CH. VI
Chapter VI
Perturbation Theory
The Hamiltonian we shall consider is in the form
H  H o  gV
(1)
with Ho is the unperturbed Hamiltonian with well-known eigen-functions and eigenvalues. gV is the
perturbation with g is a small parameter.
Time Independent Non-degenerate Perturbation Theory:
Let
H o no   no no
(2)
And
H n   n n
(3)
Substituting for H from Eq. (1), Eq. (3) becomes
H o  gV  n
with
  n n
(4)
g 0
n  no
(5)
It will be assumed that
n  n0  g n1 g 2 n2  
(6)
and
 n =  n0  g 1n  g 2 n2  
(7)
where ni and  ni represent the ith order correction to the eigenvectors and eigenvalues, respectively.
Now, substituting Eqs. (6) and (7) into Eq. (4) 
H o  gV  n0

0
n

 g n1 g 2 n2   


(8)
 g 1n  g 2 n2   n0  g n1 g 2 n2  
Collecting terms of like power of g 
H
o
n0   n0 n0

 g H
o

n1  V no  n0 n1   1n no 

g 2 H o n2  V n1  no n2  1n n1   n2 n1    0
1
(9)
CH. VI
Since Eq.(9) is to be valid for any g  every coefficient of g must equal to zero 
H o n0   n0 n0
(10)
H o n1  V no  n0 n1   1n no
(11)
H o n2  V n1  n0 n2   1n n1   n2 no
(12)
First Order Correction:
Let us make the expansion
n1   a1nj  oj
n  1,2,
(13)
j 1
Substituting Eq. (13) into Eq. (11) we get
H o  a1nj  oj  V no  n0  a1nj  oj   1n no
j 1
(14)
j 1
Now using Eq.(10), Eq. (14) reads




  0j   n0 a1nj  oj   1n  V no
(15)
j
Multiplying Eq. (15) by ko we get
 
0
j



  n0 a1nj ko  jo  ko  n1  V no
j
This gives


  0j   n0 a1nj  kj   1n kn  ko V no
(16)
j
Applying the properties of the Kronekel Delta we obtain

0
k

  n0 a1nk   1n kn  ko V no
(17)
To find the value of  n1 we set k=n 
 1n  no V no  Vnn  V
(18)
n
2
CH. VI
from Eq. (7) to first order we get
 n   n0  gVnn
(19)
To find ank, we set n  k in Eq. (17) 

0
k

  n0 a1nk   ko V no  Vkn
(20)
From Which we get
a1nk 
Vkn
 n0
(21)
  k0
From Eqs. (6) and (13) to first order we get
n  n0  g  a1nj  oj  n0  ga1nn n0  g  a1nj  oj
j 1
(22)
j n
Now substituting for a 1nj from Eq. (21) we obtain


n  1  ga1nn n0  g 
V jn
0
j n  n
  0j
 oj
(23)
Multiplying the above Eq. By its complex conjugate and dropping terms of order g2 
1  1  ga1nn
2
 1  2 ga1nn

Re a1nn  0
(24)
n  n0  g 
V jn
0
j n  n
  0j
 oj
(25)
The wave-mechanical analogue of Eq. (25) is
n  no  g 
V jn
0
j n  n
  0j
 oj
(26)
with
V jn    j V n dv
(27)
3
CH. VI
Example: Find the first order correction for the eigenvalues and the eigenfunctions of the Hamiltonian
H  H o  gx
2
p2
Ho 
2
with
1
2
 2 x 2
From Eq.(18) we have
 1n  Vnn  no V no  n x 2 n
But x 
a  a 
2 

†
Vnn 
Knowing that
Vnn 


2
x2 
2
2

aa  aa
†
 a†a  a†a†



n aa  aa †  a † a  a † a † n
n aa n  n a † a † n  0 ,


n a † a n  n , and
n aa † n  n  1

2n  1
Now from Eq.(19) we get
 n   n0  gVnn   n  12 

n  12 

g
Now from Eq.(25) we have
n  n0  g 
V jn
0
j n  n
V jn 
Knowing that
V jn 

2
 nn  1
  0j

2


j aa  aa †  a † a  a † a † n
j n  2  n  1 j n  n j n 
n  n0  g

 oj
 
 nn  1 o
n2 
 0
0
2 



n2
 n


n  1n  2
n  1n  2
 n0   n0 2

j n2




no 2 



But  n0   n02   n  12  n  2  12  2 and  n0   n0 2   n  12  n  2  12  2
n  n0 
g
4 2
 nn  1 n  2 
n  1n  2 n  2 
4

CH. VI
Second Order Correction:
Make the expansion
2
n2   anj
 oj
n  1,2,
(28)
j 1
Substituting in Eq. (28) into Eq. (12) 
2
2
H o  anj
 oj  V n1  n0  anj
 oj   1n n1   n2 no
j 1
(29)
j 1
Multiplying Eq. (29) by no and using Eq. (10) where H o  0j   0j  0j 
 anj2  oj no  oj  no V n1  n0  anj2 no  oj   1n no n1   n2 no no
j 1
j 1
Using the identity
n  j   nj
(30)
we obtain
2 o
2
a nn
 n  no V n1  n0 a nn
  n1 no n1   n2
(31)
Now substituting for n1 from Eq. (25), we get
2 o
a nn
n 

