# Time reversal for fermionic fields

I have some doubts about the way we apply time reversal to Dirac's Lagrangian in QFT. Looking for the transformed field, $$\psi^t(x)$$, I've found sources (see below) that claims:

$$\psi^t(x) = \gamma^1\gamma^3\psi(-t, \vec{x}),\quad x = (t, \vec{x}) \tag1$$

($$\gamma^\mu$$ in Weyl representation)

Nevertheless, I tried to prove it in the following way and find another result. The transformed field $$\psi^t(x)$$ is a fermion field so it has to satisfy Dirac's equation as $$\psi(x)$$ does, i.e.,

$$(i\gamma^\mu\partial_\mu - m)\psi(x)\ \Rightarrow\ (i\gamma^\mu\partial_\mu - m)\psi^t(x) \tag2$$

Also, due to left equation in (2), $$\psi(-t, \vec{x})$$ satisfies

$$(-i\gamma^0\partial_0 + i\vec{\gamma}·\vec{\partial} - m)\psi(-t, \vec{x})\ \Rightarrow\ (i\gamma^0\partial_0 - i\vec{\gamma}^{\ *}·\vec{\partial} - m)\psi^*(-t, \vec{x}) \tag3$$

(Star * denotes complex conjugation)

Now, in (2)'s right equation you can take $$\psi^t(x) = M·\psi^*(-t, \vec{x})$$ with $$M$$ unitary (the anti-unitary property of time reversal transformations is already applied over the field via complex conjugation) and such that

$$[M, \gamma^0] = 0, \quad \vec{\gamma}^{\ *}M = -M\vec{\gamma}\tag4$$

With this properties $$M$$ leads you from (4) to (3), so this $$\psi^t(x) = M\psi^*(-t, \vec{x})$$ is a fermionic field that satisfies the time reversal Dirac's equation. Moreover, the conditions in (4) force $$M = \gamma^1\gamma^3$$. Then, I obtain

$$\psi^t(x) = \gamma^1\gamma^3\psi^*(-t, \vec{x}) \tag5$$

(5) is different from (1), so am I wrong or the sources are?