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Physics 115C
Homework 2
Problem 1
Our full Hamiltonian is
p2
1
H =
+ mω 2 x2 + βx4
2m 2
= H0 + H ′
where the unperturbed Hamiltonian is our usual
H0 =
p2
1
+ mω 2 x2
2m 2
and the perturbation is
H ′ = βx4
Assuming β is small, the perturbation is small, and we can use perturbation theory to
estimate the corrections to the energy levels. From perturbation theory, these first-order
corrections are given by
En(1) = ψn(0) H ′ ψn(0)
E
(0)
where the ψn are the unperturbed eigenstates. For our H ′ , this expression becomes
En(1) = β hn| x4 |ni
E
(0)
for brevity. Now, we could turn the inner
where I’m using the usual notation |ni ≡ ψn
product into an integral and try to evaluate it explicitly; we’ll have integrals over arbitrary
Hermite polynomials, which will be pretty hideous (although doable if we were to cleverly
exploit generating functions). Instead, let’s be less masochistic and use our usual ladder
operators. Recall that we can express the position operator in terms of the ladder operators
as
r
~
x=
a† + a
2mω
1
We can therefore write
En(1) = β hn| x4 |ni
2
4
~
= β
hn| a† + a |ni
2mω
We can save ourselves some time in evaluating this expression via the following observation:
recall that the raising and lowering operators acting on energy eigenstates give
√
n |n − 1i
a |ni =
√
†
a |ni =
n + 1 |ni
Also recall that the energy eigenstates are orthonormal: hn|mi = δnm . Now, in our expression
(1)
for En , we have a whole bunch of raising and lowering operators acting on |ni; since the
4
operators are being raised to the fourth power, the state a† + a |ni will be some linear
combination of all the states from |n − 4i to |n + 4i. But since we are then taking the
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inner product with hn|, the only contibuting terms in the expression for a† + a |ni will
be those that are proportional to |ni. These terms will only appear if we have an equal
number of raising and lowering operators acting on |ni, so that after raising and lowering,
we end back up with something proportional to |ni. The conclusion is that the only terms
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we’re interested in in the expansion of a† + a are those that contain two raising and two
lowering operators; the rest we can discard. With this in mind, let’s expand:
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a† + a |ni = a† + a a† + a a† + a a† + a |ni
= a† a† aa + a† aa† a + a† aaa† + aa† a† a + aa† aa† + aaa† a† |ni
+(terms not containing |ni)
We evaluate the terms via the usual rules, e.g.
√ † †
a† a† aa |ni =
na a a |n − 1i
√ √
=
n n − 1a† a† |n − 2i
√
=
n (n − 1)a† |n − 1i
= n(n − 1) |ni
Evaluating all the terms in this way, we get
4
a† + a |ni = n(n − 1) + n2 + n(n + 1) + n(n + 1) + (n + 1)2 + (n + 1)(n + 2) |ni
+(terms not containing |ni)
= (6n2 + 6n + 3) |ni + (terms not containing |ni)
and therefore
hn| a† + a
4
|ni = 6n2 + 6n + 3
2
Our first-order energy shifts therefore become
2
4
~
(1)
En = β
hn| a† + a |ni
2mω
En(1)
=β
~
2mω
2
(6n2 + 6n + 3)
Note that since the polynomial 6n2 + 6n + 3 is always positive, the first-order energy shifts
