I Instability of hydrogen ground state if the time-reversal operator is unitary

[ergospherical]

Apparently if we try to represent the time reversal operator by a unitary operator $T$ satisfying $U(t)T = TU(-t)$, then the ground state of hydrogen (the hamiltonian of which is time-reversal invariant) is unstable. But if $T$ is anti-unitary (i.e. $\langle a | T^{\dagger} T | b \rangle = \langle a | b \rangle^*$) then the instability is avoided. Why?

 

[DrClaude]

Apparently

According to ...?

 

[vanhees71]

I guess $U(t)=\exp(-\mathrm{i} H t)$ is the time-evolution operator (for the states in the Schrödinger picture). If you want to have
$$U(-t)=\exp(\mathrm{i} t H) = T^{\dagger} U(t) T = \exp[T^{\dagger} (-\mathrm{i} t H) \hat{T}],$$
you must have
$$T^{\dagger} (-\mathrm{i} t H) \hat{T}=+\mathrm{i} t H.$$
Since $t \in \mathbb{R}$ for both $T$ unitary or antiunitary that implies
$$T^{\dagger} i H T=-\mathrm{i} H.$$
If $T$ where unitary, that would imply that $T^{\dagger} H T=-H$, which implies that for any eigenvalue $E$ of $H$ also $-E$ must be an eigenvector:
$$H T |E \rangle=T T^{\dagger} H T |E \rangle=-T H |E \rangle=-E T |E \rangle,$$
i.e., indeed $T|E \rangle$ is an eigenvector of eigenvalue $(-E)$. Since now the energy spectrum for a free particle (or that of an electron in presence of a Coulomb potential) has a lower bound, $T$ cannot be a symmetry operator if realized as a unitary operator, which implies that time-reversal symmetry must be realized by an anti-unitary symmetry operator.