Mr. A. Square’s Quantum Mechanics Part 1. Bound States in 1 Dimension.
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Transcript of Mr. A. Square’s Quantum Mechanics Part 1. Bound States in 1 Dimension.
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? Energy levels
2. 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 3-dimensional 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. Potential Energy will depend on position, not time
2. We will deal with relativity later.
3. We will ensure that the wavefunction, , and its first derivative with respect to position are continuous.
4. will have reasonable values at the extremes
Mathematically,
1. ( , , )
2. ( ) ( )
3. ( ) ( )
( ) ( )
4. ( ) ( )
a b
a b
a b
V f x y z or equivalent
x x
d dx x
dx dxx x
and
Re-writing Schroedinger
A more easily form of the Schroedinger equation is as follows:
2 2
2
2 2
2
2
2 2
( )( ) ( ) ( )
2
( )( ( )) ( )
2
( ) 2( ( ) ) ( )
xV x x E x
m x
xE V x x
m x
x mV x E x
x
Infinite Square Well
2
2 2
22
2
0Potential Definition: ( )
,
In order to ensure that is well behaved, we
will let 0 in the region where V=
In the region from (-a, a)
( ) 2(0 ) ( )
2Let
( )
a x aV x
otherwise
x mE x
xmE
k
x
x
22
22
2
( )
( )( ) 0 (obviously sinusoidal function)
( ) cos( ) sin( )
k x
xk x
xx A kx B kx
Apply Boundary conditions
At x=-a, must be zero in order to satisfy our continuity condition; also, at x=a.
cos( ) sin( ) 0
cos( ) sin( ) 0
cos( ) sin( ) 0
cos( ) sin( ) 0
Add these two equations:
2 cos 0
Subtract these two equations:
2 sin 0
A ka B ka
A ka B ka
or
A ka B ka
A ka B ka
A ka
B ka
Even Solutions
Assume A <>0, then B is zero
These are called “even” functions since (-x)=(x)
2 cos 0
3 5, ,
2 2 2
A ka
ka
Odd Solutions
Assume B <>0, then A is zero
These are called “odd” functions since (-x)=-(x)
Evenness or oddness has to do with “parity”
2 sin 0
2 4 6, ,
2 2 2
A ka
ka
Writing a general function for k
2 2 2 2 2
2
, 1, 2,3,2
, 1,2,3,2 8
n
n
nk n
athen
k nE n
m ma
Wavefunctions for Even and Odd
cos , 1,3,5,2
sin , 2,4,6,2
even n
odd n
nA x n
a
nB x n
a
Finding An and Bn
2
2 2
2
2
2 2 2
2
1
cos ?2
1 1sin sin(2 )
2 41 1
cos sin(2 )2 4
cos 12
1
a
a
a
na
a
n na
n
dx
n xA dx
a
x dx x x
x dx x x
n xA dx A a
a
and
B a
1
1
n
n
Aa
and
Ba
Wavefunctions for Even and Odd
1cos , 1,3,5,
2
1sin , 2,4,6,
2
even
odd
nx n
aa
nx n
aa
Wavefunctions
-0.2
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
0.2
-60 -40 -20 0 20 40 60
N=1
N=2
N=3
In the momentum representation
0
1( ) ( ( )) ( )
2
1( ) 2 cos( )cos( )
22
1 1( ) sin sin
2 2 22 2
1( ) sin
2 22
ikxeven even even
a
even
even
odd
k FT x e x dx
n xk kx dx
a
a n nk ka ka
n nka ka
a nk i ka
nka
1sin
22
nka
nka
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
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 , 1,3,5...
m
m nm x n x dx
n
2 22
m , 2,4,6...
1sin cos
2 2
sin sin2 24
a
a
m
so
m x n xm x n x dx
a a a
m n m n
am x n
m n m n
Several Transitions
2
2 2
2 2
4 402 5
1 1 0 441
4 8 4 162 1 1 2 4 1
9 225
4 24 4 482 3 4 3
25 49
ax
x n x n
a ax x x
a ax x
If I write the wavefunctions as
0 0
1 0
0 02 and 4
0 1
0 0
Then I can write x as a m x n matrix
2
8 160 0 09 2258 240 09 25
240 04 25ˆ16 0 0225
0 0
0
ax
The momentum operator is derived in a similar manner but is imaginary and anti-symmetric
840 0 03 154 120 03 5
120 05ˆ8 0 015
0 0
0
ip
a
Finite Square
The finite square well can be made more realistic by assuming the walls of the container are a finite height, Vo.
