The potential $-x^4$ is non-hermitian? I am currently reading a paper by Carl Bender about non-hermitian Hamiltonians. In this paper he says that $$H=p^2-x^4$$ was a non-hermitian Hamiltonian whereas $$H=p^2+x^4$$ was hermitian. It is probably really simple but I just don't see the reason for that. Can anyone explain to me why the one with a minus is not hermitian?
 A: Writings of Carl Bender while full of very imaginative ideas often don't bother with mathematical strictness. Well, most of the physicist usually do the same. However strictly speaking the Hermiticity of the operator $(Hx,y)=(x,Hy)$ usually doesn't guarantee its self-adjointness $H=H^\ast$ because of the domain issues. The self-adjointness which is required for the unitary evolution and spectral decomposition is actually a very non-trivial property. Yet even if the Hermitian operator is not self-adjoint it may have so-called self-adjoint extensions obtained by a suitable enlargement of the domain.
The classic example is $p=-i\partial_x$ operator on a segment $[0,a]$ that's not self-adjoint - you can see that from the fact that if you try to construct $U=e^{i\alpha p}$ that shifts the wavefunction $U\psi(x)=\psi(x+\alpha)$ it can't be uniquely defined because you should also specify what happens with wavefunction at the boundaries.
For Hamiltonian operators of the form $H=-\Delta+V(x)$ there exist a number of conditions that guarantees self-adjointness. What's relevant here is that if $V(x)<0$ as $x\rightarrow \pm\infty$ it shouldn't drop too fast i.e. $V(x)>-C|x|^2$. If it goes faster self-adjointness is not guaranteed and you have to build a self-adjoint extension. You can suspect that something wrong in a similar way to $p$ operator as for such potential classical particle goes to infinity in a finite time. So your self-adjoint extension should specify what exactly happens with wavefunction at infinities.
However for Bender's case this all is irrelevant actually. What he considers is a class of operators $p^2+x^2(ix)^\epsilon$ which are generally non-Hermitian but symmetric under $PT$-symmetry understood as,
\begin{equation}
x\mapsto -x,\quad p\mapsto p,\quad \quad i\mapsto -i
\end{equation}
If you consider non-singular wavefunctions decreasing at infinity these Hamiltonians actually possess a discrete real positive spectrum of such wavefunctions $\phi_n$ that happen to be orthogonal under special inner product. On the span of this wavefunctions those Hamiltonian can actually be considered as self-adjoint operators in terms of some special norm.
