12. Approx. Methods for Time Ind. Systems
12A. The Variational Principle
A Great Way to Find the Ground State
Given an arbitrary time-independent Hamiltonian H, we want to approximate
• The eigenstates |i
• The eigen-energies Ei
Idea behind the variational principle:
• The ground state is the state with the lowest energy
• The first excited state is the lowest energy state orthogonal to the ground state
• Etc.
The method:
• Choose a large number of state vectors
• Measure their energy
• Pick the one with the lowest energy
The Ground State Has the Lowest Energy
•
•
•
•
•
Imagine we knew the eigenstates and energies of the Hamiltonian H i  Ei i
These states are assumed to be complete and orthonormal E  E  E 
1
2
3
Assume the energies are ordered:
   ci i
For an arbitrary state |, we use completeness to write:
i
Let’s find the expectation value of the Hamiltonian for |:
 H    c*j  ci  j H i   c*j ci Ei  j i   c*j ci Ei ij   ci Ei
2
j
i
 E1  ci   ci
2
i 1
i
• It follows that
j
2
 Ei  E1 
 H   E1  ci
j
i
i
i
2
i
*

c



c
c


• Consider the inner product:
  i j i  j  ci ji   ci
*
j
j
• We therefore have:
 H
 E1

i
j
i
i
2
The Variational Principle
 H
Select a wide range of trial vectors |
 E1
Calculate the expectation values at right

Choose the one with the lowest ratio at right
Use this ratio as an estimate of the ground state energy
Use the corresponding state vector | (normalized) as an estimate of
ground state vector |1
The variational part:
• We want as many state vectors as possible, ideally infinitely many
• Best way to do this is to make | have
   1 ,  2 , ,  n 
one or more variational parameters:
• Calculate the energy as a function of these parameters
 α  H  α 
E α  
• Find the value of min that minimizes E()
 α   α 
• Then the estimate of the energy and wave function is:
•
•
•
•
•
E1  E  α min 
1 
1
  α min    α min 
  α min 
Theory vs. Practice
For homework problems:
• Pick a set of trial vectors described by a small
    
number ( 2) of parameters
• Find E() analytically

E α  α  0
• Find the minimum using derivatives
min
 i
• Substitute back in min
For research problems:
• Pick a set of vectors described by a large number (100’s) of parameters
• Find E() numerically
• Find the minimum using multi-dimensional search
 α  H  α 
E α  
algorithms (simplex method, for example)
 α   α 
• Substitute back in min
E1  E  α min 
1 
1
  α min    α min 
  α min 
Good Trial Wave Functions
• How do you pick a good trial wave function?
• Discontinuous functions will have infinite derivative
– Can show this yields infinite P2 and hence infinite H
– Don’t use discontinuous functions
No!
Danger!
• Non-smooth functions are okay, but can be tricky to evaluate
• Discontinuous first derivatives have infinite second
derivative at a point
2
2
3
2

P



d
r

r

 r 



• This will contribute non-trivially to
• Unless  vanishes at the non-smooth point
Fine!
• When in doubt, you can avoid this problem with:
 P   P
2
2

2
 d r 
3
2
Sample Problem (1)
A particle of mass m in 1D lies in potential
V = m2x2/2. Estimate the ground-state energy.
• The potential rises suddenly
a 2  x 2 x  a
• Try a function that
  x  
x a
 0
disappears suddenly
• No need to normalize in this formalism
• We now need
 H
1  1
1
2
2
2
E
a


P

m

X


to calculate

   2m
2
• Work out the pieces,
a
2
2
2 2
one at a time

      x  dx    a  x  dx
a
2
P
2
  P   P
2
X
2
2
d
  i
  x  dx 
dx
  x   x  dx   x  a  x
2
2
a
a
2
2

2 2
2

a
a
dx
 2 x  dx
2
P2 


5
 16
a
15
8
3
2
a3
16
X 2  105
a7
Sample Problem (2)
2

1
m

2
2 
A particle of mass m in 1D lies in potential
E a 
P 
X 

2
2
   2m
2
V = m x /2. Estimate the ground-state energy.

