Download Physics 505 Homework No. 9 Solutions S9

Physics 505
Homework No. 9
Solutions
S9-1
1. As promised, here is the trick for summing the matrix elements for the Stark effect for
the ground state of the hydrogen atom.
Recall, we need to calculate the correction to the ground state energy to second order
in the perturbation due to an external field. This correction is
∆E1 = e2 E 2
∞
X
| hm, 1, 0 | z | 1, 0, 0i |2
.
E
−
E
1
m
m=2
To simplify the notation, lets call |1, 0, 0i = |0i, the ground state with energy E0 and
call |m, 1, 0i = |ni with energy En and n ≥ 1. So, we want to compute
∞
X
| hn | z | 0i |2
∆E0 = e E
.
E
−
E
0
n
n=1
2
2
(a) Suppose we had an operator A such that
z |0i = (AH0 − H0 A) |0i ,
where H0 is the unperturbed Hamiltonian for the hydrogen atom,
e2
p2
−
,
2m
r
where m is the reduced mass of the electron and proton. Show that
H0 =
hn | z | 0i = (E0 − En ) hn | A | 0i .
Also show that
∞
X
| hn | z | 0i |2
= h0 | zA | 0i − h0 | z | 0i h0 | A | 0i = h0 | zA | 0i .
E0 − En
n=1
Solution
hn | z | 0i = hn | AH0 | 0i − hn | H0 A | 0i = (E0 − En ) hn | A | 0i ,
Since H0 is Hermitian. Also,
∞
X h0 | z | ni (E0 − En ) hn | A | 0i
X
| hn | z | 0i |2
=
E0 − En
E0 − En
n=1
n6=1
X
=
h0 | z | ni hn | A | 0i
n6=1
=
X
n
h0 | z | ni hn | A | 0i − h0 | z | 0i h0 | A | 0i
= h0 | zA | 0i − h0 | z | 0i h0 | A | 0i
= h0 | zA | 0i ,
c 2012, Edward J. Groth
Copyright (1)
Physics 505
Homework No. 9
Solutions
S9-2
where the second term has been dropped, since h0 | z | 0i = 0.
End Solution
(b) So, if we knew A, we could get the answer just by calculating one matrix element. If
we assume A is a function only of coordinates, then equation (1) is an inhomogeneous
differential equation for A. If you’re really good at differential equations, you could
solve it. The result is
ma r
A=− 2
+a z.
2
h̄
Show that this expression does, in fact, solve equation (1). (Note that the normalization of |0i cancels out, so you can just take |0i = exp(−r/a).)
Solution
We need to calculate
2
h̄2
h̄2
e2
∂
2 ∂
∂
1 ∂
− −
−
−
+
sin θ
2m ∂r 2
r ∂r
2mr 2 sin θ ∂θ
∂θ
r
2
ma r
−r/a
.
+ ar cos θ e
− 2
2
h̄
This is the second term on the right hand side of equation (1). There may be some ways
to simplify this, but in the end, it appears brute force is required, so, we’ll just evaluate
each term. First of all, the second derivative term is
2
a ∂2
a ∂
1 r2
r
−r/a
−r/a
− cos θ
= − cos θ
(r + a) −
+ ar e
+ ar
e
2 ∂r 2
2
2 ∂r
a 2
2
1 r2
a
1 − (r + a) + 2
+ ar
e−r/a
= − cos θ
2
a
a
2
r2
a r
e−r/a .
+ −
= cos θ
2 2 4a
The first derivative term is
2
a ∂
r
a
1 r2
−r/a
− cos θ
= − cos θ
(r + a) −
+ ar e
+ ar
e−r/a
r ∂r
2
r
a 2
r a2
e−r/a .
−
= cos θ
2
r
The angular derivative term is
2
2
a
r
1 ∂
∂
a
r
−r/a
− 2
cos θ
sin θ
+ ar e
= 2 cos θ
+ ar e−r/a
2r
sin θ ∂θ
∂θ
2
r
2
a a2
e−r/a .
