228
Appendix E: Solutions
p + | ˆ l x |p 0 =
√
2
2 . ˆ l x p + =
1
2 ( ˆ l + + ˆ l − )p + =
1
2 (0 +
√
2p 0 ) =
√
2
2 p 0 and analogous for
ˆ l x p − =
√
2
2 p 0 . Then p 0 | ˆ l x |p + ==p 0 | ˆ l x |p − =
√
2
2 .
Exercise 2.11 The eigenvalues of the matrix
⎛
⎝
ab0
b −2ab
0 ba
⎞
⎠ are E 1 = a and E 2,3 =
−
1
2 a ±
√
9a 2 + 8b 2 /2. Shifting the diagonal elements by +2a, the matrix becomes
⎛
⎝
3ab 0
b 0 b
0 b 3a
⎞
⎠ with eigenvalues E ′
1 = 3a and E ′
2,3 = (3a/2) ±
√
9a 2 + 8b 2 /2. The
constant difference of 2a between E i and E ′
i is the same quantity by which the
diagonal matrix elements have been shifted.
Problem 2.1 (1) see Appendix C.( 2 )
Ψ ′
1 |
Ψ ′
1 =0.97211913;;
Ψ ′
1 |
Ψ ′
2 =0.0000
0457 + 0.00001946i ==
Ψ ′
2 |
Ψ ′
1 ∗ ;;
Ψ ′
1 |
Ψ ′
3 =0.00004069 − 0.00000414i
==
Ψ ′
3 |
Ψ ′
1 ∗ ;;
Ψ ′
2 |
Ψ ′
2 =0.97480639;;
Ψ ′
2 |
Ψ ′
3 =0.00010103 − 0.00078090i =
Ψ ′
3 |
Ψ ′
2 ∗ ;;
Ψ ′
2 |
Ψ ′
2 =0.98401064. (3) 1, 1| ˆ
H eff |1, 1=6.5359;;1, 1| ˆ
H eff |1, 0=
1, 0| ˆ
H eff |1, 1 ∗ =− 4.6142 − 2.2944i;;1, 1| ˆ
H eff |1, −1== 1, −1| ˆ
H eff |1, 1 ∗ =
5.3478 − 0.7801i;;1, 0| ˆ
H eff |1, 0=35.9344;;1, 0| ˆ
H eff |1, −1==1, −1| ˆ
H eff |1, 0 ∗
= 4.6142 + 2.2944i;;1, −1| ˆ
H eff |1, −1=6.5359. (4) D 11 = 23.315016; D 12 =
D 21 = 0.780050; D 13 = D 31 =− 6.525412; D 22 = 12.619358; D 23 = D 32 =
3.244771; D 33 =− 11.431250. After diagonalization: D 11 = 24.503124; D 22 =
13.048843; D 33 =− 13.048843. Using Eq. 2.16, D =− 31.8c m −1 and E = 5.8
cm −1 . Using the energies (Eq. 2.22): D =
1
2 (11.54 + 0) − 37.55 =−31.8cm −1 and
E =
1
2 (11.54 − 0) = 5.8cm −1 .
Problem 2.2 (1) see inset Fig. 2.1 (2) Since there is only one energy difference, only
one effective anisotropy parameter can be derived, under the assumption E = 0,
this parameter can be considered to be D. Unless the orientation of the magnetic
axes frame is known, the sign of D cannot be determined. (3) see Appendix C.(4)
3
2 ,
3
2 | ˆ
H eff |
3
2 ,
3
2 ==
3
2 , −
3
2 | ˆ
H eff |
3
2 , −
3
2 =3.634038;;
3
2 ,
3
2 | ˆ
H eff |
3
2 ,
1
2 ==
3
2 ,
1
2 | ˆ
H eff |
3
2 ,
3
2 ∗ = 0.173864 + 9.819256i;;
3
2 ,
3
2 | ˆ
H eff |
3
2 , −
1
2 ==
3
2 , −
1
2 | ˆ
H eff |
3
2 ,
3
2 ∗ =
2.856287 + 0.412321i;;
3
2 ,
3
2 | ˆ
H eff |
3
2 , −
3
2 ==
3
2 , −
3
2 | ˆ
H eff |
3
2 ,
3
2 =0;;
3
2 ,
1
2 | ˆ
H eff |
3
2 ,
1
2 ==
3
2 , −
1
2 | ˆ
H eff |
3
2 , −
1
2 =28.831949;;
3
2 ,
1
2 | ˆ
H eff |
3
2 , −
1
2 ==
3
2 , −
1
2 | ˆ
H eff |
3
2 ,
1
2 =0;;
3
2 ,
1
2 | ˆ
H eff |
3
2 , −
3
2 ==
3
2 , −
3
2 | ˆ
H eff |
3
2 ,
1
2 ∗ = 2.856287 + 0.412321i;;
3
2 , −
1
2 |
ˆ
H eff |
3
2 , −
3
2 ==
3
2 , −
3
2 | ˆ
H eff |
3
2 , −
1
2 ∗ =−0.173864−9.819255i. D 11 = 10.177529;
D 12 = D 21 =− 0.238054; D 13 = D 31 = 0.100381; D 22 = 6.879372; D 23 =
D 32 =−5.669150; D 33 =−4.070506. Diagonalization leads to D 11 = 10.246963;
D 22 = 9.216256; D 33 =− 6.476824. From this D =− 16.21 cm −1 and E = 0.51
cm −1 .NegativeD indicates that the wave function of the lowest level is dominated
by M S =±
3
2 contributions (see
Ψ ′
1,2 ), hence, the molecule exhibits easy-axis magnetism.