1
• Put the pieces together:
5
   16
a
15
P2 
8
3
2
a3
m 2 a 2
15  2 8 3 m 2 16 7  5 2

E a 
 a 

a 
2
5 
14
16a  2m 3
2 105  4ma
2
2
d

5
m

a
• Minimize with respect to a 0 
E a 

3
da
2ma
7
• Substitute back in to get E(a), an estimate of the energy:
16
X 2  105
a7
amin
2
m 2 amin
5 2
5 2 m 2 m 2
35
E  amin  



 
2
4mamin
14
4m
35 14 m 2
 
5
14
E1  0.598 

1
2
35

m 2
5
14
 12
5
14

Some comments on how we did
• We picked a terrible trial wave function
• We still did pretty well (20% error)
• The error in the energy is caused by the square of the amount of bad
2
wave functions in the wave function
 H   E1  ci   ci
– Small things squared are very small
i
i 1
• The state vector has first order errors
– Not as reliable
• We got an overestimate of the energy
– This will always happen
 H
 E1

• With more parameters, you can do much better
• This method is powerful; most realistic problems are solved with
(advanced) variational approaches
2
 Ei  E1 
Sample Problem – Hydrogen Estimate (1)
A particle of mass m in 3D lies in potential
V = - kee 2/r. Estimate the ground-state energy.
Do this in class
Trial wave functions:
• First function tried:   r   r 1
– This is not normalizable, | = 
• Second function tried:
  r   e r
– This looks very promising
1
1
2
2
• Things we need to find:
 H 
P  ke e R
2m


2 r 3
    e d r  0 e2 r 4 r 2 dr  4 2! 3  3
 2  
 P2   P 
R
1
 r e
1 2 r
2
  d 3r i 

d r  4 0 e
3
2 r
2

2
rdr  4
3
 r
ˆ
d
r
r

e

1!
 2 
2

 2

and

2

r  2 2  e 2 r d 3r 

2
Sample Problem – Hydrogen Estimate (2)
 r   e
 r
1
1
2
2
 H 
P  ke e R
2m
•
•
•
•
•
•
A particle of mass m in 3D lies in potential
V = - kee 2/r. Estimate the ground-state energy.

  3

P
2
 2


R
1

 2

2 2
2
3
2

H

 1 
 ke e     k e2

We now calculate
E   


e

2 
the energy function:
2
m

  2m 
 
2
d
E      ke e 2
Find the minimum: 0 
  a01
  ke e 2 m 2
d
m
2
2 4
2 4
2
2
2
Substitute in to get
k
e
m
1
 ke e m 


k
k
e
m


e
2
ee m
e
  1 E1  
2
2
the minimum energy: E1  2m  2   ke e  2 
2
2






We got it exactly right!
3
The wave function is
1

 r


r

e
g  
Also exactly right


Can We Get Beyond the Ground State?
Is there a way to get states beyond the ground state?
• Yes, if we pick states orthogonal to the ground state
Removing the approximate ground state:
• Assume you have an estimate of the true normalized
 1  1
ground state found by variational method
 2 α 
• Create a set of states that you estimate are close to the next excited state
• Remove the portion of these states
 2  α    2  α    1  1  2  α 
in the approximate ground state
• This state is, by construction,
 1  2  α    1  2  α    1  2  α   0
orthogonal to |1
 2  α  H  2  α 
• Calculate the energy for this state: E  α  
 2  α   2  α 
• Minimize this energy and find min
• Then we approximate the first excited state energy as
E2  E  αmin 
Comments on Excited State(s)
?
Is the resulting energy guaranteed to be an overestimate?
E  α min   E2
• In general no:
– The state is guaranteed to contain none of the estimated ground state |1 
– But it could contain a small mixture of the actual ground state |1 
Does this work as well for excited states as it did for the ground state?
• In general no:
– Error from previous step gets compounded with this step
– Over many steps, errors accumulate
An exception where it does work
• Suppose the problem has some symmetry
• Then all eigenstates can be classified by their eigenvalues under this symmetry
• Choose trial state vectors that have this symmetry eigenvalue
• The energies you find will be true overestimates of the energy
E  α min   E2
of the lowest state with each symmetry eigenvalues
Sample Problem (1)
A particle of mass m in 1D lies in potential V = m2x2/2.
Estimate the first excited state energy. Is it an overestimate?
•
•
•
•
•
The potential is symmetric under parity
The ground state, previously discussed, has even parity
Let’s find the lowest energy odd parity state
a 2 x  x 3
  x  
Try an odd wave function
0


