Mr. A. Square’s Quantum
Mechanics
Part 1
Bound States in 1 Dimension
The Game’s Afoot

Primarily, the rest of this class is to
solve the time independent
Schroedinger Equation (TISE) for
various potentials
1.
What do we solve for?

2.
Energy levels
Why?

Because we can measure energy most easily
by studying emissions and absorptions
2) What type of potentials?
We will start slow– with 1-dimensional
potentials which have limited physicality
i.e. not very realistic
 We then take a detour into the Quantum
Theory of Angular Momentum
 Using these results, we then study 3dimensional motion and work our way up
to the hydrogen atom….

Some of these potentials seem a
little … dumb.

While a particular potential may not
seem realistic, they teach how to solve
the harder, more realistic problems. We
learn a number of tips and tricks.
The Big Rules
1.
2.
3.
4.
Potential Energy will depend on
position, not time
We will deal with relativity later.
We will ensure that the wavefunction,
y, and its first derivative with respect
to position are continuous.
y will have reasonable values at the
extremes
Mathematically,
1. V  f ( x, y , z ) or equivalent
2. y a ( x )  y b ( x )
3.
d
d
y a ( x)  y b ( x)
dx
dx
y a ( x )  y b ( x )
4. y ( )   and y ( )  
Re-writing Schroedinger

A more easily form of the Schroedinger
equation is as follows:
 y ( x)

 V ( x)y ( x)  Ey ( x)
2
2m x
2
2
 y ( x)

 ( E  V ( x))y ( x)
2
2m x
 2y ( x) 2m
 2 (V ( x)  E )y ( x)
2
x
2
2
Infinite Square Well
0  a  x  a
Potential Definition: V ( x)  
, otherwise
In order to ensure that y is well behaved, we
will let y  0 in the region where V=
In the region from (-a, a)
 2y ( x) 2m
 2 (0  E )y ( x)
2
x
2mE
Let k 2  2
 2y ( x)
  k 2y ( x)
2
x
 2y ( x)
 k 2y ( x)  0 (obviously sinusoidal function)
2
x
y ( x)  A cos( kx)  B sin(kx)
Apply Boundary conditions

At x=-a, y must be zero in order to satisfy
our continuity condition; also, at x=a.
A cos(ka )  B sin( ka )  0
A cos(  ka )  B sin(  ka)  0
or
A cos(ka )  B sin( ka )  0
A cos(ka )  B sin( ka )  0
Add these two equations:
2 A cos ka  0
Subtract these two equations:
2 B sin ka  0
Even Solutions

Assume A <>0, then B is zero
2 A cos ka  0
ka 

 3 5
,
2 2
,
2
These are called “even” functions since

y(-x)=y(x)
Odd Solutions

Assume B <>0, then A is zero
2 A sin ka  0
2 4 6
ka 
,
,
2 2 2

These are called “odd” functions since


y(-x)=-y(x)
Evenness or oddness has to do with
“parity”
Writing a general function for k
n
kn 
, n  1, 2,3,
2a
then
k
n
En 

, n  1, 2,3,
2
2m
8ma
2
2
2
2
2
Wavefunctions for Even and Odd
 n 
y even  An cos 
x  , n  1,3,5,
 2a 
 n 
y odd  Bn sin 
x  , n  2, 4,6,
 2a 
Finding An and Bn

a
a
y 2 dx  1
2
2  n x 
A
cos
 a n  2a  dx  ?
1
1
2
 sin x dx  2 x  4 sin(2 x)
1
1
2
 cos x dx  2 x  4 sin(2 x)
a
2
2  n x 
2
A
cos
dx

