Download Sample Chapter 9

C HA PT E R
9
Time-Independent Perturbation
Theory
In theory, the Schrödinger equation allows us to solve any quantum mechanical
system exactly. We simply insert the potential V and solve for the wave function ψ
and the energy E. Unfortunately, there are very few potentials, such as the infinite
square well or the Coulomb potential of the hydrogen atom, for which a simple
exact solution exists. In order to make any further progress, we need to develop
some techniques for finding approximate solutions to the Schrödinger equation.
This chapter and Chapter 11 are devoted to a very important set of these techniques
called perturbation theory.
The basic idea of perturbation theory rests on a simple general argument. Suppose we begin with a potential for which we can solve the Schrödinger equation
exactly, such as the infinite square well of width a; recall from
√ Chapter 4 that the
ground-state energy and wave function are given by ψ = 2/a sin(π x/a) and
E = h̄2 π 2 /2ma 2 . Now suppose we make a tiny change in V such as a small notch
in the center of the potential (Figure 9.1). We cannot solve the Schrödinger equation for this new potential, but intuition suggests that a small change in V ought
to produce a small change in ψ and in E. This intuition is correct. The reason is
that the Schrödinger equation is a special kind of differential equation: it is linear,
i.e., ψ and its derivatives are taken only to the first power. Linear differential equations like the Schrödinger equation have the property that small changes in the
parameters produce small changes in the solution. The fact that a small change in
V produces a small change in ψ and E is the fundamental idea of perturbation
theory.
9.1
DERIVATION OF TIME-INDEPENDENT PERTURBATION THEORY
Now we will calculate mathematically the change in E produced by an arbitrary
small change in the Hamiltonian H . [Although we talk generically about a change
in the Hamiltonian, this usually amounts to a change in the potential.] Assume we
have a Hamiltonian H for which we can solve the Schrödinger equation exactly.
We need to consider two possibilities for a small change in H : either the change
in H is constant in time, or it varies as a function of time. If the change in H is
constant in time, we are dealing with time-independent perturbation theory, which
is the subject of this chapter. If, on the other hand, the change in H varies with time,
we have time-dependent perturbation theory, which is discussed in Chapter 11.
195
196
Chapter 9 Time-Independent Perturbation Theory
'
E
E'
0
a
0
a
FIGURE 9.1√ The infinite square well on the left has the ground-state wave function and
energy ψ = 2/a sin(πx/a) and E = h̄2 π 2 /2ma 2 . A small change in the potential, shown
on the right, produces a small change in ψ and E.
We will now derive what happens if we have a small, time-independent change
in the Hamiltonian. Assume that we have a Hamiltonian H0 , for which we can find
all of the eigenstates |ψn with energies En :
H0 |ψn = En |ψn Note that this is just shorthand for an infinite set of equations: H0 |ψ1 = E1 |ψ1 ,
H0 |ψ2 = E2 |ψ2 , and so on. For example, the |ψn could correspond to the hydrogen wave functions, the spin eigenfunctions of an electron in a magnetic field,
or any other set of wave functions that are exact solutions of the Schrödinger
equation. Now add a small perturbation λH to the Hamiltonian:
H = H0 + λH (9.1)
Here λ is taken to be a dimensionless small number, λ 1, so that the perturbation
λH is small compared to the original Hamiltonian H0 . We would like to solve the
new Schrödinger equation:
H |ψ = E|ψ
(9.2)
In this equation H is the new (perturbed) Hamiltonian of Equation (9.1), |ψ is
the new wave function after we have added the perturbation to the Hamiltonian,
and E is the new energy. Of course, we cannot solve this equation exactly (or
we wouldn’t be bothering to develop perturbation theory), but we can use some
mathematical techniques to see how the change in the Hamiltonian changes the
wave functions and energies.
The first step is to write the new energy and wave function in Equation (9.2) as
a power series in the small number λ that appears in Equation (9.1):
E = En + λE [1] + λ2 E [2] + · · ·
(9.3)
|ψ = |ψn + λ|φ1 + λ |φ2 + · · ·
(9.4)
2
In these equations, |ψn and En are the original eigenfunction and energy before
we apply the perturbation; since Equation (9.2) has an infinite number of solutions (corresponding to a small perturbation applied to any of the eigenfunctions
9.1
197
Derivation of Time-Independent Perturbation Theory
of H0 ) we have to pick a particular eigenfunction |ψn to perturb. The energies
E [1] , E [2] , . . . and the wave functions |φ1 , |φ2 are unknown; our goal is to solve
for them.
Recall that we are interested in the small change in E which results from our
small change λH in the Hamiltonian. The first term in Equation (9.3) gives us
the energy before we apply the small perturbation. The rest of the terms give us
the small change in the energy due to the small change in H . But if λ is tiny,
only the first of these terms really matters. For instance, if we take λ = 10−6 , then
λ2 = 10−12 , λ3 = 10−18 , and so on. So the third term in Equation (9.3) is tiny
compared to the second term, and the fourth term is tiny compared to the third
term. That means the change in E is essentially λE [1] , and we can ignore all of the
other terms in Equation (9.3). [Exception: if, for a particular perturbation, λE [1] is
exactly zero, we will have to go further and solve for λ2 E [2] .]
