1
PHY4605
Problem Set 6
Solutions
1. Atomic configurations. Griffiths Problem 5.12.
(a) H: 1s; He: 1s2 ; Li: 1s2 2s; Be: 1s2 2s2 ; B: 1s2 2s2 2p; C: 1s2 2s2 2p2 ; N: 1s2 2s2 2p3 ;
O: 1s2 2s2 2p4 ; F: 1s2 2s2 2p5 ; Ne: 1s2 2s2 2p6
(b) H: 2 S1/2 ; He: 1 S0 ; Li: 2 S1/2 . There is only one possibility in each of these cases
for the term symbol, because ` = 0, so the triangle rule allows only one value for
j, i.e. s. For the other cases one must be careful to determine first the possible
total `, then the possible total s, then the possible j’s which can result. B: the
one extra p electron means ` = 1, s = 1/2, so the possibilities for j are 1/2, 3/2:
2
P1/2 , 2 P3/2 . C: two 2p electrons can have spins which add to 1 or 0, while their
orbital a.m. can be 2, 1, or 0. This makes ten possible combinations: 1 S0 , 3 S1 ,
1
P1 , 3 P0 , 3 P1 , 3 P2 , 1 D2 , 3 D1 , 3 D2 , 3 D3 . For N there are 3 ` = 1 momenta which
can add to 0,1,2, or 3, and 3 spins 1/2 which can add to 1/2, or 3/2: 4 S3/2 ,
2
S1/2 ,4 P1/2 , 4 P3/2 , 4 P5/2 , 2 P1/2 , 2 P3/2 , 4 D1/2 ,4 D3/2 , 4 D5/2 ,4 D7/2 , 2 D3/2 , 2 D5/2 ,
4
F3/2 ,4 F5/2 ,4 F7/2 ,4 F9/2 , 2 F7/2 ,2 F5/2 .
2. Yukawa Hydrogen Griffiths Problem 7.14.
q
Take φ = πb13 e−r/b , b a variational parameter. The expectation value of the kinetic
energy hT i has the same form as for usual Hydrogen, but with a0 → b, i.e.
hT i →
~2
.
2mb2
The expectation value of the Yukawa potential in φ is
hV i =
=
=
≈
Z
−µr
e2 4π ∞
2 −2r/b e
−
dr
r
e
4π²0 πb3 0
r
Z ∞
2
e 4
=−
dr r e−(µ+2/b)r
4π²0 b3 0
e2 4
1
e2
1
−
=
−
4π²0 b3 (µ + 2/b)2
4π²0 b(1 + µb/2)2
2
e 1
−
(1 − µb + 3µ2 b2 /4),
4π²0 b
2
where the last expansion follows for sufficiently small µ. The expectation value of H
is then
~2
e2
−
(1 − µb + 3(µb)2 /4)/b
2
2mb
4π²0
µ
¶
1
∂hHi
~2
e2
2
=0 = − 3 +
− 3µ /4
∂b
mb
4π²0 b2
hHi =
⇒
This is a cubic equation, but we only need the perturbative solution. The leading
order solution, if we neglect the µ2 term, gives
b ≈
4π²0 ~2
= a0 ,
me2
so write b = a0 + δ and expand for small δ:
µ
¶
1
1
1
2
0 = −
+
− 3µ /4
(a0 + δ)3 a0 (a0 + δ)2
µ2 1 3
δ
−3
(a + ...) ⇒
0 ≈ −1 + 1 +
a0
4 a0 0
δ
≈ 3µ2 a20 /4
a0
So b ≈ a0 (1 + 3µ2 a20 /4). Inserting into the expression for hHi, we find
hHi = −R + e2 µ/(4π²0 ) + O(µ2 ).