A
na 1
 a n  2a 
and
a
Bn2 a  1
1
An 
a
and
1
Bn 
a
Wavefunctions for Even and Odd
1
 n 
y even 
cos 
x  , n  1,3,5,
a
 2a 
1
 n 
y odd 
sin 
x  , n  2, 4,6,
a
 2a 
Wavefunctions
0.2
N=1
N=2
N=3
0.15
0.1
0.05
0
-60
-40
-20
0
-0.05
-0.1
-0.15
-0.2
20
40
60
In the momentum representation
1
even (k )  FT (y even ( x)) 
2
1
even (k )  2 
2
a

e

 cos(kx) cos(
0

a 
1

even (k ) 
n
2  
ka



2

y even ( x)dx
 ikx
n x
) dx
2a

n 
1
n  



sin  ka 
sin  ka 


n  
2  
2 
 

 ka 


2 




a 
1

odd (k )  i
n
2  
ka


2


n 
1
n  



sin  ka 
sin  ka 


n  
2  
2 
 
ka





2 



Matrix Elements of the Infinite
Square Well


The matrix
elements of
position, x, are of
interest since they
will determine the
dipole selection
rules.
The matrix
element is defined
as

m x n   y m  x y n dx

This integral vanishes except for states of opposite parity
It is convenient to specify that n is even parity
m is odd parity
Symbolically using the parity operator, 
 n  n , n  1,3,5...
 m   m , m  2, 4, 6...
so
m x n 
1 a
 m x 
 n x 
x
sin
cos



 dx


a
a
 2a 
 2a 
   m  n
 sin 
2
4a  
m x n  2
   m  n 2



  m  n
sin


2


2
m  n






Several Transitions
1 x 1 0 n x n
 4a  8
2 x 1  2   1 x 2
  9
 4a  24
2 x 3   2 
   25
 4a  40
2 x 5  2 
   441
 4a  16
4 x 1   2 
   225
 4a  48
4 x 3  2 
   49
If I write the wavefunctions as
0
0
 
 
1
0
 
 
0
0
2    and 4   
0
1
 
 
 
 
0
0
Then I can write x as a m x n matrix
8
16
 0
0
9
225

 8
24
0
0
9
25


24
0
0
4a 
25
xˆ  2 
 16
0
0

225

 0



0
0











0

The momentum operator is derived in a similar
manner but is imaginary and anti-symmetric
 0
4
0
3

 4
12
0
5
 3

12
0
0
i 
5
pˆ  
a 8
0
 15

 0



8
15
0
0
0
0











0

Finite Square

The finite square well can be made more
realistic by assuming the walls of the
container are a finite height, Vo.
V(x)
V0
Region 2
Region 1
x= -a
Region 3
x=a
x
The Schroedinger Equations
 2y ( x) 2m
Region 1:
 2 (V0  E )y ( x), x  a
x 2
 2y ( x) 2m
Region 2:
 2 (0  E )y ( x),  a  x  a
2
x
 2y ( x) 2m
Region 3:
 2 (V0  E )y ( x), x  a
2
x
2mE
Let  2  2
Let  2 
2m(V0  E )
2
Region 1: y 1 ( x)  Aex
Region 2: y 2 ( x)  B cos( x)  C sin( x)
Region 3: y 3 ( x)  Ae x
The Boundary Conditions
y1(-a)=y2(-a)
 y2(a)=y3(a)


y1’(-a)=y2’(-a)
 y2’(a)=y3’(a)

The only way to satisfy all
four equations is to have
either B or C vanish
If C=0, then we have states of even
parity
Let  
2
2mE
2
and  
2
2m(V0  E )
Region 1: y 1 ( x)  Ae x
Region 2: y 2 ( x)  B cos( x)
Region 3: y 3 ( x)  Ae x
at x  a
B cos  a  Ae a
 B sin  a  Ae a
Dividing these two:
 tan  a  
2
If C=0, then we have states of even
parity
Let  
2
2mE
and  
2
2
Equation 1:    
2
2
2m(V0  E )
2
2mV0
2
Equation 2:  tan  a  


If these 2 equations could be solved simultaneously
for  and , then E could be found.
Two options:


Numerically (a computer)
Graphically
Even Parity Solutions
9
8
7
Each
intersection
represents a
solution to the
Schroedinger
equation
6
1
k=6.8
K= 4.2
5
2
3
4
5
4
6
7
8
3
a
b
2
c
1
0
0
1
2
3
4
5
6
7
8
9
Odd Parity Solutions
9
1
8
2
3
Each
intersection
represents a
solution to the
Schroedinger
equation
4
7
5
6
6
7
8
a
5
b
c
4
3
2
1
0
0
1
2
3
4
5
6
7
8
9
Inspecting these graphs

