Shifting momentum by a constant in the Schrodinger Equation

Question

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.

Answer 1

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.

Answer 2

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.