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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.

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1 Answer 1

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You should do what you should always do in such a situation:

  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 $$ \frac1N\sum_{k=0}^{N-1} e^{ik(n-n')} = \delta_{n,n'}\ . $$

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