In class, we have define the following operator:

$$\Pi_{\pm} = \frac{1 \pm \gamma^0}{2} \tag1$$

Where, $\gamma^0$ is the usual first gamma matrix in Weyl representation.

Applying it to a 4-momentum in rest frame, amin

$$p_* = \begin{pmatrix}m\\ 0\\ 0\\ 0 \end{pmatrix} \tag2$$

You get,

$$\Pi_{\pm}(p_*) = \frac{m \pm (p_*)_\mu \gamma^\mu}{2m} \tag3$$

Here, my professor said that this expression was Lorentz invariant due to the product $(p_*)_\mu \gamma^\mu$, but I don't understand this because gamma matrices do not transform, so what's happening here? Actually, because of this supposed invariance he wrote that in any other reference frame (where you have 4-momentum $p$), you could write:

$$\Pi_{\pm}(p) = \frac{m \pm p_\mu \gamma^\mu}{2m} \tag4$$

  • $\begingroup$ The expression is not Lorentz invariant. It is, at best, Lorentz covariant. $\endgroup$ – AccidentalFourierTransform Nov 27 '18 at 2:54
  • $\begingroup$ Then, how can I get Eq. (4) by Eq. (3)? $\endgroup$ – Vicky Nov 27 '18 at 2:55

Your Answer

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

Browse other questions tagged or ask your own question.