are all positive: this is not surprising, since our perturbation is positive everywhere.
3
Problem 2
(a) The Hamiltonian for a particle with spin in a magnetic field is (ignoring the spatial part)
H = −γB · S
Our magnetic field is B = B0 k̂ + B î, so
H = −γ(B0 Sz + BSx )
We can solve for the energy eigenvalues of this Hamiltonian in two different ways. The
first is to use a brute-force approach: using the matrix representations of the spin matrices for spin-1/2,
~ 1 0
Sz =
2 0 −1
~ 0 1
Sx =
2 1 0
we can write the Hamiltonian as
1
B0 B
H = − ~γ
B −B0
2
Finding the energy eigenvalues simply means finding the eigenvalues of the above 2 × 2
matrix. This is a very simple and straightforward procedure, but I’ll instead opt for a
more physically intuitive approach.
Note that the magnetic field is just a constant vector; in particular, since its x and zcomponents are B and B0 , respectively, its magnitude is
q
|B| = B02 + B 2
This field is pointing in a direction given by some unit normal n̂ (which, if we wanted,
we could express in terms of the unit vectors
p î and k̂ and B0 and B, but nevermind - we
don’t need to). Thus we can write B = B02 + B 2 n̂, so the Hamiltonian becomes
q
H = −γ B02 + B 2 Sn
where Sn is the component of the spin along the axis n̂. Now, recall from last quarter
(or whenever you learned about spin) that we can diagolize any component of the spin
that we want; in particular, we can diagonalize the component of the spin along any
axis whatsoever, not just the x, y, z axes. By convention, we usually choose to diagonalize along the z axis (i.e. we diagonalize Sz ), but in this problem it seems natural
to diagoanlize the component of the spin along the n̂ axis (i.e. Sn ). Recall that for
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spin-1/2, the eigenvalues of the components of the spin are ±~/2, and therefore Sn has
eigenvalues ±~/2. Since H is proportional to Sn , we thus find that the eigenvalues of H
are
q
~
2
2
E = −γ B0 + B ±
2
q
1
E± = ± γ~ B02 + B 2
2
(b) Let’s write the Hamiltonian as
H = H0 + H ′
where H0 = −γB0 Sz is the unperturbed Hamiltonian and H ′ = −γBSx is the perturbation. Since the eigenvalues of Sz are ±~/2, the eigenvalues of the unperturbed
Hamiltonian are
1
(0)
E± = ± ~γB0
2
E
E
(z)
(z)
and the unperturbed spin states are the eigenstates of Sz , χ∓ (with χ− corre(0)
sponding to the energy E+ and vice versa). The first-order shifts in the energy are
thus
D
E
(z) (1)
′ (z)
E± = χ∓ H χ∓
E
D
(z)
(z) = −γB χ∓ Sx χ∓
This expression can be evaluated by either expressing Sx in terms of the ladder operators
via Sx = E(S+ + S− )/2, or by using the matrix and column vector representations of Sx
(z)
and χ∓ . I’ll do the latter: we have
(z)
χ−
and thus
(1)
E+
(1)
E−
~ 0 1
Sx =
2 1 0
1
0
(z)
χ+ =
=
0
1
0
1
= − ~γB 0 1
1
2
0
1
= − ~γB 1 0
1
2
0
=0
1
1
1
=0
0
0
1
0
Evidently, the first-order corrections are zero, and we need to move on to second-order
perturbation theory to get the first non-vanishing corrections. The second-order energy
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shifts are given by
