Another side issue is whether the exchange “constants” that work well above the Curie temperature also work well below the Curie temperature. Since the development of the Heisenberg Hamiltonian was only phenomenological, this is a sensible question to ask. It is particularly sensible since J depends on R and R increases as the temperature is increased (by thermal expansion). Charap and Boyd9 and Wojtowicz10 have shown for EuS (which is one of the few “ideal” Heisenberg ferromagnets) that the same set of J will fit both the low-temperature specific heat and magnetization and the high-temperature specific heat.
We have made many approximations in developing the Heisenberg Hamiltonian. The use of the Heitler–London method is itself an approximation. But there are other ways of understanding the binding of the hydrogen atoms and hence of developing the Heisenberg Hamiltonian. The Hund–Mulliken11method is one of these techniques. The Hund–Mulliken method should work for smaller R, whereas the Heitler–London works for larger R. However, they both qualitatively lead to a Heisenberg Hamiltonian.
We should also mention the Ising model, where H = −∑Jijσizσjz, and the σ are the Pauli spin matrices. Only nearest-neighbor coupling is commonly used. This model has been solved exactly in two dimensions (see Huang [7.32 p341ff]). The Ising model has spawned a huge number of calculations.
The Heisenberg Hamiltonian and its Relationship to the Weiss Mean Field Theory (B)
We now show how the mean molecular field arises from the Heisenberg Hamiltonian. If we assume a mean field γM then the interaction energy of moment μk with this field is
Ek = −μ0γM μk .
(7.89)
Also from the Heisenberg Hamiltonian
Ek = −∑i′ Jik Si Sk − ∑ j′ Jkj Sk S j ,
and since Jij = Jji, and noting that j is a dummy summation variable
Ek = −2∑i′ Jik Si Sk .
(7.90)
9See [7.10].
10See Wojtowicz [7.70].
11See Patterson [7.53 p176ff].
378 7 Magnetism, Magnons, and Magnetic Resonance
In the spirit of the mean-field approximation we replace Si by its average S¯i = S since the average of each site is the same. Further, we assume only nearestneighbor interactions so Jik = J for each of the Z nearest neighbors. So
Ek −2ZJS Sk .
(7.91)
But
μk −
gμB Sk
(7.92)
(with μB = |e| /2m), and the magnetization M is
M −
NgμB S
,
(7.93)
where N is the number of atomic moments per unit volume (≡ 1/Ω, where Ω is the atomic volume). Thus we can also write
Ek
−2ZJ ΩM μk 2 .
(7.94)
(gμB )2
Comparing (7.89) and (7.94)
μ γ (gμ
B
)2
J =
0
.
(7.95)
2ZΩ 2
This not only shows how Heisenberg’s theory “explains” the Weiss mean molecular field, but also gives an approximate way of evaluating the parameter J. Slight modifications in (7.95) result for other than nearest-neighbor interactions.
RKKY Interaction12 (A)
The Ruderman, Kittel, Kasuya, Yosida, (RKKY) interaction is important for rare earths. It is an interaction between the conduction electrons with the localized moments associated with the 4f electrons. Since the spins cause the localized moments, the conduction electrons can mediate an indirect exchange interaction between the spins. This interaction is called RKKY interaction.
We assume, following previous work, that the total exchange interaction is of the form
H exTotal = −∑i,α J x (ri − Rα )Sα Si ,
(7.96)
12 Kittel [60, pp 360-366] and White [7.68 pp 197-200].
7.2 Origin and Consequences of Magnetic Order
379
where Sα is an ion spin and Si is the conduction spin. For convenience we assume the S are dimensionless with absorbed in the J. We assume Jx(ri − Rα) is short range (the size of 4f orbitals) and define
J = ∫ J x (r − Rα )dr .
(7.97)
Consistent with (7.97), we assume
J x (ri − Rα ) = Jδ (r) ,
(7.98)
where r = ri − Rα and write
H ex = −JSα Siδ (r)
for the exchange interaction between the ion α and the conduction electron. This is the same form as the Fermi contact term, but the physical basis is different. We can regard Siδ(r) = Si(r) as the electronic conduction spin density. Now, the interaction between the ion spin Sα and the conduction spin Si can be written (gaussian units, μ0 = 1)
−JSα Siδ (r) = −(−gμB Si ) Heff (r) ,
so this defines an effective field
Heff
= −
JSα
δ(r) .
