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 $\bar\psi\psi$?

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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).

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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