My book states that if we perturb a given Hamiltonian for the Schrödinger Equation

$$ H = \frac{p^2}{2m} +V(x) $$


$$ H' = \frac{p^2}{2m} + V(x) + \frac{\lambda p}{m} $$

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.

  • $\begingroup$ It is certainly straight-forward to show that the new functions are eigenfunctions of the shifted Hamiltonian: just plug them into the Schrodinger equation with the new Hamiltonian and show that they work. In so doing, you'll see what the energy eigenvalues are, too. I'm not entirely sure how to straight-forwardly show that this is all the eigenfunctions, but I think that's pretty apparent. $\endgroup$
    – march
    Dec 21, 2015 at 2:05
  • $\begingroup$ I suppose I would prefer a formal derivation that shows how they are obtained besides guess-and-check; indeed, it is clear that they work from plugging them in. $\endgroup$
    – Loonuh
    Dec 21, 2015 at 2:11

2 Answers 2


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.


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.

  • $\begingroup$ I think there is an error in this problem actually. If what you say is true about the fourier transform via substitution, then I think since $\psi(x)$ are the inverse fourier transform of $\psi(p)$ it should be the case that $\psi(p +\lambda) \to \psi(x)e^{i\lambda x/\hbar}$, where the minus sign in the exponent has now been made positive. $\endgroup$
    – Loonuh
    Dec 21, 2015 at 7:38
  • 3
    $\begingroup$ @Loonuh: Hmmm...yes, there might be a sign error somewhere, but that depends on your exact definition of the Fourier transform. It depends on which sign is in your exponential for the direction $p\to x$, I can't answer that for you. $\endgroup$
    – ACuriousMind
    Dec 21, 2015 at 7:59
  • $\begingroup$ Plugging their eignefunctions in directly it seems like they do have the correct form of them. $\endgroup$
    – Loonuh
    Dec 21, 2015 at 8:13

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