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

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


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

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

I'm confused. Thanks for your answers.

  • 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 Nov 18 '12 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. Nov 18 '12 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 Nov 18 '12 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. Nov 18 '12 at 1:49

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?

  • $\begingroup$ 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. Nov 18 '12 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 Nov 18 '12 at 20:03

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