En(2) =
D
E
(0) ′ (0) 2
X ψ m H ψ n (0)
(0)
En − Em
m6=n
In our case, the Hilbert space is two-dimensional,so nEandm only
E take on the values ±.
(z)
(0)
Using the fact that our unperturbed states were ψn = χ± , we get
(2)
E±
=
D
E
(z) ′ (z) 2
χ
H
χ
X m ± (0)
(0)
E± − Em
D
E
(z) ′ (z) 2
χ∓ H χ± m6=±
=
(0)
(0)
E± − E∓
D
E
(z) (z) 2
χ ∓ Sx χ ± = γ 2B2
(0)
(0)
E± − E∓
From our expressions for the unperturbed energies, we have
(0)
(0)
E± − E∓ = ±~γB0
and thus
γB 2 D (z) (z) E2
χ ∓ Sx χ ± ~B0
Finally, the matrix elements we need are
E
D
0 1
1
1
(z)
(z) = − ~ 0 1
χ − Sx χ +
0
1
0
2
1
= − ~
2
E
D
0 1
1
0
(z)
(z) = − ~ 1 0
χ + Sx χ −
1
1 0
2
1
= − ~
2
(2)
E± = ±
Using these expressions, our second-order energy shifts becomes
2
γB 2 1
(2)
~
E± = ±
~B0 2
~γB 2
= ±
4B0
6
(2)
E±
1
= ± ~γB0
4
B
B0
2
The energy levels up to second order in the perturbation B are therefore
(0)
(2)
(1)
E± = E± + E± + E± + · · ·
"
3 #
2
1
1
B
B
= ± ~γB0 + ~γB0
+O
2
4
B0
B0
"
2
3 #
1
B
1 B
E± = ± ~γB0 1 +
+O
2
2 B0
B0
(c) Let’s go back to our exact energy expressions from part (a) and rewrite them in terms
of the small parameter B/B0 :
s
2
1
B
E± = ± γ~B0 1 +
2
B0
If we Taylor-expand the square root, we get
"
4 #
2
1 B
B
1
+O
E± = ± γ~B0 1 +
2
2 B0
B0
which agrees precisely with the expression we obtained in part (b) above using perturbation theory.
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Problem 3
(a) This is a standard freshman E&M problem. Because the charged sphere is spherically
symmetric, we expect the electric field to only point in the radial direction. If we draw a
spherical Gaussian surface of radius r concentric with the charged sphere and containing
it, then the total enclosed charge is e (remember that our “charged sphere” is the nucleus
of a hydrogen atom, which just has charge e). Then by Gauss’ law, the total flux through
the surface must be proportional to the enclosed charge:
I
e
E · dA =
ǫ0
But the electric field at the surface is uniform and normal to the surface, so the flux
through the surface is just
I
E · dA = EA = 4πr2 E
where A = 4πr2 is the area of the surface. Thus outside the sphere, we have
E=
e 1
4πǫ0 r2
(r > R)
If we instead consider a spherical Gaussian surface smaller than the sphere (i.e. of
radius r < R), then the enclosed charge will be the volume enclosed by the surface times
the charge density ρ of the sphere:
4
Qenclosed = πr3 ρ
3
But the charge density is just the total charge of the sphere divided by its volume,
so ρ = e/(4/3)πR3 , and thus
r 3
4 3
3e
Qenclosed = πr
=e
3
4πR3
R
Thus inside the sphere, Gauss’ law gives
I
Qenclosed
E · dA =
ǫ0
r 3
e
4πr2 E =
ǫ0 R
e r
E=
(r < R)
4πǫ0 R3
In short,
(
1
r≥R
e
r2
E(r) =
r
4πǫ0 R3 r < R
8
To find the potential, we need to take the line integral of the electric field:
Z r
V (r) − V (r0 ) =
E · dr
r0
where r0 is some arbitrary reference position. It is conventional to take the potential to
be zero at infinity, so let’s choose r0 = ∞ and V (r0 ) = 0. Since the electric field points
radially outward, if we integrate radially inward from infinity, we have that E · dr =
−E dr, and thus
Z
r
V (r) = −
E(r) dr
