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Consider the four $2\times 2$ matrices $\{\sigma_\mu\}$, with $\mu = 0,1,2,3$, which are defined as follows $$ \sigma_0 =\left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right) $$ $$ \sigma_1 =\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right) $$ $$ \sigma_2 =\left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right) $$ $$ \sigma_3 =\left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right) $$ i.e. the identity matrix and the three Pauli matrices. For the trace of the product of any two matrices $\sigma_\mu$ one has the identity $\text{tr}(\sigma_\mu \sigma_\nu)= 2 \delta_{\mu \nu}$. I was wondering if a similar identity can be derived for the product of three sigma matrices, $$ \text{tr}(\sigma_\mu \sigma_\nu \sigma_\lambda)= \;? $$

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First I note a few things about the defined matrices:

1) Commutation: $[\sigma_\mu,\sigma_\nu] = 2 i \epsilon_{0 \mu \nu \rho} \sigma_\rho$

2) Anti-commutation: $\{ \sigma_\mu ,\sigma_\nu \} = 2 \delta_{\{\mu \nu} \sigma_{0\}} - 4 \delta_{\mu 0} \delta_{\nu 0} \sigma_0$

Where $a_{ \{b c} d_{i\}} = a_{bc} d_i + a_{ib} d_{c} + a_{ci}d_b$

This gives: $\sigma_\mu \sigma_\nu = i \epsilon_{0 \mu \nu \rho} \sigma_\rho + \delta_{\{\mu \nu} \sigma_{0\}}-\delta_{\mu 0} \delta_{\nu 0} \sigma_0$

Thus: \begin{eqnarray} \text{tr}(\sigma_\mu \sigma_\nu \sigma_\rho) &=& \text{tr}(i \epsilon_{0 \mu \nu \alpha} \sigma_\alpha\sigma_\rho + \delta_{\{\mu \nu} \sigma_{0\}}\sigma_\rho-2\delta_{\mu 0} \delta_{\nu 0} \sigma_\rho)\\ &=& 2 i \epsilon_{0\mu\nu\rho}+2\delta_{\{\mu \nu}\delta_{0\}\rho}-4\delta_{\mu0}\delta_{\nu 0} \delta_{\rho 0 } \end{eqnarray} where I have used the trace identity sketched out in the problem statement.

We can check the results of this expression by confirming that if one of the three, say $\mu=0$ then we get back the original identity. We can also look at what happens if $\mu ,\nu,\rho \ne0$. In this case we would get $2i\epsilon_{0\mu \nu \rho}$ this matches the other answer.

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You can check that $\sigma_1\sigma_2\sigma_3 = i I$ so $\mathrm{tr}(\sigma_1\sigma_2\sigma_3) = 2i$. By cyclic invariance of the trace, we also have $$\mathrm{tr}(\sigma_1\sigma_2\sigma_3) = \mathrm{tr}(\sigma_2\sigma_3\sigma_1) = \mathrm{tr}(\sigma_3\sigma_1\sigma_2) = 2i$$ Since $\sigma_r \sigma_s = - \sigma_s \sigma_r$ for distinct $r,s \in \{1,2,3\}$, we also get $$\mathrm{tr}(\sigma_2\sigma_1\sigma_3) = \mathrm{tr}(\sigma_3\sigma_2\sigma_1) = \mathrm{tr}(\sigma_1\sigma_3\sigma_2) = -2i$$ On the other hand, if two of $i,j,k \in \{1,2,3\}$ are equal then, from the the commutation relation mentioned above and the fact that $\sigma_r^2 = I$ for $r \in \{1,2,3\}$, you can conclude that $\sigma_i \sigma_j \sigma_k \in \{ \pm \sigma_1 ,\pm \sigma_2 , \pm \sigma_3\}$ and, in particular, $\mathrm{tr}(\sigma_i \sigma_j \sigma_k) = 0$.

This can be summarized by saying that, for $i,j,k \in \{1,2,3\}$, $$\mathrm{tr}(\sigma_i \sigma_j \sigma_k) = \varepsilon_{ijk} 2i$$ where $\varepsilon_{ijk}$ is the Levi-Cevita symbol given by $$\varepsilon_{ijk} = \begin{cases} 1 && \text{ if } ijk \text{ is an even permutation of } 123 \\ -1 && \text{ if } ijk \text{ is an odd permutation of } 123 \\ 0 && \text{ if } ijk \text{ is not a permutation of } 123 \\ \end{cases}$$

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  • $\begingroup$ What about adding in the $\sigma_0$ to the mix? $\mathtt{tr}(\sigma_0 \sigma_i \sigma_i)$? $\endgroup$ – Kitter Catter Nov 2 '16 at 18:17
  • $\begingroup$ @Kitter Catter: Since $\sigma_0$ is just the identity, I don't think it's worth trying to include it in considerations. You might argue that would mean we lose information in the case of two matrices, where we had $\mathrm{tr}(\sigma_i \sigma_j) = 2 \delta_{ij}$ for $i,j \in \{0,1,2,3\}$, but I tend to think of that a strange way of combining two facts about $\sigma_1,\sigma_2,\sigma_3$ into a single statement. Instead, I would break that down into (i) $\mathrm{tr}(\sigma_i \sigma_j) = 2 \delta_{ij}$ for $i,j \in \{1,2,3\}$ and (ii) $\mathrm{tr}(\sigma_i) = 0$ for $i \in \{1,2,3\}$. $\endgroup$ – Mike F Nov 2 '16 at 18:28
  • $\begingroup$ Wouldn't it make sense to at least comment on that in your answer then? Otherwise it seems like you aren't fully addressing the question $\endgroup$ – Kitter Catter Nov 2 '16 at 18:36
  • $\begingroup$ @KitterCatter: Yeah it would have been good to mention. Anyway, the cat is out of the bag here in the comment section! $\endgroup$ – Mike F Nov 2 '16 at 18:46

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