Proving lorentz invariance of Dirac bilinears

[GreyBadger]

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

[tex]\bar{\psi}^\prime\psi^\prime = \psi^{\prime\dagger}\gamma_0\psi^\prime[/tex]
[tex]= \psi^{\dagger}S^\dagger\gamma_0 S\psi[/tex]
[tex]= \psi^{\dagger}\gamma_0\gamma_0S^\dagger\gamma_0 S\psi[/tex]
[tex]= \psi^{\dagger}\gamma_0 S^{-1} S\psi[/tex]
[tex]= \psi^{\dagger}\gamma_0\psi[/tex]
[tex]= \bar{\psi}\psi[/tex]

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

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

[tex]\bar{\psi}^\prime\gamma^\mu\psi^\prime = \psi^{\prime\dagger}\gamma_0\gamma^\mu\psi^\prime[/tex]
[tex]=\psi^\dagger S^\dagger\gamma_0\gamma^\mu S\psi[/tex]
[tex]=\psi^\dagger\gamma_0\gamma_0S^\dagger\gamma_0\gamma^\mu S\phi[/tex]
[tex]=\psi^\dagger\gamma_0S^{-1}\gamma^\mu S\phi[/tex]

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

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

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

Cheers!

 

[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 [tex]\Lambda_{\frac{1}{2}}[/tex]. And then you can stick their formula (3.29) into your equation and that immediately gives you the correct answer.

 

[GreyBadger]

I could well have got my notes confused... I have a copy of P&S on my desk, so will have a look when I'm in work (Can't find a preview on Google Books). Cheers!