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Perturbation
1. WELCOME TO YOU ALL
PHYSICAL
CHEMISTRY
PRESENTATION
M.SC.-II –
(SEM. III )
2013–
2014
PERTURBATION
Presented by :–
Dharmendra R. Prajapati
RAMNIRANJAN JHUNJHUNWALA
COLLEGE
3. Perturbation
Original Equation
X 2 − 25 = 0
Y = X 2 + ε X − 25
Perturbed equation
X 2 + ε X − 25 = 0
Y vs X
15
10
0 ≤ ε <1
5
0
-8
-6
-4
-2
0
2
4
6
8
-5
Y
Epsilon=0.8
-10
Epsilon=0.5
Epsilon=0.0
-15
-20
-25
-30
X
4. Perturbation
Perturbed equation
X 2 + ε X − 25 = 0
Change in result (absolute values) vs Change in equation
Perturbation in the result
Perturbation in the result (root)
Root -1
-2
-0.15
-0.15
Simple (Regular)
Perturbation
0.06
ε=0.1
ε=0.1
0.05
ε=-0.1
ε=-0.1
0.04
0.03
Answer can be
in the form
0.02
-0.1
-0.1
0.01
0.01
ε=-0.01
ε=-0.01
0
0
-0.05
0
-0.05
0
ε=0.01
ε=0.01
0.05
0.05
Epsilon (perturbation)
Epsilon (perturbation)
0.1
0.1
0.15
0.15
X = X 0 + ε φ ( X 0 ) + ε 2 ...
5. Y = εX + X − 5
2
Y vs X
Perturbation
Original Equation
X −5 = 0
Perturbed equation
ε X 2 + X −5 = 0
40
35
30
25
Epsilon=0
Two roots instead
15
Y
20
Epsilon=0.8
one
10
Epsilon=1
5
0
-6
-4
-2
-5 0
-10
X
2
4
6
Roots are not
close to the
original root
of
6. Perturbation
Y = εX + X − 5
Change in result (absolute values) vs Change in equation
2
Other root varies
from the original
root dramatically,
as epsilon
approaches zero!
Root-1
Root-2
3.5
1200
Perturbation
Perturbation in Resultin Result
3
2.5
1000
2
800
1.5
600
1
0.5
400
0
200
0
0.2
0.4
0.6
0.8
1
1.2
Epsilon
0
0
0.2
0.4
0.6
Epsilon
0.8
1
1.2
Singular
perturbation
Answer may NOT
be in the form
X = X 0 + ε φ ( X 0 ) + ε 2 ...
7. dy
=a
dx
Solution
Differential Equations
a <1
y =ax
Perturbation-1
Solution
y = 0, at x = 0
dy
= a +ε y
dx
(
y = 0, at x = 0
)
a εx
a
ε 2 x2
y = e −1
= 1 + ε x +
+ ... − 1
ε
ε
2
a
ε 2 x2
= ε x +
+ ...
ε
2
ε → 0, y → ax
εx
= ax1 + + ...
2
Regular Perturbation
8. § Time-Independent Perturbation
Theory
FIRST ORDER PERTURBATION:•
We are often interested in systems for which we
could solve the Schreodinger equation if the
potential energy were slightly different.
•
Consider a one-dimensional example for which
we can write the actual potential energy as
Vactual(x) = V(x) + υ(x)
(12.1)
• where υ(x) is a small perturbation added to the
unperturbed potential V(x).
9. •
The Hamiltonian operator of the “unperturbed”
(i.e. exactly solvable) system is
• and the Schreodinger equation of that system
therefore is
H0ψl = Elψl (12.3)
with a known set of eigenvalues El and
eigenfunctions ψl.
•
The Schreodinger equation of the “perturbed”
system contains the additional term υ(x) in the
Hamiltonian:
10. [H0 + υ(x)]ψn’ = En’ψn’
(12.4)
•
• which may make it impossible to solve directly for
the eigenfunctions ψn’ and eigenvalues En’.
•
However, Postulate 3 tells us that any
acceptable wave function may be expanded in a
series of eigenfunctions of the unperturbed
Hamiltonian.
Therefore we may write each function ψn’ as a
•
series of the functions ψl with constant coefficients:
• (The letter l is simply an index, having nothing to do
with the angular momentum quantum number; this
11. •
Substitution into Eq.(12.4) gives
•
The technique used in finding coefficients in a
Fourier series may be employed here.
