H Solutions to Problems
259
tween the complex and real triplet basis, as given in Eq. (7.39). One obtains:
0|H Ze |+1=−
1
√
2
A 1 |H|E x + iE y =
1
√
2
−a + d + i(−b − c)
0|H Ze |−1=
1
√
2
A 1 |H|E x − iE y =
1
√
2
a + d + i(b − c)
±1|H Ze |±1=
1
2
x|H|x++y|H|y±i
x|H|y−−y|H|x
=±f
From these equations the parameters may be identified as follows:
a = 0
b =−g ⊥ B y
c = 0
d = g ⊥ B x
e = 0
f = g || B z
The Zeeman Hamiltonian does not include the zero-field splitting between the
A 1 and E states. This can be rendered by a second-order spin operator, which
transforms as the octahedral E g θ quadrupole component:
H ZF =
D
3 2
2 ˜
S
2
z − ˜
S
2
x − ˜
S
2
y
=
D
2
˜
S
2
z −
1
3
˜
S
2
One then obtains
D = 3
7.5 The action of the components of the fictitious spin operator on the Γ 8 basis is
dictated by the general expressions for the action of the spin operators on the
S =
3
2 basis functions. It is verified that the spin-Hamiltonian that generates the
J p part of the matrix precisely corresponds to
H p = J p B · ˜
S
The fictitious spin operator indeed transforms as a T 1 operator and has the
tensorial rank of a p-orbital. However, as we have shown, the full Hamiltonian also includes a J f part, which involves an f -like operator. To mimic
this part by a spin Hamiltonian, one thus will need a symmetrized triple product of the fictitious spin, which will embody an f -tensor, transforming in the
octahedral symmetry as the T 1 irrep. These f -functions can be found in Table 7.1 and are of type z(5z 2 − 3r 2 ). But beware! To find the corresponding spin operator, it is not sufficient simply to substitute the Cartesian variables by the corresponding spinor components, i.e., z by ˜
S z , etc.; indeed,
while products of x,y, and z are commutative, the products of the corresponding operators are not. Hence, when constructing the octupolar product
259
tween the complex and real triplet basis, as given in Eq. (7.39). One obtains:
0|H Ze |+1=−
1
√
2
A 1 |H|E x + iE y =
1
√
2
−a + d + i(−b − c)
0|H Ze |−1=
1
√
2
A 1 |H|E x − iE y =
1
√
2
a + d + i(b − c)
±1|H Ze |±1=
1
2
x|H|x++y|H|y±i
x|H|y−−y|H|x
=±f
From these equations the parameters may be identified as follows:
a = 0
b =−g ⊥ B y
c = 0
d = g ⊥ B x
e = 0
f = g || B z
The Zeeman Hamiltonian does not include the zero-field splitting between the
A 1 and E states. This can be rendered by a second-order spin operator, which
transforms as the octahedral E g θ quadrupole component:
H ZF =
D
3 2
2 ˜
S
2
z − ˜
S
2
x − ˜
S
2
y
=
D
2
˜
S
2
z −
1
3
˜
S
2
One then obtains
D = 3
7.5 The action of the components of the fictitious spin operator on the Γ 8 basis is
dictated by the general expressions for the action of the spin operators on the
S =
3
2 basis functions. It is verified that the spin-Hamiltonian that generates the
J p part of the matrix precisely corresponds to
H p = J p B · ˜
S
The fictitious spin operator indeed transforms as a T 1 operator and has the
tensorial rank of a p-orbital. However, as we have shown, the full Hamiltonian also includes a J f part, which involves an f -like operator. To mimic
this part by a spin Hamiltonian, one thus will need a symmetrized triple product of the fictitious spin, which will embody an f -tensor, transforming in the
octahedral symmetry as the T 1 irrep. These f -functions can be found in Table 7.1 and are of type z(5z 2 − 3r 2 ). But beware! To find the corresponding spin operator, it is not sufficient simply to substitute the Cartesian variables by the corresponding spinor components, i.e., z by ˜
S z , etc.; indeed,
while products of x,y, and z are commutative, the products of the corresponding operators are not. Hence, when constructing the octupolar product