+
= cos θ
2
r
c 2012, Edward J. Groth
Copyright Physics 505
Homework No. 9
Solutions
S9-3
The potential term is
e2
−
r
ma
h̄2
r
r2
−r/a
= cos θ − − a e−r/a .
+ ar cos θ e
2
2
The first term on the right of equation (1) is considerably easier to evaluate,
2
e
AH0 |0i = A −
e−r/a
2a
2 ma r 2
e
−r/a
− 2
+ ar cos θ e
= −
2a
2
h̄
2
r
r
e−r/a .
+
= cos θ
4a 2
We now add up the results of the last 5 calculations to find
a r
r2
r
a2
a a2
r
r2
r
(AH0 − H0 A) |0i = cos θ
+ −
+ −
+ +
− −a+
+
2 2 4a 2
r
2
r
2
4a 2
= cos θ re−r/a
e−r/a
= z |0i ,
which is what was to be shown!
End Solution
(c) Calculate the Stark effect energy shift for the ground state of hydrogen to second
order in the applied field.
Solution
We need to evaluate h0 | zA | 0i. Here, the normalization of |0i matters, so we use
|0i = 2(a)−3/2 (4π)−1/2 exp(−r/a). Also, zA is proportional to z 2 . As far as evaluating
the matrix element, we can use symmetry to replace z 2 by r 2 /3. So,
2
e h0 | zA | 0i = −
9
= − a3 .
4
Z
0
∞
r2 4
+a
e−2r/a r 2 dr
3
2
3 a
r
Finally, we need to include E 2 in the energy shift, so
9
∆E1 = − a3 E 2 .
4
End Solution
c 2012, Edward J. Groth
Copyright Physics 505
Homework No. 9
Solutions
S9-4
2. A particle in a 2D box. (Based on a problem from Merzbacher.) A particle is confined
to a square box, 0 ≤ x ≤ L and 0 ≤ y ≤ L. We are not interested in the z-motion, so this
is a 2D problem.
(a) Obtain the energies and eigenfunctions. What is the degeneracy of the few lowest
levels?
Solution
The wave function must vanish at the boundaries of the box. This means
|nmi =
2
sin(nπx/L) sin(mπy/L) ,
L
where n and m are integers greater than 0. The energy is
Enm
h̄2 π 2
2
2
=
n
+
m
.
2mL2
The degeneracy has to do with how many ways one can choose n and m to give the same
n2 + m2 . The ground state is non-degenerate with n = m = 1. The first excited state
has a degeneracy of 2 with n = 2, m = 1 or n = 1, m = 2. The next excited state
is non-degenerate with n = m = 2. The third excited state is doubly degenerate with
n = 3, m = 1 or n = 1, m = 3. That’s enough!
End Solution
(b) A small perturbation V = Cxy, where C is a constant, is applied. Find the energy
change for the ground state and the first excited state in the lowest non-vanishing
order. Construct the appropriate eigenfunctions in the case of the first excited state.
Solution
We calculate
Z L
Z
2
n1 πx
n2 πx
2L π
x sin
sin
dx = 2
sin n1 x sin n2 x x dx
L
L
L
π 0
0
Z
L π
(cos((n1 − n2 )x) − cos((n1 + n2 )x)) x dx
= 2
π 0
The integrals can be evaluated with an integration by parts. If n1 6= n2 , the result is
Z
0
L
n1 πx
n2 πx
L
2
sin
dx = 2
x sin
L
L
L
π
(−1)n1 −n2 − 1 (−1)n1 +n2 − 1
−
(n1 − n2 )2
(n1 + n2 )2
.
If n1 and n2 are both odd or both even, the result is 0. If one is odd and the other is even,
the result is
Z L
2
n1 πx
n2 πx
8L n1 n2
x sin
sin
dx = − 2 2
.