We multiply each side of Eq.(12.6) by ψm*—the
•
complex conjugate of a particular unperturbed
eigenfunction ψm. We then integrate both sides over
all x, to obtain
∞
∞
∞
∞
−∞
l =1
−∞
l =1
*
'
*
ψ m [ H 0 + v( x)] ∑ anlψ l dx = ∫ Enψ m ∑ anlψ l dx
∫
12. •
Integrating each side of Eq.(12.7) term by term
and removing the space-independent factors anl and
En’ from the integrals. we obtain
•
Because the functions ψ are normalized and
orthogonal, the right-hand side of Eq.(12.8) reduces
to the single term anmEn’ for which l = m, and we
have
•
We can rewrite the left-hand side of Eq.(12.8) by
using the fact that H0ψl = Elψl [Eq.(12.3)]; this
13. •
•
∞
∞
l =1
−∞
*
'
anl ∫ψ m [ El + v( x)]ψ l dx = anm En
∑
or
∞
∞
∞
∞
*
*
'
anl El ∫ψ mψ l dx + ∑ anl ∫ψ m [ v( x)]ψ l dx = anm En
∑
•
•
(12.9)
Again, because the functions ψ are normalized
and orthogonal, the first term on the left reduces to
anmEm.
•
The second term can be written in abbreviated
form as ∑∞l=1 anlυml, where υml is an abbreviation for the
integral ∫∞-∞ψ*mυ(x)ψl dx, called the matrix element of
the perturbing potential υ(x) between the states m
and l.
l =1
−∞
l =1
−∞
14. •
(This term can also be written in Dirac notation as
<m|υ(x)|l>.)
• Substitution into Eq.(12.9) now yields
• and after rearranging terms,
•
Equation (12.10) is exact. No approximations
have been used in deriving it. However, it contains
too many unknown quantities to permit an exact
solution in most cases.
15. •
The most important of the unknown quantities is
the perturbed energy of the nth level, En’,or more
precisely,Δ En = En’ - En . So we let m=n in Eq.(12.10)
to obtain
∞
'
anl vnl = ann ( En − En )
•
∑
(12.10a)
l =0
•
To find a first approximation to this energy
difference, we make the arbitrary assumption that
anl = 1 if l = n and anl = 0 if l ≠ n. In that case, the lefthand side of Eq.(12.10a) collapses to a single term:
vnn.
• Thus Eq.(12.10) is reduced to the approximation
16. •
This is a first-order approximation to the
perturbed energy En’. You can recognize this
integral from the formula for the expectation value
of an operator.
•
This formula is not exact because the integral
contains the wave function of the unperturbed
system rather than the actual wave function.
17. Applications of perturbation theory
Perturbation theory is an important tool for describing
real quantum systems, as it turns out to be very difficult
to find exact solutions to the Schrödinger equation
for Hamiltonians of even moderate complexity.
The Hamiltonians to which we know exact solutions, such
as the hydrogen atom, the quantum harmonic
oscillator and the particle in a box, are too idealized to
adequately describe most systems.
Using perturbation theory, we can use the known
solutions of these simple Hamiltonians to generate
solutions for a range of more complicated systems.
We will start with the mathematical background. And here is the reason.
Many books start with the Navier Stokes equation and then say, ‘based on order of magnitude analysis, we throw away these terms and keep these terms. Then using the following transformation, we can solve this equation’.
Why do you have to do that way? If you are given an equation, can you figure out this is what you have to do , out of the blue?
What are the methods that are normally tried for these kind of problems and why do we choose this method? I would like you to get some idea of these issues and hence we start with some mathematical background.
A perturbation is a ‘disturbance’, usually a small disturbance. Consider an equation, or a system, and say you know the solution of the system. If the equation changes very slightly (perturbed), perhaps the solution will also change slightly. In that case, you can find the solution of the ‘perturbed equation’ to be very close to the solution of the unperturbed equation. Look at the example above.
In the case considered here, if the perturbation is smaller and smaller (i.e. equation becomes closer to the original eqn), then the solution approaches the original solution. These kind of perturbations are called simple perturbations or regular perturbations.
If the perturbation parameter (in the original equation) is epsilon, then the solution can be written as “original solution + epsilon * something…”
Let us consider another system, where the situation is not so simple. Here a small perturbation alters the solution drastically.
Even if the perturbation tends to zero, the solution does NOT tend to the original solution. We cannot write the solution like we did in the previous case. These are called ‘singular perturbation’.
We can look at similar examples in differential equations. An example of regular perturbation is given here.