Region 1 Region 3
x= -a x=a x
V(x)
V0
Region 2
The Schroedinger Equations
2
02 2
2
2 2
2
02 2
22
2 02
1
2
( ) 2Region 1: ( ) ( ),
( ) 2Region 2: (0 ) ( ),
( ) 2Region 3: ( ) ( ),
2Let
2 ( )Let
Region 1: ( )
Region 2: ( ) cos( ) sin( )
R
x
x mV E x x a
x
x mE x a x a
x
x mV E x x a
xmE
m V E
x Ae
x B x C x
3egion 3: ( ) xx Ae
The Boundary Conditions
1(-a)=2(-a)
2(a)=3(a)
1’(-a)=2’(-a)
2’(a)=3’(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
2
2 2 02 2
1
3
2 ( )2Let and
Region 1: ( )
Region 2: ( ) cos( )
Region 3: ( )
cos
sin
Dividing these two:
tan
x
x
a
a
m V EmE
x Ae
x B x
x Ae
at x a
B a Ae
B a Ae
a
If C=0, then we have states of even parity
2 2 02 2
2 2 02
2 ( )2Let and
2Equation 1:
Equation 2: tan
m V EmE
mV
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
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6 7 8 9
1
2
3
4
5
6
7
8
a
b
c
Each intersection represents a solution to the Schroedinger equation
k=6.8K= 4.2
0
1
2
3
4
5
6
7
8
9
0 1 2 3 4 5 6 7 8 9
1
2
3
4
5
6
7
8
a
b
c
Odd Parity Solutions
Each intersection represents a solution to the Schroedinger equation
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
2 2
2 2 21 1
2 2
kk m
mso
if V kx m x
TISE for SHO
2
2 2
22 2
2 2
22 2
2 2
2 2 22
2 2 2
( ) 2( ( ) ) ( )
( ) 2 1( ) ( )2
( ) 2 1( ) ( ) 0
2
( ) 2( ) ( ) 0
x mV x E x
x
x mm x E x
x
x mE m x x
x
x mE mx x
x
More Definitions
2 22
2 2
2 2 22
2 2 2
22 2
2
2
( ) 2( ) ( ) 0
( )( ) ( ) 0
Let
m mEand
so
x mE mx x
x
xx x
x
Change of Variables
22
2
2
( )( ) ( ) 0
Let
d dq x
dx dq
E
qq q
q
Let q>>
2 2
2
22
2
2 2
2
( )( ) 0
( )
However, ( ) ( ) 0 implies that B=0
so
( )
A
q qA
q
qq q
q
then
q Ae Be
q Ae
Constructing a trial solution
2
2 2
2 2 2 2 2
2
2 2
222 2 2 2 2
2
( ) ( )
( )'( ) ( )
( )'' '
q
q q
q q q q q
Let
q e H q
d qe H q qe H q
dq
qe H qe H e H qe H q e H
q
Adding terms
2
2 2 2 2 2 2 2
2
2 2 2
22
2
2 22 2 2 2 2 2 2
2
( ) ( )
( )( ) ( )
'' ' 0
0
'' 2 ' 1 0
q
q q q q q q q
q
Let
q q q e H q
so
qq q q
q
e H qe H e H qe H q e H e H q e H
Obviously e so
H qH H
Sol’ns of H are a power series
1 2
0
2
' '' ( 1)
'' 2 ' 1 0
( 1) 2 1 0
Each of these sums must vanish separately. On the
first term, I will shift everything by +2 so
( 1)
j j jj j j
j
j j jj j j
jj
H a q H ja q H j j a q
so
H qH H
becomes
j j a q ja q a q
j j a q
22
2
( 2)( 1)
( 2)( 1) 2 1 0
jj
j j j
j j a q
Now
j j a ja a
Finding aj’s
2
2
2
( 2)( 1) 2 1 0
For a given ,
( 2)( 1) 2 1 0
1 2
( 2)( 1)
j j j
j j j
j j
j j a ja a
j
j j a ja a
ja a
j j
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 vanishThis occurs when -1-2j=0
Making j equal to n
2 1
22 1
1
2
0,1,2,3,...