Work out the pieces we
a
2
3 2
     a x  x  dx
need, one at a time
7
a
16



a
105
a
2
2
2
8 2 5
2
2
2 2
P 5 a
P  P     a  3x  dx
x a
x a
a
X
2
  x   x  dx   x  a x  x
2
2
a
a
2
2

3 2
dx
16
X 2  315
a9
 H
1  1
1
2
2
2

P

m

X
E a 
   2m
2

2 5
2 9

105
4
a
8
m

a 

 16a 7  5m  315 



2
2 2
21
m a


2
4ma
6
Sample Problem (2)
21 2 m 2 a 2
E a 

2
4ma
6
• Minimize the energy:
d
21 2 m 2 a
0
E a  

3
da
2ma
3
2
min
a
63 3
7


m 2 m 2
• Substitute it back in:
21 2 m 2 m 2 3
7
E  amin  



 
4m 3
7
6 m 2

1
2
7
2
 12
7
2


• Since our state is odd, the lowest odd energy state must be lower
than this value
• Actual coefficient is 1.5 (25% off)
7
2
 1.87 
12B. The WKB Approximation
A Method For High-Energy States
• The variational method is good for ground state and other low energy states
• The WKB method is good for highly excited states
Idea behind the WKB approximation
• If the energy is large, the wave function will be quickly oscillating
• The potential will then look like it’s slowly varying
2
•
•
•
•
•
•
d2
  x   V  x   x 
Start from Schrödinger’s equation in 1D: E  x   
2
2m dx
2
2
Define: k  x   2m  E  V  x  


d2
 2   x   k 2  x   x 
dx
Then Schrödinger’s equation becomes:
If we think of k(x) as a constant, this solution would be like eikx
This suggests breaking  into a magnitude and a phase   x   A  x  ei  x 
A(x) and (x) are real functions
Breaking it into two equations
d2
 2   x   k 2  x   x 
dx
• Substitute in (denote
derivatives with primes):
  x   A  x  ei  x 
  Aei

 i
 i
k 2 Ae i    Aei    Ae  iAe  
  Aei
2iAei  iAei   Aei 2
k 2 A   A 2iA  iA   A 2
• Match real and imaginary parts: k 2 A   A  A 2 and 0  2 A   A 
• Multiply second equation by A
0  2AA   A2    A2  
N
2
A

• Anything with a zero derivative is a constant A    constant


• Rename  as W, then
2
 N
N
k
N
N


x
N
2
A



W




exp

i
W
x
dx
 



W
W
W
W


W  x


0th Order WKB Approximation

•
•
•


2
k
N
 N  N
2
exp i  W  x  dx
 

W

W  x
W
W
W


Expand out that second derivative

N
3N
 N   N
2




W

W



W
5/2
 2W 3/2
  2W 3/2
4
W

 W 
2
2




W
3
W
W
3
W
2
2
Substitute it back in k 2 


W
W

k


2
2W 4W
2W 4W 2
Warning! This expression apparently wrong in the notes!
N
x
k 2  x   2m  E  V  x  
• If energy is large, so is k(x)
• To zeroth order, assume that k2 dominates the other two terms
x
N

exp i  k  x  dx
W  x  k  x
k  x


2
Bound Problems with Steep Boundaries
N


• Suppose we want to consider bound states

exp i  k  x  dx
k  x
• For now, assume the potential goes to infinity
suddenly at x = a and x = b
E
• For bound states, we generally prefer real wave functions:
– Sines, cosines, or some   N sin x k  x  dx  
V(x)
a
linear combination
k  x
x
• We have two additional constraints:
– Wave function must vanish at x = a
a
b
– Wave function must vanish at x = b
• Vanishing at x = a requires that   0 b
• Then vanishing at x = b requires that  k  x  dx    n  1
a
– Where n = 0, 1, 2, …
2
2
k
x