Note that the ground state always has even
parity (i.e. intersection at 0,0), no matter what
value of Vo is assumed.
 The number of excited bound states increases
with Vo (the radius of the circle) and the states
of opposite parity are interleaved. If Vo
becomes very large, the values of  approach
n/2a

This asymptotic approach agrees which the
quantization of energy in the infinite square well
The Harmonic Oscillator

It is useful in describing the vibrations of atoms
that are bound in molecules; in nuclear
physics, the 3-d version is the starting point of
the nuclear shell model
 We will solve this problem in two different
ways:



Analytical (integrating as we have done before)
Using Operators
But first we need some definitions
Angular frequency
k
 
m
so
2
if
 k  m
2
1 2 1
2 2
V  kx  m x
2
2
TISE for SHO
 y ( x) 2m
 2 (V ( x)  E )y ( x)
2
x
 2y ( x) 2m 1
2 2
 2 ( m x  E )y ( x)
2
x
2
 2y ( x) 2m
1
2 2

(
E

m

x )y ( x)  0
2
2
x
2
 2y ( x) 2mE m 2 2 2
 ( 2  2 x )y ( x)  0
2
x
2
More Definitions
Let
 
2
m
2
2
2
and

2mE
2
so
 2y ( x) 2mE m 2 2 2
 ( 2  2 x )y ( x)  0
2
x
 2y ( x)
2 2

(



x )y ( x)  0
2
x
Change of Variables
Let
q  x 
d
d
 
dx
dq
 2E
 


 2y ( q)
2
 (  q )y ( q)  0
2
q
Let q>>
A
 2y (q)
2
 q y (q)  0
2
q
then
A
y (q)  Ae
q2

2
 Be
q2
2
However, y ()  y ()  0 implies that B=0
so
y (q)  Ae
q2

2
Constructing a trial solution
Let
y ( q)  e
q2

2
dy ( q)
e
dq
q2

2
 y ( q)
e
2
q
2
H ( q)
H '( q)  qe
q2

2
H '' qe
q2

2
q2

2
H ( q)
H e
q2

2
H  qe
q2

2
q2

2
2
H ' q e
H
Adding terms
Let
q 2y ( q)  q 2e

q2
2
H ( q)
so
 2y ( q)
2

y
(
q
)

q
y ( q) 
2
q
e

q2
2
H '' qe

q2
2
Obviously e
H e
q2

2

q2
2
H  qe
 0 so
H '' 2qH '    1 H  0

q2
2
H ' q 2e

q2
2
H  e

q2
2
H  q 2e

q2
2
H 0
Sol’ns of H are a power series

H   a jq j
j 0
H '   ja j q j 1
H ''   j ( j  1)a j q j  2
so
H '' 2qH '    1 H  0
becomes
 j ( j  1)a q
j 2
j
 2 ja j q j     1  a j q j  0
Each of these sums must vanish separately. On the
first term, I will shift everything by +2 so

j ( j  1)a j q j  2   ( j  2)( j  1)a j  2 q j
Now
 ( j  2)( j  1)a
j2
 2 ja j     1  a j  0
Finding aj’s
 ( j  2)( j  1)a
j2
 2 ja j     1  a j  0
For a given j ,
( j  2)( j  1)a j  2  2 ja j     1 a j  0
a j2 
   1  2 j 
( j  2)( j  1)
aj
Note: These equation connects terms of the same parity i.e 1,3,5 or 0,2,4
How to generate aj’s
For odd parity i.e. q1,q3,q5, start with
ao=0 and a1<>0
 For even parity i.e. q0,q2,q4, start with
ao<>0 and a1=0
 We will learn an established procedure in
a few pages

Too POWERful!
We have a problem, an infinite power
series has an infinite value
 So we know that the series at large
values must approximate exp(-q2/2)
 So we must truncate the power series
(make it finite)
 The easiest way is to make aj+2 vanish


This occurs when -1-2j=0
Making j equal to n
  2n  1

2E

 2n  1
1

E  n  
2

n  0,1, 2,3,...
If
  0 then
no solution because
j0

This is a big result,
since you know from
Modern Physics that
we had to get this
result.
 For a 3-d SHO,
E=(n+3/2), 1/2 
for each degree of
freedom
Summary so far:
y e
q2

