






Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
harmonic oscillator, the quantum rotator, or the hydrogen atom and small deformation
Typology: Lecture notes
1 / 10
This page cannot be seen from the preview
Don't miss anything!
So far we have concentrated on systems for which we could find exactly the eigenvalues and
eigenfunctions of the Hamiltonian, like e.g. the harmonic oscillator, the quantum rotator, or
the hydrogen atom.
However the vast majority of systems in Nature cannot be solved exactly, and we need
to develop appropriate tools to deal with them.
Perturbation theory is extremely successful in dealing with those cases that can be mod-
elled as a “small deformation” of a system that we can solve exactly.
Let us translate the above statement into a precise mathematical framework. We are
going to consider systems that have an Hamiltonian:
0
where
0 is the Hamiltonian of the unperturbed system, is a small parameter, and
is the potential describing the perturbation. We shall assume that the perturbation V is
independent of time.
Let us also assume that we can solve the time-independent Schr¨odinger equation for
0
i.e. that we know its eigenvalues and eigenfunctions:
0
ψ
(n)
(x) = E
(n)
ψ
(n)
(x). (17.2)
For simplicity we start by considering the case where all the unperturbed levels E
(n) are not
degenerate.
Let us discuss the solution of the time-independent Schr¨odinger equation for the full Hamil-
tonian H. The eigenvalue equation reads:
Hψ(x) = Eψ(x). (17.3)
Since is a small parameter, we shall expand the solution of Eq. (17.3) as a Taylor series
in :
ψ(x) = ψ 0 (x) + ψ 1 (x) +
2
ψ 2 (x) +... , (17.4)
0
1
2
E 2
Plugging Eqs. (17.4) and (17.5) into Eq. (17.3), we obtain:
0
ψ 0
(x)+ψ 1
(x) +
2
ψ 2
(x) +...
0
1
2
E 2
ψ 0
(x) + ψ 1
(x) +
2
ψ 2
(x) +...
We can now solve Eq. (17.6) order by order in .
Using Dirac’s notation, we can rewrite the solution above as:
|ψ = |ψ
(n)
+
m =n
ψ
(m) |
V |ψ
(n)
(n) − E
(m)
|ψ
(m)
+ O(
2
). (17.16)
Note that to first order in the solution in Eq. (17.16) is already normalized:
ψ|ψ = ψ
(n)
|ψ
(n)
+
m =n
ψ
(m) |
V |ψ
(n)
(n) − E
(m)
ψ
(n)
|ψ
(m)
+ c.c.
2
) (17.17)
2
). (17.18)
Example A particle moves in the 1-dimensional potential
V (x) = ∞, |x| > a, V (x) = V 0 cos(πx/ 2 a), |x| ≤ a
Calculate the ground-state energy to first order in perturbation theory.
Here we take the unperturbed Hamiltonian,
0 , to be that of the infinite square well, for
which we already know the eigenvalues and eigenfunctions:
(n)
=
π
2
2 n
2
8 ma
2
, u
(n)
=
a
cos
sin
nπx
2 a
; n
odd
even
The perturbation
is V 0 cos(πx/ 2 a), which is small provided V 0
(2)
− E
(1)
.
To first order, then,
1
11
∞
−∞
u
(1) ˆ H
u
(1)
dx =
0
a
a
−a
cos
3
πx
2 a
dx
Evaluating the integral is straightforward and yields the result
0
3 π
0
Iterative solution At order
L the eigenvalue equation yields:
0
0
ψ L
1
ψ L− 1
L
K=
K ψ L−K
Taking the same scalar products described above, we find:
L = ψ 0
V |ψ L− 1
which yields the correction of order
L
to the unperturbed energy level.
Following the computation above we also obtain:
ψ
(m)
|ψ L
ψ
(m) |V |ψ L− 1
(n) − E
(m)
(n) − E
(m)
L− 1
K=
K ψ
(m)
|ψ L−K
Using Eq. (17.21) for L = 2 we find the second-order correction to the n-th energy level:
2 = ψ 0
V |ψ 1
m =n
ψ
(n)
|
V |ψ
(m)
ψ
(m)
|
V |ψ
(n)
(n) − E
(m)
Equation (17.15) shows that the correction to the energy eigenfunctions at first order in
perturbation theory is small only if
ψ
(m) |
V |ψ
(n)
(n) − E
(m)
If the energy splitting between the unperturbed levels is small compared to the matrix element
in the numerator, then the perturbation becomes large, and the approximation breaks down.