∞
Plugging in our expression for E(r), we get
(R r
1
dr
e
r2
R∞
Rr
V (r) = −
R 1
4πǫ0
dr
+
2
∞ r
R
r
R3
dr
r≥R
r<R
Evaluating the integrals, we get
e
V (r) = −
4πǫ0
(
− 1r
− R1 +
e 1
V (r) =
4πǫ0 R
(
r2
1
2 R3
r≥R
r<R
− 12 R1
R
r
2
3
− 12 Rr 2
2
r≥R
r<R
Note that in the limit R → 0, we recover the usual potential for a point charge, as we
should.
(b) Our first step is to find the perturbing Hamiltonian that corresponds to our correction of
treating the nuclear as an extended object. Recall that the unperturbed (point-charge
nucleus) Hamiltonian was
p2
H0 =
− eVpoint (r)
2m
where Vpoint (r) is the Coulomb potential of a point charge. Our Hamiltonian now is
p2
− eVsphere (r)
2m
where Vsphere (r) is the potential of our uniformly charged sphere. But the perturbing
Hamiltonian is simply the difference between the unperturbed and exact Hamiltonians: H ′ = H − H0 = −e(Vsphere − Vpoint ). Using the expression for Vsphere that we found
above, we have
! (
r≥R
e 1 Rr
e 1
Vsphere − Vpoint =
−
2
4πǫ0 R 23 − 12 Rr 2 r < R
4πǫ0 r
(
r≥R
e 1 0
=
2
1
r
R
3
4πǫ0 R 2 − 2 R2 − r r < R
H=
9
and thus the perturbing Hamiltonian is
(
0
α
H′ =
2
R 21 Rr 2 − 23 +
R
r
r≥R
r<R
where α ≡ e2 /4πǫ0 . Now, finding the first-order energy shift of the ground state is
straightforward: by the usual perturbation theory, we have
(1)
E0 = h100| H ′ |100i
where |100i is the usual (unperturbed) ground state of the Hydrogen atom. Using the
functional form
1 −r/a
ψ100 (r) = √
e
πa3
we have
Z
(1)
E0 =
H ′ (r)|ψ100 (r)|2 d3 r
Z ∞
= 4π
H ′ (r)|ψ100 (r)|2 r2 dr
0
Z 3 R
α R 1 r2
1 −2r/a 2
− +
e
r dr
= 4π
2
R 0
2R
2
r πa3
Z R 4
4α
3 2
1r
= 3
− r + Rr e−2r/a dr
2
aR 0
2R
2
2 Z R 4
3 r2
r
dr
1r
4α R
−
+
e−2r/a
=
4
2
a
a
2R
2R
R
R
0
Now, actually computing this integral isn’t too bad, but here’s where our approximation
comes in: the reason H ′ is only a perturbation is because the radius of the nucleus R is
very small; in particular, R/a ≪ 1. Thus in this case, lowest order in the “perturbation”
means lowest order in the parameter R/a. Because r < R everywhere in the region of
integration, inside the integral r/a is also very small; this prompts us to expand the
exponential:
2r
e−2r/a = 1 −
+ ···
a
(1)
But notice that the entire expression for E0 is already of order (R/a)2 (because of the
overall prefactor in front of the integral), so in fact, to lowest order in R/a, we only need
the zeroth order term in the expansion of the exponential; i.e. we can take
e−2r/a ≈ 1
10
Changing variables to x = r/R, the integral is now simply
(1)
E0
2 Z 1 R
1 4 3 2
x − x + x dx
a
2
2
0
2 1
4α R
1 1
≈
− +
a
a
10 2 2
2
2α R
≈
5a a
4α
≈
a
Explicitly keeping track of the powers of R/a we neglected, and writing the (unper(0)
turbed) hydrogen ground-state energy as E0 = −α/2a, we get
(1)
E0
4 (0)
= − E0
5
3
2
R
R
+O
a
a
(c) Since the Bohr radius is a ≈ 5.3 × 10−11 m, we have R/a ∼ 2 × 10−5 , and therefore the
lowest-order correction to the energy due is
4
(1)
E0 ∼ (13 eV)(2 × 10−5 )2 ∼ 4 × 10−9 eV
5
This correction is really tiny: the hyperfine structure of hydrogen (which comes from
interactio between the spins of the electron and proton) gives a correction on the order
