MAGNETIC STRUCTURES - Rare Earth Magnetism: Structures and Excitations - page 104

F Z = , are included explicitly in F ( θ 0 ,φ 0 ) and its deriva-

tives. Introducing the expression (2.2.17) for the free energy, in the
two cases where the moments are either parallel or perpendicular to the
c -axis, we find

− Ngµ B H · J

(3 κ 2 +10 κ 4 +21 κ 6 ) / ( σJ ) 2

1 /χ xx =1 /χ yy = − ; θ 0 =0 , (2 . 2 . 19 a )

or

1 /χ xx =(3 κ 2 − 1 2 κ 4 + 10 8 κ 6 +6 κ 6 | ) / ( σJ ) 2 |

π
2 ,

; θ 0 =

1 /χ yy =36 |κ 6 |/ ( σJ ) 2

(2 . 2 . 19 b )
which must be positive if the structure is to be stable. In order to deter-
mine the higher derivatives of the free energy, a transverse field greater
than that corresponding to the linear regime described by the (zero-field)
susceptibility must be applied. The application of a large magnetic field
perpendicular to the magnetization axis, in a strongly anisotropic sys-
tem, creates a large mechanical torque, which may cause practical prob-
lems with maintaining the orientation of the crystal. If the experimental
facilities do not allow the determination of the higher derivatives, the
different temperature dependences of the various anisotropy parameters
may yield a rough separation of their contributions to the total axial
anisotropy. However the Callen-Callen theory is an approximation, the
corrections to which are important if the anisotropy is large, and there
are other contributions to the free energy than those which we have
considered above.
The results derived above are only valid if the anisotropy energies
are small compared to the exchange energy. In order to demonstrate the
kind of modifications which may appear in higher order, we shall consider
the simplest possible case, where only B 2 is non-zero, and we shall only
calculate the free energy at zero temperature in the MF approximation,
i.e. the ground-state energy of a single site subjected to the exchange
field h ex = J z J ( 0 ), with J z = σJ . In this case, the MF Hamiltonian

(2.1.16) is

( J z − 2

H h ( J z cos θ + J x sin θ )

+ B 2 3 J z cos 2 θ +3 J x sin 2 θ + 2

− σJ ) σJ J −

= ( 0 )

J ( J +1) ,
(2 . 2 . 20)
in an applied field h along the ζ -axis. With the J z -eigenstates as the
basis, the leading-order ground-state energy is

( J z J x + J x J z )sin2 θ −

− 2

E 0 = <J σ ) σJ 2 hJ cos θ + B 2 J (2) (3 cos 2 θ |H| J> = − (1 J ( 0 ) − − 1) .