L
L
L
π (n1 − n22 )2
0
c 2012, Edward J. Groth
Copyright Physics 505
Homework No. 9
Solutions
S9-5
If n1 = n2 , the result is L/2.
So, the ground state changes energy by
∆E11 = h11 | Cxy | 11i = CL2 /4 ,
since the expectation value is the product of two of the integrals just discussed with
n1 = n2 = m1 = m2 = 1.
The first excited state is degenerate, so we need to choose a basis which diagonalizes
the perturbation. We calculate the matrix elements for all the states:
1/4
256/81π 4
h21 | V | 21i h21 | V | 12i
2
.
= CL
256/81π 4
1/4
h12 | V | 21i h12 | V | 12i
The eigenvalues of this matrix are
∆Efirst excited = CL
with eigenvectors
2
256
1
±
4 81π 4
,
1
|first excited± i = √ (|21i ± |12i) .
2
End Solution
3. Hyperfine splitting of the hydrogen ground state. As you know, the spatial part of
the hydrogen ground state is very simple: ψ100 (r, θ, φ) = exp(−r/a) a(−3/2) π (−1/2) . Since
there is no orbital angular momentum, there is no spin orbit effect. However, the ground
state has a degeneracy of 4 since both the proton and the electron have spin 1/2. The
spins can align, giving a triplet state, or anti-align, giving a singlet state. Since there are
magnets associated with the spins, we expect that there should be a difference in energy
between the triplet and singlet states. The nuclear spin is often denoted by I and produces
a magnetic moment
egp
I,
µp =
2mp c
where µp is the magnetic moment of the proton, gp is its g-factor and mp is the mass of
the proton. (Note: to consider other nuclei, we would use the appropriate Z, g, and m.)
We take the proton as fixed at the origin and it produces a magnetic field,
B(r) =
3er (er · µp ) − µp
8π
+
µp δ(r) ,
r3
3
(Jackson, Classical Electrodynamics, 2nd ed., p. 184). The interaction energy of this field
and the magnetic moment of the electron is
HHF = −µe · B = −
3(er · µe )(er · µp ) − µe · µp
8π
−
(µe · µp )δ(r) .
3
r
3
c 2012, Edward J. Groth
Copyright Physics 505
Homework No. 9
Solutions
S9-6
Aside: If we were to consider other than s-states, the hyperfine Hamiltonian would also
include a spin orbit term due to the interaction of the magnetic moment of the nucleus
with magnetic field produced by the moving electron(s).
Evaluate the hyperfine Hamiltonian above for the ground state of hydrogen. How does
it depend on the proton and electron spins? Or, what is the energy difference between
the singlet and triplet states? Which is the actual ground state: triplet or singlet? What
are the wavelength and frequency of the radiation emitted or absorbed in the transition
between these states? Hint: can you show that the first term in HHF vanishes for s-states?
Solution
The first term in the hyperfine Hamiltonian is zero by the following argument. The
numerator times r 2 is 3xi µei xj µpj − r 2 µe · µp . When we take the expectation value with
an s-state, the angular integral will eliminate cross terms xi xj with i 6= j. The squared
terms averaged over angles become r 2 /3. This leaves 3µe · µp r 2 /3 − r 2 µe · µp = 0.
The δ-function in the second term is a three dimensional delta function and it just
picks out the value of the integrand at the origin. So,
h100 | −8πµe · µp /3 | 100i = −8µe · µp /3a3 .
The product of the magnetic moments is
−ege egp
S ·I
2me c 2mp c
ge gp
1
2
2
2
= − 2 µB µN
,
F −S −I
2
h̄
µe · µp =
where µB and µN are the Bohr and nuclear magnetons, and F is the total spin. S 2 = I 2 =
3h̄2 /4. For the singlet state F = 0, F 2 = 0 and the quantity in brackets is −3h̄2 /4. For a
triplet state, F = 1, F 2 = 2h̄2 . and the quantity in brackets is +h̄2 /4.
