0% found this document useful (0 votes)
67 views37 pages

كمي ٢

Uploaded by

r4g9mnh8dt
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
0% found this document useful (0 votes)
67 views37 pages

كمي ٢

Uploaded by

r4g9mnh8dt
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
You are on page 1/ 37

Quantum Mechanics

Approximation Methods
Time-Independent Perturbation Theory
Fourth Stage Students,
Department of Physics, College of Science,
University of Baghdad
Prof. Dr. Ali Abdulateef Alzubadi
2023-2024
Time-Independent Perturbation Theory (TIPT)
TIPT concerned with finding the changes in the energy and
the eigen function of a system when a small disturbance is
applied.
Separating the TI Hamiltonian H of a system in two parts:
Hˆ  Hˆ 0   Hˆ  (1)
where

Hˆ : the perturbed Hamiltonian operator.


Hˆ 0 : the unperturbed Hamiltonian operator.
Hˆ : a small perturbation term, such as Hˆ   H 0 .
: a parameter 0    1 and used to distinguish between
the various order of perturbation. As   0, Hˆ  H 0

Prof. Dr. Ali A.Alzubadi QM_Lectures


Perturbation of Non-Degenerate States
In non-degenerate case, there is only one e.f. corresponding to each
energy level.

Prof. Dr. Ali A.Alzubadi QM_Lectures


Prof. Dr. Ali A.Alzubadi QM_Lectures
Prof. Dr. Ali A.Alzubadi QM_Lectures
n 1
Degeneracy    2 l  1
l 0

Prof. Dr. Ali A.Alzubadi QM_Lectures


Perturbation of Non-Degenerate States
TISE (unperturbed): Hˆ 0  0n  E n0  0n (2)
where  0n is the eigen state of Hˆ 0

TISE (perturbed): Hˆ  n  E n  n
or Hˆ 0   Hˆ   n  E n  n (3)
This means, when   0  n =  0n
 n and E n are the w.f. and energy of nth level for perturbed case
respectively, written in a series:
 n   0n    (n1)   2  (2)
n  ... (4)
E n  E n0   E n(1)   2 E n(2)  ... (5)
Zero- 1st order 2nd order
order correction correction

Prof. Dr. Ali A.Alzubadi QM_Lectures


Substitute Eqs (4) and (5) into (3) one gets:
 H 0   H    0n    H 0   H    (1)
n   2
 H 0   H    (2)
n 

E 0
n 
  E n(1)   2 E n(2 )   0n    (1)
n   2
 ( 2)
n 
or H 0  0n   H   0n
  H 0  (1)
n   2
H   (1)
n

  2 H 0  (2)
n   3
H   (2)
n

 E n0  0n   E n0  (1)
n   2
E 0
n  (2)
n

+ E n(1)  0n   2 E n(1)  (1)


n   3 (1)
E n  (2)
n

+ 2 E n(2)  0n   3 E n(2)  (1)


n   4
E (2)
n  ( 2)
n

Equating the coefficients of equal power of λ we get:

Prof. Dr. Ali A.Alzubadi QM_Lectures


Equating the coefficients of equal power of λ:

 0 terms: H 0  0n  E n0  0n (6)
 1 terms: H   0n  H 0  (1)
n  E 0
n  (1)
n  E (1)
n  0
n

or
H 0  E 0
n   (1)
n    H   E (1)
n   0
n
(7)
H 0  E 0
n   (1)
n  E (1)
n  H    0n

 2 terms: H   (1)
n  H 0  (2)
n  E 0
n  (2 )
n  E (1)
n  (1)
n  E (2 )
n  0
n

H 0  E 0
n   ( 2)
n    H   E (1)
n   (1)
n  E (2)
n  0
n (8)

Prof. Dr. Ali A.Alzubadi QM_Lectures


First order correction
The unknown e.f.  (1)
n
can be expanded in terms of known function
 0m such as:

 (1)
n  C nm  0m (9)
m 1
where Cnm are the linear sum coefficients.
Substitute (9) into (7) one gets:

  0 n  nm m  n
H
m 1
 E 0
C  0
 E (1)
 H   n
 0

  m n  nm m  n
E
m 1
0
 E 0
C  0
 E (1)
 H   n
 0
(10)

Prof. Dr. Ali A.Alzubadi QM_Lectures


E
m 1
0
m  En0  Cnm  0m   En(1)  H    0n

Multiplying (10) by the conjugate  0k and integrating, one gets


E
m 1
0
m  E n0 C nm  0k  0m   0k  E n(1)  H    0n

Taking  0k  0m   km  1 for k=m

then E 0
k  E n0 C nk  E n(1)  kn   0k H   0n (11)
1
To calculate E n ;
set k  n ,   kn  1
then
0  E n(1)   0k H   0n

