2
$\begingroup$

I am studying time-independent perturbation theory in quantum mechanics and I found myself not understanding how to associate degenerate eigenstates their specific Hamiltonian eigenvalue correction.

Suppose that the full Hamiltonian has the form $$ \widehat{H}= \widehat{H_0} + \lambda \widehat{V} $$ and the unperturbed Hamiltonian $\widehat{H_0}$ has $k$ degenerate eigenstates, $\left| 1 \right\rangle \cdots \left| k \right\rangle $.

The eigenvalues of the degenerate subspace matrix $$ V_{ij}= \left\langle i \right| \widehat{V} \left| j \right\rangle \quad , \quad i,j \in \left[ 1,k \right] $$ represent the first order Hamiltonian eigenvalue corrections in $\lambda$ .

My question is, how is it possible to associate the new, perturbed energy values to the original unperturbed eigenstates?

$\endgroup$
  • $\begingroup$ I'm not sure I understand your question. Are you asking "How do we compute perturbed eigenvalues in terms of unperturbed eigenvalues and states?" or "why should it possible to do this?" or are you trying to somehow associate each perturbed eigenvalues with a particular unperturbed state? or have I missed the mark entirely? $\endgroup$ – By Symmetry Jan 12 '18 at 23:35
  • $\begingroup$ Yes, I am trying to associate perturbed eigenvalues to individual unperturbed states. I don't know if it makes any sense. $\endgroup$ – Ahornach Jan 13 '18 at 9:47
  • $\begingroup$ Related. $\endgroup$ – Cosmas Zachos May 23 '18 at 1:51
0
$\begingroup$

When you first consider a degenerate subspace of states that all share the same order zero energy, you don't really know how to pick them in such a way that they become the order zero states of your perturbation series. In practice what happens is that the perturbation is the one that selects them and we can read this as "the states you would find if you turn on the perturbation but then you turn it off back again".

In practice, the basis that you use to compute $V_{ij}$ is generically just an orthonormal basis that you found for your degenerate subspace and in principle it does not have anything to do with the order zero states.

But when you compute the first order corrections as the eigenvalues of $V_{ij}$, the corresponding eigenvectors are the actual order zero states! These are the only meaningful vectors and, if the degeneracy is fully lifted, then you can use those states from then on to compute higher order corrections now using non-degenerate perturbation theory. Otherwise, you'll still have degenerate subspaces and you would have to keep using degenerate perturbation theory in higher orders, if you really need to push that far.

$\endgroup$
  • $\begingroup$ If I have understood correctly, I can take the eigenvectors of $V_{ij}$ as the unperturbed Hamiltonian eigenstates and use those as the "starting point" to find perturbed eigenstates, since both of them are suitable orthornormal bases? $\endgroup$ – Ahornach Jan 13 '18 at 9:56
  • 1
    $\begingroup$ Yes, both are suitable orthonormal bases, but at least the eigenvectors of $V_{ij}$ diagonalize for sure the first order correction and we consider them the real order zero states. At first order, they have specific energies, namely their order zero energy (which they all shared) plus their corresponding correction (eigenvalue of $V_{ij}$). So they are going to simplify your life in higher order corrections. $\endgroup$ – secavara Jan 13 '18 at 13:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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