Calculation of the 1-dimensional Magnon dispersion relationship

Magnons are exitations from the ferromagnetic ground state, where all spins are aligned parallel. The lowest excited state in a ferromagnet is the case, where the folding of one spin is evenly distributed on all the spins. The spins precess around the z-axis and neighboring spins have a phase shift of a constant angle. This leads to wavelike excitations, called spinwaves. In analogy to phonons, which are oscillations of the lattice, magnons are oscillations of the spins relative to the lattice. From this a linear spring problem can be generated. By finding the corrosponding normal modes, the dispersion relationship can be calculated

The Energy \(U\) of a 1-dimentional spin-chain, with \(N\) spins with the absolute value \(S\), can be described by the Heisenberg-model: $$ U={-2J\sum_{p=1}^{N-1} S_p S_{p+1}} $$

where \(J\) is the exchange energy. The terms that include the p-th spin are: $$ -2JS_p (S_{p-1}+S_{p+1}) $$

With the spin expressed in terms of the magnetic moment \(\mu_p=-g\mu_BS_p \), where \(\mu_B\) is the Bohr magneton and \(g\) is the g-factor, the equation above becomes: $$ \mu_p\frac{2J}{g\mu_B} (S_{p-1}+S_{p+1}) $$

and the term \((\frac{-2J}{g\mu_B}) (S_{p-1}+S_{p+1})\) can be identified as the applied Magnetic field of the neighboring spins \(B_p\) on spin \(S_p\). This magnetic field will exert a force, that aligns the magnetic moment \(\mu_p\) in the direction of \(B_p\), which is equal to the time derivative of the angular momentum of the spin \( \hbar S_p\): $$ \hbar\frac{dS_p}{dt}=\mu_p \times B_p = 2J(S_p \times S_{p-1}+S_p \times S_{p+1}) $$

This results in 3 non-linear equations. They can be linearized under the assumption, that the amplitude of the exitations is very small (\(S_p^x,S_p^y \ll S\)), by setting \(S_p^z=S\) and neglecting the products of \(S_p^x\) and \(S_p^y\). The linearized equations are: $$ \hbar\frac{dS_p^x}{dt}= 2JS (2S_{p}^y-S_{p-1}^y-S_{p+1}^y ) $$ $$ \hbar\frac{dS_p^y}{dt}= -2JS (2S_{p}^x-S_{p-1}^x-S_{p+1}^x ) $$ $$ \hbar\frac{dS_p^z}{dt}= 0 $$

The solutions to these equations are plane waves: $$ \left( \begin{array}{b} S_p^x \\ S_p^y \end{array} \right)= \left( \begin{array}{b} u_k^x \\ u_k^y \end{array} \right)\exp[i(pka-\omega t)] $$

Where \(u_k^x\) and \(u_k^y\) are constants, \(a\) is the lattice constant and \(p\). Putting the plane wave solutions into the linearized equations, results in the following system of equations: $$-i\hbar\omega u_k^x = 2JS(2-\exp(-ika)-\exp(ika)) u_k^y = 4JS(1-\cos(ka))u_k^y$$ $$-i\hbar\omega u_k^y = -2JS (2-\exp(-ika)-\exp(ika)) u_k^x = -4JS(1-\cos(ka))u_k^x$$

This homogeneous linear system of equations has solutions for \(u_k^x\) and \(u_k^y\) if the determinant of coefficients is zero: \[\begin{vmatrix} -i\hbar\omega & 4JS(1-\cos(ka)) \\ -4JS(1-\cos(ka)) & -i\hbar\omega \end{vmatrix}=0 \]

which results in the dispersion relationship for 1 dimensional magnons: $$\hbar\omega=4JS(1-\cos(ka))$$

The result is plotted in the following figure:

$\large \frac{\omega\hbar}{4JS}$

$\large ka$