Putting everything together, we wind up with
2ge gp µB µN
3a3
2ge gp µB µN
hHHF s i = −
a3
8ge gp µB µN
hHHF t i − hHHF s i = +
.
3a3
hHHF t i = +
We see that the singlet is the ground state.
c 2012, Edward J. Groth
Copyright Physics 505
µN
Homework No. 9
Solutions
S9-7
For the numerical evaluation, we use ge = 2, gp = 5.59, µB = 0.927 × 10−20 erg G−1 ,
= 0.505 × 10−23 erg G−1 , a = 0.529 × 10−8 cm. We find,
∆HHF = 9.43 × 10−18 erg = 5.89 × 10−6 eV
νHF = 1.42 × 109 Hz
λHF = 21 cm .
This transition is the famous 21 cm line of neutral Hydrogen which is seen all over the sky.
It’s one of the principal ways to study our galaxy and other galaxies with radio telescopes!
End Solution
4. Zeeman splitting. We consider an atom with a single valence electron, subject to a
magnetic field B = Bez in the z-direction. The Hamiltonian for the electron is
H = H0 + Hso + HB ,
where
P2
+ V (r) ,
2m
accounts for the dominant electric interaction of the electron (for Hydrogen, V (r) = e2 /r,
for alkali metals, V (r) takes account of the filled shells in an approximate way). The spin
orbit interaction is
1
1 dV
Hso =
L · S = f (r) L · S .
2
2
2m c r dr
The interaction with the applied magnetic field is
H0 =
HB =
eB
eB
eB
(Lz + gSz ) =
(Lz + 2Sz ) =
(Jz + Sz ) ,
2mc
2mc
2mc
where the term proportional to B 2 has been dropped. Also, some other small terms, for
example, the relativistic correction to the momentum, have been dropped since they don’t
give a splitting dependent on j, l and s. In the calculations below, we are interested in Hso
and HB ; H0 determines the zeroth order energies and states which are used in computing
expectation values of, for example, f (r), but can otherwise be ignored.
(a) Suppose the magnetic field is very weak. What are the appropriate basis states and
what are the spin-orbit and Zeeman splittings?
Solution
If the magnetic field is completely turned off, there is only the spin-orbit interaction.
The appropriate states are those of total angular momentum, z-component of angular
momentum, and orbital and spin angular momentum: |njmj lsi. These states diagonalize
the spin-orbit interaction which is assumed to be larger than the Zeeman interaction.
h̄2
l
j = l + 1/2
hHso i = hf (r)inl
×
−(l + 1)
j = l − 1/2
2
c 2012, Edward J. Groth
Copyright Physics 505
Homework No. 9
Solutions
S9-8
Using these basis states we need to calculate the matrix elements of the Zeeman term,
eB
(Jz + Sz ) njmj ls .
hHB i = njmj ls 2mc
The matrix element of Jz is just mj h̄. The matrix element of Sz requires more work.
In particular, the states |njmj lsi must be written in terms of |nlml i |sms i. Recall the
expansions given in lecture,
r
l + mj + 1/2
l − mj + 1/2
|lmj − 1/2i |↑i +
|lmj + 1/2i |↓i
|j = l + 1/2, mj , li = +
2l + 1
2l + 1
r
r
l − mj + 1/2
l + mj + 1/2
|j = l − 1/2, mj , li = −
|lmj − 1/2i |↑i +
|lmj + 1/2i |↓i .
2l + 1
2l + 1
r
With these expansions given in lecture, we find
h̄mj
hJz + Sz i =
×
2l + 1
2l + 2
2l
j = l + 1/2
.
j = l − 1/2
Both cases are covered by
hJz + Sz i = h̄mj
2j + 1
= h̄mj g ,
2l + 1
where the g-factor varies from 2 for l = 0 to 1 for l → ∞. Then the Zeeman term is
hHB i = gmj µB B .
