Why does time reversal symmetry requires a real Hamiltonian? I have some problems understanding the consequences of time reversal symmetry.
If Hamiltonian $H$ is symmetric under time reversal, it satisfies:
$$
 \mathcal{T} H \mathcal{T}^{-1} = H \quad \mathrm{with} \quad \mathcal{T} = U \mathcal{K} \\
 U H^{*} = H U \mathcal{K} \\
H^{*} = U^{-1} H U K .
$$
But how does one obtain for a general case $$ H^{*} = H~?$$
It would be very nice, if someone could help me with this problem.
 A: Your have a mistake in the second line $U \mathcal K H = U H^* \mathcal K$ (not $U H^*$). It therefore follows that
$$ U H^* \mathcal K = H U \mathcal K \Leftrightarrow H^* = U^{-1} H U. $$
This means, the Hamiltonian is not strictly real, but it is unitarily equivalent to its complex conjugate (which is the next-best thing).
See also this question on the topic.
Let me know if there are further questions or mistakes!
A: If $\ \ \mathcal{T}\mathcal {H} \mathcal{T}^{-1} = \mathcal {H}\implies \mathcal{TH}=\mathcal{HT}$ where $\mathcal{T=UK}$ with $\mathcal K$ denoting complex conjugation, then if we apply this to an arbitrary state $\psi$ $$\mathcal{TH}\psi=\mathcal{HT}\psi$$ $$\rightarrow \mathcal{UKH}\psi=\mathcal{HUK}\psi$$
$$\rightarrow \mathcal{KH}\psi=\mathcal{U^{-1}HUK}\psi\tag1$$
$$\rightarrow \mathcal{H^*}\psi^*=\mathcal{U^{-1}}\mathcal{HU}\psi^*$$
so that in the general case, $$\rightarrow \mathcal{H^*}=\mathcal{U^{-1}}\mathcal{HU}$$ so that the complex conjugate of the Hamiltonian is unitarily equivalent to the Hamiltonian.

*

*In the case where $[\mathcal H,\mathcal U]=0$ i.e., $\mathcal U$ is a symmetry of the system, we could from equation (1) write $$\rightarrow \mathcal{UKH}\psi=\mathcal{UHK}\psi$$ $$\rightarrow \mathcal{KH}\psi=\mathcal{U^{-1}UHK}\psi$$ then $$\mathcal{H^*}\psi^*=\mathcal{H}\psi^*$$ since $\mathcal{U^{-1}U}=\mathbb I$
which then implies $$\mathcal H^*=\mathcal H$$ meaning $\mathcal H$ is real.


*Example: In the case of spinless particles, $\mathcal U$ is chosen to be the identity operator $\mathbb I$ so that $\mathcal{T}=\mathcal {UK}=\mathcal K$ (or that time reversal corresponds to simple complex conjugation of the wave
function for a spinless particle) and so if the system has time-reversal symmetry: $$ [\mathcal H,\mathcal T]=0 \implies \mathcal [\mathcal H,\mathcal K]=0 \implies\mathcal {KH}\psi=\mathcal {HK}\psi$$
which leads to
$$\mathcal H^*\psi^*=\mathcal H\psi^*$$
again meaning that the Hamiltonian must be real.
