# Pauli commutation relation in k-space

Pauli commutation relation is given as

$$[\sigma^x,\sigma^y]=2i\epsilon_{xyz}\sigma^z$$

In this article, they first define a model on a lattice with $$N$$ sites, then they use the following relation for Pauli commutation in k-space

$$\frac{1}{2}[\sigma_k^z,\sigma_{k'}^x]=\frac{i}{\sqrt{N}}\sigma_{k+k'}^y$$

I was trying to prove this identity but I am not successful. Please help. My attempt is given below:

On a real space lattice, I start with commutation relation as $$[\sigma_n^z,\sigma_n^x]=2i\sigma_n^y$$ where $$n$$ show site number $$[\sigma_n^z,\sigma_n^x] =\sigma_n^z\sigma_n^x-\sigma_n^x\sigma_n^z\\ =\sum_{k,k'}\frac{1}{\sqrt{N}}e^{ikr_n}\sigma_k^z\frac{1}{\sqrt{N}}e^{ik'r_n}\sigma_{k'}^x - \sum_{k,k'}\frac{1}{\sqrt{N}}e^{ikr_n}\sigma_{k'}^x\frac{1}{\sqrt{N}}e^{ik'r_n}\sigma_{k}^z\\ =\sum_{k,k'}\frac{1}{N}e^{i(k+k')r_n}\sigma_k^z\sigma_{k'}^x-\sum_{k,k'}\frac{1}{N}e^{i(k+k')r_n}\sigma_{k'}^x\sigma_{k}^z\\ =\sum_{k,k'}\frac{1}{N}([\sigma_k^z,\sigma_{k'}^x])e^{i(k+k')r_n}$$

and $$2i\sigma_n^y=\frac{2i}{\sqrt{N}}\sum_k e^{ikr_n}\sigma_k^y$$ so, we have:

$$\frac{1}{2}\frac{1}{N}\sum_{k,k'}[\sigma_k^z,\sigma_{k'}^x]e^{i(k+k')r_n}=\frac{i}{\sqrt{N}}\sum_k e^{ikr_n}\sigma_k^y$$ How to move further? Thank you.

1. Start with the left side of what you want to show: $$[\sigma_k^z,\sigma_{k'}^x]\ .$$
2. Insert what you know: How is $$\sigma_k$$ expressed in terms of $$\sigma_n$$? Then use the commutation relation of the $$\sigma_n$$.
Finally, don't forget that $$\frac1N\sum_{k=0}^{N-1} e^{ik(n-n')} = \delta_{n,n'}\ .$$