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If I define an object:

$$S = \gamma^a A_a + \frac{i}{4}\gamma^a[\gamma^{b},\gamma^{c}] B_{abc}.$$

Is there a way (using dirac matrix formula) to seperate out the terms to either get an expression just involving $A$ or just $B$?

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  • $\begingroup$ 1. What are $A$ and $B$? Matrices? Numbers? Mystery objects? 2. What do you mean by "separating out the terms"? Perhaps by analogy: If I write down a vector contraction $v_\mu q^\mu$ for some vectors $v,q$, is there a way to "separate out $q$? $\endgroup$ – ACuriousMind Jan 30 at 18:15
  • $\begingroup$ A and B are any tensors you like. e.g. multiply some gamma matrices on the left and right of S in order to get a term just involving A or B. $\endgroup$ – zooby Jan 30 at 18:22
  • $\begingroup$ Are $S$, $A$, and/or $B$ covariantly conserved? If so you can take the divergence to vanish, giving you another equation to solve for $A$ and/or $B$. $\endgroup$ – R. Rankin Jan 30 at 21:29
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I'm guessing B should probably be fully anti-symmetric, but let's not assume that.

Without loss of generality we can assume B is anti-symmetric in $b,c$, as it's symmetric part does not contribute, and would therefore not be extractable from the equation in any way. Then we can rewrite: $$ \frac{i}{4}\gamma^a[\gamma^{b},\gamma^{c}] B_{abc}=\frac{i}{2}\gamma^a\gamma^{b}\gamma^{c} B_{abc} $$

$$ \frac{i}{2}\gamma^a\gamma^{b}\gamma^{c} B_{abc} = \frac{i}{4}[\gamma^a,\gamma^{b}]\gamma^{c} B_{abc} +\frac{i}{4}\{\gamma^a,\gamma^{b}\}\gamma^{c} B_{abc} \\= \frac{i}{4}[\gamma^a,\gamma^{b}]\gamma^{c} B_{abc} +\frac{i}{2}\gamma^{c} g^{ab}B_{abc} $$

Now the first term is only sensitive to the antisymmetric (in $a,b$) part of $B$ and the symmetric part enters the equation only through the vector $\tilde{B}_c=\frac{i}{2} g^{ab}B_{abc}$, which we can 'absorb' in the definition of $A$. In other words, the equation cannot distinguish between $A$ and 'symmetric components of $B$'. Therefore, without loss of generality we can rephrase the question in terms of a fully antisymmetric $B$ and the equation:

$$ S = \gamma^a A_a + \frac{i}{2}\gamma^a\gamma^{b}\gamma^{c} B_{abc} $$

Then using trace identities and $B$'s anti-symmetry:

$$ {\rm tr}(\gamma^a S) = 4 A^a$$ $$ {\rm tr}(\gamma^5\gamma^a S) = 2 \epsilon^{abcd} B_{bcd}$$

Notice that $\epsilon^{abcd} B_{bcd}$ contains all the info about $B$.

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