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Fall 2022 Introductory Quantum Mechanics Lecture on Pert. Theory
Perturbation Theory
Introduction
The quantum mechanics you have studied so far has entailed solving a few carefully chosen
problems exactly. Unfortunately, those problems represent a small fraction of the realistic
problems that nature presents to us, and in most cases those exactly solvable problems are
only approximations to real problems. Now we must learn to solve more realistic problems
that do not admit exact solutions. The approach we take to solving these realistic problems
is to make them look like problems we have already solved exactly, with an additional part
that represents the new, more realistic aspect of the problem. We assume that this new
part, the perturbation, is small so that we can use approximations to find the corrections
to the exact solutions. Our focus is to discover how energies and eigenstates are affected by
small additional terms in the Hamiltonian. To guide us, we will take some exactly solvable
problems, solve them, and then expand the solutions. We will compare these results with
the new perturbation methods that we learn.
Time-Independent Perturbation Theory
The Formalism
Time-independent perturbation theory is an approximation scheme that applies in the fol-
lowing context: we know the solution to the eigenvalue problem of the Hamiltonian H
o , and
we want the solution to H = H
o
′ , where H
′ is small compared to H
o in a sense to
be made precise shortly. For instance, H
o can be the Coulomb Hamiltonian for an electron
bound to proton, and H
′ the addition due to an external electric field that is weak compared
to the proton’s field at the (average) location of the electron. One refers to H
o as the un-
perturbed Hamiltonian and H
′ as the perturbing Hamiltonian or perturbation.
We proceed as follows. We assume that to every eigenket |ψ
o n〉 ≡ |n
o 〉 of H
o with eigenvalue
E
o n, there is an eigenket^ |n〉^ of^ H^ with eigenvalue^ E. We must assume here that^ |n
o 〉 is is
non-degenerate. We then assume that the eigenkets and eigenvalues of H may expanded in
a perturbation series:
|n〉 =
∣n(o)
∣n(1)
∣n(2)
En = E
(o) n +^ E
(1) n +^ E
(2) n +^...^
To find the terms in the expansions for |n〉 and En we start with the eigenvalue equation
H |n〉 = En |n〉 (2)
or
(H
o
′ )
[∣
∣n(o)
∣n(1)
∣n(2)
]
(E
(o) n +^ E
(1) n +^ E
(2) n +^.. .)^
[∣
∣n(o)
∣n(1)
∣n(2)
] (3)
Fall 2022 Introductory Quantum Mechanics Lecture on Pert. Theory
We approach these equations as we did the perturbation of the hydrogen atom: we first
consider the zeroth-order terms of Eq. 3. We get:
H
o
∣n(o)
= E
(o) n
∣n(o)
Notice that the zeroth-order quantities
∣n(o)
and E
(o) n of^ H
′ (or, equivalently, they depend on
the zeroth power of H
′ ). By assumption, this equation may be solved and the eigenvectors ∣ ∣n(o)
and eigenvalues E
(o) n determined. So we move on to the first-order terms. We get the
equation
H
o
∣n(1)
+ H
′
∣n(o)
= E
(o) n
∣n(1)
+ E
(1) n
∣n(o)
Let us dot both sides with
n
(o)
∣. Using
n
(o)
∣ (^) Ho^ =
n
(o)
∣ (^) E(o) n and^
n
(o)
∣n(o)
= 1, we get
E
(1) n =^
n
(o)
∣H′
∣n(o)
i.e., the first-order change in energy is the expectation value of H
′ in the unperturbed state.
Or as it is very well put in your textbook: “This is the fundamental result of first-order
perturbation theory; as a practical matter, it may well be the most frequently used equation
in quantum mechanics.It says that the first-order correction to the energy is the expectation
value of the perturbation, in the unperturbed state.”
