To construct a Lorentz scalar we use $\psi^{\dagger}\gamma^{0}\psi$. Could we use $\gamma^{5}$ instead of $\gamma^{0}$ seen as both are Hermitian?

Question

Both $\gamma^{0}$ and $\gamma^{5}$ are Hermitian, so could we replace $\gamma^{0}$ with $\gamma^{5}$ to construct a Lorentz scalar with the same properties as $\bar{\psi}\psi$?

Answer 1

A Lorentz boost about the $i$th spatial direction is represented as $\psi\rightarrow V\psi$ with $V=\exp\left(\theta[\gamma^0,\gamma^i]\right)$ for some real $\theta$. The problem is that $V^\dagger\neq V^{-1}$. This follows since $\gamma^0$ is Hermitian and $\gamma^i$ is anti-Hermitian, which implies that $$[\gamma^0,\gamma^i]^\dagger=[\gamma^{i\dagger},\gamma^{0\dagger}]=+[\gamma^0,\gamma^i]$$ whereas we would want a minus sign for $V$ to be unitary.

Using $\gamma^0$ in $\psi^\dagger\gamma^0\psi$ gives us that extra minus sign we need which comes from commuting $\gamma^0$ past the factor of $\gamma^i$ in the exponent. Using $\gamma^5$ instead would not work since it gives us two minus signs from commuting past both $\gamma^0$ and $\gamma^i$ in the exponent.

Answer 2

Of course you can, just a bit different from yours. The following is a Lorentz scalar and Hermitian: $$ \bar{\psi}i\gamma^{5}\psi = \psi^{\dagger}i\gamma^{0}\gamma^{5}\psi = -\psi^{\dagger}\gamma^{1}\gamma^{2}\gamma^{3}\psi. $$

It's called "pseudoscalar" mass term. See more details here.