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This is part 2 of exercise II.1.1 of Zee's QFT in a Nutshell (here's part 1).

This is what I have got:

\begin{align} \bar\psi\gamma^\lambda\psi \mapsto \bar\psi^{\,\prime}\gamma^\lambda\psi^{\,\prime} & = {\psi^{\,\prime}}^\dagger{\gamma^0}^\dagger\gamma^\lambda\psi^{\,\prime} = (S(\Lambda)\psi)^\dagger\gamma^0\gamma^\lambda(S(\Lambda)\psi) \\ & = \psi^\dagger S(\Lambda)^\dagger\gamma^0\gamma^\lambda S(\Lambda)\psi\\ & = \tag{1}\psi^\dagger e^{\frac i4\omega_{\mu\nu}{\sigma^{\mu\nu}}^\dagger}\gamma^0\gamma^\lambda e^{-\frac i4\omega_{\mu\nu}\sigma^{\mu\nu}}\psi\\ & = \psi^\dagger\left(1+\frac i4\omega_{\mu\nu}(\sigma^{\mu\nu})^\dagger+O({\omega_{\mu\nu}}^2)\right)\gamma^0\gamma^\lambda\left(1-\frac i4\omega_{\mu\nu}\sigma^{\mu\nu}+O({\omega_{\mu\nu}}^2)\right)\psi\\ & = \psi^\dagger\gamma^0\left(\gamma^\lambda+\frac i4\omega_{\mu\nu}[[\gamma^\mu,\gamma^\nu],\gamma^\lambda]+O({\omega_{\mu\nu}}^2)\right)\psi\\ & = \psi^\dagger\gamma^0\left(\gamma^\lambda+\frac 12\omega_{\mu\nu}[\sigma^{\mu\nu},\gamma^\lambda]+O({\omega_{\mu\nu}}^2)\right)\psi \end{align}

Two questions:

  • How exactly does the last line prove that $\bar\psi\gamma^\lambda\psi$ transforms as a vector under Lorentz transformations? It certainly looks like a vector transformation to me, because of the commutator between the Lorentz generators and the "vector components" $\gamma^\mu$, but how can I prove this quantitatively?
  • Then, a more general question: How can I get rid of the $O({\omega_{\mu\nu}}^2)$? I know that it can be ignored since we consider infinitesimal transformations only, so the transformation behavior is governed by first-order terms only. But what's the mathematical rigorous way to get rid of them? Do I just write a $\simeq$ sign and leave them out at the right-hand side of the $\simeq$? That would not seem right to me, because when dealing with expansions, a $\simeq$ does not imply strict equality, it just implies "equality up to a certain order".
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  • $\begingroup$ Thanks for asking these questions, I am currently reading through this book and possibly physics pragmatism will here apply re: above (aren't cutoff type techniques applied in other expansions?), but best of luck getting an answer $\endgroup$ – user81619 Nov 22 '15 at 11:07
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It gets easier if you use the result from part 1. Then you also don't have to deal with the $\mathcal O(\omega^2)$ (see my answer to your other question).

In your calculation, you transformed $\bar\psi$ and $\psi$, but not $\gamma^\lambda$. This is correct, as I will show in the end, but I will take another point of view which is really helpful here: $\gamma^\lambda$ is an object with one Lorentz index $\lambda$ and two fermionic indices, hence it should transform as $$ {\gamma'}^\lambda = {\Lambda^\lambda}_\nu (S \gamma^\nu S^{-1}) $$ (according to the general rules how tensors transform).

Because we already know from part 1 that $\bar\psi \to \bar\psi S^{-1}$, we immediately get $$ \bar\psi \gamma^\lambda \psi \to {\Lambda^\lambda}_\nu \bar\psi S^{-1} S \gamma^\nu S^{-1} S \psi = {\Lambda^\lambda}_\nu \bar\psi \gamma^\nu \psi \;, $$ this is the expected transformation behavior of a Lorentz vector.


What you did is also correct, of course, but missing one ingredient. Since $$ {\Lambda^\lambda}_\nu (S \gamma^\nu S^{-1}) = \gamma^\lambda \;, $$ the $\gamma$ matrices actually do not transform. Proving this is the trickier bit (also not too hard, though). I haven't read Zee's book, but I would guess he proves it somewhere?

(You could start by writing $\bar\psi \gamma^\lambda \psi \to \bar\psi S^{-1} \gamma^\lambda S \psi$ and then use this identity to get to the same result.)

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  • $\begingroup$ Thank you, but what exactly is a fermionic index, and why do the gamma matrices transform this way (your first equation)? Any explanation or book reference is appreciated. $\endgroup$ – Bass Nov 22 '15 at 12:12
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    $\begingroup$ For each $\mu$, $\gamma^\mu$ is a matrix, so that gamma is actually an object ${(\gamma^\mu)^A}_B$. The indices $A$ and $B$ are fermionic indices or spinor indices. $\endgroup$ – Noiralef Nov 22 '15 at 13:05
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    $\begingroup$ The definition of a tensor is that, under Lorentz transformation, each Lorentz index is contracted with a ${\Lambda^\mu}_\nu$ and each spinor index is contracted with an ${S(\Lambda)^A}_B$. $\psi = \psi^A$ is a contravariant spinor and in part 1 of the exercise you essentially show that $\bar\psi = \bar\psi_B$ is a covariant spinor. A different way to think about what I wrote above is maybe: The equation ${\gamma'}^\lambda = \gamma^\lambda = {\Lambda^\lambda}_\nu (S \gamma^\nu S^{-1})$ proves that $\gamma$ is a tensor. After showing this, we can use it in the actual calculation, like I did. $\endgroup$ – Noiralef Nov 22 '15 at 13:11
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    $\begingroup$ Thanks for your help. Judging from your comments, am I correct that the whole $\gamma$ is a tensor with two spinor indices (one contravariant, one covariant) and one contravariant "normal" index? It indeed is pretty cool :) but it's too new stuff for me to be sure about these things. $\endgroup$ – Bass Nov 22 '15 at 13:27
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    $\begingroup$ Peskin/Schroeder: Equation (3.30). Zee: Under equation (14), part 2. Weinberg: (5.4.8) $\endgroup$ – Noiralef Nov 22 '15 at 14:34

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