be (2)1/2
We can (in principle) always solve this problem
Goal: Given (r,t = 0) = (r), find
(r,t) when no potential is present
  r, t   
•Set t = 0:
d 3k
 2 
2
c
k
exp
i
k

r

i
k
t 2m 

3/2  
 r   
d 3k
 2 
ik r
c
k
e


3/2
•This just says that c(k) is the Fourier transform of (r)
d 3r
 ik r
c k    k   

r
e
 
3/2
 2 
•Substitute it in the formula above for (r,t) and we are done
•Actually doing the integrals may be difficult
Sample Problem
A particle in one dimension with no
potential has wave function at t = 0
given by   x   Nx exp   12 Ax 2 
What is the wave function at
arbitrary time?
•Need 1D versions
of these formulas:
  x, t   
c k   
dx
  x  eikx
2
dk
c  k  exp  ikx  i k 2t 2m 
2
2
2








B
B

B
 Ax  Bx
From Appendix:
x
e
dx

exp


   3/2 exp 


B  A
4
A
2
A
4
A




2


ik


N

ik

N
2
1
Find c(k): c  k  

exp  1 
x
dx
exp


3/2
2 Ax  ikx 

1
2 2  2 A
2
 4  2 A 
2
 k2 
Nik
c  k    3/2 exp  

A
 2A 
Sample Problem (2)
A particle in one dimension with no
potential has wave function at t = 0
given by   x   Nx exp   12 Ax 2 
What is the wave function at
arbitrary time?
 k2 
Nik
c  k    3/2 exp  

A
 2A 
dk
  x, t   
c  k  exp  ikx  i k 2t 2m 
2
2
2


B

B
 Ax  Bx
xe
dx  3/2 exp 
• Now find (x,t):


2
A
4
A


 k2 
dk Nik
2
  x, t    
exp  
 exp  ikx  i k t 2m 
3/2
2 A
 2A 
 Ni
 2 1 i t 

 3/2
k dk exp  k 

  ikx 

A
2
 2 A 2m 


 Niix   1 i t 
 3/2



2
A
2
m
2A
2 

3/2
2

 Nx exp   Ax 2  2  2iA t m  
ix




exp 

3/2
 4 1 2 A  i t 2m  
1

iA
t
m


2B. The Time Independent Schrödinger Eqn.
Separation of Variables
•Suppose that the potential is independent of time:
2

i
  r, t   
 2   r, t   V  r    r, t 
t
2m
•Conjecture solutions of the form:
  r,t     r   t 
2
d
i  r   t   
  t   2  r   V  r   r    t 
•Substitute it in
dt
2m
2
i d
1
2

t



 r   V r 
E


•Divide by (r)(t)
  t  dt
2m   r 
•Left side is independent of r
•Right side is independent of t
•Both sides are independent of both! Must be a constant. Call it E
Solving the time equation
i
d
 t   E
  t  dt
2
1

 2  r   V  r   E
2m   r 
•The first equation is easy to solve


i d ln   t    Edt
•Integrate both sides
i ln   t    Et
  t   eiEt
•By comparison with e-it, we see that E =  is the energy
•Substitute it back in:
  r, t     r   t     r  eiEt
  r,t     r   t 
The Time Independent Schrödinger Equation
2
1

 2  r   V  r   E
2m   r 
•Multiply the other equation by (r) again: 
  r, t     r  eiEt
2
2m
 2  r   V  r   r   E  r 
The Strategy for solving
•Given a potential V(r) independent of time, what is most general
solution of Schrödinger’s time-dependent equation?
•First, solve Schrödinger’s time-independent equation
•You should find many solutions n(r) with different energies En
•Now just multiply by the phase factor
•Then take linear combinations
  r, t    cn n  r  eiEnt
n
•Later we’ll learn how to find cn
Why is time-independent better?
2

i
  r, t   
 2   r, t   V  r, t    r, t 
t
2m

2
2m
 2  r   V  r   r   E  r 
•Time-independent is one less variable – significantly easier
•It is a real equation (in this case), which is less hassle to solve
•If in one dimension, it reduces to an ordinary differential equation
– These are much easier to solve, especially numerically
2
d2