(7.99)
gμB
The Fourier component of the effective field can be written
Heff (q) = ∫ Heff (r)e−iq r dr = −
J
Sα .
(7.100)
gμB
We can now determine the magnetization induced by the effective field by use of the magnetic susceptibility. In Fourier space
χ(q) =
M (q) .
(7.101)
H (q)
This gives us the response in magnetization of a free-electron gas to a magnetic field. It turns out that this response (at T = 0) is functionally just like the response to an electric field (see Sect. 9.5.3 where Friedel oscillation in the screening of a point charge is discussed).
We find
χ(q) =
3g 2
μB2
N
A(q / 2kF ) ,
(7.102)
8EF
V
380 7 Magnetism, Magnons, and Magnetic Resonance
where N/V is the number of electrons per unit volume and
) = 1
kF
q
2
2kF + q
A(q / 2k
F
+
1
−
ln
.
(7.103)
2
2
2q
4kF
2kF − q
The magnetization M(r) of the conduction electrons can now be calculated from (7.101), (7.102), and (7.103).
M (r) =
1
∑
q
M (q)eiq r
V
=
1
∑q χ(q)Heff (q)eiq r
(7.104)
V
J
Sα ∑q χ(q)eiq r .
= −
gμBV
With the aid of (7.102) and (7.103), we can evaluate (7.104) to find
M (r) = −
J
KG(r)S ,
(7.105)
gμB
α
where
K =
3g 2μB2
N
kF3
,
(7.106)
8EF
V 16π
and
G(r) =
sin(2kF r) − 2kF r cos(2kF r)
.
(7.107)
(kF r)4
The localized moment Sα causes conduction spins to develop an oscillating polarization in the vicinity of it. The spin-density oscillations have the same form as the charge-density oscillations that result when an electron gas screens a charged impurity.13
Let us define
F(x) = sin x − x cos x , x4
so
G(r) = 24 F(2kF r) .
F(x) is the basic function that describes spatial oscillating polarization induced by
a localized moment in its vicinity. It is sketched in Fig. 7.8. Note as x → ∞, F(x) → –cos(x)/x3and as x → 0, F(x) → 1/(3x).
13 See Langer and Vosko [7.42].
7.2 Origin and Consequences of Magnetic Order
381
F(x)
x
Fig. 7.8. Sketch of F(x) = [sin(x) – x cos(x)]/x4, which describes the RKKY exchange interaction
Using (7.105), if S(r) is the spin density,
S(r) =
M (r)
=
J
KGS .
(7.108)
(−gμB )
(gμB )2
α
Another localized ionic spin at Sβ interacts with S(r)
H αindirectand β = −JSβ S(rα − rβ ) .
Now, summing over all α, β interactions and being careful to avoid double counting spins, we have
H
RKKY
= − 1
∑
J
αβ
S
S
β
,
(7.109)
2
α,β
α
where
J
αβ
=
J 2
KG(r = r
) .
(7.110)
(gμB )2
αβ
For strong spin-orbit coupling, it would be more natural to express the Hamiltonian in terms of J (the total angular momentum) rather than S. J = L + S and within the set of states of constant J, gJ is defined so
gJ μBJ = μB (L + 2S) = μB (J + S) ,
where remember the g factor for L is 1, while for spin S it is 2. Thus, we write (gJ −1)J = S .
If Jα is the total angular momentum associated with site α, by substitution
H
RKKY
= − 1
(g
J
−1)2
∑
α,β
J
αβ
J
α
J
β
,
(7.111)
2
where (gJ − 1)2 is called the deGennes factor.
382 7 Magnetism, Magnons, and Magnetic Resonance
Magnetic Structure and Mean Field Theory (A)
We assume the Heisenberg Hamiltonian where the lattice is assumed to have transitional symmetry, R labels the lattice sites, J(0) = 0, J(R − R′) = J(R′ − R). We wish to investigate the ground state of a Heisenberg-coupled classical spin system, and for simplicity, we will assume:
a.T = 0 K
b.The spins can be treated classically
c.A one-dimensional structure (say in the z direction), and
d.The SR are confined to the (x,y)-plane
SRx = S cosϕR , SRy = S sinϕR .
Thus, the Heisenberg Hamiltonian can be written:
H= − 12 ∑R,R′ S 2J (R − R′) cos(ϕR −ϕR′) .
e.We are going to further consider the possibility that the spins will have a constant turn angle of qa (between each spin), so φR = qR, and for adjacent spins
φR = q R = qa.
