I was just wondering if there is a proof of, or an example utilizing the following relation:


where $\{i,j\}=\{x,y,z\}$ and the Greek letters are spin indices. Written explicitly,


The nontrivial-ness is that no index is summed over, and I could not find a proof of this anywhere else. This relation seems to be consistent with everything I tried so far. Of course I can just enumerate all the index values, but does anyone happen to know this, or a similar identity? Thanks in advance!

  • $\begingroup$ Believe it or not, there are people that know lots of tricks with identities but don't remember every two index and three index Pauli matrix definition. So if you write the definitions there might be more people that can help you. $\endgroup$
    – Timaeus
    Sep 10, 2015 at 0:12
  • $\begingroup$ I did assume everyone that can help you knows what a single index Pauli matrix is. Tell us the double index and the triple index and I predict more people can help you. $\endgroup$
    – Timaeus
    Sep 10, 2015 at 2:00
  • $\begingroup$ OK.. I think you misunderstood me. The "triple indices" does not imply something like a rank-three tensor. the upper index denotes x,y,z and the two lower index are just regular 2*2 matrix indices. I explicitly keep them just to emphasize that this is not a matrix product. $\endgroup$ Sep 10, 2015 at 2:28

1 Answer 1


What I have to offer is not a beautiful formal prove, but a study of the different cases (which are not that many considering the symmetries and $2\times 2$ matrices).

$i\neq j$:

$\sigma^i_{\alpha\beta}\sigma^j_{\gamma\delta} + \sigma^j_{\alpha\beta}\sigma^i_{\gamma\delta} = \sigma^i_{\alpha\delta}\sigma^j_{\gamma\beta} + \sigma^j_{\alpha\delta}\sigma^i_{\gamma\beta}$

There are 8 possibilities for the indices ($2^4=16$ however, it is symmetric in $i\leftrightarrow j$)

$\alpha=\gamma \ , \ \beta=\delta\quad$ (4 poss.) $\quad\checkmark$

$\alpha\neq\gamma \ , \ \beta=\delta\quad$ (1 poss.) $\quad\checkmark$

$\alpha=\gamma \ , \ \beta\neq\delta\quad$ (1 poss.) $\quad\sigma^i_{11}\sigma^j_{12}+\sigma^j_{11}\sigma^i_{12} = \sigma^i_{12}\sigma^j_{11}+\sigma^j_{12}\sigma^i_{11}\quad$ $\checkmark$

$\alpha\neq\gamma \ , \ \beta\neq\delta\quad$ (2 poss.) $\quad\sigma^i_{11}\sigma^j_{22}= 0 \quad$ and $\quad\sigma^i_{12}\sigma^j_{21} + \sigma^j_{12}\sigma^i_{21} = 0\quad \checkmark$


$2\sigma^i_{\alpha\beta}\sigma^i_{\gamma\delta} +\delta_{\alpha\beta}\delta_{\gamma\delta} - \sigma^k_{\alpha\beta}\sigma^k_{\gamma\delta}= 2\sigma^i_{\alpha\delta}\sigma^i_{\gamma\beta} \ , \quad$ where $k$ is summed over.

Now there are only 4 possibilities for $\{\alpha,\beta,\gamma,\delta\}$, namely

$\{1,1,1,1\}\quad \Rightarrow \quad 2A + 1 - 1 = 2A\quad \checkmark$

$\{1,1,1,2\}\quad \Rightarrow \quad 0=0 \quad \checkmark$

$\{1,1,2,1\}\quad \Rightarrow \quad 0=0 \quad \checkmark$

$\{1,2,2,1\}\quad \Rightarrow \quad \begin{cases}i=\{x,y\} \quad &2-1-1=0 \\ i=z \quad &0 -1-1= -2 \end{cases}\quad \checkmark$

I confess, it is not very elegant and I too usually dislike brute force methods. Unfortunately, that is all I can come up with right now ;)


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