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I would like to show that for a Dirac spinor $\psi$, the scalar product $\bar\psi\psi$ transforms as a scalar under a Lorentz transformation $\Lambda$, where $\bar\psi = \psi^\dagger\gamma^0$. This is exercise II.1.1 a) of Zee's QFT in a Nutshell.

This is what I have tried so far:

$\psi$ transforms as $\psi\mapsto\psi' = S(\Lambda)\psi = e^{-\frac i4\omega_{\mu\nu}\sigma^{\mu\nu}}\psi$, where $\sigma^{\mu\nu}=\frac i2[\gamma^\mu,\gamma^\nu]$ are the generators of the Lorentz Lie algebra, and $\omega_{\mu\nu}$ are the coefficients of the Lorentz transformation $\Lambda$.

So we have the transformation

\begin{align} \bar\psi\psi \mapsto \bar\psi^{\,\prime}\psi^{\,\prime} & = {\psi^{\,\prime}}^\dagger{\gamma^0}^\dagger\psi^{\,\prime} = (S(\Lambda)\psi)^\dagger\gamma^0(S(\Lambda)\psi) \\ & = \psi^\dagger S(\Lambda)^\dagger\gamma^0 S(\Lambda)\psi\\ & = \tag{1}\psi^\dagger e^{\frac i4\omega_{\mu\nu}{\sigma^{\mu\nu}}^\dagger}\gamma^0e^{-\frac i4\omega_{\mu\nu}\sigma^{\mu\nu}}\psi \end{align}

In the last line, I have a situation similar to $e^ABe^{-A}$ which is normally evaluated in an expansion (first-order) as $e^{\lambda A}Be^{-\lambda A} \simeq B+\lambda[A,B]+O(\lambda^2)$. This would make sense if the commutator $[A,B]$ vanishes, because then (1) would be equal to $\psi^\dagger\gamma^0\psi=\bar\psi\psi$, which is what we want to prove.

But in this case, this does not work, since in the first exponential we have a ${\sigma^{\mu\nu}}^\dagger$. So I tried to calculate this first:

\begin{align} {\sigma^{\mu\nu}}^\dagger & = -\frac i2(\gamma^\mu\gamma^\nu-\gamma^\nu\gamma^\mu)^\dagger\\ & = \frac i2({\gamma^\mu}^\dagger{\gamma^\nu}^\dagger-{\gamma^\nu}^\dagger{\gamma^\mu}^\dagger)\\ & = \frac i2(\gamma^0\gamma^\mu\gamma^0\gamma^0\gamma^\nu\gamma^0-\gamma^0\gamma^\nu\gamma^0\gamma^0\gamma^\mu\gamma^0)\\ \tag{2}& = \frac i2(\gamma^0\gamma^\mu\gamma^\nu\gamma^0-\gamma^0\gamma^\nu\gamma^\mu\gamma^0) \end{align}

How to proceed? How can I use (2) to transform (1) into $\bar\psi\psi$?

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    $\begingroup$ You're incredibly close! What happens to $\sigma^\dagger \gamma^0$ if you use (2)? $\endgroup$ – innisfree Nov 21 '15 at 22:50
  • $\begingroup$ What's $(\gamma^0)^2$? $\endgroup$ – innisfree Nov 21 '15 at 22:50
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    $\begingroup$ It is enough to show that $\overline{\psi}\psi$ is invariant to first order. Then because of the properties of the exponential, it will be invariant to all orders. Now that you have found ${\sigma^{\mu\nu}}^\dagger$, expand the exponentials and keep terms of at most first order. $\endgroup$ – Robin Ekman Nov 22 '15 at 1:10
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It is actually possible, and not too difficult, to prove this without expanding the exponentials to first order only.

What you are trying to prove is $S^\dagger \gamma^0 = \gamma^0 S^{-1}$, this is equivalent to $$ \gamma^0 S^\dagger \gamma^0 = S^{-1} $$ because $( \gamma^0 )^2 = 1$. Expand $S^\dagger = \sum_n \frac{1}{n!} \left( \frac i 4 \omega_{\mu\nu} \sigma^{\mu\nu\dagger} \right)^n$ and you see that it is sufficient to prove $$ \gamma^0 \left( \sigma^{\mu\nu\dagger} \right)^n \gamma^0 = \left( \sigma^{\mu\nu} \right)^n \;. $$

Now, the left hand side is equal to $\left( \gamma^0 \sigma^{\mu\nu\dagger} \gamma^0 \right)^n$, hence we only need to prove that $$ \gamma^0 ( \sigma^{\mu\nu} )^\dagger \gamma^0 = \sigma^{\mu\nu} \;. $$ This is obvious from your equation (2).

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  • $\begingroup$ Thank you! Can this also be done for the vector $\bar\psi\gamma^\mu\psi$? See this question. A hint is much appreciated (doesn't have to be a complete answer, because this is homework-and-exercises). $\endgroup$ – Bass Nov 22 '15 at 11:41

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