Substituting (in the Hamiltonian above), we find
H = −
NS 2
J (q) ,
(7.112)
2
where
J (q) = ∑R J (R)eiqR
(7.113)
and J(q) = J(−q). Thus, the problem of finding Hmin reduces to the problem of finding J(q)max.
x
φR
a
a
z
R
y
Fig. 7.9. Graphical depiction of the classical spin system assumptions
7.2 Origin and Consequences of Magnetic Order
383
q = 0, get ferromagnetism,
Note if J(q) → max for q = π/a, get antiferromagnetism,
qa ≠ 0 or π, get heliomagnetism with qa defining the turn angles.
It may be best to give an example. We suppose that J(a) = J1, J(2a) = J2 and the rest are zero. Using (7.113) we find:
J (q) = 2J1 cos(qa) + 2J2 cos(2qa) .
(7.114)
For a minimum of energy [maximum J(q)] we require
∂J
= 0 → J1 = −4J2 cos(qa)
or
q = 0 or π
,
∂q
a
and
∂
2 J
< 0 or J1 cos(qa) > −4J2 cos(2qa) .
∂q2
The three cases give:
q = 0
q = π/a
q ≠ 0, π/a
J1 > −4J2
J1 < 4J2
Turn angle qa defined by
Ferromagnetism
Antiferromagnetism
cos(qa) = −J1/4J2 and
e.g. J1 > 0, J2 = 0
e.g. J1 < 0, J2 = 0
J1cos(qa) > −4J2cos(2qa)
7.2.2 Magnetic Anisotropy and Magnetostatic Interactions (A)
Anisotropy
Exchange interactions drive the spins to lock together at low temperature into an ordered state, but often the exchange interaction is isotropic. So, the question arises as to why the solid magnetizes in a particular direction. The answer is that other interactions are active that lock in the magnetization direction. These interactions cause magnetic anisotropy.
Anisotropy can be caused by different mechanisms. In rare earths, because of the strong-spin orbit coupling, magnetic moments arise from both spin and orbital motion of electrons. Anisotropy, then, can be caused by direct coupling between the orbit and lattice.
There is a different situation in the iron group magnetic materials. Here we think of the spins of the 3d electrons as causing ferromagnetism. However, the spins are not directly coupled to the lattice. Anisotropy arises because the orbit “feels” the lattice, and the spins are coupled to the orbit by the spin-orbit coupling.
Let us first discuss the rare earths, which are perhaps the easier of the two to understand. As mentioned, the anisotropy comes from a direct coupling between the crystalline field and the electrons. In this connection, it is useful to consider
384 7 Magnetism, Magnons, and Magnetic Resonance
the classical multipole expansion for the energy of a charge distribution in a potential Φ. The first three terms are given below:
− p E(0)
1
∑i, j
∂Ej
(7.115)
u = qΦ(0)
−
Qij
+ higher-order terms.
6
∂xi 0
Here, q is the total charge, p is the dipole moment, Qij is the quadrupole moment, and the electric field is E = − Φ. For charge distributions arising from states with definite parity, p = 0. (We assume this, or equivalently we assume the parity operator commutes with the Hamiltonian.) Since the term qΦ(0) is an additive constant, and since p = 0, the first term that merits consideration is the quadrupole term. The quadrupole term describes the interaction of the quadrupole moment with the gradient of the electric field. Generally, the quadrupole moments will vary with |J, M (J = total angular momentum quantum number and M refers to the z component), which will enable us to construct an effective Hamiltonian. This Hamiltonian will include the anisotropy in which different states within a manifold of constant J will have different energies, hence anisotropy. We now develop this idea in quantum mechanics below.
We suppose the crystal field is caused by an array of charges described by ρ(R). Then, the potential energy of −e at the point ri is given by
V (ri ) = −∫
eρ
(R)dR
.
