Finding eigenvalues of $H$ Here is my question:
The dynamics of a particle moving one-dimensionally in a potential $V(x)$ is governed by the Hamiltonian $H_0 = p^2 /2m + V(x)$, where $p = -i\hbar d/dx$ is the momentum operator. Let $E^{(0)}_n$, $n =1,2,3...,$ be the eigenvalues
of $H_0$. Now consider a new Hamiltonian $H = H_0 + \lambda p/m$, where $\lambda$ is a given parameter. Given $m$ and $E^{(0)}_n$, How do I find the eigenvalues of $H$?
This is my work:
I can see that the new Hamiltonian is
$$H = H_0 + \lambda p/m = p^2/2m + \lambda p/m + V(x)$$
$$= (p + \lambda)^2 /2m + V(x) - \lambda^2/2m$$
According to my teacher, the corresponding eigenvalues are
$$E_n = E^{(0)}_n -\lambda^2/2m$$
He found the equation
$$\psi = \psi^{(0)} e^{-i\lambda x/\hbar}$$
But I didn't understand his solving method and that was so confused. 
Regards!
 A: HINT/OUTLINE: The key idea is to notice that the effect on the Hamiltonian, when the new one is rewritten as you have done, looks like a translation in momentum-space. So we look for a similar transformation in the eigenvectors: we try 
$$\psi = \psi^{(0)} e^{-i\lambda x/\hbar}$$
What is then $H\psi$? Let's look first at just the operator $p + \lambda$ which appears in it:
$$ p\psi = \left(p \psi^{(0)}\right)e^{-i\lambda x/\hbar} + \psi^{(0)}\left(-i\hbar \frac{d(e^{-i\lambda x/\hbar})}{dx}\right) = \left(p \psi^{(0)}\right)e^{-i\lambda x/\hbar} - \lambda \psi \\
\therefore (p + \lambda)\psi = p\psi + \lambda \psi = \left(p \psi^{(0)}\right)e^{-i\lambda x/\hbar}
$$
So $p + \lambda$ acts on $\psi$ in "the same way" as $p$ acts on $\psi^{(0)}$. In fact, for this it isn't even important that $\psi^{(0)}$ is an eigenvector of the original. So you can apply this one more time, with $p\psi^{(0)}$ taking the place of $\psi^{(0)}$ to see what $(p+\lambda)^2$ does. It is then straightforward to verify that any such $\psi$ is an eigenvector of the new Hamiltonian with eigenvalues given by that equation you have. By considering the reverse transformation, you could also show that any eigenvector of the new Hamiltonian must be of this form. 
