Time reversal symmetry

Remarks

If $\Theta$ commutes with $H$ and $\psi(t)$ solves $i\hbar \partial_t \psi = H\psi$, then $\phi(t) = \Theta\psi(-t)$ solves it too, because $\Theta$ turns $i$ into $-i$; this is why $\Theta$ must be antiunitary.

Example: for one spinless particle on $L^2(\R^3)$, complex conjugation $\Theta\psi = \bar{\psi}$ satisfies the clauses, since $\overline{-i\hbar\nabla\psi} = -(-i\hbar\nabla\bar{\psi})$, and $\Theta^2 = I$. The Hamiltonian $-\frac{\hbar^2}{2m}\Delta + V(x)$ with a real potential $V$ commutes with it, so this system has time reversal symmetry. For a spin one-half particle on $L^2(\R^3) \otimes \C^2$, one can take $\Theta = -i\sigma_2 K$, where $K$ is complex conjugation and $\sigma_2$ the second Pauli matrix; then $\Theta^2 = -I$.

Non-example: for a charged particle in a magnetic field with vector potential $A$, $H = \frac{1}{2m}(p - qA(x))^2$. Conjugation by $\Theta$ gives $\frac{1}{2m}(-p - qA(x))^2 = \frac{1}{2m}(p + qA(x))^2$, which differs from $H$ when the magnetic field $\nabla \times A$ is nonzero, so the system has no time reversal symmetry.