Ph235
Autumn 2008
F. Merritt
Oct 17, 2008
Lecture 10 (October 20, 2008):
(version 1.0; 20-Oct-2008::23:00)
Time Independent Perturbation Theory
[The first part of this lecture was spent finishing the Variation Method, and going
through the calculation of the helium ground state energy using this technique. The last
15 minutes was spent on introducing TIPT, following Shankar 17.1]
Introduction:
There are a fairly small number of Hamiltonians in QM where we can calculate a
complete set of exact energy eigenfunctions. The best examples are the hydrogen atom
and the simple harmonic oscillator.
But we often need to deal with problems where the Hamiltonian is altered by
additional terms; for example, how does a magnetic field alter the eigenstates and energy
levels of hydrogen? How do we find the modified eigenvectors and eigenvalues?
Suppose we have a complete set of eigenstates and eigenfunctions for a
Hamiltonian H 0 :
H 0 n0  En0 n0
with
n0 m0   nm
(9.1)
But now we need to deal with a slightly altered Hamiltonian,
(9.2)
H  H 0  H
0

where H is relatively small compared to H . We want to find the eigenvectors and
eigenvalues of the full Hamiltonian, defined by
(9.3)
H n  En n
with n m   nm
We begin by writing (9.2) as
H  H 0   H 
(9.4)
where  is a dimensionless factor which we can continuously vary over the range 0,1 .
Both the eigenvalues and eigenvectors of H  will clearly vary with  , so we can write
the eigenvalues and eigenvectors of equation (9.4) in terms of a power series in  :
n   n0   n1   2 n 2   3 n3  .....
(9.5)
En  En0   En1   2 En2   3 En3  .....
We will solve for n and En by explicitly calculating each of the coefficients in these
expansions; in practice, we only need the first few terms. Substituting these expressions
into (9.4) gives
H  n   En n  
H
0
  H    n0   n1   2 n 2  ...
(9.6)
  En0   En1   2 En2  ...  n0   n1   2 n 2  ...
The coefficients of each power of  must be the same on both sides, so we have a
separate equation for each order of  . The first few of these are:
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Ph235
Autumn 2008
F. Merritt
Oct 17, 2008
H 0 n 0  En0 n0
H 0 n1  H  n 0  En0 n1  En1 n 0
(9.7)
H 0 n 2  H  n1  En0 n 2  En1 n1  En2 n 0
H 0 n3  H  n 2  En0 n3  En1 n 2  En2 n1  En3 n 0
. Note that the sum of the upper indices is always equal to the corresponding power of
 . At this point we can dispense with the  ’s and just work with the equations.
The first equation is just (9.1), from our unperturbed Hamiltonian. For the second
equation, multiply each side by n 0 :
n 0 H 0 n1  n 0 H  n 0  En0 n 0 n1  En1 n 0 n 0
 En0 n 0 n1  n0 H  n 0  En0 n 0 n1  En1
The first terms cancel, so we are left with
En1  n0 H  n0
(9.8)
(9.9)
giving us the first term in the energy series. This is often all we need. But let’s keep
going. To find the first term in the eigenvector expansion n1 , multiply both sides of
 
(9.7b) by m0 (for m  n ):
m0 H 0 n1  m0 H  n 0  En0 m0 n1  En1 m0 n 0
 Em0 m0 n1  m0 H  n 0  En0 m0 n1
(9.10)
Solving for m 0 n1 gives
m n 
0
1
m0 H  n 0
(9.11)
En0  Em0
Since m0 form a complete set, any vector (like n1 ) can be expanded in terms of them.
The equation we have just derived gives us the expansion coefficients, so we have
 m0 H  n 0 
1
 m0
(9.12)
n  
0
0
En  Em 
m n 


It’s not difficult to continue to higher orders. For second order, take (9.7c) and
again multiply both sides by n 0 to obtain
n0 H 0 n 2  n0 H  n1  En0 n 0 n 2  En1 n 0 n1  En2
(9.13)
The first terms on each side cancel, and from (9.12) n 0 n1  0 , so we are left with
E  n H n  
2
n
0
1
m n
 m0 H  n 0
n H m 
 En0  Em0

0
0




(9.14)
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Ph235
Autumn 2008
F. Merritt
Oct 17, 2008
giving
En2  
m0 H  n0
m n
2
(9.15)
En0  Em0
We could keep going. To find the second-order correction to the wavefunction,
dot m0 into both sides of (9.7c), and substitute in the known values of the terms given
by (9.9), (9.12), and (9.15). But the expressions get increasingly complicated, and the 3
equations we have already derived are sufficient for most problems – and for all the
problems we will tackle in Ph235.
Let’s take the ISW as an example. The potential outside the well is infinite, and
the potential inside is zero, so over the interval x  [0, L] the Hamiltonian is
H0  
2
2m x 2
2
(9.16)
The eigenfunctions and eigenvalues are
2
2
 n 
 n 
sin 
x
 En0 
(9.17)


L
2m  L 
 L 
First, suppose we add a constant V0 to the potential over the entire region. Then
2
n0 
from (9.9), the first-order change to every eigenvalue is V0 (which certainly makes
sense). From (9.12), there is no change to any of the eigenfunctions. From (9.15), the
second order change is zero as well, and in fact the first-order correction term gives us
the exact solution.
Now suppose we add a delta-function at the center of the well, so
(9.18)
H    L   x  L / 2
(parmeterized so that  has the dimensions of energy). Then from (9.9) the first-order
correction to the energy of solutions with even n is zero, and the first-order correction for
odd n is
En1  2
(n odd)
(9.19)
En1  0
(n even)
From (9.12), the first-order correction to the wave-function for even n is zero, since each
term in the numerator vanishes. For odd n and odd m, we get
nm
nm
 L  2 / L 

2  n2 
1
0
1
2
(9.20)
m n 
 0 2
(1) 2
 1
En0  Em0
En  m  n2 
so
nm
 n2 
2
1
(9.21)
n  0   2
(1) 2 m0
2
En m odd,n  m  n 
So the change to the ground state wavefunction, to first order, is
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Ph235
Autumn 2008
F. Merritt
Oct 17, 2008
 2mL  2  1  3
 sin 
2 2 
   L 8  L
 11 ( x)  2 
 m L / 2    3
   2 2  sin 
   L

 1
 5
x   sin 
 24  L
 1  5
x   sin 
 3  L
 1
 7
x   sin 
 48  L
 1  7
x   sin 
 6  L


x   ...




x   ...


(9.22)
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