Lecture 14
Time-independent perturbation theory
(c) So Hirata, Department of Chemistry, University of Illinois at Urbana-Champaign. This material has
been developed and made available online by work supported jointly by University of Illinois, the
National Science Foundation under Grant CHE-1118616 (CAREER), and the Camille & Henry Dreyfus
Foundation, Inc. through the Camille Dreyfus Teacher-Scholar program. Any opinions, findings, and
conclusions or recommendations expressed in this material are those of the author(s) and do not
necessarily reflect the views of the sponsoring agencies.
Perturbation theory


What is perturbation theory?
To solve complex physical and mathematical
problems approximately



We first identify the most closely related
problem (zeroth order) with known solutions,
We then adjust the solutions by analyzing the
difference.
The main assumption is that the difference is only
a small perturbation (disturbance) and we can
obtain accurate approximate solutions more easily
than solving the problem more exactly.
Example of perturbation theory




10 people went to a bar. Nine (9) of them (Group A) had one
drink each and one (1) (Person B) had two drinks. The bill
was $100.
Each paid $10 (= $100 / 10 people) (zeroth order solution).
Person B (who had two drinks - perturbation) felt bad and
pitched in another $10 (first order perturbation solution).
There is an overage of $10 and so $1 (= $10 / 10 people) is
returned to each (second order perturbation solution). Each in
Group A paid $9 and Person B paid $19, whereas the exact
solution is, each in Group A pays $100/11 =
$9.0909090909… and Person B pays $100/11 x 2 =
$18.1818181818…
Significance of the theory


The most successful many-body theory. Most of
the accurate numerical solutions of the
Schrödinger equations (no exact analytical
solutions) are based on this theory.
Some of the physical quantities and concepts
are based on perturbation theory: dipole
moment (1st order), polarizability (2nd order),
spin-orbit coupling theory (1st order), one-photon
optical transitions (time-dependent 1st order),
two-photon optical transitions (time-dependent
2nd order), just to name a few.
Zeroth order

Suppose we want to know at least
approximately the ground-state solution of
the Schrödinger equation
ˆ
H  0  E0 0

The ground
state
Suppose we have found a similar problem
with a known solution
Known solution
(0) (0)
(0) (0)
(0)
ˆ
ˆ
H Y0 = E0 Y0 , H
ˆ
H
Partitioning of Hamiltonian

First we view H as the sum of H(0) and a
perturbation
Hˆ  Hˆ
0th-order
part
Perturbation
(0)
 Hˆ
(1)
 Hˆ
(0)
The greater the
number in
parenthesis, the
smaller the term. This
term is a small
perturbation as
compared to (0) term
  Hˆ
(1)
λ is equal to 1,
and has no effect
except it helps us
classify terms
So far no
approximation yet
Series expansions

If we “turn on” H(1) in the Hamiltonian, the
wave function and energy change – the size
of change that is linear, quadratic, cubic, etc.
to the size of H(1) (cf. Taylor expansion)
Y0 = Y
E0 = E
(0)
0
(0)
0
+ lY + l Y
(1)
0
2
+ lE + l E
(1)
0
2
(2)
0
(2)
0
+…
+…
If we carry through this infinite summation, we
get the exact solution of HΨ0=E0Ψ0. So far, no
approximation has been made.
Sizes of terms



The powers in λ indicate the relative sizes of
terms.
For example, if the perturbation H(1) is 1% of
H(0) in size, Ψ(1) is also roughly 1% of Ψ(0) and
then Ψ(2) is 1% x 1% = 0.0001 of Ψ(0).
We equate terms of roughly equal sizes. So,
for example, the terms having λ1 and λ2 are
very different in size and should not be
compared.
The Schrödinger equation

Substitute the Hamiltonian, wave function,
energy into the Schrödinger equation that we
intend to solve (not the zeroth-order one we
already know the solution of).
ˆ
H  0  E0 0
(
)(
(0)
(1)
(0)
(1)
2 (2)
ˆ
ˆ
H + lH
Y 0 + lY 0 + l Y 0 +…
(
= E0(0) + l E0(1) + l 2 E0(2) +…
)(
)
Y 0(0) + lY 0(1) + l 2 Y (2)
+…
0
)
Order-by-order equations
(
)(
(1)
2 (2)
Hˆ (0) + l Hˆ (1) Y (0)
+
l
Y
+
l
Y 0 +…
0
0
(
= E0(0) + l E0(1) + l 2 E0(2) +…

)(
)
Y 0(0) + lY 0(1) + l 2 Y (2)
+…
0
)
Equating the terms containing the same
powers of λ,
l 0 : Hˆ (0) Y 0(0) = E0(0) Y 0(0)
This is nothing but the zerothorder equation we started with.
ˆ (0) Y (1) = E (1) Y (0) + E (0) Y (1)
l 1 : Hˆ (1) Y (0)
+
H
0
0
0
0
0
0
(2)
(0)
(1)
(1)
(0)
(2)
l 2 : Hˆ (1) Y 0(1) + Hˆ (0) Y (2)
=
E
Y
+
E
Y
+
E
Y
0
0
0
0
0
0
0
First order


Repeating this process, we obtain an infinite
series of corrections the sum of which
converge at the exact solution. Truncating the
series constitutes an approximation.
For example, if we stop at a finite order and
solve the following first-order equation, then
we obtain E(0) + E(1) as an approximation to
true E.
 : Hˆ
1
(1)
(0)
 0  Hˆ
(0)
 0  E0  0  E0  0
(1)
(1)
(0)
(0)
(1)
First order

Now we expand the unknown wave
function Ψ0(1) as a linear combination of
known wave functions {Ψn(0)}. This is
always possible because of the
completeness of eigenfunctions of an
Hermitian operator. {Ψn(0)} [ground- and all
excited-state wave functions of H(0)] forms a
complete set.

