# non-negative energies

Can someone help me with the following question:

given the hamiltonian: $H=\frac{1}{2m}(iP-\hbar f(X))(-iP-\hbar f(X))$

where $f(x)$ an analytic real function. prove that the eigenenergies of the system are non-negative and then given that the ground state has $0$ energy find it's eigenstate.

all I got is that from the given hamiltionian I can write it in the form: $$H=\frac{P^2}{2m}+\frac{\hbar ^2}{2m}(f^2(X)-f'(X))$$ and if I consider some eigenstate $|\varphi\rangle$ of $H$ I try to show that $\langle \varphi |H|\varphi \rangle\geq0$ but I can't show that $\langle \varphi |f'(x)|\varphi \rangle\geq0$ for arbitrary function $f$.

• While I haven't worked this problem I would suggest that you compare this to the usual treatment for the quantum harmonic oscillator. Can you generalize from there? – dmckee --- ex-moderator kitten Jul 10 '16 at 17:42
• I thought about it but I don't quite know how – dorsh605 Jul 10 '16 at 17:44

It is best to start from the form of the Hamiltonian itself: The operator $H$ is of the form $H = \frac{1}{2m} Q^\dagger Q$ with $Q^\dagger = iP -\hbar f(x)$ so that for any $|\phi>$ $$<\phi| H |\phi> = \frac{1}{2m}<\phi| Q^\dagger Q |\phi> = \frac{1}{2m} \left( Q |\phi>, Q|\phi>\right) \, \, \ge 0$$ and is only zero when $Q |\phi> = 0$.