I have some doubt about the Bose-Einstein distribution for magnons/spin-waves.

A one-dimensional ferromagnet placed in an external magnetic field $\mathbf{B} = B\, \hat{z}$ obeys the Hamiltonian

$$H = - |J|\sum_i \mathbf{S}_i \cdot \mathbf{S}_{i+1} - B \sum_i {S}^z_i. ,$$ where we set the lattice constant $a = 1$.

By utilizing the Holstein-Primakoff transformation and the Fourier transform we can diagonalize this Hamiltonian and find the dispersion relation for the magnons $$\hbar \omega = 2 J S (1-\cos k) + B$$ where $k$ is the wavenumber of the magnon.

We know that magnons are bosons and should satisfy Bose-Einstein statistics. So the Bose-Einstein distribution for magnons is

$$n_k = \frac{1}{e^{\frac{\hbar \omega - \mu}{k_bT}}-1} = \frac{1}{e^{\frac{2 \hbar J S (1-\cos k) + B - \mu}{k_bT}}-1}.$$

However, shouldn't the distribution be normalized in the sense that if we integrate it over the Brilluin zone we should get unity $$\int_{1. BZ} n_k dk = 1?$$


1 Answer 1


Magnons are quasiparticles, so their chemical potential $\mu$ is equal to zero. Correspondingly the number of magnons is not fixed and is defined by temperature. We use the following approximation for energy levels of a ferromagnet $$ E(\{n\}) = E_0 + \sum_{k}\hbar\omega_k n_k, $$ where $n_k$ is the number of magnons with quasimomentum $k$. Hence the partition function of a ferromagnet looks like a grand partition function of a system of magnons with zero chemical potential. The mean number of magnons with quasimomentum $k$ equals to $$ \overline{n}_k = \frac1{e^{\hbar\omega_k/k_bT}-1}. $$ The overall number of magnons depends on temperature and is equal to $$ N = \sum_k \overline{n}_k $$ When $B>0$ we have $N\to0$ if $T\to0$.


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