0 
(1)

n
cn  n
(0)
First order

Substituting
Hˆ
Hˆ
(1)
(1)
(0)
 0  Hˆ
(0)
 0  Hˆ
(0)
 0  E0  0  E0  0
(1)
(1)
(0)
(0)

(0)

(1)

cn  n  E 0  0  E 0
(0)
(1)
(0)
(0)

n
Multiply

( 0 )*
0

( 0 )*
k
Hˆ
Hˆ
(1)
(1)


(0)
0
(0)
0
( 0 )*
 0 or
d 
d 

( 0 )*
0

( 0 )*
k
cn  n
(0)
n
( 0 )*
 k (k > 0) from the left and integrate over space


n
n
(0)
(0)
(1)
( 0 )*
(0)
(0)
( 0 )*
(0)
Hˆ  c n  n d   E 0   0  0 d   E 0  c n   0  n d 
Hˆ

(0)
c
n

n

(0)
n
d  E
(1)
0

( 0 )*
k

(0)
0
d  E
(0)
0
c 
n
n
( 0 )*
k
 n d
(0)
First order


From the first equation:
( 0 )*
0
Hˆ
(1)

(0)
0
d 

( 0 )*
0
Hˆ

(0)
c

n

(0)
n
d  E
(1)
0
n


Hˆ

(0)
c
n

n

(0)
n
d 
c 
n
( 0 )*
0
Hˆ

(0)
n
d 
c0 contribution cancels exactly and
it is simply repeating the λ0 equation

d  E
(0)
0
c 
n
( 0 )*
0
 n d
(0)
n
cancel
exactly

(0)
n
( 0 )*
0

(0)
0
1
cancel
exactly
( 0 )*
0
( 0 )*
0
c
(0)
n
En
n
  0  n d  c0 E 0
( 0 )*
(0)
(0)

E
(0)
0
c 
n
n
(1)
(0)
(1)
ˆ
H  0 d  E 0
( 0 )*
0
 n d  c0 E 0
(0)
(0)
First order

òY
(0)*
k
From the second equation
Hˆ Y dt + ò Y
(1)
òY
(0)*
k
(0)
0
Hˆ
(0)*
k
Hˆ
¥
(0)
¥
(0)
åc Y
n
n
(0)
n
dt = E
(1)
0
å cnY dt = å cn ò Y
n
Y dt + E
(0)
0
n
(0)
n


(0)
0
d  ck E
k
( 0 )*
ck 
Hˆ
(0)
E0
(1)
n
n
(0)*
k
Y(0)
dt
n
n
(0)
0
(1)
åc ò Y
(0)
(0)
Hˆ Y dt = å cn En(0) ò Y (0)*
Y
d
t
=
c
E
k
n
k k
(0)
E
Hˆ
¥
¥
(0)*
k
¥
( 0 )*
k
(0)
0
0
¥
(0)
n
òY
(0)*
k
(0)
k
 ck E
 0 d
(0)
 Ek
(0)
(0)
0
(0)* (0)
(0)
c
Y
Y
d
t
=
c
E
å nò k n
k 0
n
Second order
Hˆ
(1)

(1)
0
 Hˆ
(0)

(2)
0
 E
(2)
0

(0)
0
E
(1)
0

(1)
0
E
(0)
0

Multiply  (00 )* from the left and integrate over space
(0)* ˆ (1)
(1)
(0)* ˆ (0)
(2)
Y
H
Y
d
t
+
Y
H
Y
dt
0
0
ò 0
ò 0
cancel
(0)
(1)
(0)* (1)
(0)
(0)* (2)
= E0(2) ò Y (0)*
Y
d
t
+
E
Y
Y
d
t
+
E
Y
Y0 d t
0
0
0 ò
0
0
0 ò
0
1
E
(2)
0
= òY
= òY
k¹0
(1)
(0)*
(0)
Hˆ (1) Y (1)
d
t
E
Y
c
Y
dt
å
0
0 ò
0
k
k
k
¥
(0)*
0
Hˆ (1) å ck Y (0)
d t - c0 ò Y (0)*
Hˆ (1) Y (0)
dt
k
0
0
Y
ò
=å
¥
¥
(0)*
0
k
(0)*
0
(0)* ˆ (1) (0)
Hˆ (1) Y (0)
d
t
Y
k
ò k H Y 0 dt
E
(0)
0
-E
(0)
k
¥
=å
k¹0
òY
(0)*
k
Hˆ (1) Y (0)
dt
0
E0(0) - Ek(0)
2
(2)
0
First order

This looks like an expectation value of the
perturbation operator.
0
( 0 )*
(1)
(0)
(1)
Hˆ  0 d   E 0
Second order
Excited

E
(2)
0


k 0

( 0 )*
0
Hˆ
(1)

De-excited
(0)
k
d  
(0)
E0
( 0 )*
k
Hˆ
(1)

1st-order WF
(0)
0
d
 Ek
(0)
Excited - deexcited
Summary

Perturbation theory works as follows:





Find a suitable zeroth-order Hamiltonian (close to the
true Hamiltonian) whose eigenfunctions and
eigenvalues are known.
Partition the Hamiltonian into the zeroth-order part
and perturbation.
Expand the energy and wave function into series of
smaller and smaller corrections to zeroth-order
quantities.
Equate the terms according to the sizes.
Use completeness, normalization, and
orthogonality of zeroth-order eigenfunctions to obtain
general expressions for energies and wave functions.