of 10−8 eV, so our correction is one order of magnitude smaller than that.
11
Problem 4
(a) The Hamiltonian of an electron in a uniform electric field E is
H ′ = er · E
In our case, E = E0 k̂, so the perturbing Hamiltonian is
H ′ = eE0 z
Since we’re interested in evaluating the matrix elements hnℓm| H ′ |n′ ℓ′ m′ i, it will be
useful to convert the perturbing Hamiltonian to spherical coordinates. To that end, let’s
write z = r cos θ, so that
H ′ = eEo r cos θ
Now, we’re interested in calculating the matrix elements hnℓm| H ′ |n′ ℓ′ m′ i for the n = 2
states; thus all we’re really interested in are the elements h2ℓm| H ′ |2ℓ′ m′ i. From now on,
I’ll drop the n = 2 label in the kets, and denote the matrix elements as hℓm| H ′ |ℓ′ m′ i.
There are four states to consider: |00i, |11i, |10i, and |1 − 1i. These four states form a
four-dimensional degenerate subspace, so we have sixteen matrix elements to compute
(i.e. the matrix representation of H ′ in this basis is a 4 × 4 matrix). Since H ′ must
be Hermitian, we really only need to calculate ten matrix elements: the four diagonal
ones hnℓ| H ′ hnℓ| and six off-diagonal elements; the other elements are related to these
by complex conjugation: hnℓ| H ′ |n′ ℓ′ i = hn′ ℓ′ | H ′ |nℓi∗ . To begin, let’s write down the
position-space representation of the states we’re interested in (which I got from Griffiths’
Tables 4.3 and 4.7):
ψ211
ψ210
ψ21−1
1
r −r/2a
1−
e
2a
8πa3
1 r −r/2a
e
sin θeiφ
= R21 Y11 = − √
3
a
8 πa
1
r −r/2a
= R21 Y10 = √
e
cos θ
3
32πa a
1 r −r/2a
= R21 Y1−1 = √
e
sin θe−iφ
3
8 πa a
ψ200 = R20 Y00 = √
Let’s begin calculating matrix elements. A ton of the matrix elements are zero; I’ll
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highlight these by extracting the particular angular integral that integrates to zero:
Z
′
h00| H |00i = eE0 |ψ200 |2 r cos θr2 sin θ dr dθ dφ
Z π
cos θ sin θ dθ = 0
∝
0
Z
′
∗
h00| H |11i = eE0 ψ200
ψ211 r cos θr2 sin θ dr dθ dφ
Z 2π
eiφ dφ = 0
∝
0
Z
′
∗
h00| H |10i = eE0 ψ200
ψ210 r cos θr2 sin θ dr dθ dφ
Z ∞
Z 2π
Z π
1
r −r/a
1
3r
2
√
r
dφ
cos θ sin θ dθ
1−
e
dr
= eE0 √
a
2a
8πa3 32πa3 0
0
0
Z π
Z
eE0 a ∞ r 4 r −r/a dr
2
=
cos θ d(cos θ)
−
1−
e
8
a
2a
a
0
0
π
Z
1
1
eE0 a ∞ 4
−x
3
x 1 − x e dx − cos θ
=
8
2
3
0
0
Z ∞ eE0 a
1
=
x4 1 − x e−x dx
12 0
2
eE0 a
1
=
4! − 5!
12
2
= −3eE
Z 0a
∗
h00| H ′ |1 − 1i = eE0 ψ200
ψ21−1 r cos θr2 sin θ dr dθ dφ
Z 2π
e−iφ dφ = 0
∝
0 Z
h11| H ′ |11i = eE0 |ψ211 |2 r cos θr2 sin θ dr dθ dφ
Z π
∝
sin3 θ cos θ dθ = 0
0
Z
′
∗
h11| H |10i = eE0 ψ211
ψ210 r cos θr2 sin θ dr dθ dφ
Z 2π
∝
e−iφ dφ = 0
0
Z
′
∗
h11| H |1 − 1i = eE0 ψ211
ψ21−1 r cos θr2 sin θ dr dθ dφ
Z 2π
∝
e−2iφ dφ = 0
0
13
Z
h10| H |10i = eE0 |ψ210 |2 r cos θr2 sin θ dr dθ dφ
Z π
cos3 θ sin θ dθ = 0
∝
0
Z
′
∗
h10| H |1 − 1i = eE0 ψ210
ψ21−1 r cos θr2 sin θ dr dθ dφ
Z 2π
e−iφ dφ = 0
∝
0 Z
h1 − 1| H ′ |1 − 1i = eE0 |ψ21−1 |2 r cos θr2 sin θ dr dθ dφ
Z π
∝
sin3 θ cos θ dθ = 0
′
0
Whew! Those are all the independent matrix elements; the rest follow from complexconjugating the off-diagonal ones. In the basis {|00i , |11i , |10i , |1 − 1i}, the matrix
for H ′ is then


