Pauli matrices identity with no repeating indices

Question

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

$\sigma^i_{\alpha\beta}\sigma^j_{\gamma\delta}+\sigma^j_{\alpha\beta}\sigma^i_{\gamma\delta}+\delta^{ij}(\delta_{\alpha\beta}\delta_{\gamma\delta}-\vec\sigma_{\alpha\beta}\cdot\vec\sigma_{\gamma\delta})=\sigma^i_{\alpha\delta}\sigma^j_{\gamma\beta}+\sigma^j_{\alpha\delta}\sigma^i_{\gamma\beta}$.

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

$\sigma^{x}_{\alpha\beta}=\left(\begin{array}{cc}0&1\\1&0\end{array}\right)_{\alpha\beta},\sigma^{y}_{\alpha\beta}=\left(\begin{array}{cc}0&i\\-i&0\end{array}\right)_{\alpha\beta},\sigma^{z}_{\alpha\beta}=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)_{\alpha\beta}$.

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!

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 $\quad\sigma^i_{12}\sigma^j_{21} + \sigma^j_{12}\sigma^i_{21} = 0\quad \checkmark$

$i=j$:

$2\sigma^i_{\alpha\beta}\sigma^i_{\gamma\delta} +\delta_{\alpha\beta}\delta_{\gamma\delta} - \sigma^k_{\alpha\beta}\sigma^k_{\gamma\delta}= 2\sigma^i_{\alpha\delta}\sigma^i_{\gamma\beta} \ , \quad$ 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 ;)