Notice that E
(1) n is indeed proportional to the first power of^ H
′
. Let us next dot both sides
of Eq. 5 with
m
(o)
∣, m 6 = n, to get
m
(o)
∣Ho
∣n(1)
m
(o)
∣H′
∣n(o)
= E
(o) n
m
(o)
∣n(1)
or
m
(o)
∣n(1)
m
(o)
∣H′
∣n(o)
E
(o) n −^ E
(o) m
Since m 6 = n, this equation determines all the components of
∣n(1)
in the eigenbasis of
H
o , except for the component parallel to
∣n(1)
in the eigenbasis of H
o , except for the
component parallel to
∣n(o)
, let’s call it
∣n
(1) ‖
. We determine it by the requirement that |n〉
is normalized to this order. In obvious notation, we have
1 = 〈n|n〉 =
n
(o)
n
(1) ⊥
n
(1) ‖
∣n(o)
∣n
(1) ⊥
∣n
(1) ‖
which leads to
n
(o)
∣n(o)
n
(1) ‖
∣n
(o)
n
(o)
∣n
(1) ‖
or
n
(1) ‖
∣n
(o)
n
(o)
∣n
(1) ‖
Fall 2022 Introductory Quantum Mechanics Lecture on Pert. Theory
Fig. 1: Energy levels of a spin-1/2 particle in a uniform magnetic field.
eigenstate is
∣n(0)
, the perturbation mixes in orthogonal states
∣m(0)
; this mixing is directly
proportional to the matrix element
m
(o)
∣H′
∣n(o)
and inversely proportional to the energy
difference between the two levels, which measures the “rigidity” of the system. If for any
reason the above inequality is not fulfilled (say due to degeneracy, E
(o) n =^ E
(o) m ) we must
turn to an alternate formalism called degenerate perturbation theory to be described in later
lecture.
Some Examples
The Spin−
1 2
Example
To get a feel for what perturbation theory is and how it works, let’s go back to our old
standby: the spin− 1 / 2 problem. The usual Hamiltonian of a spin− 1 / 2 system is the po-
tential energy of the spin magnetic moment in an applied magnetic field. For an applied
magnetic field in the z-direction B~ = Bo ˆz, the Hamiltonian is:
H
o = −~μ · B~ = ωoSz ≡
ωo 0
0 −ωo
where we have defined the Larmor frequency ωo = eBo/me. The subscript zero on the
Hamiltonian indicates that this is the zeroth-order Hamiltonian (i.e., the Hamiltonian before
we apply a perturbation). The energy eigenstates of the zeroth-order Hamiltonian are the
spin up and down states |±〉 and the energy eigenvalues (zeroth-order energy eigenstates)
are:
E
(1) ± =^ ±
ℏωo
The goal of perturbation theory is to find the higher-order corrections to the energy eigen-
values and eigenstates caused by the application of a perturbation to the system. For this
spin-1/2 system, we will solve the problem exactly and then expand the solutions to discover
how perturbation series behave. Our exact solution should contain the zeroth-order solutions
shown in Eq 19 and small corrections.
The simplest way to perturb this spin system is to change the magnetic field. Any gen-
Fall 2022 Introductory Quantum Mechanics Lecture on Pert. Theory
Fig. 2: (a) Perturbing magnetic fields and (b) the resultant Larmor frequencies.
eral change to the magnetic field can be decomposed into an additional component along
the original field in the z-direction, and an additional component perpendicular to that, as
shown in Fig. 2-(a). We write the new total field as B~ = (Bo + B 1 )ˆz + B 2 xˆ and characterize
the two additional field components by their respective Larmor frequencies ω 1 = eB 1 /me
and ω 2 = eB 2 /me. With this notation, the new Hamiltonian is
H = −~μ · B~ = (ωo + ω 1 )Sz + ω 2 Sx ≡
ωo + ω 1 ω 2
ω 2 −ωo − ω 1
It is useful to separate the new Hamiltonian H = H
o +H
′ : into the zeroth-order Hamiltonian
H
o and the perturbation Hamiltonian that we denote H
′
. The zeroth-order Hamiltonian is
given by Eq. 18 and the perturbation Hamiltonian is
H
′
ω 1 ω 2
ω 2 −ω 1
The perturbation Hamiltonian has terms along the diagonal and terms off the diagonal.
These diagonal and off-diagonal terms play important roles in perturbation theory.
