quantum field theory - Derivation of Gordon identity from Srednicki

On srednicki page 240 (print) there is a derivation of the Gordon identity, and it starts with stating that

$$\require{cancel} \gamma^{\mu}\cancel{p} = \frac{1}{2} \big\{\gamma^{\mu},\cancel{p} \big\} + \frac{1}{2} \big[ \gamma^{\mu},\cancel{p} \big] = -p^{\mu} -2iS^{\mu\nu}p'_\nu{}$$

Where exactly does this come from?

Answer

Let $\gamma^\mu$ be a gamma matrix, i.e., $$\frac{1}{2}(\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu)=-\eta^{\mu\nu} \tag{1}$$

Let $$S^{\mu\nu}\equiv\frac{i}{4}(\gamma^\mu\gamma^\nu-\gamma^\nu\gamma^\mu) \tag{2}$$

Let us write $$\gamma^\mu\gamma^\nu=\frac{1}{2}(\gamma^\mu\gamma^\nu+\gamma^\mu\gamma^\nu) \tag{3}$$ (note that I did nothing yet)

Next, write $$0=\frac{1}{2}(\gamma^\nu\gamma^\mu-\gamma^\nu\gamma^\mu) \tag{4}$$ and add it to $(3)$: $$\gamma^\mu\gamma^\nu=\frac{1}{2}(\gamma^\mu\gamma^\nu+\gamma^\mu\gamma^\nu)+\frac{1}{2}(\gamma^\nu\gamma^\mu-\gamma^\nu\gamma^\mu)=\frac{1}{2}(\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu)+\frac{1}{2}(\gamma^\mu\gamma^\nu-\gamma^\nu\gamma^\mu) \tag{5}$$ where I just rearranged the terms.

Finally, plug $(1)$ and $(2)$ into $(5)$: $$\gamma^\mu\gamma^\nu=-\eta^{\mu\nu}-2iS^{\mu\nu} \tag{6}$$

This is all we need to prove the relation given by Srednicki: $$\gamma^\mu\not p\equiv p_\nu\gamma^\mu\gamma^\nu=p_\nu(-\eta^{\mu\nu}-2iS^{\mu\nu})=-p^\mu-2iS^{\mu\nu}p_\nu \tag{7}$$

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Note that this can be slightly generalised: let $a^\mu,b^\mu$ be any two objects (vector, matrices, operators, etc.)

With this $$a^\mu b^\nu=\frac{1}{2}\{a^\mu,b^\nu\}+\frac{1}{2}[a^\mu,b^\nu] \tag{8}$$ as can be easiliy checked by expanding the commutators. In this case, $a^\mu=b^\mu=\gamma^\mu$, so that $(6)$ follows after realising that $\{\gamma^\mu,\gamma^\nu\}=-2\eta^{\mu\nu}$ by definition of the gamma matrices, and $[\gamma^\mu,\gamma^\nu]=-4iS^{\mu\nu}$ by definition of $S^{\mu\nu}$.