3. Rubber band Helium Griffiths Problem 7.17.
√
√
(a) First invert variable change, r1 = (u + v)/ 2; r2 = (u − v)/ 2;
´ ³ 2
³ 2
2
∂ f
r12 + r22 = u2 + v 2 . Now I want to show that ∂∂xf2 + ∂∂xf2 = ∂u
2 +
1
2
x
∂2f
∂vx2
´
, also for
3
y and z. Explicitly,
¶
µ
∂ 2f
∂f ∂ux
∂f ∂ux
1
∂f
∂f
=
+
=√
+
∂x1
∂ux ∂x1 ∂vx ∂x1
2 ∂ux ∂vx
µ
¶
∂ 2f
∂f ∂ux
∂f ∂ux
1
∂f
∂f
=
+
=√
−
∂x2
∂ux ∂x2 ∂vx ∂x2
2 ∂ux ∂vx
µ
¶
∂ 2f
1 ∂
∂f
∂f
=√
+
∂x21
2 ∂x1 ∂ux ∂vx
µ 2
¶
1
∂ 2 f ∂vx
∂ 2 f ∂ux ∂ 2 f ∂vx
∂ f ∂ux
= √
+
+
+ 2
∂vx ∂x1
2 ∂u2x ∂x1 ∂ux ∂vx ∂x1 ∂vx ∂ux ∂x1
µ 2
¶
2
2
∂ f
∂ f
1 ∂ f
+2
+ 2
=
2
2 ∂ux
∂ux ∂vx ∂vx
µ
¶
2
∂ f
1 ∂
∂f
∂f
=√
−
2
∂x2
2 ∂x2 ∂ux ∂vx
¶
µ 2
∂ 2 f ∂vx
∂ 2 f ∂ux ∂ 2 f ∂vx
1
∂ f ∂ux
= √
+
−
+ 2
∂vx ∂x2
2 ∂u2x ∂x2 ∂ux ∂vx ∂x2 ∂vx ∂ux ∂x2
µ 2
¶
1 ∂ f
∂ 2f
∂ 2f
=
+
−
2
.
2 ∂u2x
∂ux ∂vx ∂vx2
Adding the two results for the two second derivatives gives the desired result.
As a consequence, we may write
~2
1
λ
H = −
(∇2u + ∇2v ) + mω 2 (u2 + v 2 ) − mω 2 2v 2
2 ¶ µ
4
µ 2m 2
¶
2
~
1
~
1
2
2 2
2
2 2
= −
∇ + mω u + −
∇ + (1 − λ)mω v
2m u 2
2m v 2
which is a sum of 3D SHO’s if λ < 1!
(b) Since the system separates into two SHO’s, we know the exact energies. In
√
particular the exact ground state energy is (3/2)~ω(1 + 1 − λ).
(c) The properly normalized ground state wavefunction of the 1D SHO is
2 /2~
ψ0 (x) = (mω/π~)1/4 e−mωx
(mω/π~)3/4 e−mωr
2 /2~
(Griffiths 2.59), so the 3D SHO ground state is
. The problem asks us simply to evaluate the energy in the
product state of two such oscillators, without the λ term (in Sec. 7.2 the
product of hydrogenic orbitals is the ground state only if the electron-electron
2
2
interaction is neglected). So we take ψ(r1 , r2 ) = (mω/π~)3/4 e−mω(r1 +r2 )/2~ , and
calculate the expectation value of the energy
3
3
λ
~ω + ~ω − h mω 2 |r1 − r2 |2 i
2
2
4
³
´3 Z
λ
λ
2
2
2
2
2 mω
−h mω |r1 − r2 | i = − mω
e−mω(r1 +r2 )/2~ (r1 − r2 )2 d3 r1 d3 r2 .
4
4
π~
hHi =
4
Now note (r1 − r2 )2 = r12 + r22 − 2r1 · r2 . The r1 · r2 term vanishes by symmetry
(the integrand for fixed r2 is odd in each component of r1 ), and the r12 and r22
terms contribute the same, so
³
´3 Z
λ
λ
2
2
2
2
2 mω
−h mω |r1 − r2 | i = − mω
2 e−mω(r1 +r2 )/2~ r12 d3 r1 d3 r2
4
4
π~
Z ∞
Z ∞
³ mω ´3
λ
2
2
2
−mωr12 /2~ 4
(4π)
= − mω
e
r1 dr1
e−mωr2 /2~ r22 dr2
2
π~
0
0
"
#" µ
#
r
¶2 r
4 5
~
8m ω 1 ~
π~
3
π~
3
= −λ
=
−
λ~ω
π~3
4 mω mω
8 mω
mω
4
So hHi = 3~ω(1 − λ/4), which must be greater than or equal to exact energy
for 0 ≤ λ ≤ 1. By expanding the square root in the exact expression (answer to
b)), you can convince yourself this is indeed the case.