In the literature (like those in quantum chaos), it seems that time-reversal symmetry implies that the Hamiltonian of the system is a real symmetric one, instead of just being complex Hermitian.
Is there any rigorous justification about this link?
[Physics] Time reversal symmetry and real symmetric Hamiltonian matrix
homework-and-exercisesoperatorsquantum mechanicstime-reversal-symmetry
Related Solutions
We can infer some general topological properties of the system for special cases. This has been used in the recent "trending topic" of condensed matter physics, topological insulators. For simplicity, I will restrict myself in the following to $2\text{D}$ systems although one can generalize everything to $3\text{D}$.
The time-reversal operator is an antiunitary operator that admits the following representation:
$\hat{\Theta} = \exp \left(i\pi\hat{S}_y/\hbar \right)K$
where $K$ means complex conjugation and $\hat{S}_y$ is the spin operator along the $\hat{y}$ axis. Consider a fermionic Hamiltonian for spin $s=1/2$ electrons. Then
$\hat{\Theta} =-\hat{1}$ [*]
In this case, Kramers theorem applies:
Let $\hat{\mathcal{H}}$ be a $T$-invariant (fermionic) Hamiltonian. Then, all the eigenstates of the Hamiltonian are twofold degenerate.
The proof of this statement is simple once you have understood [*]. As a consequence, $T$-invariant fermionic systems must have topologically protected twofold degenerate states. The $T$-invariant Hamiltonian satisfies
$ \hat{\Theta}\hat{\mathcal{H}} (\mathbf{k}) =\hat{\mathcal{H}} (-\mathbf{k}) \hat{\Theta}$
and can be classified by a new topological index, called the $\mathbb{Z}_2$ index. The $\mathbb{Z}_2$ index, $\nu$, is an integer given by the number of edge states modulo $2$ and distinguishes the $\nu = 0$ or insulating phase from the $\nu = 1$, the topological insulator. Thus, the equivalence classes of $T$-invariant Hamiltonians for insulators can be classified by its $n = 0$ Thouless-Kohmoto-Nightingaleden Nijs invariant [i.e. its $C=0$, first Chern index] and the additional index $\nu$. This gives a $\mathbb{Z} \times \mathbb{Z}_2$ symmetry for the $2\text{D}$ band structures.
After all, what can we infer from the Hamiltonian? For example, that we have time reversal invariant electronic states with a bulk electronic band gap that supports the transport of charge and spin in gapless edge states. This is exactly what we get in the so-called quantum spin Hall phase.
If we define $\mathcal{T}=i\sigma_y K$ where $K$ is complex conjugation, i.e.
$\mathcal{T}\psi_\uparrow \mathcal{T}^{-1}=\psi_\downarrow, \mathcal{T}\psi_\downarrow \mathcal{T}^{-1}=-\psi_\uparrow$,
Then naively a term like $\Delta e^{i\Phi}\psi_{\uparrow}^\dagger \psi_{\downarrow}^\dagger$ is not invariant under $\mathcal{T}$. This is basically the problem you encountered, phrased a little differently. However, it does not mean that the system actually breaks $T$, as all physical observables will be invariant under $\mathcal{T}$. The resolution is in the definition of the transformation of $\psi$ under $\mathcal{T}$. Let us modify the definition to be
$\mathcal{T}\psi_\uparrow \mathcal{T}^{-1}=\psi_\downarrow e^{-i\Phi}, \mathcal{T}\psi_\downarrow \mathcal{T}^{-1}=-\psi_\uparrow e^{-i\Phi}$.
This phase is not observable (it is essentially redefining the phases of the basis states of the second-quantized Fock space, which has no consequences on physical observables), so we are free to do so. Notice that the important algebraic relation $\mathcal{T}^2=-1$ (for the classification of TI/TSC, etc.) is not affected. Then the pairing term is invariant.
Of course, this only works when $\Phi$ does not depend on positions. Otherwise (e.g. when there is a vortex) one can not get rid of the phase, since $\nabla\Phi$ is an observable, the supercurrent.
Best Answer
A time reversal operator is an anti-unitary operator, which can be expressed as:
$\mathcal{T}=UK$
where $K$ denotes complex conjugate and $U$ is a unitary operator. In case of spinless particles, $U$ is chosen to be Identity. Thus $\mathcal{T}=K$.
If the system has $\mathcal{T}$-reversal symmetry:
$$ KH\psi=HK\psi $$
which leads to:
$$ H^*\psi^*=H\psi^* $$ meaning that the Hamiltonian must be real symmetric.