V jn
0
jn  n
  0j
2
no V  oj  n0 a nn
  n1 
V jn
0
j n  n
  0j
no  oj   n2
(32)
Using the identity of Eq. (30) we get
 n2 

V jn
0
jn  n
  0j
no V  oj 

V jn Vnj
0
jn  n
  0j


V jn
0
jn  n
  0j
And for the energy we have
 n =  n0  g n1  g 2 n2
with  n1 &  n2 are to be evaluated from Eqs. (18) and (33) .
5
2
(33)
CH. VI
Example: Find the eigenvalues of the Hamiltonian

p2 1
H
 2  2 x 2  2bx
2
a) By exact solution

b) by perturbation method
a) Let us rewrite the Hamiltonian as
H
p2 1
 2  2 x  b 2  12  2 b 2  H o  12  2 b 2
2
H

Now
H n  n n 
But
H o n   n0 n   n  12 n

0
n


o

 12  2 b 2 n   n n
 12  2 b 2 n   n n



 n   n0  12  2 b 2   n  12   12  2 b 2

p2 1
  2 x 2   2 bx  H o  gx
2 2
From Eq.(18) we have
with g    2 b
b) H 

 1n  Vnn  no V no  n x n 

Now from Eq.(33) with V  x 
 n2
 
V jn
0
j n  n
 



aa
†


2
  0j
V jn 
Knowing that
V jn 
2 
2 
n a  a† n  0
n
2

 n2 



j a  a† n
2 
j n 1 
n  1

j n 1 

2
 n  j,n1 
n  1 j ,n1 



n  1 
 
n


 0
0
0
0 
2 
  n   n1  n   n1 



But  n0   n01   n  12  n  1  12   and  n0   n01   n  12  n  1  12  
6
CH. VI
 n2 
  n n  1
1



 
2  
 
2 2

 n   n0  g 1n  g 2 n2    n0 

 2 4 b 2 
  n  12   12  2 b 2
2 
2  
Time-Independent Perturbation Theory for Degenerate Case:
Until now we have assumed that the perturbed eigenstate n differ slightly from the unperturbed eigenstate
no . When the unperturbed energy  no is degenerate, there are several eigenstates
o
nk
k  1,2,3,, d 
corresponding to this eigenvalues, and we don’t know to which state n will tend as g  0 , so the
perturbation procedure developed previously cann't be applied without modification. We will assume that the
perturbation removes the degeneracy so that n must be unique and will be approximately equal to one
particular unperturbed state (selected state)  no , i.e.,
g 0
n 
  no
(34)
The first order shift in the energy will be given by (from Eq.(19))
 n =  n0  g  no V  no
(35)
We will consider the perturbation of an d-fold degeneracy and set  no   o and
d
 no   bnk ko
n  1,2,  d
(36)
k
with
 no  no  1 
d
d
  bnj bnk  oj ko   bnj bnk  jk 1 
j 1 k
 bnk
j ,k
2
1
(37)
j ,k
It is clear from Eq.(36) that the selected state is a linear combination of all the unperturbed states.
Now rewrite Eq. (11) for  no we get
H o n1  V  no  n0 n1   n1  no
(38)
7
CH. VI
Using Eqs. (10), (13), and (36), we get for Eq. (38)
d
d
 a1nj  0j  oj   bnk V ko  n0  a1nj  oj   1n  bnk ko
j 1
j 1
k
Since  no   oj   o for n  d 
(39)
k
the above equation becomes
d
d
 a1nj  0j  oj   bnk V ko  n0  a1nj  oj   1n  bnk ko
j d
k 1
j d
Now, multiplying Eq. 40) by mo
(40)
k 1
m  1,2,3,, d  
d
 a1nj  0j mo  oj   bnk mo V ko 
j d
k 1
 n0
 a1nj
j d
mo
 oj
  1n
(41)
d
 bnk mo
k 1
ko
Using the properties of Kronekkel delta the first sum of both sides in Eq. (41) will vanish and we have
d
 bnk mo V ko   1n bnm
k 1
or
d
 Vmk bnk   1n bnm
(42)
k 1
or in matrix form
 V11  V1d  bn1 
 bn1 




1
      n   
 
V


b 
 d1  Vdd  bnd 
 nd 
(43)

Example: Find the eigenvalues of the H-atom with the application of a uniform electric field E  E kˆ (the
Stark effect)
It is known that
 
 
U   p  E  er  E  eEz

H  H o  eEz  H o  eE r cos 
For the ground state (n=1) we have no degeneracy (d=1)
V  V11  100 z 100   100 r cos  100 d 3 x
8

CH. VI
1
Knowing that 100 
V  V11 
a 3
1
a
3
e
r

a

2

0
0
0
3 2 r a
dr  0 
 sin  cos  d  d  r e
11  0
For the 1st excited state (n=2) we have 4-fold degeneracy (d=4), namely 200 , 210 , 211 , 211