  x   V  x   x   E  x 
2
2m dx
2C. Probability current
Probability Conservation
*

r
,
t


r
,
t






 r, t    r, t 
•Recall the probability density is:
2
•This can change as the wave function changes
•Where does the probability “go” as it changes?
– Does it “flow” just like electric charge does?
•Want to show that the probability moves around
•Ideally, show that it “flows” from place to place
– A formula from E and M – can we make it like this?
•To make things easier, let’s make all functions
of space and time implicit, not write them

i
  r, t   
 2   r, t   V  r, t    r, t 
t
2m
2

  r, t     J  r, t 
t
2

i

2  V 
t
2m
The derivation (1)
2

•Start with Schrödinger’s equation
i

2  V 
t
2m
2

•Multiply on the left by *:
i *   
* 2   *V 
t
2m
2

•Take complex conjugate of this i  *  
 2 *  V *
equation:
t
2m
2


 *
*
* 2
2 *
i












 



•Subtract:
t
t
2m


•Rewrite first term as a total
derivative
•Cancel a factor of i
•Left side is probability density

i
*
  
 * 2    2  * 


t
2m

i

 * 2    2  * 

t
2m
The derivation (2)

i

 * 2    2  * 

t
2m
•Consider the following expression:     *   * 
•Use product rule on the divergence   AB   A  B  A  B 
    *   *     *        *         *     *
  * 2    2  *
•Substitute this in above

i

    *   * 
t
2m
•Define the probability current j:
•Then we have:

    j
t
i
j
 *   * 

2m
Why is it called probability current?
i
j
 *   * 

2m
•Integrate it over any volume V with surface S

    j
t

3
3

r
,
t
d
r




j
r
,
t
d
 
  r


V
V
t
•Left side is P(r  V)
•Use Gauss’s law on right side
d
P  r V     n  j  r, t  dA
S
dt
•Change in probability is due to current flowing out
•If the wave function falls off at infinity (as it must) and
the volume V becomes all of space, we have
d
P  r  anywhere   0
dt
V
j
Calculating probability current
i
j
 *   * 

2m
•This expression is best when doing proofs
*
•Note that you have a real number minus its complex conjugate A  A  2i Im  A
•A quicker formula for calculation is:
j  Im   * 
•Let’s find  and j for a plane wave:
m
  r, t   N exp  ik  r  it 
     N exp  ik  r  it  N exp ik  r  it 
*
j


m
m
*
Im   *  
m

 N
2

Im N * exp  ik  r  it    N exp  ik  r  it  
Im  N * exp  ik  r  it  ikN exp  ik  r  it 
N * N Im ik  k N
m
m
2
k
j

m
Sample Problem
A particle in the 1D infinite square well has wave function
  x, t   a 1/2 sin  x a  e i1t  sin  2 x a  e i2t 
For the region 0 < x < a. Find  and j.
  *
 a 1 sin  x a  ei1t  sin  2 x a  ei2t  sin  x a  e i1t  sin  2 x a  e i2t 
 a 1 sin 2  x a   sin 2  2 x a   sin  x a  sin  2 x a   ei2t i1t  ei1t i2t 
  2  1
ei  ei  2 cos 
  a 1 sin 2  x a   sin 2  2 x a   2sin  x a  sin  2 x a  cos  t   
Sample Problem (2)
A particle in the 1D infinite square well has wave function
  x, t   a 1/2 sin  x a  e i1t  sin  2 x a  e i2t 
For the region 0 < x < a. Find  and j.
j
j

Im    
*
m

Im  sin  x a  ei1t
ma 

 * d 

  2  1
•In 1D: j  Im  
m 
dx 
d

 sin  2 x a  ei2t  sin  x a  e i1t  sin  2 x a  e i2t  
dx



i1t
i2t
 i1t
 i2t




Im
sin

x
a
e

sin
2

x
a
e
cos

x
a
e

2
cos
2

x
a
e








2



ma
sin  x a  cos  x a   2sin  2 x a  cos  2 x a  


Im 

2
i1t i2t
i2t i1t
ma
 sin  2 x a  cos  x a  e
 2sin  x a  cos  2 x a  e


 it 
it 



Im
2sin

x
a
cos
2

x
a
e

sin
2

x
a
cos

x
a
e








2

ma

Sample Problem (3)
A particle in the 1D infinite square well has wave function
  x, t   a 1/2 sin  x a  e i1t  sin  2 x a  e i2t 
j
For the region 0 < x < a. Find  and j.