2
H (q )
where
H '' 2qH '    1 H  0
H is a power series:
H n   an q n
n
and
1

E  n  
2


The solutions, Hn, to
the differential
equation are called
“Hermite”
polynomials.
 In the next few
slides, we will learn
how to generate
them
Method 1: Recursion Formula

The current convention
is that the coefficient for
the n-th degree term of
Hn is 2n


E.g. For H5, the
coefficient for q5 is 25
Also, recall for n=5,
E=(5+1/2) or 11/2 

At this eigenvalue, H=
a1q+a3q3+a5q5 where
a5=32
Previously,
(  1  2 j )
a j+2 
aj
( j  1)( j  2)
but   2n  1
a j+2
2(n  j )

aj
( j  1)( j  2)
or
( j  1)( j  2)
aj 
a j+2
2(n  j )
Method 1: Recursion Formula

The current
convention is that the
coefficient for the nth degree term of Hn
is 2n


E.g. For H5, the
coefficient for q5 is
25
Also, recall for n=5,
E=(5+1/2) or 11/2


At this eigenvalue,
H= a1q+a3q3+a5q5
where a5=32
In this case, j+2=5 and n=5, a j+2  32
( j  1)( j  2)
a j+2
2(n  j )
(3  1)(3  2)
a3 
*32=-160
2(5  3)
(1  1)(1  2)
a1 
*(-160)=120
2(5  1)
aj 
H 5  32q 5  160q 3  120q
Method 2: Formula of Rodriques
n
d  q2
H n   1 e
e
n
dq
n
q2
Method 3: Generating Function

The generating function is a function of two
variables, q and s, where s is an auxiliary
function
For Hermite polynomials, the generating function is
S ( q, s )  e

2 qs  s 2
To use this function, take the derivative n
times and set s equal to 0
Method 4: Recurrance Relations

Hn+1=2*q*Hn-2*n*Hn-1

We know that H0=1 and H1=2q
Ideally suited for computers
 Also, the derivative with respect to q is
 H’n =2*n*Hn-1

Method 5: Table Look Up
H0  q   1
H1  q   2 q
H 2  q   4q 2  2
H 3  q   8q 3  12q
H 4  q   16q 4  48q 2  12
H 5  q   32q 5  160q 3  120q
H 6  q   64q 6  480q 4  720q 2  120
Other Methods
Continued Fractions
 Schmidt Orthogonalization

Integrals with Hermite Polynomials
Recall
m
q
x
y n  Cn H n ( q ) e

q2
2

We know that  y n*y n dq  1
-

C
2
n
e
 q2
H n2 dq  1
-
d n  q2
From Rodriques formula, H n  (1) e
e
n
dq
we use 1 factor of H to subsitute this
n

1=C
2
n
e
-
 q2
d n  q2
(1) e
e H n dq 
n
dq
n
q2
q2
We know that the biggest term of Hn
is 2nqn
Hn~2nqn
 H’n~2n*n*qn-1
 H’’n~2n*n*n-1*qn-2
 Hn-’=2n*n!*qo=2n*n!

Going back and using the n-th
derivative
First integrate this

1=C
2
n
e
 q2
-
d n  q2
( 1) e
e H n dq
n
dq
n
q2
By parts for eight different times,

1=C
2
n
e
 q2
-
dn
H n dq
n
dq
d nH
and
 2n n !
n
q

1=Cn2 2 n n !  e  q dq  Cn2 2 n n ! 
2
-
Cn 
1
2n n ! 4 
Now, we have everything
Cn 
y (q) 
1
2 n! 
n
4
1
2 n! 
1
En  (n  ) 
2
n
4
H n ( q )e
q2

2
Transition Matrix Elements

m q n  CmCn  H m  q H n e

Cj 
 q2
dq
1
2 j j!4 
Using the recurrance formula
H n+1  2qH n  2nH n 1
 H n 1
  q2
m q n  Cm Cn  H m  
 nH n 1   e dq

 2

1
mq n 
n m ,n 1  n  1 m ,n 1
2



SHO via Operators (or via Algebra)