Substituting the expressions for E and |ψ from Equations (9.3) and (9.4) into
the Schrödinger equation given by Equation (9.2) gives the following rather messy
result:
(H0 + λH )(|ψn + λ|φ1 + λ2 |φ2 + · · ·)
= (En + λE [1] + λ2 E [2] + · · ·)(|ψn + λ|φ1 + λ2 |φ2 + · · ·)
(9.5)
Although it looks like this is only making matters worse, we can now apply two
ideas to simplify this equation. First, recall that if λ is small, then λ is much larger
than λ2 , which is much larger then λ3 , and so on. This means that for very small λ,
the terms in Equation (9.5) with different powers of λ do not affect each other, so
Equation (9.5) must be satisfied separately for each individual power of λ. Equating
powers of λ in this equation gives
λ0
H0 |ψn = En |ψn λ1
λH |ψn + λH0 |φ1 = λEn |φ1 + λE [1] |ψn λ
2
(9.6)
[1]
[2]
(9.7)
λ H0 |φ2 + λ H |φ1 = λ E |φ1 + λ E |ψn + λ En |φ2 2
2
2
2
2
(9.8)
Equation (9.6) is just the original unperturbed Schrödinger equation, which is
reassuring, but we cannot make further progress with Equations (9.7) and (9.8)
unless we know what happens when H0 operates on |φ1 and |φ2 . However, we
don’t even know what |φ1 and |φ2 are! Nonetheless, we can solve this problem
by applying a second idea. Recall that for any Hamiltonian H0 , we can find a set
of eigenfunctions |ψm which form an orthonormal basis, i.e., any wave function
|ψ can be written as
|ψ = c1 |ψ1 + c2 |ψ2 + · · · + cm |ψm + · · ·
So even though we don’t know |φ1 and |φ2 in Equations (9.7) and (9.8), we can
expand them out as linear combinations of eigenfunctions of H0 , and we do know
198
Chapter 9 Time-Independent Perturbation Theory
what happens when we apply H0 to these eigenfunctions. We write
ci |ψi |φ1 = c1 |ψ1 + c2 |ψ2 + · · · + cm |ψm + · · · =
(9.9)
i
and
|φ2 =
di |ψi (9.10)
i
Then when H0 is applied to the sums in Equations (9.9) and (9.10), it simply pulls
out the appropriate energy in front of each term:
ci |ψi =
Ei ci |ψi (9.11)
H0 |φ1 = H0
i
i
and
H0 |φ2 =
Ei di |ψi (9.12)
i
When H0 is applied to |φ1 in Equation (9.7), it produces the sum given in Equation
(9.11). Then we get
λH |ψn + λ
Ei ci |ψi = λEn
ci |ψi + λE [1] |ψn (9.13)
i
i
We would like to solve this equation to find E [1] . We take the inner product of ψn |
with both sides of the equation, recalling that ψn |ψn = 1:
Ei ci ψn |ψi = λEn
ci ψn |ψi + λE [1]
(9.14)
λψn |H |ψn + λ
i
i
Note that |ψn is the original eigenfunction which satisfies the unperturbed
Schrödinger equation, while the |ψi ’s are the complete set of such eigenfunctions, including |ψn as a particular case. Hence, when we take the inner product
of ψn | with |ψi , we get zero for i = n and one for i = n. This selects out i = n
from the sum in Equation (9.14) giving
λψn |H |ψn + λEn cn = λEn cn + λE [1]
so
λE [1] = ψn |λH |ψn (9.15)
where λE [1] is the dominant or lowest order (i.e., lowest power of λ) change in the
energy due to the small change λH in the Hamiltonian.
We can now calculate the second-order change in E, i.e., the term in Equation
(9.3) which is proportional to λ2 . Of course, it is reasonable to ask why we would
9.1
199
Derivation of Time-Independent Perturbation Theory
ever want to know this, since λE [1] is much larger than λ2 E [2] . Normally, the
second order perturbation is irrelevant except for one very important case: if λE [1]
is exactly zero, then λ2 E [2] gives the dominant change in the energy. To find
λ2 E [2] , we substitute both Equations (9.9) and (9.10) into Equation (9.8). When
H0 operates on the sum of eigenvectors as in Equation (9.12), we get
λ2
di Ei |ψi + λ2 H i
ci |ψi = λ2 E [1]
i
ci |ψi + λ2 E [2] |ψn + λ2 En
i
di |ψi i
As before, we take the inner product with ψn |, and now we substitute ψn |H |ψn for E [1] . Solving for λ2 E [2] , we get
λ2 E [2] = λ2
ci ψn |H |ψi − λ2 cn ψn |H |ψn i
The right-hand side is the sum over all eigenfunctions |ψi minus the particular
eignfunction |ψn , so it can be written as
λ2 E [2] = λ2
ci ψn |H |ψi (9.16)
i=n
Now all we have to do is to find the coefficients ci which first appeared in Equation
(9.9). We go back to Equation (9.13), but now instead of applying ψn | to both
sides, we apply an arbitrary eigenfunction ψm |, where ψm | = ψn |. This gives
λψm |H |ψn + λEm cm = λEn cm + 0
so that
cm =
ψm |H |ψn En − E m
(9.17)
Substituting this expression for the ci ’s in Equation (9.16) gives the final expression
for λ2 E [2] :
λ2 E [2] =
ψn |λH |ψi ψi |λH |ψn i=n
=
En − E i
|ψn |λH |ψi |2
i=n
En − E i
(9.18)
To summarize: if we start out with some Hamiltonian H0 for which we can
solve the Schrödinger equation exactly, and we begin in an eigenstate |ψn with
energy En , then after we change the Hamiltonian by the small amount λH , the
dominant change in the energy, proportional to λ, will be given by Equation (9.15),
while the next largest change, proportional to λ2 , will be given by Equation (9.18).
200
Chapter 9 Time-Independent Perturbation Theory
While we have used λ to remember which terms are larger than others, we can
now simplify our expressions by writing the change in H as
H = H0 + H1
where
H1 = λH is very small compared to the unperturbed Hamiltonian H0 . Then we write the first
and second order changes in the energy as
E (1) = λE [1]
and
E (2) = λ2 E [2]
These changes in the energy are given by
E (1) = ψn |H1 |ψn (9.19)
and
E (2) =
|ψn |H1 |ψi |2
i=n
En − E i
(9.20)
where the |ψi ’s which appear in Equation (9.20) are all of the other eigenfunctions
of H0 aside from the one being perturbed. Equations (9.19) and (9.20) are the main
result of this section; they give the first-order and second-order perturbations to
the energy from an arbitrary perturbation to the Hamiltonian. As usual, the inner
products which appear in Equations (9.19) and (9.20) can represent a variety of
different mathematical possibilities. If the wave functions and the perturbation
are functions of position (as in Example 9.1), then these inner products will be
integrals. If the eigenstates are spin states and the perturbation is a function of spin
operators (as in Example 9.2), then these inner products will be matrix products.
There is one case in which the entire argument falls apart. If the original
eigenfunction |ψn is degenerate with some other eigenfunction |ψm of H0 , i.e.,
En = Em , then the argument fails. This can be seen from the fact that both Equations (9.17) and (9.20) “blow up” in this case with zero in the denominator. Note
that Equation (9.19) might be completely well behaved in this case, and it is a very
common mistake to use Equation (9.19) in the case of this kind of degeneracy.
DON’T DO IT! Since the second-order term in this case is infinite, the entire perturbation expansion becomes inapplicable for the case of degeneracy. Perturbation
9.1
Derivation of Time-Independent Perturbation Theory
201
theory applied to degenerate eigenfunctions requires some further mathematical
machinery (degenerate perturbation theory) which is beyond the scope of this text.
Note, however, that there is one special case (which we will encounter frequently)
in which we can use Equations (9.19) and (9.20) with degenerate wave functions:
if ψn |H1 |ψi = 0 whenever En = Ei , then our expressions will be well behaved.