We now solve for the energy eigenvalues and eigenstates of the new Hamiltonian in Eq. 20
exactly by diagonalizing the matrix for the general spin-1/2 case of a magnetic field at an
angle θ to the z-axis. We find that the Hamiltonian is proportional to the spin component
Sn along the new magnetic field direction ˆn, and can be expressed in terms of the angle θ of
the new field as:
H = ωnewSn ≡
ℏωnew
cos θ sin θ
sin θ − cos θ
where
tan θ =
ω 2
ωo + ω 1
Fall 2022 Introductory Quantum Mechanics Lecture on Pert. Theory
In both energies, we identify the first term as the zeroth-order energy E
(o) ± given by Eq. 19,
and we note two additional terms. The first is linear or first order in the perturbation and
is equal to the corresponding diagonal term ±ℏω 1 / 2 in the perturbation Hamiltonian [Eq.
21]. The second additional term is quadratic or second order in the perturbation and is
proportional to the square of the off-diagonal term ℏω 2 / 2 in the perturbation Hamiltonian.
This general pattern of corrections is characteristic of perturbation theory, so we denote
perturbed energies as the series
En = E
(o) n +^ E
(1) n +^ E
(2) n +^...^ (29)
where the superscript indicates the order of the perturbation. We found in this spin-1/2 ex-
ample that the linear corrections arose from the diagonal terms in the perturbation Hamil-
tonian and the quadratic terms arose from the off-diagonal terms, another characteristic
pattern of general perturbation theory. In Eq. 28, the second-order energy correction due to
the off-diagonal terms has a factor of ωo in the denominator, and it will diverge if the energy
splitting ℏωo is zero, (i.e., if the original levels are degenerate in energy). This divergence
violates the assumption that the perturbation corrections are small, which creates a prob-
lem that is addressed later. In addition to these features of the perturbed energies, we can
also draw some conclusions about the perturbation corrections to the eigenstates from our
knowledge of the exact eigenstate solutions. The eigenstates of the full Hamiltonian in Eq.
22 are the spin up and down eigenstates |±〉n along the direction ˆn :
n
= cos
θ
|+〉 + sin
θ
|−〉n = − sin
θ
|+〉 + cos
θ
From Fig. 2-(a), it is evident that the angle θ is small for small perturbing magnetic fields,
so we can also use θ as a small parameter for a series expansion to second order in the angle
θ the new eigenstates have two correction terms: a first-order term that is orthogonal to
the original state, and a second-order term that is parallel (in a Hilbert space sense, not a
geometric sense) to the original state. If we neglect the parallel terms we get:
|+〉n
θ
n
= |−〉 −^
θ
Using the schematic in Fig. 2-(b), we express the small angle θ in terms of the Larmor
frequencies. To first order, we obtain:
θ
ω 2 √
(ωo + ω 1 )
2
ω 2
ωo
Fall 2022 Introductory Quantum Mechanics Lecture on Pert. Theory
Thus, we arrive at the perturbation series expansion for the perturbed states, to first order
|+〉n
ω 2
2 ωo
|−〉n ∼= |−〉 −
ω 2
2 ωo
We conclude that the first-order eigenstate correction depends only on the off-diagonal matrix
element, not on the diagonal elements. Note that the coefficient of the original state remains
one, which makes it appear that the state is no longer normalized. But if we check the
normalization of the perturbed state:
n
[
ω 2
2 ωo
] [
ω 2
2 ωo
]
ω 2
2 ωo
we see that it is normalized to first order in the small perturbation parameters.
The Harmonic Oscillator
Why Study the Harmonic Oscillator?
In this section I will put the harmonic oscillator in its place – on a pedestal. Not only
is it a system that can be exactly solved (in classical and quantum theory) and a superb
pedagogical tool (which will be repeatedly exploited in this text), but it is also a system of
great physical relevance. As will be shown below, any system fluctuating by small amounts
near a configuration of stable equilibrium may be described either by an oscillator or by a
collection of decoupled harmonic oscillators. Since the dynamics of a collection of nonin-
teracting oscillators is no more complicated than that of a single oscillator (apart from the
obvious N-fold increase in degrees of freedom), in addressing the problem of the oscillator
we are actually confronting the general problem of small oscillations near equilibrium of an
arbitrary system.
