4
$\begingroup$

Suppose we are doing a degenerate Rayleigh-Schodinger perturbation problem. Let's say the Hamiltonian $H_0$ is perturbed by a small perturbation $H'$, and we want corrections to the energy eigenstates/eigenfunctions (stationary states). Assume we have solved the unperturbed system.

It is always preferable to find a set of hermitian operators that commute with the perturbation $H'$ and $H_0$, so the perturbation will already be diagonal in this basis, and we don't have to diagonalise ourselves. Now my textbook says: we can just do some ordinary (non-degenerate) perturbation theory and we are done. I get this, but one thing confuses me:

Why is it not necessary that $H'$ also commutes with $H_0$?

The only reason I can think of is that we're assuming that the perturbation is sufficiently small, such that energies of different levels are not connected in perturbation theory, that is, the off-diagonal matrix elements of the degerate part, $$\tilde{H}'_{nr,n's} = \langle \chi_{nr}^{(0)}\vert H' \vert \chi_{n's}^{(0)} \rangle \,\,\,\,\,(r,s = 1,2,\dots,\alpha \text{ (degree of degeneracy}),$$ (where $\chi_{ns}^{(0)}$ are the correct zero order wave functions) vanish anyway?

Let me clarify with an example. let's say we're considering a simultaneous eigenstate of $L^2$, $L_z$ and $H$, $\vert n, l, m \rangle$. Suppose the perturbation commutes with $L^2$, $[L^2,H']=0$. We can use this to show that: $$\langle n'l'm'\vert H' \vert n l m \rangle \propto \delta_{l'l},$$ so states with different values of $l$ are not connected. How can we conclude $$\langle n'l'm'\vert H' \vert n l m \rangle \propto \delta_{n'n}$$ when we don't have that $[H,H']=0$?

$\endgroup$
9
  • $\begingroup$ Why should it be necessary ? (it's another way to say that I don't understand the question...) $\endgroup$
    – Adam
    Jun 3 '16 at 12:31
  • 1
    $\begingroup$ If the perturbation commutes with the Hamiltonian, it will just shift the different unperturbed eigenstates. However, it is usually not the case (for example, the perturbation might lift a degeneracy, en.wikipedia.org/wiki/… ) $\endgroup$
    – Adam
    Jun 3 '16 at 12:47
  • $\begingroup$ fawning - -it is not true what you want to "conclude" that $H'$ is diagonal in the basis of unperturbed $H_0$ eigenstates. Instead, perturbing means to replace $H_0$ by $H_0+H'$ and diagonalize this whole new Hamiltonian. It will have new eigenstates, close to the old ones, but not quite the same. Almost no operators in QM are diagonal in a randomly chosen basis - that's really a part of its being "quantum" mechanics. $\endgroup$ Jun 3 '16 at 12:58
  • 1
    $\begingroup$ fawning - the claim that $H'$ is diagonal in the basis of $H_0$ eigenstates is equivalent to $[H_0,H']=0$. Because the latter isn't true for the Stark effect, the former is also untrue. At most, due to the unbroken $SO(2)$ rotation symmetry of the electric field, you may see that the matrix elements vanish for $m\neq m'$ etc. but they don't vanish e.g. for $l,m=l',m'$ but $n\neq n'$. The perturbed $n$-th state surely contains an admixture of other $n'$-th states, even in the Stark effect. Not sure why you think otherwise. $\endgroup$ Jun 3 '16 at 13:21
  • 1
    $\begingroup$ @LubošMotl what I need is a better textbook. ;-) thanks for your help anyway. $\endgroup$ Jun 3 '16 at 13:38

Your Answer

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

Browse other questions tagged or ask your own question.