Although the change in the energy is usually the quantity that can be most easily
measured directly, it is also possible to calculate the change in the wave function
due to the perturbation. Returning to Equation (9.4), we see that the lowest-order
change to |ψn is given by λ|φ1 , and |φ1 has already been expressed as a sum
over the unperturbed eigenfunctions in Equation (9.9) with the ci ’s that appear in
this equation given by Equation (9.17). Substituting these values for the ci ’s from
Equation (9.17) into Equation (9.9), we obtain
λ|φ1 =
ψi |λH |ψn i=n
=
En − E i
ψi |H1 |ψn i=n
En − E i
|ψi |ψi Note that we have dropped the i = n term from the sum in Equation (9.9). We
have the freedom to do this because this term is just proportional to the original
unperturbed eigenfunction |ψn . Hence, in the original expansion of the wave
function (Equation 9.4), this term can be removed from the λ|φ1 term and absorbed
into the |ψn term. Then to first order, the new wave function in the presence of
the perturbation is
|ψ = |ψn +
ψi |H1 |ψn En − E i
i=n
|ψi Thus, the effect of the perturbation is to “mix together” all of the other eigenfunctions in the new perturbed wave function.
Example 9.1. The Anharmonic Oscillator.
In Chapter 4, we derived the solutions for the one-dimensional harmonic oscillator
potential,
V (x) =
1
Kx 2
2
The energies are
1
E = n+
h̄ω,
2
n = 0, 1, . . .
202
Chapter 9 Time-Independent Perturbation Theory
V(x)
x
FIGURE 9.2 Solid curve: the unperturbed harmonic oscillator potential V = (1/2)Kx 2 .
Dashed curve: the perturbed potential V = (1/2)Kx 2 + βx 4 .
√
where ω = K/m is the classical oscillation frequency, and the corresponding
wave functions are
1 −s 2 /2
e
π 1/4
√
2
2
ψ1 (s) = 1/4 se−s /2
π
..
.
ψ0 (s) =
with s = [(Km)1/4 /h̄1/2 ]x.
Suppose we begin in the first excited state ψ1 with energy E = (3/2)h̄ω. We
will calculate what happens to the energy of this state if we add a small anharmonic
term to the potential (Figure 9.2):
V (x) =
1
Kx 2 + βx 4
2
From Equation (9.19), the first-order perturbation is
E (1) = n = 1|βx 4 |n = 1
9.1
Derivation of Time-Independent Perturbation Theory
203
Expressing the perturbation in terms of s, we get
∗
√
∞ √
2 −s 2 /2
2 −s 2 /2
h̄2 4
(1)
E =
ds
s
β
se
se
1/4
1/4
Km
π
s=−∞ π
∞
2
h̄2
2
=√ β
s 6 e−s ds
π Km s=−∞
=
With ω =
15 h̄2
β
4 Km
√
K/m, the total energy including the perturbation is
3 15 h̄ω
E = h̄ω
+ β 2
2
4 K
Example 9.1 shows how to apply perturbation theory when the wave function
is a function of position. However, perturbation theory can also be applied to the
matrix representation of spin states.
Example 9.2. Spins in a Magnetic Field.
Suppose we begin with an electron having spin magnetic moment µs in a strong
magnetic field Bz in the z direction (Figure 9.3). Recall from Chapter 8 that the
potential for an electron with spin magnetic moment µs in a magnetic field B is
V = −µs · B
where
µs = −
gS µB
S
h̄
Hence, the Hamiltonian is
gS µB
B·S
h̄
gS
=
µB B · σ
2
H0 =
and gs = 2 for the electron. The eigenstates of H0 are just the spin up and spin
down states, | ↑ and | ↓ , with energies E+ = +µB Bz and E− = −µB Bz .
Suppose we are in the spin up state, | ↑ , and we add the small magnetic field
Bx x̂ with Bx Bz (Figure 9.3). What is the change in the energy of the electron
due to this new magnetic field? Our perturbing potential is
V1 =
gS
µ B Bx σ x
2
204
Chapter 9 Time-Independent Perturbation Theory
Bx x
^
Bz z
^
FIGURE 9.3 We begin with an electron in the spin up direction in the magnetic field Bz ẑ,
and we add the small perturbation Bx x̂.
with gs = 2, and the first-order change in the energy is
E (1) = ψn |H1 |ψn = ↑ |µB Bx σx | ↑ 0 1
1
= µB Bx ( 1 0 )
1 0
0
0
= µB Bx ( 1 0 )
=0
1
So the first-order perturbation, proportional to Bx , is equal to zero. The secondorder perturbation is
E (2) =
|ψn |H1 |ψi |2
i=n
En − E i
| ↑ |µB Bx σx | ↓ |2
E+ − E −
2
[µB Bx ]2 0 1
0 =
(1 0)
1
0
1 2µB Bz
=
=
µB Bx2
2Bz
As expected, E (2) is proportional to Bx2 . In fact, this problem can be solved exactly,
and it is instructive to see how the exact solution compares with the perturbation
theory result (see Exercise 9.1).
9.2
PERTURBATIONS TO THE ATOMIC ENERGY LEVELS
In Chapter 6 we developed an elegant model for the hydrogen atom. The energy
levels were determined by the principal quantum number n, while the other quantum numbers l, ml , and ms had no effect on the energy. As is often the case in
9.2
205
Perturbations to the Atomic Energy Levels
En1
En2
FIGURE 9.4 A given spectral line corresponds to a single transition between two different
energy levels. If two supposedly degenerate levels are slightly separated in energy, a double
line will be produced.
physics, the elegant theory is extremely accurate, but it is not exact. There are
small corrections to the theory due to internal interactions in the hydrogen atom.
With perturbation theory, we now have the tools to derive these corrections.
Fine Structure
Recall that for hydrogen, the energy is given by
En = (−13.6 eV)
1
n2
In hot hydrogen gas, a series of spectral lines are observed, each one corresponding to a particular transition with hν = En1 − En2 . However, upon examining the
spectrum closely, it is observed that some spectral lines are not really single lines
but are closely spaced double lines. This feature is called fine structure. There is
an obvious way to get such closely spaced spectral lines: they will be observed if
the degenerate energy levels are not truly degenerate but separated in energy by a
very small amount (Figure 9.4). Recall that a given energy level En corresponds
to 2n2 different states. Apparently, some sort of interaction, which we have not yet
accounted for, splits some of these states apart in energy.
Our simple model for the hydrogen atom considered only the Coulomb interaction between the proton and the electron. What we have neglected are the various
magnetic fields produced in the atom. The orbital motion of the electron sets up a
magnetic field with magnetic moment given by
µl = −
gl e
L,
2me
gl = 1
(9.21)
while the spin of the electron produces a spin magnetic moment equal to
µs = −
gs e
S,
2me
gs ≈ 2
So the orbital motion of the electron produces a magnetic field, and the electron
itself acts like a small magnet embedded in this magnetic field (Figure 9.5). The
electron will prefer to line up with its spin magnetic field in the opposite direction
to the orbital magnetic field. Hence, the “spin up” and “spin down” states of the
electron will have different energies. This is the basis of fine structure: it arises
from the spin-orbit coupling of the electron.
