# Reality condition on Hamiltonian from Time-reversal and particle-hole inversion symmetry

Starting from a Fermionic second quantized Hamiltonian as $$H = \sum_{A,B} \psi^{\dagger}_A \mathcal{H}_{A,B} \psi_B$$ Imposing time-reversal symmetry and charge-conjugation inversion symmetry, like this i.e,

$$\tau H \tau^{-1} = {H}\\c H c^{-1} = H$$ $$H$$ reduces to $$U_T^{\dagger} \mathcal{H}^* U_T = + \mathcal{H} \\ U_c^{\dagger} \mathcal{H}^* U_c = - \mathcal{H}$$ Here $$\mathcal{H}$$ is the single particle Hamiltonian obtained from the second quantized hamiltonian. Now, I don't see how these two equations impose a reality condition on $$\mathcal H$$.

I can understand that in the former equation $$\mathcal H^*$$ is the same as $$\mathcal H$$ up-to a rotation by a unitary matrix, and this is convincible to be real because of the +ve sign on the RHS.

But in the latter equation, the sign is flipped which leads me to believe that the Hamiltonian is imaginary.

To see how this can happen mathematically, consider the matrix $$A=\left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]$$ which is Hermitian and real (so in the standard basis). Changing basis means conjugating by a unitary, such as $$Q$$ where $$Q=\left[\begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\\ \frac{-i}{\sqrt{2}} & \frac{i}{\sqrt{2}} \end{array}\right]$$ and in the new basis, $$A$$ is represented by $$Q^\dagger AQ=\left[\begin{array}{cc} 0 & i\\ -i & 0 \end{array}\right]$$ which is imaginary. Same operator, two mathematical pictures, one real, the other imaginary. Now in physics the basis that leads to one of these conditions can be strange, but mathematically there is no issue.