Appendix E: Solutions
p + | ˆ l x |p 0 =
√
2
2 . ˆ l x p + =
1
2 ( ˆ l + + ˆ l − )p + =
1
2 (0 +
√
2p 0 ) =
√
2
2 p 0 and analogous for
ˆ l x p − =
√
2
2 p 0 . Then p 0 | ˆ l x |p + ==p 0 | ˆ l x |p − =
√
2
2 .
Exercise 2.11 The eigenvalues of the matrix
⎛
⎝
ab0
b −2ab
0 ba
⎞
⎠ are E 1 = a and E 2,3 =
−
1
2 a ±
√
9a 2 + 8b 2 /2. Shifting the diagonal elements by +2a, the matrix becomes
⎛
⎝
3ab 0
b 0 b
0 b 3a
⎞
⎠ with eigenvalues E ′
1 = 3a and E ′
2,3 = (3a/2) ±
√
9a 2 + 8b 2 /2. The
constant difference of 2a between E i and E ′
i is the same quantity by which the
diagonal matrix elements have been shifted.
Problem 2.1 (1) see Appendix C.( 2 )
Ψ ′
1 |
Ψ ′
1 =0.97211913;;
Ψ ′
1 |
Ψ ′
2 =0.0000
0457 + 0.00001946i ==
Ψ ′
2 |
Ψ ′
1 ∗ ;;
Ψ ′
1 |
Ψ ′
3 =0.00004069 − 0.00000414i
==
Ψ ′
3 |
Ψ ′
1 ∗ ;;
Ψ ′
2 |
Ψ ′
2 =0.97480639;;
Ψ ′
2 |
Ψ ′
3 =0.00010103 − 0.00078090i =
Ψ ′
3 |
Ψ ′
2 ∗ ;;
Ψ ′
2 |
Ψ ′
2 =0.98401064. (3) 1, 1| ˆ
H eff |1, 1=6.5359;;1, 1| ˆ
H eff |1, 0=
1, 0| ˆ
H eff |1, 1 ∗ =− 4.6142 − 2.2944i;;1, 1| ˆ
H eff |1, −1== 1, −1| ˆ
H eff |1, 1 ∗ =
5.3478 − 0.7801i;;1, 0| ˆ
H eff |1, 0=35.9344;;1, 0| ˆ
H eff |1, −1==1, −1| ˆ
H eff |1, 0 ∗
= 4.6142 + 2.2944i;;1, −1| ˆ
H eff |1, −1=6.5359. (4) D 11 = 23.315016; D 12 =
D 21 = 0.780050; D 13 = D 31 =− 6.525412; D 22 = 12.619358; D 23 = D 32 =
3.244771; D 33 =− 11.431250. After diagonalization: D 11 = 24.503124; D 22 =
13.048843; D 33 =− 13.048843. Using Eq. 2.16, D =− 31.8c m −1 and E = 5.8
cm −1 . Using the energies (Eq. 2.22): D =
1
2 (11.54 + 0) − 37.55 =−31.8cm −1 and
E =
1
2 (11.54 − 0) = 5.8cm −1 .
Problem 2.2 (1) see inset Fig. 2.1 (2) Since there is only one energy difference, only
one effective anisotropy parameter can be derived, under the assumption E = 0,
this parameter can be considered to be D. Unless the orientation of the magnetic
axes frame is known, the sign of D cannot be determined. (3) see Appendix C.(4)
3
2 ,
3
2 | ˆ
H eff |
3
2 ,
3
2 ==
3
2 , −
3
2 | ˆ
H eff |
3
2 , −
3
2 =3.634038;;
3
2 ,
3
2 | ˆ
H eff |
3
2 ,
1
2 ==
3
2 ,
1
2 | ˆ
H eff |
3
2 ,
3
2 ∗ = 0.173864 + 9.819256i;;
3
2 ,
3
2 | ˆ
H eff |
3
2 , −
1
2 ==
3
2 , −
1
2 | ˆ
H eff |
3
2 ,
3
2 ∗ =
2.856287 + 0.412321i;;
3
2 ,
3
2 | ˆ
H eff |
3
2 , −
3
2 ==
3
2 , −
3
2 | ˆ
H eff |
3
2 ,
3
2 =0;;
3
2 ,
1
2 | ˆ
H eff |
3
2 ,
1
2 ==
3
2 , −
1
2 | ˆ
H eff |
3
2 , −
1
2 =28.831949;;
3
2 ,
1
2 | ˆ
H eff |
3
2 , −
1
2 ==
3
2 , −
1
2 | ˆ
H eff |
3
2 ,
1
2 =0;;
3
2 ,
1
2 | ˆ
H eff |
3
2 , −
3
2 ==
3
2 , −
3
2 | ˆ
H eff |
3
2 ,
1
2 ∗ = 2.856287 + 0.412321i;;
3
2 , −
1
2 |
ˆ
H eff |
3
2 , −
3
2 ==
3
2 , −
3
2 | ˆ
H eff |
3
2 , −
1
2 ∗ =−0.173864−9.819255i. D 11 = 10.177529;
D 12 = D 21 =− 0.238054; D 13 = D 31 = 0.100381; D 22 = 6.879372; D 23 =
D 32 =−5.669150; D 33 =−4.070506. Diagonalization leads to D 11 = 10.246963;
D 22 = 9.216256; D 33 =− 6.476824. From this D =− 16.21 cm −1 and E = 0.51
cm −1 .NegativeD indicates that the wave function of the lowest level is dominated
by M S =±
3
2 contributions (see
Ψ ′
1,2 ), hence, the molecule exhibits easy-axis magnetism.