206
Chapter 9 Time-Independent Perturbation Theory
␮s
+
Borbital
–
FIGURE 9.5 The fine-structure splitting is produced by the interaction between the spin
magnetic field of the electron and the magnetic field produced by its orbital motion.
More precisely, the interaction energy between the spin magnetic moment µs
and the external magnetic field B (produced by the orbital motion of the electron)
is
H1 = −µs · B
(9.22)
To find B imagine that we are sitting in the rest frame of the electron watching the
proton orbit around us. In the rest frame of the electron, the electric field of the
proton,
E=
1 e
r
4π 0 r 3
transforms into a magnetic field, B = −v × E/c2 . Using L = r × p, we obtain,
for the magnetic field induced by the orbital motion of the electron,
1
e
B=
L
4π 0 me c2 r 3
Substituting our expressions for µs and B into Equation (9.22), we obtain, for the
energy of the spin-orbit interaction,
H1 =
e2
S·L
4π 0 m2e c2 r 3
Unfortunately, this expression is wrong. The problem arises because we did the
calculation in the rest frame of the electron, which is an accelerating frame of
reference, so we cannot simply transform back into the rest frame of the proton
and expect to get the right answer. When this error is corrected, an additional factor
of 1/2 is obtained (this correction is called Thomas precession after L.H. Thomas,
who explained this effect in 1926). The corrected expression is
H1 =
e2
S·L
8π 0 m2e c2 r 3
(9.23)
9.2
207
Perturbations to the Atomic Energy Levels
L
S
+
–
FIGURE 9.6 The spin-orbit interaction Hamiltonian is proportional to S · L. Classically,
S · L = SL cos θ.
From a classical point of view, this expression makes sense, since we expect the
interaction energy of the two magnetic fields to depend on the alignment of the S
and L vectors (Figure 9.6).
We can get a crude order of magnitude estimate of the size of this interaction
energy by taking S ∼ h̄, L ∼ h̄, and r ∼ 10−10 m. Then Equation (9.23) gives a
perturbation energy of the order of 10−4 eV compared to a hydrogen binding energy
of 13.6 eV. It is therefore a good approximation to treat the spin-orbit interaction
as a small perturbation.
In our expression for H1 , the interaction is proportional to S · L = Sx Lx +
Sy Ly + Sz Lz . However, this expansion is essentially useless, since the electron
is never in a state which is a simultaneous eigenstate of all three components of
angular momentum. Instead, we use the standard procedure from Chapter 8 for
dealing with dot products of operators. We define the total angular momentum
operator J to be
J=L+S
Then
J 2 = L2 + S 2 + 2L · S
which implies
L·S=
1 2
(J − L2 − S 2 )
2
208
Chapter 9 Time-Independent Perturbation Theory
and the spin-orbit perturbation becomes
H1 =
e2
(J 2 − L2 − S 2 )
16π 0 m2e c2 r 3
In Chapter 6 we wrote the hydrogen wave functions in terms of the quantum
numbers n, l, and ml , and in Chapter 8 we added the spin quantum number ms ,
but now we want to express the hydrogen wave functions as eigenfunctions of J
and Jz , i.e., in the form |n l j mj , where
J 2 |n l j mj = h̄2 j (j + 1)|n l j mj Jz |n l j mj = h̄mj |n l j mj Then the perturbation to the energy levels due to spin-orbit coupling is
(1)
= n l j mj |H1 |n l j mj Espin-orbit
= n l j mj |
e2
(J 2 − L2 − S 2 )|n l j mj 16π 0 m2e c2 r 3
= h̄2 [j (j + 1) − l(l + 1) − s(s + 1)]n l j mj |
e2
|n l j mj 16π 0 m2e c2 r 3
What is the value of n l j mj |(e2 /16π 0 m2e c2 r 3 )|n l j mj ? Note that this is
just an integral over the radial wave function, which is a function only of n and l;
hence, we can write
n l j mj |(e2 /16π 0 m2e c2 r 3 )|n l j mj = fnl
where fnl is a function only of n and l. Then our expression for the change in E
due to the spin-orbit interaction (taking s = 1/2 for the electron) becomes
3
(1)
= h̄2 fnl j (j + 1) − l(l + 1) −
Espin-orbit
4
(9.24)
From Chapter 9 recall that for l = 0, j has two possible values, either l − 1/2
or l + 1/2, while for l = 0 we have only j = 1/2. From Equation (9.24), it is clear
that the spin-orbit coupling splits each state with l = 0 into two different states with
j = l + 1/2 having the higher energy and j = l − 1/2 having the lower energy.
The expression for fnl can be evaluated to yield a final expression for the change
in E from spin-orbit coupling:
(1)
Espin-orbit
= |En |α 2
1 [j (j + 1) − l(l + 1) − 3/4]
2n
l(l + 1/2)(l + 1)
(9.25)
9.2
Perturbations to the Atomic Energy Levels
209
where
α=
e2
4π 0 h̄c
Note that α is a dimensionless number with a value of roughly 1/137. Because
of its origin in this calculation, α is called the fine-structure constant, although it
crops up in many other areas of physics. Recall that the hydrogen energy levels
En are all negative, so we take the absolute value of En in Equation (9.25) and
in Equations (9.28)–(9.29) to avoid any confusion over the sign of the change in
energy.
However, this is not the full story of the fine-structure splitting. We have applied
our standard nonrelativistic treatment to the electron, and this is an excellent approximation, since the electron in the hydrogen atom is not highly relativistic (its
kinetic energy is much smaller than its rest energy; classically, this corresponds to
an electron velocity v much smaller than the speed of light). However, now that
we are working in the realm of tiny changes in the energy levels, we have to take
into account small corrections due to relativistic effects.
In relativistic classical mechanics, the total energy of a particle with rest mass
m and momentum p is
E=
p 2 c 2 + m2 c 4
(9.26)
For now we are interested only in the case where the particle is only slightly
relativistic so that p mc. (We will relax this restriction in Chapter 15 when we
discuss relativistic quantum mechanics in more detail.) In this limit we can expand
the square root in Equation (9.26) to obtain
E ≈ mc2 +
p4
p2
−
+ ···
2m 8m3 c2
(9.27)
In the limit of small p, each term in this expression is small compared to the
preceding one. In relativity, the first term in Equation (9.27) is interpreted as the
rest energy of the particle, while the remainder of the expression corresponds to
the energy of motion. But what is the correct energy to use in the Schrödinger
equation? In the standard nonrelativistic Schrödinger equation, the Hamiltonian
operator corresponds to the kinetic energy plus the potential energy, and the rest
energy plays no role. Hence, in writing the Hamiltonian, we use the second term
in Equation (9.27) to give us the unperturbed Hamilton, while the third term gives
the lowest-order perturbation due to relativistic effects.
