1
PHY4605
Problem Set 4
Solutions
1. Finite extent of proton.
Treat proton as ball of radius r0 change density ρ0 = e/ 34 πr03 . Outside the ball the
potential energy of electron is clearly −e2 /r just as for point proton. But inside r0 , the
potentials differ! Treat small difference as perturbation:
Gauss’s Law:

ρ0 43 πr3


1
r2
E=
e
4π²0 
r < r0
r2
Z
r
0
0
dr E(r ) =
φ=−
∞
U = −eφ =
µ
Perturbation V̂ = U −
2
e
−
r
¶
⇒
(1)
r > r0

¡

−ρ0 4 π r2 −
3
2

e
r

 3e2
1 − 2r0 +
r02 ¢
2
+
e
r0
r < r0
⇒
(2)
r > r0
e2 r 2
2r03
4π²0 
− e2
r < r0
⇒
(3)
r > r0
r
2 2
er
e2 3e2
=
+ −
2r03
r
2r0
Evaluate first-order energy shift in ground state:
·
¸
1 e2 r2 e2 3e2
δE = h100|
+ −
4π²0 2r03
r
2r0
for r < r0 ,
θ(r − r)
| 0{z }
0 otherwise
|100i
(4)
(5)
i.e. only integrate r < r0
where
1
|100i = p
πa30
e−r/a0
We need hr2 i0 , h 1r i0 , and of course h1i0 e.g.
Z
Z r0 3
d r −2r/a0 2 4πa20 r0 /a0
2 0
e
r =
dxx4 e−2x
hr i =
3
πa
π
0
0
0
we used x = r/a0 and since r0 /a0 ¿ 1
¯
Z x0
5 ¯x0
4r5
4
2x ¯
2
dxx (1 − 2x + · · · ) ' 4a0 ¯ = 03
' 4a0
5 0
5a0
0
(6)
(7)
(8)
2
1
h i0 =
r
Z
Z x0
Z
d3 r −2r/a0 1
4π
4 x0
2r02
−2x
=
e
dxxe
'
dxx(1
−
2x
+
·
·
·
)
=
πa30
r
πa0 0
a0 0
a30
Z r0 3
Z
d r −2r/a0
4π x0
4r03
0
2
h1i =
e
'
dxx
(1
−
2x
+
·
·
·
)
'
πa30
π 0
3a30
0
r0
(9)
(10)
So
2
e2 2 0
3e2 0
e2 4r05
3e2 4r03
2 1 0
2 2r0
(4π²0 )δE = 3 hr i + e h i −
h1i = 2 3 + e 3 −
2r0
r
2r0
2r0 5a0
a0
2r0 3a30
µ
¶
·
¸
µ
¶
2
2
e2 r0
4 e2 r0
4
=
+2−2 =
a0 a0
10
10 a0 a0
(11)
Since binding energy of unperturbed ground state is e2 /(4π²0 · 2a0 ) = 13.6 eV≡ B,
fractional change in B due to finite extent of proton is
µ ¶2 µ −15 ¶2
δB
8 r0
10 m
'
' 10−10 small!
=
−10
B
10 a0
10 m
(12)
b)Although δEn=1 is small, it’s a good deal larger than δE in other hydrogenic states since
the electron “spends so much less time” near the origin r = 0. Find, e.g.
µ ¶4
r0
δB
1
n = 2, l = 1,
'
B
1120 a0
(13)
Due to “centrifugal barrier” for l > 0 states.
2. Bead on wire loop with “dimple”
a) We said in class that for a particle confined to a circular loop with otherwise free
motion, the Hamiltonian is that for the quantum rotor, H0 = L2z /2I, with I = mR2 the
moment of inertia of a point particle orbiting at radius R. Clearly the eigenfunctions of H0
are those of angular momentum, einθ , and since Lz = −i~∂θ , the energies are
En = ~2 n2 /(2mR2 ). Griffiths wants to express the eigenfunctions in terms of x, which he
defines as the displacement along the loop, so θ = x/R ≡ 2πx/L, where L is the
p
RL
circumference of the loop. The prefactor of 1/L is required to ensure 0 dx|ψn |2 = 1.
b) Using Griffith’s formulae for degenerate 2-state perturbation theory, with a → n,
b → −n, and noting that ψn ∗ = ψ−n , we have
Z
Z
V0 L/2
V0 ∞
V0 √
2
2
−x2 /a2
Waa = Wbb = −
dxe
dx ≈ −
dxe−x /a dx = − a π
L −L/2
L −∞
L
Z L/2
Z ∞
V0 √
V0
V0
2
2
2
2
2
dxe−x /a e−4πinx/L dx ≈ −
dxe−x /a −4πinx/L dx = − a πe−(2πna/L) ,
Wab = −
L −L/2
L −∞
L
3
where in the last step we completed the square in the exponent to factor out the
exponential term out of the Gaussian integral. Eq. 6.27 of Griffiths then gives first order
correction
E±1
´
√ V0 a ³
−(2πna/L)2
= Waa ± |Wab | = − π
1∓e
L
c) Eq. 6.22 of Griffiths ⇒ the linear combinations which diagonalize H 0 are αψn + βψ−n ,
with β = α(E−1 − Waa )/Wab = ∓α, so what Griffiths calls the “good” eigenstates (the ones
which diagonalize H 0 ) are

1
2  cos 2πnx
L
= √ (ψn ± ψ−n ) =
L  i sin 2πnx
2
L
Z L/2
´
√ V0 a ³
2
2πnx
2
2
2
= hψ+ |H 0 |ψ+ i = (−V0 )
e−x /a cos2
dx ≈ − π
1 + e−(2πna/L)
L
L
L
−L/2
Z L/2
´
√ V0 a ³
2
2πnx
2
2
2
= hψ+ |H 0 |ψ+ i = (−V0 )
1 − e−(2πna/L) ,
e−x /a sin2
dx ≈ − π
L
L
L
−L/2
r
ψ±
E+1
E−1
where I used Maple to evaluate the integrals after replacing the limits ±L/2 with ±∞.
d) Note from c) that the eigenfunctions ψ± have definite parity, so we can use the parity
operator Π to apply the theorem. Works!