Appendix E: Solutions
227
∆ =
⎛
⎝
D xx −
1
3 λ
00
0
D yy −
1
3 λ
0
00
D zz −
1
3 λ
⎞
⎠
ˆ
S∆ ˆ
S = D xx ˆ
S 2
x + D yy ˆ
S 2
y + D zz ˆ
S 2
z −
1
3 λ ˆ
S 2 = D xx ˆ
S 2
x + D yy ˆ
S 2
y + D zz ˆ
S 2
z −
1
3 D xx ˆ
S 2
x −
1
3 D xx ˆ
S 2
y −
1
3 D xx ˆ
S 2
z −
1
3 D yy ˆ
S 2
x −
1
3 D yy ˆ
S 2
y −
1
3 D yy ˆ
S 2
z −
1
3 D zz ˆ
S 2
x −
1
3 D zz ˆ
S 2
y −
1
3 D zz ˆ
S 2
z =
2
3 D xx ˆ
S 2
x −
1
3 D xx ˆ
S 2
y −
1
3 D xx ˆ
S 2
z −
1
3 D yy ˆ
S 2
x +
2
3 D yy ˆ
S 2
y −
1
3 D yy ˆ
S 2
z −
1
3 D zz ˆ
S 2
x −
1
3 D zz ˆ
S 2
y +
2
3 D zz ˆ
S 2
z to be compared with D( ˆ
S 2
z −
1
3
ˆ
S 2 ) + E( ˆ
S 2
x − ˆ
S 2
y ) = D ˆ
S 2
z −
1
3 D ˆ
S 2
x −
1
3 D ˆ
S 2
y −
1
3 D ˆ
S 2
z + E ˆ
S 2
x − E ˆ
S 2
y . After substituting the definitions of D and E: D zz ˆ
S 2
z −
1
2 D xx ˆ
S 2
z −
1
2 D xx ˆ
S 2
z −
1
3 D zz ˆ
S 2
x −
1
3 D zz ˆ
S 2
y −
1
3 D zz ˆ
S 2
z +
1
6 D xx ˆ
S 2
x +
1
6 D xx ˆ
S 2
y +
1
6 D xx ˆ
S 2
z +
1
6 D yy ˆ
S 2
x +
1
6 D yy ˆ
S 2
y +
1
6 D yy ˆ
S 2
z +
1
2 D xx ˆ
S 2
x −
1
2 D yy ˆ
S 2
x −
1
2 D xx ˆ
S 2
y +
1
2 D yy ˆ
S 2
y =
2
3 D xx ˆ
S 2
x −
1
3 D xx ˆ
S 2
y −
1
3 D xx ˆ
S 2
z −
1
3 D yy ˆ
S 2
x +
2
3 D yy ˆ
S 2
y −
1
3 D yy ˆ
S 2
z −
1
3 D zz ˆ
S 2
x −
1
3 D zz ˆ
S 2
y +
2
3 D zz ˆ
S 2
z , equal to what
is obtained from ˆ
S∆ ˆ
S.
Exercise 2.6 ˆ
S 2
z |
1
2 ,
1
2 =
1
4 |
1
2 ,
1
2 ; ˆ
S 2
z |
1
2 , −
1
2 =
1
4 |
1
2 , −
1
2 ⇒D( ˆ
S 2
z −
1
3
ˆ
S 2 )|
1
2 , ±
1
2
= D[
1
4 |
1
2 , ±
1
2 −
1
3 (
1
2 (
1
2 + 1))|
1
2 , ±
1
2 ] = D(
1
4 −
1
4 )|
1
2 , ±
1
2 =0; ˆ
S 2
x |
1
2 , ±
1
2 =
1
4 ( ˆ
S + ˆ
S + + ˆ
S − ˆ
S − + ˆ
S + ˆ
S − + ˆ
S − ˆ
S + )|
1
2 , ±
1
2 =|
1
2 , ±
1
2 , ˆ
S 2
y |
1
2 , ±
1
2 =−
1
4 ( ˆ
S + ˆ
S + +
ˆ
S − ˆ
S − − ˆ
S + ˆ
S − − ˆ
S − ˆ
S + ) =|
1
2 , ±
1
2 ⇒E( ˆ
S 2
x − ˆ
S 2
y )|
1
2 , ±
1
2 =E(
1
4 −
1
4 )|
1
2 , ±
1
2 =0.