2
m
E

V
x






• Recall


• So we have
b
 2m  E  V  xdx    n  1

a

x
Propagation in Classically Forbidden Zones
• What do we do in regions where E < V(x)?
• If the difference is large, we can still use the same
formulas, we just need to not be naïve
 x   2m  E  V  x 
 2  x   2m V  x   E 
In this case, define
k
2
•
• Then our wave
function will
look like


  x

2
exp     x  dx
x
k  x

exp i  k  x  dx
x
V  x
2
N
N

• Which solution we want will depend on whether we want growing or
damping exponentials
• Just like for E > V(x), this works well only if E – V(x) is large, and V(x) is
slowly varying
E
x

Bound State w/ Symmetric Sudden Boundaries




x
N
• Consider a potential which jumps up suddenly and
 
exp     x  dx
a
symmetrically at x = a and x = b
  x
• Divide it into the three regions:
x
N
– x < a, x > b, and a < x < b

sin  k  x  dx  
a
k  x
• Solutions in these three regions will be:
• We chose the sign of the exponentials to make sure
x
N
 
exp     x  dx
it vanishes at infinity
b
  x


• Because we made the gap symmetric, it is easy
to see that
  a   k  a     b   k  b   2m
• We now need to match the wave functions and
their derivatives at the boundaries
• Assume  and k are slowly varying
– Derivatives of them are negligible




E
V(x)
a
b
Matching at the Boundaries:
• Match < and  and their derivatives at x = a:
N
 a
N

N   a 

k a
Nk  a 
sin 
 N  N sin 
 a
k a
• Match  and > and their derivatives at x = b:
b
N
N

sin  k  x  dx  14 
a
 b
k b
b
tan  k  x  dx  14   1
 N   b 
 b

b
a

Nk  b 
k b
cos


b
a
k  x  dx  14 
k  x dx    n  
1
4
3
4
a

b
a


a
a
cos   N  N cos 

  x
exp     x  dx
x
x
tan   1   14 


N

sin   k  x  dx   
k  x
N
 
exp      x  dx 
  x
 
N

b

k  x  dx   n  12  
x





E
V(x)
a
b
Bound States with Smooth Boundaries
b
• It gets complicated with smooth boundaries
1


k
x
dx

n




2 

a
• All our approximations only work away from the
classical turning points a and b
• Must solve equations in the neighborhood of the
turning points more carefully
E
– Treat potential as linear in this neighborhood
– Solve equation using Bessel Functions (ick!)
V(x)
• Sometimes with imaginary arguments (ick ick!)
– Match to WKB solutions in the other regions
a
b
• Read Schiff Quantum Mechanics for more details
• Bottom line: we get the same quantization criterion as before:

b
a
k  x  dx   n  12  
• Substituting in the expression for k, we have
n  0,1, 2,
Three Different Cases
• We can get quantization conditions for
three types of problems:
• Hard boundaries:
n  0,1, 2,

b
a
2m  E  V  x dx   n  1 
E
• Soft boundaries:
V(x)
x
• One hard, one soft:
a
V  x
b
E
E
V(x)
a

b
a
b
2m  E  V  x  dx   n 
x
a
3
4


b
a
b
2m  E  V  x  dx   n  12  
How to use these formulas
• Given a potential V(x), can we estimate the energy
eigenstates?
• Pick the energy E
• Find the classical turning points a and b, the
points where the particle would stop classically
V  a   V b   E
• Write the relevant integral, usually
• Do the integral
• Solve for En as a function of n
V  x
E
a

b
a
b
2m  E  V  x  dx   n  12  
n  0,1, 2,
x
Sample Problem
Estimate the eigenenergies of a particle of mass m
in the 1D Harmonic oscillator with V(x) = m2x2/2
E
• We pick an energy E
• We solve for the
2E
2 2
1
x