In particular, if there are degenerate levels, the denominator is singular, and the solution is
not applicable.
Let us see how we can deal with a g 0 -fold degenerate level of the unperturbed Hamiltonian.
We shall denote P the projector onto such level, and Q the projector orthogonal to this level.
The first-order equation:
0
0
ψ 1
1
ψ 0
can be projected using P onto the space spun by the degenerate states:
1
ψ 0
Choosing a basis for the space of degenerate levels, we can write ψ 0
as:
ψ 0
g 0
i=
c i φ i
and then rewrite Eq. (17.25):
φ i
V |φ j c j
1 c i
i.e. E 1
is an eigenvalue of the matrix V ij
= φ i
V |φ j
. This equation has g 0
roots (not
necessarily distinct), and generalizes Eq. (17.10) to the case of degenerate levels. If the
eigenvalues are indeed all distinct, then the degeneracy is completed lifted. If some of the
eigenvalues are equal, the degeneracy is only partially lifted.
We can now attempt to incorporate the effect of the inter-electron Coulomb repulsion by
treating it as a perturbation. We write the Hamiltonian as
0
where
0
1
2
and
=
e
2
4 π 0
|r 1
− r 2
The ground state wavefunction that we wrote down earlier is an eigenfunction of the
unperturbed Hamiltonian,
0
Ψ(ground state) = u 100 (r 1
) u 100 (r 2
) χ 0 , 0
To compute the first order correction to the ground state energy, we have to evaluate the
expectation value of the perturbation,
, with respect to this wavefunction;
1
e
2
4 π 0
u
∗
100
(r 1
) u
∗
100
(r 2
) χ
∗
0 , 0
r 12
u 100 (r 1
) u 100 (r 2
) χ 0 , 0 dτ 1 dτ 2
The scalar product of χ 0 , 0 with its conjugate = 1, since it is normalised. Putting in the
explicit form of the hydrogenic wavefunction from Lecture 10
u 100
(r) =
π
(Z/a 0
3 / 2
exp(−Zr/a 0
thus yields the expression
1
e
2
4 π 0
3
πa
3
0
2
r 12
exp{− 2 Z(r 1
)/a 0
} dτ 1
dτ 2
Amazingly, this integral can be evaluated analytically. See, for example, Bransden and
Joachain, Introduction to Quantum Mechanics, pp 465-466. The result is
1
Z Ry =
Ry = 34 eV
giving for the first-order estimate of the ground state energy
1
= − 108 .8 + 34 eV = − 74. 8 eV = − 5. 5 Ry
to be compared with the experimentally-measured value of − 78. 957 eV.
Classically, an electron of mass M and charge −e moving in an orbit with angular momentum
L would have a magnetic moment
μ = −
e
suggesting that in the quantum case,
μ ˆ = −
e
L and μˆ z
e
z
The eigenvalues of ˆμ z
are thus given by
e
m
≡ −μ B
m
where the quantity μ B
is known as the Bohr magneton.
Similarly, there is a magnetic moment associated with the intrinsic spin of the electron;
μ ˆ z
g s
e
z
where the constant, g s
, cannot be determined from classical arguments, but is predicted to
be 2 by relativistic quantum theory and is found experimentally to be very close to 2.
The interaction between the orbital and spin magnetic moments of the electron introduces
an extra term into the Hamiltonian of the form
S−O = f (r)
where
f (r) =
2 c
2 r
dV (r)
dr
We can attempt to treat this extra term by the methods of perturbation theory, by taking
the unperturbed Hamiltonian to be
0
ˆp
2
ˆp
2
Ze
2
4 π 0
r
Cautionary Note In our derivation of the first-order formula for the shift in energy in-
duced by a perturbation, we assumed that there were no degeneracies in the energy eigenvalue
spectrum and noted that the method could break down in the presence of degeneracies.
necessary to use degenerate state perturbation theory, which we briefly discussed above.
As usual, we summarize the main concepts introduced in this lecture.