ma 2
Im  2sin  x a  cos  2 x a  e  it   sin  2 x a  cos  x a  eit  

sin  t   sin  2 x a  cos  x a   2sin  x a  cos  2 x a  
ma
a
•After some work …

j
2
2
3
sin
t


sin


 x a 
2
ma
ma 2 j h
2D. Reflection from a Step Boundary
The Case E > V0: Solutions in Each Region
•A particle with energy E impacts a step-function barrier
I
from the left:
incident
0
x

0
,

2
transmitted
d 2
V  x  
E  
 V  x 
reflected
V0 x  0 .
2

II
2m dx
Solve the equation in each of the regions
2
•Assume E > V0
2 2
d 2
k
 ikx
E





e
E
•Region I
I
2
2m dx
2m
2 2
2
2

k
 ik x
d

 II  e
•Region II
E  V0 
 E  V0   
2
2m
2m dx
•Most general solution:
ikx
 ikx

x

Ae

Be


I
– A is incident wave
– B is reflected wave
 II  x   Ceik x  Deik x
– C is transmitted wave
– D is incoming wave from the right: set D = 0
Step with E > V0: The solution
 I  Aeikx  Beikx
2 2
k
I
k2
E

V

E
0
ik x
incident
2m
 II  Ce
2m
transmitted
reflected
•Schrödinger’s equation: second derivative finite
II
•(x) and ’(x) must be continuous at x = 0
 continuous  A  B  C ,
  continuous  ikA  ikB  ik C  ik   A  B    k  k  A   k  k  B
k  k
k  k
A C  2k A
B
A C  A B  A
k  k
k  k
k  k
2
•We can’t normalize wave functions
2
•Use probability currents! j  k  m
2
jA  k A m
jB  k B
2
jC  k  C
2
m
m
 k  k 
R


jA  k  k  
jB
2
jC
4kk 
T

j A  k  k  2
Summary: Step with E > V0
2
k2
E
2m
k 2
E  V0 
2m
 k  k 
R


jA  k  k  
jB
jC
4kk 
T

j A  k  k  2
2
2
 E  E  V0
R
 E  E V
0

T

4 E  E  V0 
E  E  V0





V0 
I
2
incident
reflected
2
8
9
E
transmitted
II
Step with E < V0
•What if V0 > E?
2 2
 ikx
E

k 2m I incident
II
•Region I same as before  I  e
 II  e x V0  E  2 2 2m
•Region II: we have
evanescent
reflected
2
d 2
 V0  E 
2
2m dx
 I  x   Aeikx  Beikx
•Most general solution:
2
2
– A is incident wave
 x
x
B
 II  x   Ce  De
ik  
R 2 
– B is reflected wave
ik  
A
– C is damped “evanescent” wave
– D is growing wave, can’t be normalized
•(x) and ’(x) must be continuous at x = 0:
R 1
 continuous  A  B  C ,
•No transmission since
evanescent wave is damped
  continuous  ikA  ikB   C    A  B 
ik  
T 0
ik


A

ik


B
B

A

 

ik  
Step Potential: All cases summarized
•For V0 > E, all is reflected
•Note that it penetrates, a little
bit into the classically
forbidden region, x > 0
•This suggests if barrier had
finite thickness, some of it
would bet through
V0  2 E
•Reflection probability:
 E  E  V
0

R   E  E  V0


1





T
R
2
if V0  E
if V0  E
V0 E
T  1 R
2E. Quantum Tunneling
Setting Up the Problem
V(x)
 0 x  12 d
• Barrier of finite height and width: V  x   
V0
1
• Solve the equation in each of the regions V0 x  2 d
III
II
I
• Particle impacts from left with E < V0
• General solution in all three regions:
- d/2
+ d/2 x
ikx
ikx
 x
x
ikx
 I  x   Ae  Be
 II  x   Ce  De  III  x   Fe
2 2
k
E