0
0
n
En
E n
n
If then
no solution because
j
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:
2
2 ( )
'' 2 ' 1 0
is a power series:
1
2
q
nn n
n
e H q
where
H qH H
H
H a q
and
E n
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
j+2 j
j+2 j
j j+2
Previously,
( 1 2 )a a
( 1)( 2)
but 2 1
2( )a a
( 1)( 2)
( 1)( 2)a a
2( )
j
j j
n
n j
j j
or
j j
n j
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
j+2
j j+2
3
1
5 35
In this case, j+2=5 and n=5, a 32
( 1)( 2)a a
2( )
(3 1)(3 2)a *32=-160
2(5 3)
(1 1)(1 2)a *(-160)=120
2(5 1)
H 32 160 120
j j
n j
q q q
Method 2: Formula of Rodriques
2 2
1n
n q qn n
dH e e
dq
Method 3: Generating Function
The generating function is a function of two variables, q and s, where s is an auxiliary function
To use this function, take the derivative n times and set s equal to 0
22
For Hermite polynomials, the generating function is
( , ) qs sS q s e
Method 4: Recurrance Relations
Hn+1=2*q*Hn-2*n*Hn-1
We know that H0=1 and H1=2q
Ideally suited for computersAlso, the derivative with respect to q isH’n =2*n*Hn-1
Method 5: Table Look Up
0
1
22
33
4 24
5 35
6 4 26
1
2
4 2
8 12
16 48 12
32 160 120
64 480 720 120
H q
H q q
H q q
H q q q
H q q q
H q q q q
H q q q q
Other Methods
Continued FractionsSchmidt Orthogonalization
Integrals with Hermite Polynomials
2
2
2 2
2 2 2
2
*
-
2 2
-
2
-
Recall
( )
We know that 1
1
From Rodriques formula, ( 1)
we use 1 factor of H to subsitute this
1= ( 1)
q
n n n
n n
qn n
nn q q
n n
nq n q q
n nn
mq x C H q e
dq
C e H dq
dH e e
dq
dC e e e H dq
dq
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
2 2 2
2
2
2
-
2
-
2 2
-
4
First integrate this
1= ( 1)
By parts for eight different times,
1=
and 2 !
1= 2 ! 2 !
1
2 !
nq n q q
n nn
nq
n nn
nn
n
n q nn n
n n
dC e e e H dq
dq
dC e H dq
dq
d Hn
q
C n e dq C n
Cn
Now, we have everything
2
4
2
4
1
2 !
1( ) ( )
2 !1
( )2
n n
q
nn
n
Cn
q H q en
E n
Transition Matrix Elements
2
2
4
n+1 1
11
, 1 , 1
1
2 !
Using the recurrance formula
H 2 2
2
11
2
qm n m n
j j
n n
qnm n m n
m n m n
m q n C C H q H e dq
Cj
qH nH
Hm q n C C H nH e dq
m q n n n
SHO via Operators (or via Algebra)
While the analytical solution is one way to solve this problem, using operators
1. Teaches us the operator method which will be required when we study the quantum theory of angular momentum as well as quantum field theory
2. Gives us another tool in our toolbox for problem solving.
Many Slides ago,
22
2
2
( )( ) ( ) 0
Let
d dq x
dx dq
E
qq q
q
Which is good Schroedinger notation but what about Dirac?
In Dirac notation,
22
2( ) 0q n
q
Where |n> is the nth state of the oscillator
Even More Definitions,
22
2
Now, using the commutator,
[q,p]=i
Proof: where ( )
[ , ]
[ , ]
[ , ]
and furthermore,
p
dp i
dq
f q
d d d dq p qp pq i q q iq i q i
dq dq dq dq
q p i
q p i
d
dq
A new way to TISE
22
2
2 2
2 2
( ) 0
0
dq n
dq
p q n
p q n n
This is TISE in operator form
A necessary aside
2 2
We can write;
if [ , ] 0
i i
Gosh, more definitions?