V 



200 z 200
200 z 210
200 z 211
210 z 200
210 z 210
210 z 211
211 z 200
211 z 210
211 z 211
211 z 200
211 z 210
211 z 211

200 z 211 

210 z 211 
211 z 211 

211 z 211 

Since z  r cos is independent of  and noting that  Ylm
'Ylm d  N mm' ,all terms with different m vanish. 
V13  200 z 211  0 ,
V14  200 z 211  0
V23  210 z 211  0 ,
V24  210 z 211  0
V31  211 z 200  0 ,
V32  211 z 210  0
V34  211 z 211  0
V41  211 z 200  0 ,
V42  211 z 210  0
V43  211 z 211  0
So we conclude that there is no splitting associated with the states 211
Knowing that 200  14
r   r 2a

2

,

e
a
2a 3 
1
210  14
r  r 2a
e
cos  and
2a 3 a
211  18
1 r  r 2a
e
sin  e i
3 a
a
1
and from the oddness property of the  integral we also conclude

2

2
V11  200 z 200 
1
16
r  r 3 r a

cos

sin

d

d

2

e
dr  0



 
a a
2a 3 0
0
0
V22  210 z 210 
1
16
3
 r  r r a
 cos  sin  d  d   a  a e dr  0
2a 3 0
0
0 
1
1

1
V33  V44  211 z 211  16
2
3
2
3
2 
 r  r r a
3
cos

sin

d

d


   a  a e dr  0 
a 3 0
0
0 
1


2
9
CH. VI
 0 V12 

V  
V21 0 

1
V12  V21  200 z 210  210 z 200  16

Using the result of  x n e ax dx 
0
n!
a n1
2 
r  r 4 r a

2
sin

cos

d

d

2


   a  a e dr

2a 3 0
0
0

a 0, n is  ve integer
 3a 
 0

V  
0 
  3a

V12  V21  3a
,

1
Now from Eq.(43) we obtain
 3a  b1 
b 
 0

    12  1  
0  b2 
  3a
 b2 
 3a    12
 0


0   0
  3a
 

1 2
2
 9a 2

 12  3a
0 
0
 12 

 2   20  g 12   20  3aeE
1
For  12  3a we have b   N   , and for  12  3a we have
  1
2  N  200  210

1
b   N  
1
and 2  N  200  210
 from Eq.(36)


2
1
200
210
211
100
211
3aeE

3aeE
100
E. Field Off
E. Field On
10
CH. VI
The Variational Method:
The variational method is the other main approximate method used in quantum mechanics. Compared to
perturbation theory, the variational method can be more efficient in situations when there is no small
dimensionless parameter in the problem, or the system is not exactly solvable when the small parameter is sent
to zero.
The basic idea of the variational method is to guess a ``trial'' wavefunction for the problem, which consists of
some adjustable parameters called ``variational parameters.'' These parameters are adjusted until the energy of
the trial wavefunction is minimized.
Why would it make sense that the best approximate trial wavefunction is the one with the lowest energy? This
results from the Variational Theorem, which states that the energy of any trial wavefunction E is always an
upper bound to the exact ground state energy. This can be proven easily by the following Lemma.
Lemma: Let  be an arbitrary normalized ket and 1 be the ground state energy of the Hamiltonian H. Then
the following inequality always holds
 H   1
(44)
Proof:
Let    ai  i
With H i   i i
(45)
i
Using Eq.(45), the L.H.S of Eq.(44) becomes

 
L.H .S     i ai  H   a j  j
 i
  j
L.H .S   ai a j  j  ij    j ai
i, j

   ai a j  j  i  j
 i, j


2
i
But since 1 is the ground state energy
1   i 

2
L.H .S    j ai   1  ai
i
Since  is normalized and using Eq.(45) we get  ai
2
1

2
i
L.H .S   1
i
The variational method is applied by choosing a reasonable trial wave-function which depends on a set of Nparameters 1 , 2 , N , i.e.,
trial  trial 1 , 2 , N 

 H    trial
Htrial dv  I 1 , 2 ,   N 

(46)
11
CH. VI
I 1 , 2 , N    1 
Now from the lemma we have
To find the ground state energy we have to minimize the energy function I 1, 2 ,N  with respect to the
parameters , i.e.,
I
0 

 var  I min
(47)
Example: Let us sole the H-atom by a adopting a trial function of the form trial 
1

3
e
r

2
2

 
 p  e trial dv
Htrial dv   trial
Now from Eq.(46) we have I     H    trial
 2 r 


After some calculation we obtain
I   

2
22
2
 e2

a
e2



I      2
e 2 




  22 d 
  I min 
2
 2 
2  2 
e 


2

e2
 2 


  e2 


Let us now try another trial wavefunction (Gaussian) trial
I   

3 2
22
3  2
2 2 e 2

2 2 e2
 

   I min  

2
3

 2 
 2 
  
2

e2
2
 e4
2 2
r
3
e
0

which is the exact result.
2
2

3 2 2 2e 2
I     3 2 2 2 e 2 





0

  22
  
3
 2
8  e4
 0.85 1 which a good answer within 15%.
3 2 2
12
