Potential must be real for Hamiltonian to be Hermitian?

Question

I have seen a few proofs specify for finite wells, step functions, and harmonic oscillators, that $V$ must be real for $H$ to be Hermitian. Why is that?

If we're solving the Schrodinger equation, we have $$i\hbar \frac{\partial}{\partial t}\psi=-\frac{\hbar^2}{2m}\nabla^2\psi+V\psi$$ and we know that $\psi$ being complex will have a complex spatial derivative. So for a real energy value to be observed, could the complex part cancel out with the complex part of $V$? That would still be hermitian, right?

Why does a electric Potential have to be real, but not a Potential in quantum mechanics? says that potentials in QM can be complex, but I want to know why we'd talk about complex potentials when they're not physically achievable and they cause operators to be non-hermitian and hence useless.

Answer 1

If $V$ is just a function of position then the requirement of being hermitian is equivalent to the requirement of reality. This is not true for different systems: suppose an electron is subject to an uniform magnetic field. The interaction with it's spin with the magnetic field will break the degeneracy associated with the spin states: spin up states will have a different energy than spin down states. This is effectively a 2 state system now, where we can represent the ground state as $$ |1\rangle = \begin{pmatrix} 1\\ 0 \end{pmatrix} $$ and the excited state as $$ |2\rangle = \begin{pmatrix} 0\\ 1\end{pmatrix}. $$ Suppose now we introduce another potential $V$ to this system. The most general form of $V$ will be represented by a 2x2 matrix $$ V = \begin{pmatrix} V_{11} & V_{12} \\ V_{21} & V_{22} \end{pmatrix}. $$ As you can easily deduce, the hermitian requirement ($V^\dagger = (V^*)^T=V$) allows for off-diagonal complex elements. And complex potentials may be achievable, one example I can think of is the Rabi oscillations potential that models a time-dependent 2 state system.