# Hermitian operator ?? is this possible

we know that the operator

$H= - \hbar ^{2} \frac{d^{2}}{dx^{2}}+ V(x)$ is hermitian isn't it ??

however what would happen if the potential were still real but it depends on the Wave function, for example huypothetically

$V(x) = |\Psi (x)|^{2}$ or $V(x) = arg \Psi (x)$

since the functions $|x|$ and $arg (x+iy)$ are ALWAYS real, then the hamiltonian with potentials (1) and (2) should be Hermitian but they depend on the solution so i am not sure about the Hermiticity of a Hamiltonian in the form $H =p^{2}+ |\Psi (x) |^{2}$ i believe that a real potential makes the Hamiltonian hermitian even in the case that we do not exactly know what potential is

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That would give a nonlinear equation. Quantum mechanics on the other hand assumes that all observables (and hence also the time, and space evolution operators) be linear operators on the space of physical states. – user10001 Jul 16 '12 at 15:20
One problem could be that by restricting your operator to be real you can't differentiate it with respect to $\Psi(x)$ and thus also not with respect to $x$. – Laar Jul 16 '12 at 15:35

If $V(x)$ is a real function, then it's hermitian as you say. If the Hamiltonian contains any function $|\Phi(x)|^2$, then this is a priori just the previous case. If you mean this to be the respective wave function, then the operator is more of a functional. In that case the associated differential equation is certainly not linear, so you leave the framework in the standard terminology. Although the operator as such will still behave hermitean w.r.t. to the scalar product, so the answer is a cautious yes. As a remark, then function involving the argument of the wave function is pretty unphysical as the phase is exactly the quantity, which should be undetectable.