Then we have, for our perturbation,
H1 = −
p4
8m3e c2
210
Chapter 9 Time-Independent Perturbation Theory
and the lowest-order change in the hydrogen energy levels is
(1)
Erelativistic
=−
1
n l j mj |p 4 |n l j mj 8m3e c2
This expression can be evaluated for the hydrogen wave functions, yielding a final
result of
(1)
Erelativistic
= −|En |α 2
1
n2
2n
3
−
2l + 1 4
(9.28)
(1)
Since Erelativistic
is a function of n and l but not a function of j , the relativistic
correction does not contribute anything to the splitting of the energy levels, which
is determined entirely by the spin-orbit interaction, but it does change the overall
dependence of the energy levels on n and l. Since l ≤ n − 1, the term in brackets in
Equation (9.28) is always positive, so the relativistic correction always decreases
the energy levels.
(1)
(1)
Note that Espin-orbit
and Erelativistic
are roughly equal in magnitude; both of them
2
are approximately α En . Hence, neither contribution to the fine structure can be
neglected. We therefore add Equations (9.25) and (9.28) to get the total change in
energy due to both the spin-orbit coupling and relativistic effects:
(1)
Efine
structure
(1)
(1)
= Espin-orbit
+ Erelativistic
n
1 3
−
= |En |α 2 2
n 4 j + 1/2
(9.29)
The dependence on l has cancelled out, so that the total change in energy is a
function only of j and n. The net effect of the fine structure is to split the l − 1/2 and
l + 1/2 states (due to the spin-orbit coupling) and to decrease the energies of both
states relative to the unperturbed hydrogen energy levels (see Exercise 9.10). The
fine-structure perturbation (both the decrease in energy relative to the unperturbed
energy levels and the splitting between the j = l + 1/2 and the j = l − 1/2 states)
is of order α 2 En ≈ 10−4 En , since the other factors in Equation (9.29) are all of
order unity.
We now introduce a standard if somewhat arcane notation to describe the angular
momentum states of the hydrogen atom. In this notation, the different l states of
the hydrogen atom are written with different capital letters: the l = 0 state is
called the S state, the l = 1 state is called the P state, the l = 2 state is called
the D state, the l = 3 state is called the F state, and then the sequence continues
alphabetically (G, H, I, . . .). (The origin of these abbreviations is buried in the
history of spectroscopy; there is nothing particularly logical about them.) The
standard way of writing the various j states is to indicate the value of j as a
subscript, e.g., P1/2 is the notation for l = 1, j = 1/2. [We will see an additional
twist to this notation in Chapter 13.] The different hydrogen energy levels written
9.2
211
Perturbations to the Atomic Energy Levels
D5/2
n=3
S1/2
P3/2
D3/2
P1/2
P3/2
n=2
n=1
S1/2
P1/2
S1/2
l=0
l=1
l=2
FIGURE 9.7 The energy levels of hydrogen showing the fine structure (not drawn to
scale).
Sp
+
Se
–
FIGURE 9.8 The spin-spin interaction between the proton and electron produces the
hyperfine splitting.
in this notation are shown in Figure 9.7. The S states (l = 0) do not split since they
correspond to a single value of j , while the l = 0 states split into the j = l + 1/2
state and the j = l − 1/2 state, with the former having higher energy than the
latter.
Finally, note that we have blithely applied first-order perturbation theory
to degenerate states, ignoring the warning in the previous section. However,
it is all right to use nondegenerate perturbation theory in this case, since
n l ml ms |H1 |n l ml ms = 0 if l = l or ml = ml or ms = ms .
There is a second internal magnetic interaction in the hydrogen atom with a
much smaller effect. The proton has a spin magnetic moment given by
µp =
gp e
S
2mp
(9.30)
with gp ≈ 5.6. So the electron also feels this magnetic field and is perturbed by
it (Figure 9.8). However, a comparison of Equation (9.30) with Equation (9.21)
shows that the ratio of the spin magnetic field of the proton to the magnetic field
produced by the orbital motion of the electron is roughly me /mp ≈ 6 × 10−4 .
Hence, we expect the splitting from the spin-spin interaction to be much smaller
than the effect of the spin-orbit interaction.
212
Chapter 9 Time-Independent Perturbation Theory
Nonetheless, this spin-spin interaction does produce a splitting in energies.
Since it is so much smaller than the fine structure, it is called hyperfine splitting.
In the ground state of hydrogen, for example, the triplet (S = 1) state has a higher
energy than the singlet (S = 0) state; the energy difference is E = 5.9 × 10−6 eV.
Although this energy difference is tiny, hyperfine splitting has an importance out of
proportion to its magnitude. The universe contains clouds of neutral hydrogen gas;
this gas radiates by dropping from the triplet into the singlet state. The frequency
of this radiation is ν = E/ h = 1420 MHz, corresponding to a wavelength of
λ = 21 cm: the famous “21-centimeter line.”
The Lamb Shift
The fine-structure calculations in the previous section predict that the hydrogen
energy levels do not depend on l. Hence, two states with the same n and j quantum
numbers but different values of l should be degenerate in energy. In 1947 Willis
Lamb and his student, R.C. Retherford, showed experimentally that this was not
the case. Specifically, they measured a splitting between the n = 2, S1/2 state and
the n = 2, P1/2 state.
This splitting, now called the Lamb shift, cannot be explained in the context
of quantum mechanics, but arises from the more esoteric area of quantum field
theory (which was, in part, motivated by Lamb’s experimental result). Quantum
field theory is beyond the scope of this book; here we will simply use one of the
predictions of the theory.
In quantum field theory, the vacuum is no longer simply empty space; it is
literally seething with activity. Virtual particles, such as electron-positron pairs,
can pop into existence and disappear. As long as the energy of the particles E and
their lifetime t satisfy Et < h̄/2, then these particle-antiparticle pairs cannot be
detected directly. [This is a rather crude explanation, which is made much more
precise within the framework of quantum field theory.]
These particle-antiparticle pairs produce an effect called vacuum polarization.
Consider a dielectric surrounding a point positive charge. The point charge polarizes the dielectric, attracting negative charge inward and repelling positive charge
outward. This tends to cancel the electric field produced by the point charge, leading to a reduced electric field inside the dielectric (Figure 9.9).
Now consider the same positive charge in a vacuum. The production of virtual
electron-positron pairs tends to cancel the charge, just as in a physical dielectric.
