286
11 Spintronics
11.2.1 Application to Fe
3+
: TiO 2
The five unpaired electrons in Fe
3+ are each in the 3d state, meaning that the ion
is in an S-state, with a spin, S = 5/2. The total number of spin-states, therefore, is
N s = 2S + 1 = 6. The spin-Hamiltonian, H 0 , for the Fe
3+
: TiO 2 complex is [96]
H 0 = gβH 0 · S+D
S
2
z −35/12
+E
S
2
x −S
2
y
+(a/6)
S
4
x +S
4
y +S
4
z −707/16
+ (7/36)F
S
4
z − (95/14)S
2
z + 81/16
,
(11.6)
where the nominal values of the derived constants are g = 2.0, D =
20.35 GHz, E = 2.21 GHz, a = 1.1 GHz, F = −0.5 GHz, and H 0 is the dc
magnetic field. S x , S y , and S z are 6 × 6 Pauli spin-matrices. This Hamiltonian gives
us frequency directly, rather than energy. The D term has axial symmetry (about
the z-axis), and corresponds to the ion having an electric quadrupole moment, that
is acted upon by the crystalline electric fields. The E term represents and additional
nonaxially symmetric anisotropy in the xy plane, and corresponds to the ion’s
possessing an electric moment of higher order than quadrupolar. These are the main
terms, as the size of D and E would suggest; the remaining terms are due to the fact
that S > 2 and that the crystal symmetry is complicated. Clearly, these latter terms
are less important, but must be included for completeness.
The eigenvalue equation that determines the unperturbed energy levels (or
frequencies in this case) is
H 0 u = Eu ,
(11.7)
and when this equation is solved as a function of H 0 = H a z , we get the six curves
shown in Fig. 11.1. The zero-field energies occur in pairs (Kramers’ doublets), as
is typical of a system with an odd number of electrons in an electric field (the
crystalline field).
Consider the system at H = 1.78 kilogauss; the eigenvalues of H 0 are
E 1 = −58.20 × 10 9 h E 2 = −54.15 × 10 9 h E 3 = −19.60 × 10 9 h
E 4 = −5.64 × 10 9 h E 5 = 56.14 × 10 9 h E 6 = 81.05 × 10 9 h
,
(11.8)
from which we derive the resonant frequencies (in GHz)
ω 012 = 4.05 ω 023 = 34.55 ω 034 = 13.96 ω 045 = 61.78 ω 056 = 24.91
ω 013 = 38.60 ω 024 = 48.51 ω 035 = 75.74 ω 046 = 86.69
ω 014 = 52.56 ω 025 = 110.29 ω 036 = 100.65
ω 015 = 114.34 ω 026 = 135.20
ω 016 = 139.25
,
(11.9)
11 Spintronics
11.2.1 Application to Fe
3+
: TiO 2
The five unpaired electrons in Fe
3+ are each in the 3d state, meaning that the ion
is in an S-state, with a spin, S = 5/2. The total number of spin-states, therefore, is
N s = 2S + 1 = 6. The spin-Hamiltonian, H 0 , for the Fe
3+
: TiO 2 complex is [96]
H 0 = gβH 0 · S+D
S
2
z −35/12
+E
S
2
x −S
2
y
+(a/6)
S
4
x +S
4
y +S
4
z −707/16
+ (7/36)F
S
4
z − (95/14)S
2
z + 81/16
,
(11.6)
where the nominal values of the derived constants are g = 2.0, D =
20.35 GHz, E = 2.21 GHz, a = 1.1 GHz, F = −0.5 GHz, and H 0 is the dc
magnetic field. S x , S y , and S z are 6 × 6 Pauli spin-matrices. This Hamiltonian gives
us frequency directly, rather than energy. The D term has axial symmetry (about
the z-axis), and corresponds to the ion having an electric quadrupole moment, that
is acted upon by the crystalline electric fields. The E term represents and additional
nonaxially symmetric anisotropy in the xy plane, and corresponds to the ion’s
possessing an electric moment of higher order than quadrupolar. These are the main
terms, as the size of D and E would suggest; the remaining terms are due to the fact
that S > 2 and that the crystal symmetry is complicated. Clearly, these latter terms
are less important, but must be included for completeness.
The eigenvalue equation that determines the unperturbed energy levels (or
frequencies in this case) is
H 0 u = Eu ,
(11.7)
and when this equation is solved as a function of H 0 = H a z , we get the six curves
shown in Fig. 11.1. The zero-field energies occur in pairs (Kramers’ doublets), as
is typical of a system with an odd number of electrons in an electric field (the
crystalline field).
Consider the system at H = 1.78 kilogauss; the eigenvalues of H 0 are
E 1 = −58.20 × 10 9 h E 2 = −54.15 × 10 9 h E 3 = −19.60 × 10 9 h
E 4 = −5.64 × 10 9 h E 5 = 56.14 × 10 9 h E 6 = 81.05 × 10 9 h
,
(11.8)
from which we derive the resonant frequencies (in GHz)
ω 012 = 4.05 ω 023 = 34.55 ω 034 = 13.96 ω 045 = 61.78 ω 056 = 24.91
ω 013 = 38.60 ω 024 = 48.51 ω 035 = 75.74 ω 046 = 86.69
ω 014 = 52.56 ω 025 = 110.29 ω 036 = 100.65
ω 015 = 114.34 ω 026 = 135.20
ω 016 = 139.25
,
(11.9)