2
$\begingroup$

I have a problem but I don't understand the question. It says:

"Show that, to first order in energy, the eigenvalues ​​are unchanged." What does it mean? It means that if the Hamiltonian has the form

$$H=H^{(0)}+\lambda H^{(1)}$$

Where $H^{(0)}$ is the Hamiltonian of the unperturbed system, $H^{(1)}$ is the perturbation and $\lambda$ is a small parameter, then if

$$E_{n}=E_{n}^{(0)}+\lambda E_{n}^{(1)}$$

Where

$$E_{n}^{(1)}=\left\langle \psi_{m}^{(0)}|H^{(1)}|\psi_{m}^{(0)}\right\rangle $$

I have to show that $$E_{n}^{(1)}=0$$ ?

I'm confused. Thanks for your answers.

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ That can't be right. If $H^{(1)}=1$ is scalar, i.e. $H=H^{(0)}+\epsilon 1$, the eigenvectors are unchanged, and the eigenvalues are shifted by $\lambda=\epsilon$ : $E_n=E_n^{(0)}+\epsilon$, so $E_n^{(1)}=\epsilon$. $\endgroup$
    – Emilio Pisanty
    Commented Nov 18, 2012 at 1:06
  • $\begingroup$ I think I understand your comment but I think that's not my question, or I don't get it. With your supposition you conclude that the eigenvalues are shifted, but not unchanged. $\endgroup$
    – Ana S. H.
    Commented Nov 18, 2012 at 1:21
  • $\begingroup$ Exactly. The counter-example shows that your understanding of the question is incorrect. Whatever specific result you set out to prove must hold in the specific case I mentioned for it to have a chance of holding in general. $\endgroup$
    – Emilio Pisanty
    Commented Nov 18, 2012 at 1:34
  • $\begingroup$ Oh yes, but $H^{(1)}$ is not constant, in fact $H^{(1)}=-Fx$ And $H^{(0)}$ is the hamiltonian of the harmonic oscillator. $\endgroup$
    – Ana S. H.
    Commented Nov 18, 2012 at 1:49

1 Answer 1

Reset to default
4
$\begingroup$

As I mentioned in the comments, the assertion that $E_n^{(1)}\equiv0$ cannot hold in general since a scalar perturbation does not obey it.

For the particular case you mention, a linear perturbation on a harmonic oscillator, however, it does hold. The simplest way to see this is that the perturbation can be included in the oscillator potential to give another, displaced, harmonic oscillator: $$ \frac{1}{2}m\omega^2 x^2-Fx=\frac{1}{2}m\omega^2\left(x-\frac{F}{m\omega^2}\right)^2-\frac{F^2}{2m\omega^2}. $$ This is displaced in position, which is irrelevant for these purposes (though it definitely affects the eigenfunctions!), and it is displaced in energy by $-\frac{F^2}{2m\omega^2}\propto F^2$. Thus there will not be any first-order shifting in energy.

As far as your question is concerned, however, you need a perturbation-theoretic argument that will prove this, and that is on you to build. The essential point here is to think parity: in the expectation value $$E_n^{(1)}=\left\langle \psi_{m}^{(0)}|H^{(1)}|\psi_{m}^{(0)}\right\rangle$$ the eigenfunctions have definite parity, as does the perturbation. What does this entail?

Improve this answer
$\endgroup$
2
  • $\begingroup$ By the way, using parity of eigenfunctions, it's obvious that the integrand of $E_{m}^{(1)}=\left\langle \psi_{m}^{(0)}|H^{(1)}|\psi_{m}^{(0)}\right\rangle$ is odd (because $H^{(1)}\propto x$), and because of the symmetric limits of integration, the result is that $E_{m}^{(1)}=0$ and then we conclude that, to first order we have $E_{n}=E_{n}^{(0)}$, i.e. the eigenvalues are unchanged. Thanks! $\endgroup$
    – Ana S. H.
    Commented Nov 18, 2012 at 17:29
  • $\begingroup$ There's an even simpler parity argument. Because $H^{(0)}$ is even, the dependence of the perturbed energies on $F$ should be exactly the same for the parity-reversed problem. This is equivalent to the same perturbation with a reversed field, which means that the perturbed energies' dependence on $F$ must be even. Then the linear term must vanish. $\endgroup$
    – Emilio Pisanty
    Commented Nov 18, 2012 at 20:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.