Time reversal
Definition: Time reversal
Time reversal $T$ is a symmetry transformation $$T: (t,\vec{x}) \to (-t,\vec{x}).$$ It is part of the Lorentz group, since it leaves $s^2$ invariant. However it is not smoothly connected to the identity.
Since $T\cancel{\partial}=T(\partial_0,\partial_1,\partial_2,\partial_3)\to (-\partial_0,\partial_1,\partial_2,\partial_3)$, to leave the kinetic Term in the Dirac Lagrangian $\bar{\psi}\cancel{\partial}\psi$ invariant we need the spinors to transform under $T$ as $$T\bar{\psi}\gamma^0\psi= -\bar{\psi}\gamma^0\psi,$$ which means $T\psi^\dagger\psi\to -\psi^\dagger\psi$. This is problematic, since $\psi^\dagger\psi$ is a positive definite quantity (Like scalar product in quantum mechanics) and no linear transformation $\psi\to\Gamma\psi$ can do this ($\Gamma^\dagger\Gamma=-1$ has no solution).
# Simple $\hat{T}$
A possible non-linear transformation of $\psi$, which would fix this is $$\hat{T}(\psi)\to \Gamma\psi^*,\hat{T}(\psi^{\dagger})=(\Gamma\psi^*)^{\dagger}=\psi^T\Gamma^{\dagger}$$ We then have $$\psi^\dagger\psi\to \psi^T\Gamma^\dagger\Gamma\psi^* = \psi_\alpha\Gamma^\dagger_{\alpha\beta}\Gamma_{\beta\gamma}\psi^*_\gamma=-\psi^*_\gamma\Gamma^T_{\gamma\beta}\Gamma^{\dagger T}_{\beta\alpha}\psi_\alpha=-\psi^\dagger(\Gamma^\dagger\Gamma)^T\psi,$$ where we used that $\psi_\alpha\psi^_\gamma = -\psi^\gamma\psi\alpha$ due to the anti-commuting nature of the fermion fields. Therefore, $\hat{T}\psi^\dagger\psi=-\psi^\dagger\psi$ as required to leave $\bar{\psi}\cancel{\partial}\psi$ invariant, iff $\Gamma^\dagger\Gamma=1$, which means $\Gamma$ is unitary.
- To leave the full term $\bar{\psi}\psi$ invariant we need ${\Gamma,\gamma_0}=0$.
- To leave $\bar{\psi}\gamma_i\psi$ invariant we need $[\Gamma,\gamma_1]=[\Gamma,\gamma_3]={\Gamma,\gamma_2}=0$.
The unique Matrix which fulfills all this is $\Gamma=\gamma_0\gamma_2$.
Therefore $\hat{T}$ acts as:
- Spinors $$\hat{T}(\psi)=\gamma_0\gamma_2\psi^*(-t,\vec{x})$$
- Vectors $$\hat{T}(A_0)=-A_0(-t,\vec{x}),\qquad \hat{T}(A_i)=A_i(-t,\vec{x})$$
We can consider the action of all discrete transformations $CP\hat{T}$: $$CP\hat{T}(\psi)=-i\psi(-t,-\vec{x})$$ $$CP\hat{T}(\bar{\psi}\gamma^\mu\psi)=\bar{\psi}(-t,-\vec{x})\gamma^\mu\psi(-t,-\vec{x})$$ $$CP\hat{T}(A_\mu)=A_\mu(-t,-\vec{x})$$ $$CP\hat{T}(\bar{\psi}\psi)=\bar{\psi}\psi$$ $$CP\hat{T}(\bar{\psi}\cancel{\partial}\psi)=\bar{\psi}\cancel{\partial}\psi$$ so that the Lagrangian in invariant under $CP\hat{T}$.