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
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Perturbation of Non-Degenerate States
In non-degenerate case, there is only one e.f. corresponding to each
energy level.
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Prof. Dr. Ali A.Alzubadi QM_Lectures
Prof. Dr. Ali A.Alzubadi QM_Lectures
n 1
Degeneracy 2 l 1
l 0
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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
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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)
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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
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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:
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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
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