Shifting momentum by a constant in the Schrodinger Equation My book states that if we perturb a given Hamiltonian for the Schrödinger Equation 
$$
H = \frac{p^2}{2m} +V(x)
$$ 
to 
$$
H' = \frac{p^2}{2m} + V(x) + \frac{\lambda p}{m}
$$ 
then we can rewrite the perturbed Hamiltonian in the form
$$
H' = \frac{(p+\lambda)^2}{2m} + V(x) - \frac{\lambda^2}{2m} = \frac{p'^2}{2m} + V(x)  - \frac{\lambda^2}{2m}.
$$
Furthermore, it goes on to say that if we know the eigenvalues and eigenfunctions of the unperturbed Hamiltonian, $E_n^{(0)}$ and $\psi_n^{(0)}$ respectively, we can readily use the fact that the wave number is now $k' = \frac{p'}{\hbar} = \frac{(p+\lambda)}{\hbar}$ to say that the eigenfunctions of the perturbed Hamiltonian must be
$$\psi_n = \psi^{(0)}_n e^{-i\lambda x/\hbar}$$
and thus the new energies are
$$E_n = E_n^{(0)} - \frac{\lambda^2}{2m}.$$
Can someone please explain how the energy eigenfunctions can be so easily obtained considering the momentum operator has been shifted by a constant? I believe that the answer has to do with the fact that momentum space representation of $\psi$ is the Fourier Transform of $\psi(x)$, but I'm not sure how to prove this. 
 A: You can write
\begin{eqnarray*}
H^{\prime } &=&H^{\prime \prime }--\frac{\lambda ^{2}}{2m} \\
H^{\prime \prime } &=&\frac{1}{2m}(p+\lambda )^{2}+V(x)=\exp [-i\lambda
x]H\exp [i\lambda x]
\end{eqnarray*}
so $H$ and $H^{\prime \prime }$ are unitarily equivalent, so their
eigenvalues coincide.. In the coordinate representation let
\begin{equation*}
H\psi (x)=\mu \psi (x)
\end{equation*}
so
\begin{eqnarray*}
\exp [i\lambda x]H^{\prime \prime }\exp [-i\lambda x]\psi (x) &=&\mu \psi (x)
\\
H^{\prime \prime }\exp [-i\lambda x]\psi (x) &=&\mu \exp [-i\lambda x]\psi
(x)
\end{eqnarray*}
The shift by the number $-\frac{\lambda ^{2}}{2m}$ is trivial.
A: If $\phi_n(p)$ is an eigenfunction of $H$ in momentum space with eigenvalue $E_n$, then $\phi_n(p+\lambda)$ is an eigenfunction of $H'$ with eigenvalue $E_n -\frac{\lambda^2}{2m}$ (and vice versa), since $H$ and $H'$ differ from each other only by the linear redefinition of the momentum and adding a constant:The notation in the following is not optimal, but I hope it is clear what I mean.
\begin{align} H'\phi_n(p+\lambda) & = \left(\frac{(p+\lambda)^2}{2m}+V(\partial_{p}) \right)\phi_n(p+\lambda) -\frac{\lambda^2 }{2m}\phi_n \\ & = \left(\frac{p^2}{2m}\phi_n +V(\partial_p)\phi_n\right)(p+\lambda)-\frac{\lambda^2 }{2m}\phi_n\end{align}
where the last equality holds because $\partial_p(\phi(p+\lambda)) = (\partial_p \phi)(p+\lambda)$ by the chain rule and we know that $\left(\frac{p}{2m}\phi_n +V(\partial_p)\phi_n\right) = H\phi_n = E_n\phi$.
By a general property of the Fourier transform, if $\psi_n(x)$ is the Fourier transform of $\phi_n(p)$, then $\mathrm{e}^{-\mathrm{i}\lambda x}\psi_n(x)$ is the Fourier transform of $\phi_n(p+\lambda)$. This property just follows from integration by substitution in the defining integral of the Fourier transform.