However, unlike a dielectric, we can never remove the positive charge from the
vacuum polarization to measure its true charge: the charge we measure has already
been cancelled by the effect of the vacuum polarization. This means that the “bare”
charge, which cannot be measured directly, is much larger than we thought; in fact,
it is mathematically infinite!
The upshot of all of this is that the electric field of a point charge must be
modified at the origin (where the charge is “infinite”), but everywhere else in
9.2
213
Perturbations to the Atomic Energy Levels
Dielectric
Vacuum
+
e+
–
e–
+
–
–
+
+
+
e–
e+
e–
–
e+
+
FIGURE 9.9 In a dielectric, polarization reduces the electric field produced by a point
charge. Vacuum polarization produces the same effect in a vacuum.
space the charge has already been cancelled by the effects of vacuum polarization,
so the electric field is unchanged. The result for V (r), derived from quantum field
theory, is
V (r) = −
αe2 h̄2 3
e2 1
−
δ (r)
4π 0 r
15π 0 m2e c2
(9.31)
where δ 3 (r) is the three-dimensional Dirac delta function discussed in Chapter 7.
The second term in Equation (9.31) is the perturbation to the Hamiltonian, so the
first-order shift in the energy is
E (1) = n l ml ms |H1 |n l ml ms αe2 h̄2 3
3
∗
= d r ψnlml (r) −
δ (r) ψnlml (r)
15π 0 m2e c2
=−
αe2 h̄2
|ψnlml (0)|2
15π 0 m2e c2
Recall that the hydrogen wave functions are all identically zero at the origin, except
for the l = 0 states. Thus, the effect of the Lamb shift is to reduce the energy of the
l = 0 states relative to the corresponding l = 0 states. The effect is smaller than
the fine-structure splitting, e.g., for the n = 2 states, the splitting between the l = 0
and l = 1 state is about 10−7 eV. As bizarre as all of this sounds, it is important to
remember that it is based on solid experimental evidence.
214
Chapter 9 Time-Independent Perturbation Theory
9.3
THE ATOM IN EXTERNAL ELECTRIC OR MAGNETIC FIELDS
In the previous section, we discussed perturbations which are intrinsic to the atom.
We will now examine what happens when the atom is placed in an external electromagnetic field. Since the atom consists of charged particles, and the electons
produce both a spin and orbital magnetic moment, any external electric or magnetic
field will perturb the energy levels of the atom. The effect produced by an external
electric field is called the Stark effect, while the effect of an external magnetic field
is the Zeeman effect.
The Atom in an Electric Field: The Stark Effect
We will first examine the effect of a uniform electric field with magnitude E on
the ground state of hydrogen. Recall that the ground-state wave function is
ψ100 = √
1
π a03
e−r/a0
where the “100” subscript denotes the n l ml quantum numbers. We can ignore the
spin state of the electron, since the spin interacts only with magnetic fields through
the electron’s spin magnetic moment. (Of course, we will have to consider spin in
the next section when we discuss external magnetic fields.) We take the electric
field to be uniform, static, and pointing in the z direction.
Since the ground state of hydrogen is nondegenerate, we can use the perturbation
theory expressions from Section 9.1. Classically, the potential energy of a charge
e in an electric field E is V = eEz, so the perturbation H1 produced by the electric
field is
H1 = eEz
and the first-order change in the energy of the hydrogen atom is, from Equation
(9.19),
E (1) = ψ100 |eEz|ψ100 (9.32)
Taking z = r cos θ and writing Equation (9.32) in spherical coordinates, we get
1
1
(1)
−r/a0
−r/a0
(eEr cos θ ) r 2 dr sin θ dθ dφ
e
e
E =
3
3
π a0
π a0
But the integral over θ vanishes:
π
1
cos θ sin θ dθ = sin2 θ|π0 = 0
2
θ=0
so E (1) = 0.
9.3 The Atom in External Electric or Magnetic Fields
215
Since the first-order perturbation vanishes, we must use second-order perturbation theory to calculate the change in the energy due to the external electric field.
Equation (9.20) gives
E (2) =
|ψ100 |eEz|ψnlm |2
l
E
−
E
1
n
n,l,m
(9.33)
l
where En = 13.6 eV/n2 . Recall from Chapter 6 that every hydrogen wave function
can be written as the product of a radial wave function Rnl (r) and the spherical
harmonic Ylm (θ, φ). Then the inner product which appears in Equation (9.33) can
be written in the form
∗
ψ100 |eEz|ψnlml = eE R10
(r)Y00∗ (θ, φ) r cos θ Rnl (r)Ylm (θ, φ) d 3 r (9.34)
√
√
√
But now recall that Y00 = 1/ 4π , and Y10 = 3/4π cos θ, so Y00 cos θ = (1/ 3)Y10 .
This allows us to write Equation (9.34) as
eE
∗
ψ100 |eEz|ψnlml = √
(r)Rnl (r)r 3 dr Y10∗ (θ, φ)Ylm (θ, φ) sin θ dθ dφ
R10
3
But the Ylm ’s are orthonormal, so
0∗
Y1 (θ, φ)Ylm (θ, φ) sin θ dθ dφ = 1 (l = 1, m = 0)
= 0 (l = 1 or m = 0)
Hence, in the sum in Equation (9.33), only the l = 1, m = 0 terms are nonzero
giving
√ |(eE/ 3) R ∗ (r)Rn1 (r)r 3 dr|2
10
(2)
E =
(9.35)
E
1 − En
n
The integral under the sum in Equation (9.35) can be evaluated exactly for all
values of n, and the terms in the series decrease rapidly with n:
9
E (2) = −(4π 0 )a03 E 2 (1.48 + 0.20 + 0.066 + · · ·) = − (4π 0 )a03 E 2
4
The change in energy is negative, since the hydrogen atom becomes polarized and
aligns itself so as to partially cancel the external electric field (Figure 9.10; see
also Exercise 9.2). Since the change in energy is proportional to the square of the
applied electric field, this effect is called the quadratic Stark effect.
Our use of nondegenerate perturbation theory breaks down for the excited states
of hydrogen, since these states are degenerate. Using degenerate perturbation theory, it is possible to show that the change in energy for these excited states is
proportional to E rather than E 2 . Hence, the change in energy when an electric
field is applied to the excited states of hydrogen is called the linear Stark effect.
216
Chapter 9 Time-Independent Perturbation Theory
+
external
atomic
–
FIGURE 9.10 A classical picture of the quadratic Stark effect: the hydrogen atom is
polarized by the external electric field, and the field produced by the polarized atom is in
the opposite direction to the external field.
