A hamiltonian for one dimensional axis for a particle of mass m is given by:

$$H_0 = \frac{p^2}{2m} +V(x)$$

Let we apply a term to the hamiltonian: $$H = \frac{p^2}{2m} +V(x) + \frac{\lambda}{m} p$$ where p is momentum operator and $\lambda$ , $p$ are constants. Therefore the New hamiltonian would become :

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

I have the solution where they write the perturbed eigenstates as $$\bar{\psi_n } (x)= e^\frac{-ix\lambda}{\hbar} \psi_n (x)$$


It seems that they considered the unperturbed hamiltonian as $$H_0= \frac{(p+ \lambda)^2 }{2m} +V(x)$$

Why is that? I mean why did they include the $\lambda$ term?

  • $\begingroup$ The expression $(x)\exp(-ix \lambda)$ doesn't make sense. $\endgroup$
    – DanielSank
    Dec 23, 2016 at 4:31
  • $\begingroup$ Sorry, i have edited the typo. $\endgroup$
    – user58143
    Dec 23, 2016 at 4:34
  • $\begingroup$ I think you completed the square incorrectly in the "new Hamiltonian". $\endgroup$
    – DanielSank
    Dec 23, 2016 at 4:42
  • $\begingroup$ You have another typo. The $p$ should be removed from the $-\lambda^2 p / 2m$ term in the second expression for $H$. $\endgroup$ Dec 23, 2016 at 4:43
  • $\begingroup$ The constant term $-\lambda^2 / 2m$ does not affect the eigenstates. The new eigenstates can thus be obtained from the old by sending $p \to p-\lambda$. Such a momentum translation is implemented on position space wave functions with the plane wave factor $e^{-i x \lambda / \hbar}$ $\endgroup$ Dec 23, 2016 at 4:46

1 Answer 1


We start with the Hamiltonian $$ H = H_0 + \delta H $$ where $$ H_0 = \frac{p^2}{2m} + V(x) $$ and $$ \delta H = \frac{\lambda}{m} p \ . $$ Moreover, we assume that we know the eigenstates and eigenenergies of $H_0$. Denote them $\psi_n(x)$ and $E_n$ respectively. As the questioner has pointed out, we may complete the square to write the Hamiltonian in the form $$ H = \frac{(p + \lambda)^2}{2m} + V(x) - \frac{\lambda^2}{2m} \ . $$

The claim, which we will check, is that the eigenstates of $H$ in a position space basis have the form $\tilde \psi_n (x) = e^{-i x \lambda / \hbar} \psi_n(x)$. Recall that in a position space basis, we may write the operator $p = \frac{\hbar}{i} \frac{\partial}{\partial x}$. First consider how $p+\lambda$ acts on the shifted wave function: $$ (p+ \lambda) \tilde \psi_n(x) = \left(\frac{\hbar}{i} \frac{\partial}{\partial x} + \lambda \right) e^{-i x \lambda/ \hbar} \psi_n(x) = e^{-i x \lambda / \hbar} p \psi_n(x) \ . $$ Next how $(p+\lambda)^2$ acts: $$ (p+ \lambda)^2 \tilde \psi_n(x) = (p+\lambda) e^{-i x \lambda / \hbar} p \psi_n(x) = e^{-i x \lambda / \hbar} p^2 \psi_n(x) \ . $$ It should then be clear that $$ H \tilde \psi_n(x) = e^{-i x \lambda / \hbar} \left(H_0 - \frac{\lambda^2}{2m} \right) \psi_n(x) = \left( E_n - \frac{\lambda^2}{2m} \right) \tilde \psi_n(x) \ . $$ Thus the eigenenergies of $H$ are the shifted quantities $$ \tilde E_n = E_n - \frac{\lambda^2}{2m} \ . $$

To motivate the factor $e^{-i \lambda x / \hbar}$, it's probably simpler to think about momentum space wave-functions. Consider the Fourier transform $$ \phi(p) = \int dx \, e^{-i x p / \hbar} \psi(x) \ . $$ In momentum space, clearly if $\phi(p)$ is an eigenstate of $H_0$, then $\tilde \phi(p) = \phi(p + \lambda)$ is an eigenstate of $H$. But then we get the chain of equalities $$ \int dx \, e^{-i x p/ \hbar} \tilde \psi(x) = \tilde \phi(p) =\phi(p+\lambda) = \int dx\, e^{-i x (p+\lambda) / \hbar} \psi(x) \ . $$ Multiplying both sides by $e^{i x' p / \hbar}$ and integrating over $p$, we find $\tilde \psi(x) = e^{-i \lambda x / \hbar} \psi(x)$.

  • $\begingroup$ I understand that $\tilde \psi_n (x) = e^{-i x \lambda / \hbar} \psi_n(x)$ ? satisfies the perturbed eigenstate. But how to guess quick this type of plane wave? Could you help me out of this. I understand the rest of the procedure. $\endgroup$
    – user58143
    Dec 23, 2016 at 6:02

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