While the analytical solution is one way
to solve this problem, using operators
1.
2.
Teaches us the operator method which will
be required when we study the quantum
theory of angular momentum as well as
quantum field theory
Gives us another tool in our toolbox for
problem solving.
Many Slides ago,
Let
q  x 
d
d
 
dx
dq
 2E
 


2
 y ( q)
2
 (  q )y ( q)  0
2
q
Which is good
Schroedinger
notation but
what about
Dirac?
In Dirac notation,
 
2 
 2  (  q )  n  0
 q

2
Where |n> is the nth state of the oscillator
Even More Definitions,
d
dq
Now, using the commutator,
p  i
[q,p]=i
Proof: where   f (q)
 d
d 
d
d
[q, p]    qp  pq    i  q
 q    iq
 i q  i 
dq
dq
 dq dq 
[ q, p ]   i 
[ q, p ]  i
and furthermore,
d2
p  2
dq
2
A new way to TISE
 d
2 
 2  (  q )  n  0
 dq

2
 p    q  n  0
p q  n  n
2
2
2
2
This is TISE in
operator form
A necessary aside
We can write;
  i   i   
if [ ,  ]  0
2

2
Gosh, more definitions?
Let
q  ip
a
2
q  ip
†
a 
2
It turns out that
Let

1 2
aa  q  p 2  i[q, p ]
2
but [ q, p]  i
†
so




1 2
aa  q  p 2  1
2
1 2
†
a a  q  p2 1
2
†

Let’s solve for q2+p2




1 2
2
aa  q  p  1
2
1 2
†
a a  q  p2 1
2
Adding these together, we get
†
a † a  aa †  q 2  p 2
Subtracting these , we get
aa  a a  1
†
†
A crazy form of TISE
 a a  aa  n
†
†
 n
Theorem 8
a+ is
called the
“raising operator”
Proof of Thm 8
( p 2  q 2 ) n  ( a †a  aa † ) n
( p  q )a n  ( a aa  aa a ) n
2
2
†
†
†
† †
Recall [a, a † ]  1 i.e. aa †  a †a  1
aa †  a †a  1
( p 2  q 2 )a † n  ( a †aa †   a †a  1 a † ) n
( p  q )a n  ( a ( aa  aa  1)) n
2
2
†
†
†
†
( p 2  q 2 )a † n  ( a † ( aa †   a †a  1  1)) n


( p 2  q 2 )a † n  (a † (aa †  a † a  1  1)) n
( p 2  q 2 )a † n  (a † (aa †  a † a  2)) n
p2  q2
( p 2  q 2 )a † n  (a † ( p 2  q 2  2)) n  (a † ( p 2  q 2 ) n )  a † (2) n
n
( p  q )a n  a (  2) n  (  2)a n
2
2
†
†
but
( p 2  q 2 ) n  1  (  2) n  1
Thus
a† n  n  1
†
Theorem 9
a is
called the
“lowering operator”
Proof of Thm 9

See “Proof of Thm 8” and put minus signs
in the appropriate places

a n    2  n  q  p
2
2
 n 1
Theorem 10
1
Proof
First: n n  0
n 0
a†a
n


1 2 2
q  p 1
2

1 2
2
n  q  p  1 n  0

2  

so
1
  1  0
2
  1 QED
Consequences

We could start with state |n> and lower it
until we reached a ground state i.e.
a(a|n>)=(-4)|n> etc.
Theorem 11
In the ground state, =1
Proof
First: a 0  0
a†a 0  0
0 0
a†a


1 2 2
q  p 1
2
1
 o  1 0  0
2
o  1
Now, let’s find the wavefunction of the ground state
(hint: it must agree with what we found earlier
Hoexp(-q2/2)
a 0 0
 q  ip 

 0  0 or  q  ip  0  0
 2 
d
Use our definition of p=-i
dq

d 
q

0 0


dq 

d
0   q 0 (Let me go to S. notation) A is found by
0
dq
normalizing the
d
y 0   qy 0
wavefunction and this is
dq
dy 0
y0
  q dq  y 0  A0e
 q2
2
exactly in agreement with
the analytical method.
Now other states, use a+ on |0> and
raise, raise, raise

d 
1  A1  q   e
dq 

or
q2

2
n

d 
n  An  q   e
dq 

q2

2
An is found by
normalization
Calculating Transition Matrix
Elements
Let
a † n  AR n  1
a n  AL n  1
where A R and A L are just normalization constants
 