The Atom in a Magnetic Field: The Zeeman Effect
Now consider what happens when we apply an external magnetic field B to the
hydrogen atom. Assume that the magnetic field has magnitude B and is pointing
in the z direction so that
B = B ẑ
9.3 The Atom in External Electric or Magnetic Fields
217
S
S
J
␮
L
FIGURE 9.11 The magnetic moment µ of the hydrogen atom is proportional to L + 2S,
while the total angular momentum J is proportional to L + S, so µ and J are not, in general,
parallel.
The potential energy of a magnetic dipole µ in a magnetic field B is just V =
−µ · B, so the perturbation produced by the magnetic field is
H1 = −µ · B
(9.36)
In Equation (9.36), there are two contributions to the atomic magnetic moment: the
contribution from the orbital magnetic moment µl and the contribution from the
spin magnetic moment of the electron µs . (In principle, we should also include the
spin magnetic moment of the proton, but this is much smaller and can be ignored.)
Hence, the total magnetic moment is
µ = µl + µs
g s µB
gl µB
L−
S
=−
h̄
h̄
Recall that gl = 1 and gs ≈ 2, so the expression for µ becomes
µ=−
µB
[L + 2S]
h̄
(9.37)
Note that the magnetic moment of the atom is not proportional to the total angular
momentum operator J, which is L + S. In classical terms, the angular momentum
vector J and magnetic moment vector µ are not parallel (Figure 9.11). This has
important consequences for the Zeeman effect. (You are already familiar with a
much larger classical system in which the angular momentum and magnetic dipole
are not parallel: the earth!)
218
Chapter 9 Time-Independent Perturbation Theory
We can use Equation (9.37) to rewrite the perturbation in Equation (9.36) as
H1 = B
µB
[Lz + 2Sz ]
h̄
Applying this perturbation to the hydrogen state |n l ml ms gives the first-order
change in energy,
E (1) = n l ml ms |(BµB /h̄)(Lz + 2Sz )|n l ml ms = BµB (ml + 2ms )
(9.38)
The problem with this result is that it requires the atom to be in a state of definite
ml and ms (or, equivalently, an eigenstate of Sz and Lz ). However, we saw in our
discussion of fine structure in Section 9.2 that the spin-orbit coupling drives the
atom into an eigenstate of J 2 , which does not commute with Sz and Lz . Hence,
the atom is in a state of definite j and mj rather than ml and ms , so our argument
would appear to be invalid.
To clarify this situation, we can write the full Hamiltonian as
H = H0 +
e2
µB
[Lz + 2Sz ]
S·L+B
8π 0 m2e c2 r 3
h̄
(9.39)
where the second term is the perturbation due to the spin-orbit interaction (given
in Equation 9.23), and the third term is the perturbation from the external magnetic
field.
Now consider two possible cases: for very strong magnetic fields (B 1 T), the
third term in Equation (9.39) dominates the second term, while for weak magnetic
fields (B 1 T), the second term dominates the third. Consider the case of strong
magnetic fields first. For this case we simply ignore the effect of the spin-orbit
coupling; the strong magnetic field overwhelms the spin-orbit coupling and drives
the atom back into a state of definite ml and ms . Therefore, for the strong magnetic
field case, the expression we derived for E (1) in Equation (9.38) is correct:
E (1) = BµB (ml + 2ms )
This regime of the Zeeman effect is called the strong-field Zeeman effect or the
Paschen-Back effect. An illustration of this perturbation in the energy levels is
shown in Figure 9.12 for the case l = 1.
Now consider the opposite regime in which spin-orbit coupling dominates the
effect of the external magnetic field. In this case the atom is in a state of definite
j and mj rather than ml and ms , and the perturbation must be written as
E (1) = n l j mj |(BµB /h̄)(Lz + 2Sz )|n l j mj 219
9.3 The Atom in External Electric or Magnetic Fields
ml = 1,
ms = 1/2
ml = 0,
ms = 1/2
ml = –1, ms = 1/2
ml = 1, ms = –1/2
l=1
ml = –1, 0, 1
ms = –1/2, +1/2
ml = 0,
ms = –1/2
ml = –1, ms = –1/2
ml + 2ms
2
1
0
–1
–2
FIGURE 9.12 The strong-field Zeeman effect for the energy levels of an l = 1 state in
hydrogen.
This can be partially simplified by using the fact that Jz = Lz + Sz :
E (1) = n l j mj |(BµB /h̄)(Jz + Sz )|n l j mj = BµB mj +
BµB
n l j mj |Sz |n l j mj h̄
(9.40)
In order to further simplify this expression, the state |n l j mj must be written as a
linear combination of the |n l ml ms states. From Chapter 8, we know that s = 1/2
and a given value of l can couple to give either j = l + 1/2 or j = l − 1/2, while
mj = ml + ms . The actual linear combination is
l + 1/2 + mj 1/2
|ml = mj − 1/2, ms = 1/2
|j = l + 1/2, mj =
2l + 1
l + 1/2 − mj 1/2
+
|ml = mj + 1/2, ms = −1/2
2l + 1
l + 1/2 − mj 1/2
|j = l − 1/2, mj =
|ml = mj − 1/2, ms = 1/2
2l + 1
l + 1/2 + mj 1/2
−
|ml = mj + 1/2, ms = −1/2
2l + 1
We can use these equations to solve for n l j mj |Sz |n l j mj . For j = l + 1/2,
220
Chapter 9 Time-Independent Perturbation Theory
we get
l + 1/2 + mj 1/2
ml = mj − 1/2, ms = 1/2|
2l + 1
l + 1/2 − mj 1/2
+
ml = mj + 1/2, ms = −1/2|
2l + 1
l + 1/2 + mj 1/2
× Sz
|ml = mj − 1/2, ms = 1/2
2l + 1
l + 1/2 − mj 1/2
+
|ml = mj + 1/2, ms = −1/2
2l + 1
h̄ l + 1/2 + mj
h̄ l + 1/2 − mj
=
−
2
2l + 1
2
2l + 1
n l j mj |Sz |n l j mj =
=
mj h̄
2l + 1
Similarly, for j = l − 1/2, we obtain
n l j mj |Sz |n l j mj = −
mj h̄
2l + 1
Combining the results for j = l + 1/2 and j = l − 1/2, we get
n l j mj |Sz |n l j mj =
mj h̄
2(j − l)
2l + 1
and substituting this result into Equation (9.40) yields
2j + 1
E (1) = BµB mj
2l + 1
In analogy with the gs factor for the electron spin and gl for the orbital angular
momentum, we can write
g=
2j + 1
2l + 1
where this g is called the Landé g factor. In terms of the Landé g factor, the energy
shift becomes
E (1) = gBµB mj
In contrast to gs and gl which are constant, the Landé g factor is not constant,
but rather is a function of j and l. The reason for this is the fact, already alluded
to, that µ is not parallel to J, since the operator which determines µ is L + 2S
221
Exercises
mj
+3/2
g = 4/3
+1/2
P3/2 (l = 1, j = 3/2)
–1/2
–3/2
g = 2/3
+1/2
P1/2 (l = 1, j = 1/2)
–1/2
FIGURE 9.13 The splitting of the P3/2 and P1/2 states in the weak-field Zeeman effect.
while J = L + S. Hence, the ratio between µ and J can depend on the relative
orientation of L and S (Figure 9.11), so µ is not a fixed multiple of mj . The effect
of the weak-field Zeeman effect is to split the energies of the individual mj levels,
with a magnitude which depends on both the magnitude of the magnetic field and
the value of the Landé g factor (Figure 9.13).