A concrete example of a single harmonic oscillator is a mass m coupled to a spring of force
constant k. For small deformations x, the spring will exert the force given by Hooke’s law,
F = −kx, (k being its force constant) and produce a potential V =
1 2
kx
2
. The classical
Hamiltonian for this system is
Hclassical = T + V =
p
2
2 m
mω
2 x
2 (35)
Where ω =
k/m is the classical frequency of oscillation. Any Hamiltonian of the above
form, quadratic in the coordinate and momentum, will be called the harmonic oscillator
Hamiltonian. Now, the mass-spring system is just one among a family of systems described
by the oscillator Hamiltonian.
Fall 2022 Introductory Quantum Mechanics Lecture on Pert. Theory
Fig. 3: Perturbation of the harmonic oscillator. (a) Shifts of the first three energy levels.
(b) Dependence of the shift of the ground state energy on the perturbation strength to first
order (dotted), second order (dashed), and exact (solid).
The expectation value of the perturbation is
E
(1) n =^ ε
ℏω
n
(o)
∣a†a†^ + a†a + aa†^ + aa
∣n(o)
The operators a
† a
† and aa contribute zero because they raise or lower the state
∣n(o)
twice
and produce a new state that is orthogonal to
∣n(o)
. The remaining terms are calculated
using a |n〉 =
n |n − 1 〉 and a
† |n〉 =
n + 1 |n + 1〉; we then obtain:
E
(1) n =^ ε
ℏω
n +
The resultant energy of level n to first order in the perturbation is
En = E
(o) n +^ E
(1) n
= ℏω
n +
ℏω
n +
= ℏω
n +
ε
Each state is shifted upwards, with the shift larger for larger states. The original and
first-order perturbed energy levels are shown in Fig. 3-(a).
Now consider the second-order energy correction
E
(2) n =^
m 6 =n
m
(o)
∣H′
∣n(o)
2
E
(o) n −^ E
(o) m
Fall 2022 Introductory Quantum Mechanics Lecture on Pert. Theory
This looks like an infinite sum, which would be problematic, but we plow ahead and find
that the sum is reduced for the harmonic oscillator case. For a given energy level n, only
two terms in the sum (m 6 = n) contribute, yielding
E
(2) n =
[
1 4
εℏω
n − 1
n
] 2
E
(o) n −^ E
(o) n− 2
[
1 4
εℏω
n + 2
n + 1
] 2
E
(o) n −^ E
(o) n+
εℏω
) 2 [
n(n − 1)
2 ℏω
(n + 1)(n + 2)
− 2 ℏω
]
ε
2 ℏω
n +
Note that the second-order contribution is negative. Only the two levels m = n + 2 and
m = n− 2 contribute to the energy correction in Eq. 46. They each have the same magnitude
energy denominators, but the matrix element is larger for the m = n + 2 state above the
state of interest, so the level is pushed down. The resultant energy of level n to second order
in the perturbation is:
En = E
(0) n +^ E
(1) n +^ E
(2) n
= ℏω
n +
ε −
ε
2
The perturbed energies to first and second order and the exact result are plotted in Fig.
3-(b) as a function of the perturbation strength.
In this example, we can find the exact answer, so we can check the perturbation result
and confirm that perturbation theory works. The exact Hamiltonian is
H = Ho + H
′
pˆ
2
2 m
mω
2 xˆ
2
mω
2 ˆx
2
pˆ
2
2 m
mω
2 xˆ
2 (1 + ε)
pˆ
2
2 m
mω
2 p xˆ
2
where we have defined a new perturbed harmonic frequency
ωp = ω
1 + ε (49)
This new Hamiltonian has the same form as the original harmonic oscillator problem, but
with a new characteristic frequency. Hence, we know the energy eigenvalues exactly. They
are
En =
n +
ℏωp =
n +
ℏω
1 + ε (50)
