Perturbation method & eigenvalues I have a problem but I don't understand the question. It says:
"Show that, to first order in energy, the eigenvalues ​​are unchanged."
 What does it mean?
 It means that if the Hamiltonian has the form
$$H=H^{(0)}+\lambda H^{(1)}$$
Where $H^{(0)}$ is the Hamiltonian of the unperturbed system, $H^{(1)}$ is the perturbation and $\lambda$ is a small parameter, then if
$$E_{n}=E_{n}^{(0)}+\lambda E_{n}^{(1)}$$
Where
$$E_{n}^{(1)}=\left\langle \psi_{m}^{(0)}|H^{(1)}|\psi_{m}^{(0)}\right\rangle $$
I have to show that 
$$E_{n}^{(1)}=0$$
?
I'm confused. Thanks for your answers.
 A: As I mentioned in the comments, the assertion that $E_n^{(1)}\equiv0$ cannot hold in general since a scalar perturbation does not obey it.
For the particular case you mention, a linear perturbation on a harmonic oscillator, however, it does hold. The simplest way to see this is that the perturbation can be included in the oscillator potential to give another, displaced, harmonic oscillator:
$$
\frac{1}{2}m\omega^2 x^2-Fx=\frac{1}{2}m\omega^2\left(x-\frac{F}{m\omega^2}\right)^2-\frac{F^2}{2m\omega^2}.
$$
This is displaced in position, which is irrelevant for these purposes (though it definitely affects the eigenfunctions!), and it is displaced in energy by $-\frac{F^2}{2m\omega^2}\propto F^2$. Thus there will not be any first-order shifting in energy.
As far as your question is concerned, however, you need a perturbation-theoretic argument that will prove this, and that is on you to build. The essential point here is to think parity: in the expectation value
$$E_n^{(1)}=\left\langle \psi_{m}^{(0)}|H^{(1)}|\psi_{m}^{(0)}\right\rangle$$
the eigenfunctions have definite parity, as does the perturbation. What does this entail?