†
Let
2
2
q ipa
q ipa
It turns out that
† 2 2
† 2 2
† 2 2
Let
1[ , ]
2but [ , ]
so
11
21
12
aa q p i q p
q p i
aa q p
a a q p
Let’s solve for q2+p2
† 2 2
† 2 2
† † 2 2
† †
11
21
12
Adding these together, we get
Subtracting these , we get
1
aa q p
a a q p
a a aa q p
aa a a
A crazy form of TISE
† †a a aa n n
Theorem 8
a+ is called the “raising operator”
Proof of Thm 8
2 2 † †
2 2 † † † † †
† † †
† †
2 2 † † † † †
2 2 † † † †
2 2 † † † †
( ) ( )
( ) ( )
Recall [ , ] 1 i.e. 1
1
( ) ( 1 )
( ) ( ( 1))
( ) ( ( 1 1))
p q n a a aa n
p q a n a aa aa a n
a a aa a a
aa a a
p q a n a aa a a a n
p q a n a aa aa n
p q a n a aa a a n
2 2
2 2 † † † †
2 2 † † † †
2 2 † † 2 2 † 2 2 †
2 2 † † †
2 2
†
( ) ( ( 1 1))
( ) ( ( 2))
( ) ( ( 2)) ( ( ) ) (2)
( ) ( 2) ( 2)
( ) 1 ( 2) 1
1
p q
n
p q a n a aa a a n
p q a n a aa a a n
p q a n a p q n a p q n a n
p q a n a n a n
but
p q n n
Thus
a n n
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
2 22 1a n n q p n
Theorem 10
1
2 2
†
11
2
2 2
Proof
First: 0
0
11 0
2
11 0
21 QED
q p
n n
n a a n
n q p n
so
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
2 2
†
†
11
2
Proof
First: 0 0
0 0
0 0
11 0 0
21
q p
o
o
a
a a
a a
Now, let’s find the wavefunction of the ground state (hint: it must agree with what we found earlier Hoexp(-q2/2)
2
0 0
0 20 0
0
0 0
0 0 or 0 02
dUse our definition of p=-i
dq
0 0
0 0 (Let me go to S. notation)
q
a
q ipq ip
dq
dq
dq
dq
dq
dq
dq dq A e
A0 is found by
normalizing the wavefunction and this is exactly in agreement with the analytical method.
Now other states, use a+ on |0> and raise, raise, raise
2
2
21
2
1q
n q
n
dA q e
dq
or
dn A q e
dq
An is found by normalization
Calculating Transition Matrix Elements
†
R L
††
† 2
† 2
† 2
2 2
2 2
2 2
Let
1
1
where A and A are just normalization constants
Since
1
Adding them, we get our old friend ( )
2 1
1
Adding these relation
R
L
R
R
R
L
R L
R L
a n A n
a n A n
a a
n a A n
n a a n A
n aa n A
n a a n A
q p
A A n
A A
2 2
ships
1R LA n A n
So the matrices look like
, 1
†, 1
†
1
0 1 0 0 0 0
0 0 2 1 0 0
0 0 0 0 2 0
m n
m n
m a n n
m a n n
so
a a
The matrices of q and p are
† †
2 2
0 1 0 0 1 0
1 0 2 1 0 21
2 20 2 0 0 2 0
a a a aq p i
so
iq p
What about matrices of q2 and p2 ?
