# Operators, gamma matrices and Lorentz invariance

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:

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

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