# What does the Hermiticity have to do with the conservation of energy?

Naomichi Hatano's Non-Hermitian quantum mechanics (link here) published in PTEP 12, 2020 says that Hermitian operators are used when the energy is conserved but when it isn't then non-Hermitian operators are suitable.

How does Hermiticity assure the conservation of energy?

The Hamiltonian of the whole universe may be Hermitian according to von Neumann, but a part of it, for example, a radioactive nuclide, a quantum dot, or whatever is connected to the rest of the macroscopic universe, does not conserve energy, and hence can be described by an effective non-Hermitian Hamiltonian after eliminating the environmental degrees of freedom

And what is meant by "effective non-Hermitian Hamiltonian"?

Let's assume a Hamiltonian which is not time-dependent. If $$H$$ is Hermitian, then the time evolution operator $$U(t)=\mathrm e^{-\mathrm i \frac{t}{\hbar}H}$$ is unitary. So then the expected energy is constant:
\begin{align}\langle H(t)\rangle&=\langle\psi(t)\mid H\mid\psi(t)\rangle\\ &=\langle\psi(0)\mid U^\dagger(t)HU(t)\mid\psi(0)\rangle\\ &=\langle\psi(0)\mid U^\dagger(t)U(t)H\mid\psi(0)\rangle\\ &=\langle \psi(0)\mid H\mid\psi(0)\rangle\\ &=\langle H(0)\rangle \end{align}
You need to use that the time evolution operator commutes with the Hamiltonian (because it's an exponential of the Hamiltonian) and that it's unitary. So at least one of these assumptions must be broken if we want to model a system without conservation of energy. Now, $$U(t)$$ must commute with $$H$$, unless we throw out the Schrödinger equation, too. So we have to make $$U(t)$$ non-unitary. But if $$H$$ is Hermitian, then $$U(t)$$ is automatically unitary, so $$H$$ can't be Hermitian for this to work. So the Hamiltonian of a system without conservation of energy must be non-Hermitian.