Perturbation Examples
Perturbation Examples
Perturbation Examples
2 2
0
2
2
d
m dx
=
H ,
0
2
N
N x
sin
a a
= and
2
2
0
2
N
N
E
m a
=
=
along the x axis the charged particle experiences an additional potential energy V q
and so the perturbed problem is
Fx
N
( )
0
N N
= =
Exercise: Evaluate
0
sin N sin K d
J. F. Harrison Michigan State University 9/19/2006 1
B. Harmonic Oscillator with a cubic perturbation
Suppose we are interested in estimating the eigenvalues and eigenvectors associated with
the Hamiltonian
2 2
2 3 0 3
2
1
2 2
d
H kx x H
m dx
x = + + = +
The first order correction to the energy of the state
0
n
is then zero by parity arguments.
(1) 0 3 0
0
n n n
E x = =
The second order correction is
2 2
2
0 0 0 3 0 3
(2)
0 0
( ) ( )
n p n p
n
n p
p n p n p n
V x n x p
E
h n p h n p
E E
= = =
So we need to evaluate matrix elements of
3
x between various harmonic oscillator states.
Using recursion relationships between the Hermite polynomials many texts show that
, 1 , 1
1
( )
2 2
j i j i ij
j j
i x j x
+
+
= + =
and this result allows us to evaluate matrix elements of
l
x for any positive integer value
of using the resolution of the identity. Since the set of eigenfunctions is complete
we may write
l
0
n
0 0
0 0
1
n n
n n
n n
= =
= =
So
2 2
0
( )
ij
n
i x j x i xx j i x n n x j
=
= = =
=
=
Using the above result for ( )
ij
x the sum can be evaluated and we obtain
J. F. Harrison Michigan State University 9/19/2006 2
2
, 2 , 2 2 2
( 1) 2 1 ( 1)( 2)
( )
4 2 4
ij j i ij j i
i i i i i
x
+
+ + + +
= + +
Note the systematics. The matrix element ( )
ij
x is zero unless i & j differ by 1, say j is
equal to i . The matrix element 1& 1 i +
2
( )
ij
x is zero unless j equals i i . The
matrix element (
2, , 2 i +
3
)
ij
x will vanish unless 3, 1, 1& 3 i i i j i = + + whereas
4
( )
ij
x will
vanish except for , and so on. Returning to the perturbation sum
for
4, 2, , 2, i i i i = + + 4 j i
(2)
n
E we see
3 2 3 2 3 2 3 2 3 2
2 2
, 3 , 1 , 1 , 3 (2)
( ) ( ) ( ) ( ) ( )
( ) 3 1 1 3
np n n n n n n n n
n
p n
x x x x x
E
h n p h
+ +
= = + +
x
Now
3 2
, 3 , , 3
0
( ) ( ) ( )
n n n k k n
k
x x
+ +
=
=
and since ( ) vanishes unless
, 3 k n
x
+
2 k n or n+4 = + we have
3 2 2
, 3 , 4 4, 3 , 2 2, 3
( ) ( ) ( ) ( ) ( )
n n n n n n n n n n
x x x x x
+ + + + + +
= +
+
and since ( ) , we have
2
, 4
0
n n
x
+
=
3 2
, 3 , 2 2, 3 3
( 2)( 1)( 3
( ) ( ) ( )
(2 )
n n n n n n
n n n
x x x
+ + + +
) + + +
= =
In a similar fashion we find
3
3
, 1 3
( 1)
( ) 3
(2 )
n n
n
x
+
+
=
3
3
, 1 3
( ) 3
(2 )
n n
n
x
=
3
, 3 3
( 1) 2)
( )
(2 )
n n
n n n
x
=
Assembling these components we find
J. F. Harrison Michigan State University 9/19/2006 3
2
(2) 2
3
30
( 11/ 30)
(2 )
n
E n n
h
= + +
The first order correction to the wavefunction depends on the same matrix elements so
3 0 3 0 3 0 3 0 3 0
, 3 3 , 1 1 , 1 1 , 3 3 (1)
( ) ( ) ( ) ( ) ( )
( ) 3 1 1 3
np p n n n n n n n n n n n n
n
p n
x x x x x
h n p h
+ + + +
= = + + +
C. Harmonic Oscillator in a Constant Electric Field
Consider a one dimensional harmonic oscillator in a constant electric field , and let the
charge on the oscillator be q. If the oscillator is on the x axis, the Hamiltonian is
F
2 2
2
2
1
( )
2 2
d
H kx
m dx
= + +
q x
In one dimension
d
Fx
dx
x
2 2
d
H kx
m dx
= +
qFx
We can solve this problem exactly and then compare the perturbation and exact results.
Lets solve it exactly.
First note that we can complete the square on the potential
2 2
2 2
1 2
2 2 2
k qF k qF qF
kx qFx x x x
k k
= =
k
and then define the new variable
qF
x
k
= , resulting in the Schrodinger equation
( )
2
2 2
2
2
1
( ) ( ) ( )
2 2 2
qF
d
H k
m d k
E
= + + =
or
J. F. Harrison Michigan State University 9/19/2006 4
( )
2
2 2
2
2
1
( ) ( )
2 2 2
qF
d
k E
m d k
+ = +
This is the harmonic oscillator equation, so, as we have seen above
2
1/ 2
( ) ( )
n n n
N H e
= and ( )
2
1
( )
2
n
2 E h n qF k = +
Note that all of the levels have been lowered by the same amount and the wavefunctions
have all been shifted along the x axis so that they are centered at
qF
x
k
= . If q and F are
both positive the equilibrium point is shifted in the x + direction as expected.
The explicit dependence of the first two wavefunctions on the electric field is
( )
( )
( ) ( )
( )
1/ 4
0
1/ 4
3
1
2
2
/ 2
/ 2
( )
4
( )
qF
x
k
qF
x
k
x
qF
x x
k
e
e
=
=
Now lets use perturbation theory to solve the problem.
The perturbation is and the first order correction to the energy is zero by parity.
The second order correction is then
qFx
( )
2 2
2
2 pn ( )
n
p n p n
x
p qFx n
( qF )
E
hv( n p ) h ( n p )
= =
From above the only terms that will appear in the summation are 1 p n = so we have
2 2 2
2 1 1
1 1
( ) n,n n.n
n
( x ) ( x ) ( qF )
E
h
+
= +
2 2
1 1
1
2 2
n,n n,n
n n
( x ) & ( x )
+
+
= =
so
( )
2
2
2
( )
n
qF
E
k
=
J. F. Harrison Michigan State University 9/19/2006 5
Note that because the first order correction to the energy vanishes, the third order
correction has the form
(3)
0 0
np pt tn
n
np nt
p n t n
x x x
E
E E
=
= = + =
+
Exercise:
We have seen that the exact solution for the ground state of the harmonic oscillator in a
constant electric field is
( )
( )
1/ 4
0
2
/ 2
( )
qF
x
k
x e
= while perturbation theory tells us that
(1) 0
1 0
2
qF
h
= Show that these two results are consistent. Hint: expand the exact
solution in a power series in the electric field.
J. F. Harrison Michigan State University 9/19/2006 6