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".
  • $\begingroup$ 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, 2015 at 11:07

1 Answer 1


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.)

  • $\begingroup$ 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, 2015 at 12:12
  • 2
    $\begingroup$ 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, 2015 at 13:05
  • 1
    $\begingroup$ 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, 2015 at 13:11
  • 1
    $\begingroup$ 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, 2015 at 13:27
  • 1
    $\begingroup$ Peskin/Schroeder: Equation (3.30). Zee: Under equation (14), part 2. Weinberg: (5.4.8) $\endgroup$
    – Noiralef
    Nov 22, 2015 at 14:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.