Proving lorentz invariance of Dirac bilinears

1. Oct 18, 2009

I'm trying to work through the proof of the Lorentz invariance of the Dirac bilinears. As an example, the simplest:

$$\bar{\psi}^\prime\psi^\prime = \psi^{\prime\dagger}\gamma_0\psi^\prime$$
$$= \psi^{\dagger}S^\dagger\gamma_0 S\psi$$
$$= \psi^{\dagger}\gamma_0\gamma_0S^\dagger\gamma_0 S\psi$$
$$= \psi^{\dagger}\gamma_0 S^{-1} S\psi$$
$$= \psi^{\dagger}\gamma_0\psi$$
$$= \bar{\psi}\psi$$

Where the following have been used: $$\gamma_0\gamma_0=\textbf{I}$$, $$S^{-1} = \gamma_0 S^\dagger\gamma_0$$.

Now, attempting this for the vector current, I get stuck:

$$\bar{\psi}^\prime\gamma^\mu\psi^\prime = \psi^{\prime\dagger}\gamma_0\gamma^\mu\psi^\prime$$
$$=\psi^\dagger S^\dagger\gamma_0\gamma^\mu S\psi$$
$$=\psi^\dagger\gamma_0\gamma_0S^\dagger\gamma_0\gamma^\mu S\phi$$
$$=\psi^\dagger\gamma_0S^{-1}\gamma^\mu S\phi$$

The problem being I don't know the commutation relation $$[\gamma^\mu,S]$$. Given the expression for $$S(a)$$:

$$S(a)=\exp\left( \frac{i}{4\sigma_{\mu\nu}}(a^{\mu\nu} - g^{\mu\nu}) \right)$$,

I could compute the commutator explicitly in the infinitesimal limit ($$e^x = 1 + x$$), but this seems a bit annoying... Are there any tricks?

Cheers!

Last edited: Oct 18, 2009
2. Oct 19, 2009

hamster143

Your expression for S(a) is a bit strange. Take a look at Peskin & Schroeder, page 42. (The relevant page is available on books.google.com if you don't have the physical book.) There's a bit of a mismatch in notation - what you call S, P&S call $$\Lambda_{\frac{1}{2}}$$. And then you can stick their formula (3.29) into your equation and that immediately gives you the correct answer.

Last edited: Oct 19, 2009
3. Oct 19, 2009