How to identify non-hermitian Hamiltonians? Can anyone explain how can I prove that hamiltonians like $H=p^2+ix^3$ or $H=p^2-x^4$ are non-hermitian.
I think in first case, as the conjugate of $H$ is not equal to $H$, we can say it is non-hermitian. Is it a valid argument?
And what about the 2nd case?
Please help.
 A: Technically you'd need to specify the domain on which these operators act and the inner product...  Assuming the domain is the real line and the usual inner product
$$
\langle \phi\vert\psi\rangle = \int_{-\infty}^\infty dx\,\phi^*(x)\psi(x) <\infty
$$
then the first clearly isn't because $\hat x$ and $\hat p$ are Hermitian so that
\begin{align}
\hat H^\dagger &=\left(p^2 +i x^3\right)^\dagger=(p^2)^\dagger -i (x^3)^\dagger\, ,\\
&= (p^\dagger)^2 -i (x^\dagger)^3 = p^2  -i x^3 \ne \hat H
\end{align}
since $\hat p^\dagger=\hat p$ and $\hat x^\dagger=\hat x$ on the real line with $\lim_{x\to\infty}\psi(\pm x)\to 0$.
The second would be Hermitian since
\begin{align}
\hat H^\dagger &=(p^2)^\dagger - (x^4)^\dagger\, ,\\
&= (p^\dagger)^2 - (x^\dagger)^4 = p^2 - x^4  = \hat H
\end{align}
under the same conditions.
Note that to show that $\hat{\cal O}$ is hermitian
\begin{align}
\langle \phi\vert \hat{\cal O}\psi\rangle &= 
\langle \hat{\cal O}\phi\vert\psi\rangle\, ,\\
&= \int_{-\infty}^\infty \phi^*(x) \left(\hat{\cal O} \psi(x)\right)
= \int_{-\infty}^\infty \left(\hat{\cal O} \phi^*(x)\right) \psi(x)
\end{align}
so that $\hat p\to -i\hbar\frac{\partial }{\partial x}$ is hermitian using
integration by parts and the boundary conditions on the wavefunctions at $\infty$.
This conclusion might change if you are working with different assumptions on the domain of your functions or a different definition of the inner product.