To summarize, for weak magnetic fields, the hydrogen atom can be taken to
be in a state of definite j and mj , and the magnetic field separates the energies of
the individual mj states. As the magnetic field is increased, it eventually becomes
stronger than the internal magnetic fields of the atom. In this limit, the magnetic
field drives the hydrogen atom into a state of definite ml and ms , and the perturbation
in energy is just proportional to ml + 2ms .
EXERCISES
9.1 (a) In Example 9.2, the energy of the system can be calculated exactly. Take B =
Bx x̂ + Bz ẑ, and calculate the exact energies. [Hint: Feel free to use a different
coordinate system; the energy levels cannot depend on the choice of the coordinate
system].
(b) Take the answer in part (a) and expand it out in powers of Bx , remembering
that Bx Bz . Show that the terms proportional to Bx and Bx2 correspond to the
answers derived in Example 9.2.
222
Chapter 9 Time-Independent Perturbation Theory
9.2 A particle is in a potential V0 in its ground state |ψ0 . A small perturbation H1 is
applied to the particle. Suppose that the first order perturbation to the energy is zero:
E (1) = ψ0 |H1 |ψ0 = 0. Show that the lowest-order effect of H1 is to decrease the
energy of the ground state.
9.3 A particle of mass m is confined to move in a one-dimensional square well with infinite
potential barriers at x = 0 and x = a, with V = 0 for 0 ≤ x ≤ a. The particle is in the
ground state. A perturbation H1 = λδ(x − a/2) is added, where λ is a small constant.
(a) What units does λ have?
(b) Calculate the first-order perturbation E (1) due to H1 .
(c) Calculate the second-order perturbation E (2) . The answer may be expressed as
an infinite series.
9.4 A particle of mass m is confined to move in a narrow, straight tube of length a which
is sealed at both ends with V = 0 inside the tube. Treat the tube as a one-dimensional
infinite square well. The tube is placed at an angle θ relative to the surface of the
earth. The particle experiences the usual gravitational potential V = mgh. Calculate
the lowest-order change in the energy of the ground state due to the gravitational
potential.
9.5 A particle of mass m is in the ground state in the harmonic oscillator potential
V (x) =
1
Kx 2
2
A small perturbation βx 6 is added to this potential.
(a) What are the units of β?
(b) How small must β be in order for perturbation theory to be valid?
(c) Calculate the first-order change in the energy of the particle.
9.6 In the hydrogen atom, the proton is not really a point charge but has a finite size.
Assume that the proton behaves as a uniformly-charged sphere of radius R = 10−15 m.
Calculate the shift this produces in the ground-state energy of hydrogen.
9.7 The photon is normally assumed to have zero rest mass. If the photon had a small mass,
this would alter the potential energy which the electron experiences in the electric field
of the proton. Instead of
e2 1
4π 0 r
(9.41)
e2 e−r/r0
4π 0 r
(9.42)
V (r) = −
we would have
V (r) = −
where r0 is a constant with units of length. Assume r0 is large compared to the size
of the hydrogen atom, so the potential energy given in Equation (9.42) differs only
slightly from the standard one given by Equation (9.41) in the vicinity of the electron.
Calculate the change in the ground state energy of hydrogen if the correct potential is
given by Equation (9.42) instead of Equation (9.41).
Exercises
223
9.8 Suppose that that the proton had spin 0 instead of spin 1/2.
(a) How would this alter the fine structure of the energy levels of the hydrogen atom?
(b) How would this alter the hyperfine structure of the energy levels of the hydrogen
atom?
9.9 We have seen that the spin-orbit interaction splits the l = 0 states in the hydrogen
atom into j = l + 1/2 states (with slightly higher energy) and j = l − 1/2 states
(with slightly lower energy). Suppose that the electron had spin 1. How many different
energy levels would the spin-orbit interaction produce, and what would their relative
energies be? Be sure to consider how the answer would depend on the value of l.
9.10 Equation (9.29) gives the fine-structure energy shift.
(a) Show that the j = l + 1/2 state has a higher energy than the j = l − 1/2 state.
(b) Show that the change in energy, Ef(1)ine structure , is always negative.
9.11 An electron is in the ground state in a three-dimensional rectangular box given by
0 ≤ x ≤ a, 0 ≤ y ≤ b, and 0 ≤ z ≤ c, where V = 0 inside the box, and there are
infinite potential barriers at all of the walls. A homogeneous, static electric field with
magnitude E is applied in the x direction. What is the lowest-order change in the
energy of the electron?
9.12 A hydrogen atom in its ground state is placed in a homogeneous, static electric field
with magnitude E in the x direction.
(a) Show that the first-order perturbation E (1) is 0.
(b) Show that the second-order perturbation E (2) is the same as if the field was
pointing in the z direction. [This is obvious from symmetry, but calculate E (2)
using perturbation theory and show it explicitly.]
9.13 A hydrogen atom is in its ground state. A proton is fixed in space a distance R from
the nucleus of the hydrogen atom, where R a0 . Calculate the perturbation to the
energy of the hydrogen atom due to the electric field of this proton.
9.14 The electron in a hydrogen atom is in a D state. A homogenous, static magnetic field
is applied in the z direction.
(a) Draw a diagram showing the splitting of the energy levels in the weak-field limit.
Calculate the value of g for each energy level.
(b) Draw a diagram showing the splitting of the energy levels in the strong-field limit.
9.15 (a) A particle is in a state |ψ which is an eigenfunction of the Hamiltonian H0 with
energy E. A perturbation H1 is applied such that H1 |ψ = 0. Show that the energy
of the system is completely unchanged by this perturbation.
(b) In the ground state of the helium atom, both electrons are in the l = 0 state, and
the spin wave function for the two electrons is the singlet spin state (s = 0 and
ms = 0). [This is a consequence of the Pauli exclusion principle, which will be
discussed in Chapter 13.] A homogeneous, static magnetic field is applied in the
z direction. Show that the energy of the ground state of helium is completely
unaffected by this magnetic field. [Ignore the magnetic moment of the nucleus.]
What is the physical reason for this?