Time reversal

If the hamiltonian is real and does not depend on time, then inverting the time in the Schrödinger equation is equivalent to taking its complex conjugate, and therefore the eigenstates have the property:

$$\psi({\bf r},t) = \psi^*({\bf r},-t).$$ (12.17)

It follows that both $\psi$ and $\psi^*$ satisfy the Schrödinger equation with the same eigenvalue. Consider a Bloch state identified by a vector ${\bf k}$ in the BZ. For any lattice vector $\tau$, this state has the property:

$$\psi_{\bf k}({\bf r} + \tau) = e^{i {\bf k}\cdot \tau} \psi_{\bf k}({\bf r}).$$ (12.18)

Now take the complex conjugate of :

$$\psi^*_{\bf k}({\bf r} + \tau) = e^{-i {\bf k}\cdot \tau} \psi^*_{\bf k}({\bf r}).$$ (12.19)

We see that $\psi^*_{\bf k}$ and $\psi_{\bf k}$ do not satisfy the same translational property. Let us re-write  it in terms of ${\bf -k}$:

$$\psi^*_{\bf -k}({\bf r} + \tau) = e^{i {\bf k}\cdot \tau} \psi^*_{\bf -k}({\bf r}).$$ (12.20)

and taking the complex conjugate:

$$\psi_{\bf -k}({\bf r} + \tau) = e^{-i {\bf k}\cdot \tau} \psi_{\bf -k}({\bf r}).$$ (12.21)

Comparing  with  we see that both $\psi_{\bf -k}$ and $\psi^*_{\bf k}$ satisfy the same translational property and so they are both eigenstates of $\hat{T}_\tau$ with eigenvalue $e^{-i \tau \cdot {\bf k}}$, meaning that $\psi_{\bf -k}$ and $\psi^*_{\bf k}$ must belong to the same degenerate manifold of eigenstates of the hamiltonian. It follows that the solution of the Schrödinger equation is only needed at k, as the one at -k can be obtained from the complex conjugate of $\psi_{\bf k}$.