,
• Match  and ’ at x = -d/2 and x = d/2
2m
 ikd 2
ikd 2
 d 2
d 2
1
   2 d  : Ae
 Be
 Ce
 De
2 2

V0  E 
 ikd 2
ikd 2
 d 2
d 2
1
    2 d  : ik  Ae
 Be     Ce
 De 
2m
  12 d  : Ce d 2  De d 2  Feikd 2
   12 d  :   Ce d 2  De d 2   ikFeikd 2
• Solve for F in terms of A
Why didn’t I include e-ikx
in III? Why did I skip
letter E?
Skip this Slide – Solving for F in terms of A
•Multiply 1 by ik and add to 2
•Multiply 3 by  and add to 4
•Multiply 3 by  and subtract 4
•Multiply 5 by 2 and substitute
from 6 and 7
2
4ik Aeikd 2    ik  Fe d eikd 2
   ik  Fe d eikd 2
2
  2  k 2  e  d  e d  Feikd 2
2 ik  e  d  e d  Feikd 2
1: Ae ikd 2  Beikd 2  Ce  d 2  De d 2
2 : ik  Ae ikd 2  Beikd 2     Ce  d 2  De d 2 
3 : Ce d 2  De  d 2  Feikd 2
4 :   Ce d 2  De  d 2   ikFeikd 2
5: 2ikAeikd 2   ik    Ce d 2  ik    De d 2
6 : 2 Ce d 2    ik  Feikd 2
7 : 2 De d 2    ik  Feikd 2
 2 Feikd 2  k 2   2  sinh  d   2 ik cosh  d 
2ik eikd A
F 2
 k   2  sinh  d   2ik cosh  d 
Barrier Penetration Results
2ik eikd A
F 2
2
k



 sinh  d   2ik cosh  d 
 I  x   Aeikx  Beikx  III  x   Feikx
2 2
k2

E
, V0  E 
2m
2m
2
• We want to know transmission probability
2
jF
F
4k 2 2
T
 2 
2
2
2
2
2 2
2
jA
A
k


sinh

d

4
k

cosh


 d 



k
2


2 2
• For thick barriers,
4 E V0  E 
T 2
V0 sinh 2  d   4 E V0  E 
4k 2 2
sinh 2  d   4k 2 2
sinh  d  
• Exponential suppression
of barrier penetration
1
2
e
d
 e  d   12 e d
E  E  2 d
T  16 1   e
V0  V0 
Unbound and Bound State
• For each of the following, we found solutions for any E
– No potential
– Step potential
– Barrier
• This is because we are dealing with unbound states, E > V()
• Our wave functions were, in each case, not normalizable
• Fixable by making superpositions:
  x, t    dk c  k  k  x  exp   i k 2t 2m 
•
•
•
•
•
We will now consider bounds states
These are when E < V()
There will always only be discrete energy values
And they can be normalized
Usually easier to deal with real wave functions
2F. The Infinite Square Well
Finding the Modes
2
d2
E  
  V
2
2m dx
0 0  x  a
V  x  
 otherwise
V(x)
• Infinite potential implies wave function must vanish there
  x  0    x  a   0
•
•
•
•
0
a
2
2
d
In the allowed region, Schrödinger’s equation is just E  

2
2m dx
The solution to this is simple:
2 2
k
E
  A cos  kx   Bsin  kx 
2m
Because potential is infinite, the derivative is not necessarily continuous
But wave functions must still be continuous:
2 2 2
  nx 

n


B
sin
  0  0  A  0

 E
 a 
2ma 2
  a   0  B sin  ka   0  ka   n
x
Normalizing Modes and Quantized Energies
• We can normalize this wave function:

1    dx  B
2

2

a
0
sin 2  nx a  dx  B
2   nx 
 n  x   sin 
 for 0  x  a
a  a 
  Bsin  nx a 
2 1
2
a
En 
B 2 a
 2 2 n2
2ma 2
• Note that we only get discrete energies in this case
• Note that we can normalize these
• Most general solution is then

  x, t    cn n  x  e iEnt
n 1
 i 2 n2t 
2 
  nx 

cn sin 

 exp  
2 
a n1
a
2
ma




The 3D Infinite Square Well
 0 if 0  x  a , 0  y  b , 0  z  c
E  
   V
V  x, y , z   
otherwise
2m