0 0 1 0
 0 0 0 0

hℓm| H ′ |ℓ′ m′ i = −3eE0 a 
 1 0 0 0
0 0 0 0
(b) Recall that in degenerate perturbation theory, the first-order energy shifts are simply
the eigenvalues of the perturbing Hamiltonian H ′ ; that is, we need to diagonalize H ′
(the “good” states to use in perturbation theory are then the eigenstates of H ′ ). Our
first-order energy shifts are therefore simply the eigenvalues of the matrix hℓm| H ′ |ℓ′ m′ i
that we found above. We find these as usual: defining β ≡ 3eE0 a, the characteristic
equation is
0 = det(H − λI)
−λ 0 −β 0 0 −λ 0
0 = −β 0 −λ 0 0
0
0 −λ
= λ2 (λ2 − β 2 )
= λ2 (λ + β)(λ − β)
Thus the eigenvalues of H ′ (in our degenerate basis) are λ = 0 and λ = ±3E0 ae (with
zero a double eigenvalue). These eigenvalues are just the first-order energy shifts, so the
shifts are

3E0 ae

(1)
En=2 = 0


−3E0 ae
14
Thus the applied electric field has broken the four-fold degeneracy into three distinct
energies, leaving just a two-fold degeneracy (at this order, anyway).
(c) Remember that our goal was to find the states that diagonalize the perturbation H ′ ;
these states are the “good” states that we use in perturbation theory. The “good” states
can therefore be represented by the eigenvectors of our matrix hℓm| H ′ |ℓ′ m′ i. From our
expression for this matrix above, we see right away that two of the eigenvectors are
 
 
0
0
0
1

 
|1i = 
0 |2i = 0
1
0
Both of these eigenvectors correspond to the (double) eigenvalue λ = 0 that we found.
The other two eigenvectors are slightly more complicated: because of the mixing between
the |00i and |10i states (as indicated by the fact that the matrix elements h00| H ′ |10i
and h10| H ′ |00i are nonzero), the other two eigenvectors will be some nontrivial linear
combination of |00i and |10i; their column vector representation will therefore be of the
form
 
a
0
 
b
0
We could find the coefficients a and b by brute force computation, but the form of
our matrix for H ′ is so simple that we can simply read off what these remaining two
eigenvectors must be: they must be proportional to the vectors
   
1
1
0  0 
   
1 −1
0
0
√
with a proportionality constant of 1/ 2 for normalization. Thus our remaining two
eigenvectors are
 
 
1
1



1 0
1
0

√
|4i
=
|3i = √ 
2 1
2 −1
0
0
The state |3i corresponds to the eigenvalue −3E0 ae, and the state |4i corresponds to
the eigenvalue +3E0 ae. Remembering that these column vectors are representations in
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the basis {|00i , |11i , |10i , |1 − 1i}, we can write the “good” states as
|1i = |11i |2i = |1 − 1i
(E (1) = 0)
1
|3i = √ (|00i + |10i)
(E (1) = −3E0 ae)
2
1
|4i = √ (|00i − |10i)
(E (1) = +3E0 ae)
2
In particular, we notice that the applied electric field breaks the degeneracy between the
states with quantum number m = 0, but doesn’t affect the states with m = ±1.
16