Advanced Solid State Physics

Figure 77: (a) Ground state of a linear ferromagnetic chain at zero temperature, (b) excited state
where one spin is flipped, (c) excite state where the spin flip is equally distributet among
all available spins resulting in a spin wave. [from Kittel]

To calculate the dispersion relation for Magnons we examine thekthspin in equation 135

Uk=− 2 JS~k·(S~k− 1 +S~k+1) (141)

. Inserting the relation between magnetic moment and spin~μ=γS~=−gμBS~into this relation we
obtain

Uk=−~μk·(

− 2 J

gμB

(S~k− 1 +S~k+1)) (142)

. Because the potential energy of a magnetic dipole isU=−~μ·B~we may interpret

B~k=−^2 J

gμB

(S~k− 1 +S~k+1) (143)

as the effective magnetic field on spin k due to the nearest neighbors. The rate of change of angular
momentum is the torqueT~ =d
~S
dt =~μ×
B~. Using the torque we set up equations of motion for the
spin

d

dt

S~k = −gμB

~

S~k×B~k (144)

=

2 J

~

(S~k×S~k− 1 +S~k×S~k+1) (145)

. The cartesian components of this relation are

d

dt

S~xk =^2 J

~

(Sky(Szk− 1 +Skz+1)−Szk(Sky− 1 +Sky+1)) (146)

d

dt

S~ky =^2 J

~

(Skz(Skx− 1 +Skx+1)−Skx(Skz− 1 +Skz+1)) (147)

d

dt

S~kz =^2 J

~

(Skx(Syk− 1 +Syk+1)−Sky(Sxk− 1 +Skx+1)) (148)

. One way to solve these nonlinear set of equations it is to linearize them. This is valid in the low
energy limit since we already know that low-lying excitations have a form of 77 (c) and therefore