2 2
2 2
1 0 2 1 0 2
0 3 0 0 3 01 1
2 22 0 5 2 0 5
(2 1)
q p
p q n I I
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 ( ) 1x dx
The Dirac Delta cont’d
If it is at x=a (x-a)=0 if x<>a (x-a)= if x=a
( ) 1
( ) ( ) ( )
x a dx
f x x a dx f a
TISE for the Dirac Delta
2
2 2
2
2 2
( )
( ) 2( ( ) ) ( )
( ) 2( ( ) ) ( ) 0
Let V A x
x mA x E x
x
x mA x E x
x
Some Boundary Conditions
At x=||, =0The big problem lies at x=0
It turns out that is continuousBut ’ is discontinuous
X=0
Technically, this is infinitesimally thick. So the derivatives change from + to – in an infinitesimal distance
We could integrate over a small space
X=0
X=+X=-
So the idea is to integrate from – to and then let go to zero
Mathematically
2
2 20
Let ' represent the discontinuity
2lim ( ) ( ) ( ) 0
The differential integrates to the first derivative,
the energy term goes to zero under limit condition
( )
mdx V x x dx E x dx
x
V x
2
2
2
2
2( ) (0)
'
2' (0)
mAx dx
dxx
so
mA
Using an ansatz
Let
( ) for x<0
( ) for x>0
x
x
x
x
x Ce
x Ce
dCe
dxd
Cedx
2 2
2 22
2 4
in the limit as goes to zero
' 2
at the origin,
2 22 (0)
d dC C
dx dxC C C
mA mAC C
mA m A
Plugging into the TISE
22
2
22
2
2
2 20 for x 0
2 22 2
22
Let
( ) for x<0
( ) for x>0
No matter which I choose
( ) 2( ( ) ) ( ) 0
2 20
In any case,
20
x
x
x
x
x x x x
x Ce
x Ce
dCe
dx
dCe
dx
x mA x E x
x
mE mECe Ce Ce Ce
mE
Solving for E
22
2 22
4
2 2
4 2
2
2
20
Previously,
20
2
mE
m A
m A mE
mAE
Note that energy is a single value; based on the potential amplitude, A
Find
2
2
2
2
2
For compactness, we write
( )
( ) 1
1
( )
2
x
mAx
x Ce
x dx
C mAC
so
mAx e
mAE
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
2
2 2
2
2 2
( ) 2( ) ( )
( ) 2( ) ( ) 0
Let V Gx
x mGx E x
x
x mGx E x
x
A change of variables3
2
32
32
22 2
32 2 2
2
2 2
22
32 2 2 2
32
2( )
2
2
2
( ) 2( ) ( ) 0
2 2 20
2
E mGLet z x
GE z
xG mG
d d dz mG d
dx dz dx dz
d mG d
dx dz
so
x mGx E x
xbecomes
mG d mG E z mE
dz G mG
2 2 32
32 2 2
2
2
2 20
0 which is called Airy's function
mG d mGz
dz
dz
dz
Airy’s Functions
More about Airy
32
Ai( )
(0) 0
For large z, Ai( )
Now at x=0 (where the floor is), =0
E 2at x=0, z=-
G
z
o
C z
z e
mGz
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
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
32
2n n
mGE G
Graphically,
General Features of 1-D bound states
1. vanishes at |x|=2. 1-d Bound states are non-degenerate
By degenerate, 2 states have equal energy
3. Wave functions for a 1-d bound state can be constructed so that it is real
4. For a 1-d bound state, <p>=05. If H is symmetric, the wave functions are
eigenfunctions of the parity operator6. The Schroedinger Equation can be
converted into an integral equation.
Proof of #2: Nondegeneracy of 1-d bound states
1 2
21
1 12 2
22
2 22 2
1 22
1 2
Let 2 states ( and ) have same energy E
2( ) ''
2( ) ''
'' ''2( )
d mV E
dx
d mV E
dxmV E
1 2 1 2
1 2 1 2
1 2 1 2 1 2 1 2
1 2 1 2
1 2 1 2
1 2 1 2
'' '' 0
Add a zero, ' ' ' '
'' ' ' ' ' '' 0
( ' ) ' ( ') ' 0
' ' a constant, C
As x goes to zero, so does the
first derivative of both functions so C=0
' ' 0
1 21 2
1 2
' '* these states merely differ
by a constant
D
Proof of #3: Wavefunctions can be constructed to be real
2
2 2
2 ** *
2 2
*
*
- -
2( ) ''
2( ) ''
by proof of #2,
Since
1 1
d mV E
dx
d mV E
dx
D
dx D dx D
Proof of #4: <p>=0
*
- -
2
- -
-
But ( ) 0
0
d dp dx dx
i dx i dxd d
p dx dxi dx i i dx
dp dx p
i dxso
p
Proof of #5: Wavefunctions are eigenfunctions of parity operator
1
1 2
1
( ) ( ) where is the parity operator
since the Hamiltonian is symmetric and
1
But the nondegeneracy rule means that
since E is the same for both and
where
x x
H H
H E
H E
k
2 2
1
k is some constant
1 which are the only eigenvalues for parity
k
k
Expansion on #6:
2
If there is some value of x (say x=a) where
both (a) and '(a) are known, then
2( ) ( ) '( ) ( )
where y and z are dummy variables.
yx
a a
mx a x x V z E dz dy