1
PHY4605
Problem Set 5
Solutions
1. Strong field Zeeman Splitting. Griffiths Problem 6.23. The zeroth-order Bohr
energy is the same for all states, E2 = −R/22 = −3.4eV. The Zeeman contribution is
2nd term in Griffiths Eq. 6.79, µB B(m` + 2ms ). The fine structure energy shift is
h
i
`(`+1)−m` ms
given by Eq. 6.82: Ef1s = (R/23 )α2 { 83 − `(`+1/2)(`+1)
} = 1.7 eVα2 {...}. We’re
supposed to make a table of the energy levels, within the strong field Zeeman effect
where the Zeeman energy is treated exactly and the fine structure effects perturb
these levels. The table below contains the results for the 8 states |2`m` ms i, with the
final answer for the energy −3.4eV[1 − (α2 /2){...}] + (m` + 2ms )µB B in the last
column.
State
m` + 2ms {...}
Energy
|200 12 i
1
-5/8
−3.4 eV[1 + (5/16)α2 )] + µB B
|200 − 12 i
-1
-5/8
−3.4 eV[1 + (5/16)α2 )] − µB B
|211 12 i
2
-1/8 −3.4 eV[1 + (1/16)α2 )] + 2µB B
|21 − 1 − 21 i
-2
-1/8 −3.4 eV[1 + (1/16)α2 )] − 2µB B
|210 12 i
1
-7/24 −3.4 eV[1 + (7/48)α2 )] + µB B
|210 − 12 i
-1
-7/24 −3.4 eV[1 + (7/48)α2 )] − µB B
|211 − 12 i
0
-11/24
−3.4 eV[1 + (11/48)α2 )]
|21 − 1 12 i
0
-11/24
−3.4 eV[1 + (11/48)α2 )]
2. Feynman-Hellman Theorem. Griffiths problem 6.33.
(a)
4me3
4
∂En
= −
= En
2 2
2
2
∂e
32π ²0 ~ (jmax + ` + 1)
e
∂H
2e 1
= −
,
∂e
4π²0 r
so applying the Feynman-Hellman theorem,
4
e
En = −
e
2π²0
¿ À
1
r
2
so
¿ À
1
8π²0
8π²0 R
e2 m 1
1
= − 2 En = 2 2 =
= 2 ,
2
2
r
e
en
4π²0 ~ n
n a0
where I used the standard expression for the Bohr radius, a0 = 4π²0 ~2 /me2 .
(b) Recall n = (jmax + ` + 1) from our first solution to H-atom (Griffiths Eq. 4.67).
So
∂En
2me4
2En
=
=−
2 2
2
3
∂`
32π ²0 ~ (jmax + ` + 1)
n
2
∂H
~
(2` + 1),
=
∂`
2mr2
so
¿
1
r2
À
4mEn
4mR
= 3
2
n(2` + 1)~
n (2` + 1)~2
1
= 3
n (` + 12 )a20
= −
3. Variational theorem for SHO. Griffiths Problem 7.2. Normalize:
Z
hφ|φi =
hφ|T |φi =
=
=
hφ|V |φi =
∞
1
π
dx = 3
2
2
+b )
2b
−∞
µ
¶
2 Z ∞
2
1
1
~
d
−
dx
2m −∞ x2 + b2 dx2 x2 + b2
Z ∞
1
~2
2(3x2 − b2 )
−
·
2m −∞ x2 + b2 (x2 + b2 )3
µ
¶
~2 −π
−
2m 4b5
Z ∞
1
1
x2
π
2
= mω 2
mω
2
2
2
2
2
2b
−∞ (x + b )
(x2
so
2
−~
hφ|H|φi
= 2m
hφ|φi
¡ −π ¢
4b5
π
+ 12 mω 2 2b
π
2b3
=
~2
1
+
mω 2 b2
2
4mb
2
√
Minimize ∂(hHi/hi)/∂b = 0 ⇒ b2 = ~/( 2mω). At minimum,
√
hHimin = ~ω/ 2 > ~ω/2
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