This shows that both diagonal and off-diagonal matrix elements are zero. This means
that the ZFS model Hamiltonian cannot remove the degeneracy.
Exercise 2.7 (a) Taking E (0) as zero of energy, the exponent in the denominator
becomes 1, limiting the sum over the 2S + 1 M S -sublevels of the ground state, the
denominator simplifies to 2S + 1. Quintet:
M S
M 2
S =[ (−2) 2 + (−1) 2 + 0 +
1 2 + 2 2 ]=10; S(S + 1)(2S + 1)/3 = 2(2 + 1)(2 · 2 + 1)/3 = 10. (b) C =
N A (µ B g e ) 2 S(S + 1)/3k with N A µ 2
B /3k ≈ 1/8 ⇒ C = g 2
e S(S + 1)/8. Taking g e = 2,
C = S(S + 1)/2; 3/8 (0.375), 1, 15/8 (1.875), 3, 35/8 (4.375), 6, 63/8 (7.875).
Exercise 2.8 Equation 2.44 gives S z =(−S(S + 1)/3k B T ) · (µ B g e H − nJS z ) ⇒
S z −(−S(S + 1)/3k B T )nJS z =(−S(S + 1)/3k B T )µ B g e H ⇒⇒S z (3k B T − S(S +
1)nJ) =−S(S + 1)µ B g e H, which directly leads to Eq. 2.45.
Exercise 2.9 First for α: ψ
(0)
i | ˆ
L · ˆ
S|ψ
(0)
0 ==ψ
(0)
i | ˆ
L z · ˆ
S z |ψ
(0)
0 +
1
2 ψ
(0)
i | ˆ
L + · ˆ
S − |
ψ
(0)
0 +
1
2 ψ
(0)
i | ˆ
L + · ˆ
S − |ψ
(0)
0 == ψ
(0)
i | ˆ
L z |ψ
(0)
0 α| ˆ
S z |α+
1
2 ψ
(0)
i | ˆ
L + |ψ
(0)
0
α| ˆ
S − |α+
1
2 ψ
(0)
i | ˆ
L − |ψ
(0)
0 α| ˆ
S + |α=
1
2 ψ
(0)
i | ˆ
L z |ψ
(0)
0 .N o wf o rβ: ψ
(0)
i | ˆ
L · ˆ
S|
ψ
(0)
0 == ψ
(0)
i | ˆ
L z · ˆ
S z |ψ
(0)
0 +
1
2 ψ
(0)
i | ˆ
L + · ˆ
S − |ψ
(0)
0 +
1
2 ψ
(0)
i | ˆ
L + · ˆ
S − |ψ
(0)
0 =
ψ
(0)
i | ˆ
L z |ψ
(0)
0 β| ˆ
S z |α+
1
2 ψ
(0)
i | ˆ
L + |ψ
(0)
0 β| ˆ
S − |α+
1
2 ψ
(0)
i | ˆ
L − |ψ
(0)
0 β| ˆ
S + |α=
1
2 ψ
(0)
i | ˆ
L + |ψ
(0)
0 .
Exercise 2.10 ˆ l + |l, m l =
√
l(l + 1) − m l (m l + 1)|l, m l + 1; ˆ l + p 0 =
√
2p + and
ˆ l − p 0 =
√
2p − From this ˆ l x |p 0 =
1
2
√
2(p + + p − . Then p − | ˆ l x |p 0 =
√
2
2 and