Electric Field-Controlled Magnetic Anisotropy …
17
The magnetic anisotropy type, i.e., PMA or IMA, can be established in two ways.
When M||x the longitudinal component,χ
zz
= 0, and the sign of the MAE is determined by χ
0 . On the other hand, for M||z the volume susceptibility, χ
0
= 0, so the
sign of the MAE is decided by the competition between the intraband and interband
contributions to the spin susceptibility χ
xx .
Conditions favoring PMA are determined by the magnetic anisotropy constants.
When M||z and t
<< J , from (29) it follows that
χ
xx
intra ≈
( ˆ
b(k))
2
d k
(2π) 2
n=±
δ(E F − E n (k)) =
(t
)
2
2J 2 ρ(E F ),
(31)
where ρ(E F ) is the total density of states at the Fermi energy. The average of the
matrix element was simplified by assuming that the exchange fields are much stronger
than the spin–orbit fields, so that |b(k)| ≈ −M(t
)
2
/2J
3 . Combining these expressions leads to an approximate form, K 2 ≈ (M − Jρ(E F )) for the uniaxial anisotropy
constant. The scale of anisotropy constant K ref = (t
)
2
/J is a useful figure of merit
for MAE. PMA is likely to be stable when the density of states at the Fermi level is
small: since 0 ≤ M ≤ 1, it can expect PMA if Jρ(E F ) ≤ 1. The states at the Fermi
level are the ones affected by SOC in the most important way, in energetic terms. A
large DOS at the Fermi level then translates to a large number of single-particle states
that gain the most energy from SOC once the magnetization is tilted away from the
perpendicular direction, which explains why this contribution favors IMA.
4 PMA for the Gapped Half-Filled Band Spectrum
In the case, in which the ferromagnetic exchange splitting is large enough to produce
a gape for the half-filled ferromagnetic insulator, the Fermi level lies in this gape.
The majority band is full, f + (k) = 1, and the minority band is empty, f − (k) = 0.
Starting from (8), (20), and (27), the internal energy for this case is
U =
d k
(2π )
2
(E 0 (k) − |b(k)|),
(32)
where only the second term in the integrand contains information about the
orientation of the ferromagnetic background, given by the angles θ and ϕ.
The expansion of the spin splitting |b(k)| gives its explicit dependence on the θ
and ϕ, which have the form
|b(k)| = b 0 (k)
∞
n=0
1/2
n
(cos γ (k))
n
,
(33)
17
The magnetic anisotropy type, i.e., PMA or IMA, can be established in two ways.
When M||x the longitudinal component,χ
zz
= 0, and the sign of the MAE is determined by χ
0 . On the other hand, for M||z the volume susceptibility, χ
0
= 0, so the
sign of the MAE is decided by the competition between the intraband and interband
contributions to the spin susceptibility χ
xx .
Conditions favoring PMA are determined by the magnetic anisotropy constants.
When M||z and t
<< J , from (29) it follows that
χ
xx
intra ≈
( ˆ
b(k))
2
d k
(2π) 2
n=±
δ(E F − E n (k)) =
(t
)
2
2J 2 ρ(E F ),
(31)
where ρ(E F ) is the total density of states at the Fermi energy. The average of the
matrix element was simplified by assuming that the exchange fields are much stronger
than the spin–orbit fields, so that |b(k)| ≈ −M(t
)
2
/2J
3 . Combining these expressions leads to an approximate form, K 2 ≈ (M − Jρ(E F )) for the uniaxial anisotropy
constant. The scale of anisotropy constant K ref = (t
)
2
/J is a useful figure of merit
for MAE. PMA is likely to be stable when the density of states at the Fermi level is
small: since 0 ≤ M ≤ 1, it can expect PMA if Jρ(E F ) ≤ 1. The states at the Fermi
level are the ones affected by SOC in the most important way, in energetic terms. A
large DOS at the Fermi level then translates to a large number of single-particle states
that gain the most energy from SOC once the magnetization is tilted away from the
perpendicular direction, which explains why this contribution favors IMA.
4 PMA for the Gapped Half-Filled Band Spectrum
In the case, in which the ferromagnetic exchange splitting is large enough to produce
a gape for the half-filled ferromagnetic insulator, the Fermi level lies in this gape.
The majority band is full, f + (k) = 1, and the minority band is empty, f − (k) = 0.
Starting from (8), (20), and (27), the internal energy for this case is
U =
d k
(2π )
2
(E 0 (k) − |b(k)|),
(32)
where only the second term in the integrand contains information about the
orientation of the ferromagnetic background, given by the angles θ and ϕ.
The expansion of the spin splitting |b(k)| gives its explicit dependence on the θ
and ϕ, which have the form
|b(k)| = b 0 (k)
∞
n=0
1/2
n
(cos γ (k))
n
,
(33)