# Bose-Einstein distribution and magnons

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 $$\int_{1. BZ} n_k dk = 1?$$

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 $$T\to0$$.