E  V  x   2 m x
2
turning points:
m

• Substitute into the integral: 2 E m 2
2 2
1
1


2
m
E

m

x
dx

n


2
2 
2



 2 E m
• Make a trigonometric substitution:
 n  
1
2

1
2
 12 
x  sin  2 E m 2
2E

2 2E
2 
1
2m  E  2 m
sin  
cos  d
2
2
m

 m
En    n  12 
1
2E 
E
2
E
2E 12 
2
2
2


cos  d

2mE 1  sin   cos  d 
1

2  1 

 2

 2
m 2
• By coincidence, we got it exactly right
Probability Density in WKB and Classical
• Consider the wave function in the classically
allowed region in the WKB approximation:
N2
• The probability density is:  
sin 2
k  x
2

N
k  x
sin
  k  x dx   
x
a
  k  x dx   
x
a
• WKB works for high energy, k(x) large
• Rapidly oscillating sine function: sin 2  12
1
• So we have
2
  k  x 
• k is like momentum, which is like velocity
• Classically, fraction of time spent in region
of size dx is given by
2
2 dt
2dt

dx

dx
  x  dx 
vT
T dx
T

2
 v  x 
1
  x   v  x  
1
12C. Time Independent Perturbation Theory
A Series Expansion
• Consider a situation where the Hamiltonian can be
H  H0  W
split into a large, easily solved piece, and a small piece:
H  H 0  W
• Replace W by W, where  = 1
W  W
 1
• The large part is assumed to have known eigenvalues
H0 n   n n
and complete, orthonormal eigenvectors:
• Let |n be the exact eigenstates of H with energies En H  n  En  n
• In the limit   0, the |n’s will be |n’s and the En’s will go to n’s
• It makes sense, therefore, to imagine a series expansion in terms of the
parameter  for |n and |n
         2     3   
n
n
n
n
En   n   n   2 n   3 n 
n
The Idea Behind The Method
 n  n   n   2 n   3 n 
H  n  En  n
En   n   n   2 n   3 n 
• We write out Schrödinger’s equation as follows:

2
3









 n 
n
n
n
 
n
H  H 0  W
En  n   H0  W   n
  n   2 n   3 n 
  H 0  W   n   n   2 n   3 n 
• Expand to some order:
 n n     n n   n n    2  n n   n n   n n 


  3   n n   n n   n n   n n  
 H 0 n   W n  H 0 n    2 W n  H 0 n    3 W n  H 0 n  
• Since this must be true for all , the coefficients of every power of  must match
 n p  n   n p 1 n 
  n n p   W n p 1  H 0 n p 
A Small Ambiguity Problem
• Schrödinger’s time-independent equation does not
H  n  En  n
completely determine the eigenstates |n
– We can always multiply by an arbitrary complex number c:  n  c  n
– Therefore these states are slightly ambiguous
n  n  n n  1
• In the limit W = 0, we know that
• One way to resolve the ambiguity is to simply demand n  n  1
n n p   0
• The problem: this means final states |n are not normalized
 n  n   n  n  n   n  n  n  
 1  n n  n n  n n 
• Can always be normalized later
n 
• Irrelevant for energy computation
• Only relevant in 2nd or higher order state vector
1
n n
n
The Procedure
 p
 p 1
 n n   n
n 
 p
  n n
 p 1
 W n
n n p   0
 p
 H 0 n
• For each order (p):
 n p   n W n p 1  n H 0 n p 
– Act on the left with n|
– Let H0 act on n|:
 n n
– Act on the left with m| for m  n:
 n p 1 m n 
 n p   n W n p 1
 m m
  n m n p   m W n p 1  m H 0 n p 
 p
 p 1
 p 1
 p  2








W







m n 
 n m m n
m
n
n
m
n
n
– Reconstruct |n(p)
using completeness
• Iterate order by order
• When done, reconstruct
|n and En using  = 1:
 p
n
  m m n
 p
 p
n
  m m n
m
 n  n  n  n  n 
En   n   n   n   n 
  n m n p 1
 p
m n
First and Second Order:
n p    m m n p 
 n p   n W n p 1
m n
 p
 p 1
 p 1
 p  2