2
• In allowed region: E  
 2
2m
• Guess solution:
c
  nx x    ny y    nz z 
  x, y, z   N sin 
 sin 

 sin 
a
b
c



 

• Normalize it:
b
a
– This is product of 1D functions
2
2
8
  nx x    ny y    nz z 
  x, y, z  
sin 
 sin 

 sin 
abc
a
b
c


 
 
• Energy is
– This is sum of 1D energies
Enx ,ny ,nz 
 2 2 nx2
2ma
2

 2 2 ny2
2mb
2

 2 2 nz2
2mc 2
2G. The Double Delta-Function Potential
Finding Bound States
V  x      x  a     x  a  
1
2
1
2
-a/2
I
• First, write out Schrödinger’s Equation:
V(x)
d2
1
1


E  x   

x



x

a


x






2
2 a    x 
2

2m dx
2
• Bound states have E < V() = 0
d2
 E 

2
• Within each region we have:
2m dx
II
a/2
III
2
  e  x
E
2
2
2m
• General solution (deleting the parts
that blow up at infinity):
 I  x   Ae  x
 II  x   Be  x  Ce   x
 III  x   De   x
x
Dealing with Delta Functions
2
d2
1
1


E  x   

x



x

a


x






2
2 a    x 
2

2m dx
• To deal with the delta functions, integrate Schrödinger’s
equation over a small region near the delta function:
– For example, near x = +a/2
E
a 2 
a 2 
  x  dx  
2
a 2 
2m a 2
V(x)
-a/2
I
II
E
a 2 
III
a 2 
d2
1
1



x
dx



x

a


x






2
2 a    x  dx
2


a
2


dx
• Do first term on right by fundamental theorem of calculus
• Do second term on right by using the delta functions
a 2 
a/2
  x  dx  
2
   12 a       12 a       12 a 
2m
• Take the limit   0
– Left side small in this limit
0
2
  12 a   II  12 a     12 a 
 III
2m
Simplifying at x = ½a
  12 a   II  12 a   
 III
• Since there is a finite
discontinuity in ’,  must be
continuous at this boundary
 III  12 a    II  12 a 
2m
2
  12 a 
I
On the right side of
the equation above, is
that I, II, or III?
V(x)
-a/2
II
a/2
III
 I  x   Ae  x
 II  x   Be  x  Ce   x
• Write these equation out
De  a 2  Be a 2  Ce  a 2
 III  x   De   x
explicitly:
2m   a 2
   De  a 2   Ce  a 2   Be a 2
• Substitute first into second:  2 De
2m
 2  Be  a 2  Ce   a 2   2 Be  a 2
Be  a 2  Ce  a 2 
 a 2
Be
m
2
2

 
a
C e 
 1 B
 m

Repeating at x = – ½a
2

 
a
C e 
 1 B
 m

 II   12 a   I   12 a   
 I   12 a    II   12 a 
2m
2
   12 a 
 I  x   Ae  x
 II  x   Be  x  Ce   x
 III  x   De   x
• Repeat the steps
we did, this time Ae  a 2  Be  a 2  Ce  a 2
at x = –½a
2m   a 2
 a 2
a 2
 a 2

Ae


Be


Ce


Ae
• Note these
2
equations are
2

 
a
nearly identical:  2m  Be  a 2  Ce  a 2   2 Ce  a 2
Be 
 1 C
2
 m

• The only numbers
2
equal to their
 a 2
 a 2
a 2
Be
 Ce

Ce
reciprocal are 1
m
2
2

C B
 

a
 e 
 1  1
 1  e  a
B C
m
 m

Graphical Solution
2
• Right side is two curves, left side is a

 a

1

e
straight line
2 2
m

E
2m
• Black line always crosses red curve, sometimes
crosses green curve, depending on parameters
– Sometimes two solutions, sometimes one
• Normalize to finish the problem
• Note one solution symmetric, one anti-symmetric
Right side, plus
Right side, minus
Left side