Each spin-orbit level is split into 2j + 1 equi-spaced levels by the Zeeman effect.
End Solution
(b) Now suppose the magnetic field is very strong so the Zeeman term is larger than
the spin-orbit term. What are the appropriate states and what are the Zeeman and
spin-orbit splittings?
Solution
The Zeeman term is diagonal in the basis |lml i |ms i and
hHB i =
eB
eB
hLz + 2Sz i =
h̄(ml + 2ms ) = (ml + 2ms )µB B .
2mc
2mc
In this basis, the spin-orbit term is easy to evaluate,
hL · Si = hLx Sx + Ly Sy + Lz Sz i = hLz Sz i = h̄2 ml ms .
c 2012, Edward J. Groth
Copyright Physics 505
Homework No. 9
Then
Solutions
S9-9
hHso i = ml ms h̄2 hf (r)inl .
Note that in this limit there are some degeneracies. For example, if l = 1, then states
ml = −1, ms = +1/2 and ml = +1, ms = −1/2 are degenerate.
End Solution
(c) Suppose that neither the spin-orbit nor the Zeeman effect is appreciably larger than
the other. How would you determine the level splittings in this case? (This is a short
essay question, no calculations are required!)
Solution
Pick a convenient basis for the levels that are degenerate in the absence of the spinorbit and Zeeman interactions. Calculate the matrix elements in this basis. The eigenvalues
of this matrix are the energy shifts in the states that correspond to the eigenvalues.
End Solution
5. Virial theorem for a particle in a fixed potential. (See Schwabl, chapter 12.) Consider
x · p and a Hamiltonian H = p2 /2m + V (x).
(a) Show that
[H, x · p] = −ih̄
p2
− x · ∇V (x) .
m
Solution
2
We consider p first:
[p2 , x · p] = pj pj xi pi − xi pi p2
= pj (xi pj − ih̄δij )pi − xi pi p2
= pjxi pj pi − ih̄p2 − xi pi p2
= (xi pj − ih̄δij )pj pi − h̄p2 − xi pi p2
= xi pi p2 − 2ih̄p2 − xi pi p2
= −2ih̄p2 .
Now V (x):
[V (x), xipi ] = V (x)xi pi − xi pi V (x)
h̄ ∂
V (x) + V (x)pi
= V (x)xi pi − xi
i ∂xi
= +ih̄x · ∇V (x) .
So,
[H, x · p] = −ih̄
p2
− x · ∇V (x)
m
c 2012, Edward J. Groth
Copyright .
Physics 505
Homework No. 9
Solutions
End Solution
(b) If |ψi is a stationary state of H, H |ψi = E |ψi, show that
2 p ψ − hψ | x · ∇V (x) | ψi = 0 ,
ψ m and therefore, for the Coulomb potential,
2 hψ | H | ψi + ψ
Ze2 r ψ =0.
Solution
hψ | [H, x · p] | ψi = hψ | Hx · p − x · pH | ψi
= hψ | Ex · p − x · pE | ψi
= E hψ | x · p − x · p | ψi
=0.
With the Coulomb potential, we have
Ze2
Ze2 x
Ze2
x·∇ −
=x· + 3
=+
.
r
r
r
So,
p2 ψ − hψ | x · ∇V (x) | ψi
m Ze2 p2 ψ
ψ − ψ m r 2 2 p ψ − 2 ψ Ze ψ + ψ
2m r Ze2 ψ .
= 2 hψ | H | ψi + ψ r 0 = ψ = ψ =2 ψ
End Solution
(c) Determine h1/rinl for the hydrogen atom.
Solution
From the above, we have
1
En = −
2
and we know, En = −Z 2 e2 /2an2 , so
Ze2
r
,
1
Z
=
.
r nl
an2
End Solution
c 2012, Edward J. Groth
Copyright Ze2 r ψ
S9-10