Since a †
†
a
n a  AR n  1
 n aa
†

n  AR2
n aa † n  AR2
n a † a n  AL2
Adding them, we get our old friend ( q 2  p 2 )
AR2  AL2    2n  1
AR2  AL2  1
Adding these relationships
AR2  n  1 AL2  n
So the matrices look like
m a n  n m ,n 1
m a † n  n  1 m ,n 1
so
0

0
a
0


1
0
0
0
2
0

 0


 †  1
 a 

 0




0
0
0
0
2
0







The matrices of q and p are
a  a†
q
2
a†  a
pi
2
so
 0

1  1
q

2 0


1
0
2
0
2
0







 0  1
0

0
 2
i  1
p

2 0
2
0









What about matrices of q2 and p2 ?
 1 0

1 0 3
2
q  
2 2 0


p 2  q 2  (2n  1) I
2
0
5
I







 1

1 0
2
p  
2  2


0
 2
3
0
0
5







The Dirac Delta

The Dirac delta, (x),
is used to represent
a point particle



(x)=0 if x<>0
(x)= if x=0
But what keeps it
finite, is the
normalization



 ( x) dx  1
The Dirac Delta cont’d

If it is at x=a


(x-a)=0 if x<>a
(x-a)= if x=a






 ( x  a ) dx  1
f ( x)  ( x  a ) dx  f (a )
TISE for the Dirac Delta
Let V   A ( x)
 y ( x) 2m
 2 ( A ( x)  E )y ( x)
2
x
2
 y ( x) 2m
 2 ( A ( x)  E )y ( x)  0
2
x
2
Some Boundary Conditions
X=0
Technically, this is
infinitesimally thick. So the
derivatives change from + to
– in an infinitesimal distance
At x=||, y=0
 The big problem lies at x=0

It turns out that y is continuous
 But y’ is discontinuous

We could integrate over a small
space
So the idea is to integrate from – to  and then let  go to zero
X=-
X=0
X=+
Mathematically
Let y ' represent the discontinuity




   2y
2m 
dx
)
x
(
y
E

dx
)
x
(
y
)
x
(
V

dx
lim  
0
2  


  x 2
 0


The differential integrates to the first derivative,
the energy term goes to zero under limit condition



V ( x)y ( x) dx 
2mA
 2y
 x 2 dx  y '
so

y '  
2mA
2
y (0)
2
y (0)
Using an ansatz
Let
y  ( x )  Ce x for x<0
 x
y  ( x )  Ce for x>0
dy 
  Ce x
dx
dy 
  Ce   x
dx
in the limit as  goes to zero
dy 
dy 
 C
  C
dx
dx
y '   C     C   2  C
at the origin,
2 C  

2mA
mA
2
2
y (0)  
 
2
2mA
m 2 A2
4
2
C
Plugging y into the TISE
Let
y  ( x )  Ce x for x<0
y  ( x )  Ce   x for x>0
d 2y 
  2Ce x
2
dx
d 2y 
2
 x


Ce
dx 2
No matter which I choose
 2y ( x ) 2m
 2 ( A  ( x )  E )y ( x )  0
x 2
0 for x 0
 2Ce   x 
2mE
2
In any case,
2 
2mE
2
0
Ce   x  0   2Ce x 
2mE
2
Ce x
Solving for E
 
2
2mE
2
0
Previously,
2 
2
m A
4
m 2 A2
4
2

2mE
mA2
E 2
2
2
0
Note that energy
is a single value;
based on the
potential
amplitude, A
Find y
For compactness, we write
y ( x)  Ce   x

y
2
( x)dx  1

C2

1 C   
so
y ( x) 
mA
mA2
E 2
2

e
mA
2
x
mA
The Uniform Force Field

V(x)=G*x, x>0
 V(x)=0, x<0
 G is a positive constant equal to the gradient
of the potential
 Function has several physical examples:


An electric charge in an uniform field near an
impermeable plate
A tennis ball dropped down an elevator shaft
(hence the name of the quantum bouncer)
TISE
Let V  Gx
 y ( x ) 2m
 2 (Gx  E )y ( x )
2
x
2
 y ( x ) 2m
 2 (Gx  E )y ( x )  0
2
x
2
A change of variables
Let
x
z  (x 
E

G
3
E 3 2mG
)
2
G
z
2mG
2
 2mG  d 2y  2mG  2 3
  2  zy  0
 3 2 
2



 dz
d 2y
 zy  0 which is called Airy's function
dz 2
2
dy dy dz 3 2mG dy


2
dx
dz dx
dz
d 2y  3 2mG

2
dx 2 
so
2
 d 2y

2
 dz
 2y ( x) 2m
 2 (Gx  E )y ( x)  0
x 2
becomes
 2mG
 3 2




2

d y 2mG  E
 2  
dz 2
G


2
3

z 
2mE
y  2 y  0
2mG 

2

Airy’s Functions
More about Airy
y  C  Ai( z )
y (0)  0
For large z, Ai(z )  e
z
Now at x=0 (where the floor is), y =0
E
at x=0, z=G
3
2mG
2
 zo
The root of the matter



The “roots” of a function are the
values where a function is
equal to zero
We solve the previous equation
for E
We set zo equal to the roots of
the Airy function, n
En  nG 3
2mG
2
n Root
1 2.3381
2 4.08794
3 5.52055
4 6.7867
5 7.94413
6 9.02265
7 10.04017
8 11.00852
9 11.93601
10 12.82877
Graphically,
General Features of 1-D bound
states
1.
2.
y vanishes at |x|=
1-d Bound states are non-degenerate
By degenerate, 2 states have equal energy
3.
4.
5.
6.
Wave functions for a 1-d bound state can be
constructed so that it is real
For a 1-d bound state, <p>=0
If H is symmetric, the wave functions are
eigenfunctions of the parity operator
The Schroedinger Equation can be
converted into an integral equation.
Proof of #2: Nondegeneracy of 1-d
bound states
Let 2 states (y 1 and y 2 ) have same energy E
d 2y 1 2m
 2 (V  E )y 1  y 1 ''
2
dx
d 2y 2 2m
 2 (V  E )y 2  y 2 ''
2
dx
y 1 '' y 2 ''
2m
(
V

E
)


2
y1
y2
y 1 ''y 2 y 1y 2 ''  0
Add a zero, y 1 'y 2 'y 1 'y 2 '
y 1 ''y 2 y 1 'y 2 'y 1 'y 2 'y 1y 2 ''  0
(y 1 'y 2 ) ' (y 1y 2 ') '  0
y 1 'y 2 y 1y 2 '  a constant, C
As x goes to zero, so does the
first derivative of both functions so C=0
y 1 'y 2 y 1y 2 '  0
y1 ' y 2 '

 y 1  D *y 2  these states merely differ
y1 y 2
by a constant
Proof of #3: Wavefunctions can be
constructed to be real
d 2y 2m
 2 (V  E )y  y ''
2
dx
d 2y * 2m
*
*

(
V

E
)
y

y
''
2
2
dx
by proof of #2,
y *  Dy
Since



y y dx   Dy y dx  1  D  1
-
*
-
Proof of #4: <p>=0

dy
dy
p   y
dx   y
dx
i -
dx
i - dx


dy
dy
2 
p   y
dx  y
  y
dx



i
dx
i
i
dx
But y ()  0

*
dy
p   y
dx   p

i
dx
so

p 0
Proof of #5: Wavefunctions are
eigenfunctions of parity operator
y ( x )  y (  x ) where  is the parity operator
H  1  H since the Hamiltonian is symmetric and
 1     2  1
H y  E y
1
H y  E y
But the nondegeneracy rule means that y  ky
since E is the same for both y and y
where k is some constant
  y    2 y  k 2y
1
k  1 which are the only eigenvalues for parity
Expansion on #6:
If there is some value of x (say x=a) where
both y (a) andy '(a) are known, then
x y
y ( x)  y (a)  xy '( x)   
2m
2
V ( z )  E  dz dy
a a
where y and z are dummy variables.