In my library's (old -- 1996) copy of Peskin and Schroeder, there's an identity I'm struggling to prove. In my copy it occurs on page 42, between equations 3.28 and 3.29, but I don't know how well this matches newer editions.

Previously they've introduced the matrices $$ \left( \mathcal{J}^{\mu\nu} \right)_{\alpha\beta} = i \left( {\delta^\mu}_{\alpha} {\delta^\nu}_{\beta} - {\delta^\mu}_{\beta} {\delta^\nu}_{\alpha} \right) $$ and the commutator matrices $$ S^{\mu\nu} = \frac{i}{4} \left[ \gamma^\mu , \gamma^\nu \right], $$ and asserted the identity $$ \left[ \gamma^\mu , S^{\rho\sigma} \right] = {\left( \mathcal{J}^{\rho \sigma} \right)^\mu}_\nu \gamma^\nu, $$ which I'm quite happy with.

The next line says this is equivalent to $$ \left( 1 + \frac{i}{2} \omega_{\rho\sigma} S^{\rho\sigma} \right) \gamma^\mu \left( 1 - \frac{i}{2} \omega_{\rho\sigma} S^{\rho\sigma} \right) = {\left( 1 - \frac{i}{2} \omega_{\rho\sigma} \mathcal{J}^{\rho\sigma} \right)^\mu}_\nu \gamma^\nu. $$

There's immediately at least one error in this from the multiple pairs of matching indices on the LHS. I imagine that this should be rectified by changing the $\rho\sigma$ indices in one of the brackets to some other pair.

But when I try to verify this identity, I find that they only match to first order in $ \omega $. Is this correct, or is there an algebraic trick I'm missing?

  • $\begingroup$ I would call it sloppy, not an error, to sum like this. $\endgroup$
    – mvw
    Dec 15 '14 at 11:25

It states on page 40 that "$\omega$ ... gives the infinitesimal angles", and it states right after your equation on page 42 that it is an "infinitesimal version" of an equation which follows it. So basically yes, we would only expect them to match to first order in $\omega$.

  • 1
    $\begingroup$ I'm prepared to accept this as a valid answer, but since appealing to 'infinitesimals' seems thoroughly unnecessary, presumably easily avoidable by simply adding something like $ + O( \left\Vert \omega \right\Vert ^2 ) $ to the equation, it's still rather unsatisfactory. $\endgroup$
    – JCW
    Dec 19 '14 at 14:13

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