W







m n 
 n m m n
m
n
n
m
n
n
• First order:  n  n W n
 n   m  m n
 m W n
• Second order:  n  n W n
 n   m  m n

in
n   m
m n
 m W n   n m n
m W i i W n
m W n n W n

n  i
n  m
n  
m n
m W n
n  m
 n   n W m
mn
  n m n p 1
1
m W n
n  m
1
 n  
m W n
m n  n   m

m W i i W n
m W n n W n 

m 


2






 n   m 


 i  n  n i  n m 

2
Third Order Energy and Summary:
n p    m m n p 
 n p   n W n p 1
m n
• Third order energy:

m W i i W n
m W n n W n 


 n  n W n   n W m 

2






m n
 n   m 


 i  n  n i  n m 

• Put it all together: En   n   n   n   n
 n  n  n  n
En   n  n W n  
mn
  n W m
m n
 n  n   m
m n
m W n
n  m
2

m W i i W n
m W n n W n 



2
 n   m 
 i  n   n   i   n   m 

 m W n
m W i i W n
m W n n W n 




2
i  n   n   i   n   m 
 n   m 
  n   m

Sample Problem
Find the ground state energy to second order and eigenstate to first order for a
particle in 1D with mass m and potential V(x) = m2x2/2 + x4 when  is small.
1 2 1
P  2 m 2 X 2 , W   X 4
• Split Hamiltonian into H0 and perturbation W: H 0 
2m
• Find eigenstates and energies of H0:
n ,  n    n  12 
• We now need to find
E0   0  0 W 0  
m n
2
mW 0
mW 0
m
0  m
0  0  
0  m
m n
4
2
2


4


†
†
† 3
W 0  X 0   
a  a  0 
aa  0 
aa  1

2 2 
2 2 
4m 
4m 
 2m

4


2
4m 

2

aa
2 
2
4m 
2
2


† 2
2 2  0



2
4m 
2
a  a 
†
2


6 3 2 1  1
2
24 4  3 2 2  3 2 2  3 0  4m2 2


24 4  6 2 2  3 0

Sample Problem (2)
Find the ground state energy to second order and eigenstate to first order for a
particle in 1D with mass m and potential V(x) = m2x2/2 + x4 when  is small.

W 0 

2
24 4  6 2 2  3 0
4m2 2
• We therefore have:
3 2   2 
E0   0 
 2 2 
2 2
4m   4m  

n   n 
2
3 2
2 4



2 2
2 4m  16m4 4
0


 
2


2
 24 2
6 2 



 0  4 0  2 


 24 72 
 4   2  
 24

6 2
 0 
4 
2 
2 2 

4m    0   4
0  2 

1
2
E0   0  0 W 0  
m n
0  0  
m n
mW 0
2
0  m
mW 0
m
0  m
3 2 21 2 3
E0 


2 2
2 4m 
8m4 5
0  0 


4m 
2
3

1
2
6 4 3 2 2

Validity of Perturbation Theory
• Perturbation theory can only be trusted if subsequent terms get smaller
2
1
• Let’s compare first and second     W  ,   
m W n

n
n
n
n
order energy contribution:
m n  n   m
• Let  be smallest gap
 n   m   , all m.
between n and m:
• Then we have
1
1
2
1
 n  
m W n   n W m m W n   n W m m W n
 m
 mn
m n  n   m
1
 n W 2 n
• Compare to 'n

1
2
• Roughly, perturbation theory works if n W n  n W n

• So we need   W
12D. Degenerate Perturbation Theory
The Problem and Its Solution
• Look at the second order expression for energy or first order for state vector:
2
m W n
m W n
 n  n   m
En   n  n W n  
n  m
n  m
m n
mn
• If two states have the same unperturbed energy, we will get in trouble
• The problem disappears if
m W n  0 for  m   n , m  n
we get “lucky” by having
• If we have such states, then the unperturbed eigenstates are in fact ambiguous
H0 1   1
and H0 2   2
 H0      
for
   1 1   2 2
• We can use this ambiguity to change basis for our unperturbed states
• Thereby ensuring the problem goes away
• Then proceed with perturbation theory as normal
Procedure for Degenerate Perturbation Theory