(7.116)
4πε
0
r − R
i
If we further suppose ρ(R) is outside the ion in question, then in the region of the ion, V(r) is a solution of the Laplace equation, and we can expand it as a solution of this equation:
V (ri ) = ∑l,m BlmrlYlm (θ,φ) ,
(7.117)
where the constants Blm can be computed from ρ(R). For rare earths, the effects of the crystal field, typically, can be adequately calculated in first-order perturbation theory. Let |v be all states |J, M , which are formed of fixed J manifolds from |l, m , and |s, ms where l = 3 for 4f electrons. The type of matrix element that we need to evaluate can be written:
v ∑iV (ri ) v′ ,
(7.118)
summing over the 4f electrons. By (7.117), this eventually means we will have to evaluate matrix elements of the form
lm Y m′
lm′
,
(7.119)
i l′
i
and since l = 3 for 4f electrons, this must vanish if l′ > 6. Also, the parity of the functions in (7.119) is (−)2l+l′ the matrix element must vanish if l′ is odd since
7.2 Origin and Consequences of Magnetic Order
385
2l = 6, and the integral over all space is of an odd parity function is zero. For 4f electrons, we can write
V (ri ) = ∑l6′=0
∑m′ Blm′ ′rl′Ylm′ ′(θ,φ) .
(7.120)
(even)
We define the effective Hamiltonian as
H A = ∑i V (ri )
doing radial integrals only .
If we then apply the Wigner–Eckhart theorem [7.68 p33], in which one replaces (x′/r), etc. by their operator equivalents Jx, etc., we find for hexagonal symmetry
H
A
= K J 2
+ K
2
J 4
+ K
3
J 6
+ K
4
(J 6
+ J 6 ) ,
(J
±
= J
x
± iJ
y
) . (7.121)
1 z
z
z
+
−
We now discuss the anisotropy that is appropriate to the iron group [7.68 p57]. This is called single-ion anisotropy. Under the action of a crystalline field we will assume the relevant atomic states include a ground state (G) of energy ε0 and appropriate excited (E) states of energy ε0 + . We will consider only one excited state, although in reality there would be several. We assume |G and |E are separated by energy .
The states |G and |E are assumed to be spatial functions only and not spin functions. In our argument, we will carry the spin S along as a classical vector. The argument we will give is equivalent to perturbation theory.
We assume a spin-orbit interaction of the form V = λL S, which mixes some of the excited state into the ground state to produce a new ground state.
G → GT = G + a E ,
(7.122)
where a is in general complex. We further assume G|G = E|E = 1 and E|G = 0 so GT|GT = 1 to O(a). Also note the probability that |E is contained in |GT is |a|2. The increase in energy due to the mixture of the excited state is (after some algebra)
ε1 =
GT H GT −ε0
= aE + G H aE + G
−ε0,
GT GT
1+ a 2
or
ε1 =
a
2 .
(7.123)
In addition, due to first-order perturbation theory, the spin-orbit interaction will cause a change in energy given by
ε2 = λ GT L GT S .
(7.124)
We assume the angular momentum L is quenched in the original ground state so by definition G|L|G = 0. (See also White, [7.68 p43]. White explains that if
386 7 Magnetism, Magnons, and Magnetic Resonance
a crystal field removes the orbital degeneracy, then the matrix element of L must be zero. This does not mean the matrix element of L2 in the same state is zero.) Thus to first order in a,
ε2 = λa* E L G S + λa G L E S .
(7.125)
The total change in energy given by (7.123) and (7.125) is ε = ε1 + ε2. Since a and a* are complex with two components we can treat them as linearly independent, so ∂ε/∂a* = 0, which gives
a = − E λL G S .
Therefore, after some algebra ε = ε1 + ε2 becomes
ε = −a 2 = − E λL G S 2
< 0 ,
a decrease in energy. If we let
A = E λL G ,
then
ε = −A SA*S = −∑μ,ν Sμ Bμν Sν ,
where Bμν = AμAν*. If we let S become a spin operator, we get the following Hamiltonian for single-ion anisotropy:
H spin = −∑μ,ν Sμ Bμν Sν .
(7.126)
When we have axial symmetry, this simplifies to
H spin = −DSz2 .
For cubic crystal fields, the quadratic (in S) terms go to a constant and can be neglected. In that case, we have to go to a higher order. Things are also more complicated if the ground state has orbital degeneracy. Finally, it is also possible to have anisotropic exchange. Also, as we show below, the shape of the sample can generate anisotropy.
Magnetostatics (B)
The magnetostatic energy can be regarded as the quantity whose reduction causes domains to form. The other interactions then, in a sense, control the details of how the domains form. Domain formation will be considered in Sect. 7.3. Here we will show how the domain magnetostatic interaction can cause shape anisotropy.