The first order


E n(1)   0k H   0n (12)
correction

Prof. Dr. Ali A.Alzubadi QM_Lectures


To obtain Cnk
Assume k  n , then Equ .(11) becames

E 0
k  E n0 C nk    0k H   0n

 0k H   0n  0k H   0n (13)
C nk   
E k0  E n0 E n0  E k0

Substitute (13) into (9) one gets:

 0m H   0n
 (1)
n  
m n E n0  E m0
 (0)
m

Prof. Dr. Ali A.Alzubadi QM_Lectures


Example
The unperturbed wave functions for the infinite square well potential are

Suppose we perturb the system by simply raising the floor of the well a constant
amount V0. Find the first order correction to the energies.

E n(1)   0n H   0n

The total energy corrected to first order is


Exact solution:

Prof. Dr. Ali A.Alzubadi QM_Lectures


 2
d 2   2m 

 2m dx 2 V 0   E   0  multplying by  2 
   

 2m 
where k   2   E V 0 
 
Compare with equation (1), shows that
the energy corrected to first order is the
So
exact solution, i.e no higher order
corrections

Prof. Dr. Ali A.Alzubadi QM_Lectures


2m 2m
2
E n a2  2
V 0 a 2  n 2 2

2m 2m
2
E n a 2  n 2 2  2
V 0 a2

Prof. Dr. Ali A.Alzubadi QM_Lectures


Example
If the perturbation extends only half way across the well
2  n 
 0n  x   sin  x
a  a 

Prof. Dr. Ali A.Alzubadi QM_Lectures


Example
Obtain the first order relativistic correction to the ground state of Hydrogen atom.
p2
Nonrelativistic: E  V (r )
2m

The Taylor series of the square root

we get:

p4
 p 2
 p 4
with H   
So H  V (r )   3 2
 H0  H  8m 3c2
 2 m  8 m c

Prof. Dr. Ali A.Alzubadi QM_Lectures


Prof. Dr. Ali A.Alzubadi QM_Lectures
Prof. Dr. Ali A.Alzubadi QM_Lectures
Quantum Mechanics
Approximation Methods
Time-Independent Perturbation Theory
(part two)
Fourth Stage Students,
Department of Physics, College of Science,
University of Baghdad

Prof. Dr. Ali Abdulateef Alzubadi


2023-2024
Second-order correction
recall that
 0 n n
H  E 0
 (2)
   H   E n 
(1)
 (1)
n  E (2)
n  0
n (1)
Scalar product with  (0)
n

 (0)
n  H 0  E 0
n   (2)
n    (0)
n H   (1)
n  E (1)
n  (0)
n  (1)
n

 E n(2)  (0)
n  0
n (2)

Substitute the value of  (1)
n  C nm  0m we get
m 1

n  nm
0    (0)
n H   (1)
n  E (1)
C  (0)
n  (0)
m  E (2)
n (3)
m 1

E n(2)   (0)
n H   (1)
n (4)
Prof. Dr. Ali A.Alzubadi QM_Lectures
We know that
 0m H   0n
 (1)
n  
m n E n0  E m0
 (m0) (5)

Sub Equ. (5) in Equ (4) we get


 (0) H   (0)
 (0)
H   (0)
(6)

n m m n
E n(2) 
m n E n0  E m0

2
 (0)
H  (0)


n m
E (2)
n  (7)
m n E n0  E m0

Prof. Dr. Ali A.Alzubadi QM_Lectures


Example:
A diatomic molecule such as NO (Nitric oxide) has a permanent dipole moment d
along the direction connecting the two atoms. If such a molecule is placed in a uniform
electric field E , find the energy shift of the ground state. Treat the unperturbed molecule
as a rigid rotator with the Hamiltonian
Electric E
d
field

dipole
moment

Prof. Dr. Ali A.Alzubadi QM_Lectures


The unperturbed energy

The first order correction is given by

H   d E cos 

Prof. Dr. Ali A.Alzubadi QM_Lectures


The second order correction to the energy of the ground state is given by

We know that:

Prof. Dr. Ali A.Alzubadi QM_Lectures


m
From the orthonormality of Y l , the above matrix elements becomes

E d 4
 (0)
0 H   l(0)  l 1 m 0
4 3
So, only l  1 and m  0 in the summation
2
E d 4  1 
E (2)
0   (0) (0) 
4 3 E
 0  E 1 

Prof. Dr. Ali A.Alzubadi QM_Lectures

You might also like