• Suppose we have a set of degenerate states
1 , 2 , , g
• Define the reduced W matrix:
 1 W 1
• Find g normalized eigenvectors
1 W 2

and eigenvalues for this matrix:
 2 W 1
2 W 2
†
Wv i  wi v i
vi v j  ij
W 
• Change

i    vi m m

basis to:
 g W 1
 g W 2
m

• Then in the new basis, we have:
,
H 0 i   i
1 W g 

2 W  g 



g W g 
i  j   m  vi m   v j n n    nm  vi m  v j n    vi *m  v j   v†i v j  ij
m
*
*
• And:
m
m
n
n
i W  j   m  vi mW   v j n n
*
m
n
m
   vi m Wmn  v j 
*
m
   vi m w j  v j   w j vi† v j  w j ij
m
*
m
• So we can use these as our new basis states!
n
n
Working in the New Basis States
• In the new basis, the matrix elements of W vanish
for any pair of distinct states with the same energy
• This gets rid of problem terms
En   n  n W n 

 
m n
m W n
n  m
m W n  0,  m   n , m  n
2
 n  n 

 
m n
m W n
m
n  m
• Note that the basis states we start with are determined partly by the perturbation
– Even to leading (0th) order, we need to include perturbation theory
– The perturbation breaks the degeneracy and determines our states
• Note that the first correction to the energies are just
n W n   j  vn  j  j W i  vn i i   j i  vn  j W ji  vn i  vn†Wvn  wn vn†vn  wn
*
*
• First order energy easy to find just from eigenvalues of W
En   n  wn
Sample Problem
A particle of mass m in two dimensions has Hamiltonian as H  1 P 2  P 2
 x y
given at right, where b is small
2m
(a) What are the eigenstates and energies in the limit b = 0?  1 m 2  X 2  4Y 2   bYP 2
x
2
(b) For the first pair of degenerate states, determine the
eigenstates to leading order and energies to first order in b.
1 2 1
1 2 1
2
2
2
• H0 is two harmonic oscillators
H0 
Px  2 m X 
Py  2 m  2  Y 2
2m
2m
– The x-direction has frequency 
W  bYPx2
– The y-direction has frequency 2
• The unperturbed states and energies are: nm ,  nm    n  12   2   m  12 
• The first two states are nondegenerate
00 ,  00  32  10 , 10  52 
• The second excited states are degenerate
20 , 01 ,  20   01  72 
• Need to use degenerate perturbation theory!
Sample Problem (2)
(b) For the first pair of degenerate states, determine the
eigenstates to leading order and energies to first order in b.
• We place our degenerate states in some order:
 20 , 01   20   01  72 
• We then calculate the perturbation matrix
Px  i
1
2
m  ax†  ax 

20 W 01  b  i
20 W 20  b i
  14 b
1
2
1
2

m 
m
2
2
a

2m  2 
*
  14
 20 W 20
W 
 01 W 20
†

a
y
y
 a
 4m  20  a  a  a
 4m 
20  a y  a
y
m 20 a y ax†2 01   14 b
01 W 20  20 W 01
01 W 01  0
Y
2 m
1
2
2
H
P

P

x
y 
2m
 12 m 2  X 2  4Y 2   bYPx2
†
y
†
x
 ax  20  0
†
y
†
x
 ax  01
20 W 01 

01 W 01 
2
2
2 m
W  b
1
4
0 1
2 m 

1
0


Sample Problem (3)
(b) For the first pair of degenerate states, determine the
eigenstates to leading order and energies to first order in b.
• We now find eigenstates and eigenvalues of this matrix:
 1
1
1
     ,    2   , w  14 b
 1
 1
• The energies and eigenstates are therefore:
1
2
 
1
2
 20
 01  , E 
7
2

1
4
b
 20 , 01 
 20   01  72 
W  b
0 1
2 m 

1
0


2 m
2 m
• Note that to find the energies, all we needed was the eigenvalues
1
4