Skip to main content
9 events
when toggle format what by license comment
Sep 21, 2018 at 4:37 comment added Ian Let us continue this discussion in chat.
Sep 21, 2018 at 4:30 comment added Ruben Verresen @Ian I am not sure what you mean by eigenvalue of the resolvent... The poles of the resolvent are the physically meaningful quantity. I thought you were asking whether the resolvent has a pole at $z= \pm i \Gamma$. I don't think it makes sense to ask about eigenvalues of the resolvent; how would you even define that? $\hat G(z) |\psi_z\rangle = \lambda_z |\psi_z\rangle$?... Seems pretty meaningless to me!
Sep 21, 2018 at 4:21 comment added Ian My (new) knowledge is that for an operator H, the resolvent of H can in general have additional poles that aren't the eigenvalues of H. My question though is whether these poles are necessarily eigenvalues of the resolvent! I'll keep digging.
Sep 21, 2018 at 3:27 comment added Ruben Verresen @Ian Good point! I'm stumped; a fun question to think about.
Sep 21, 2018 at 3:06 comment added Ian If you define the resolvent in that way, it's not clear to me how the poles appear without the c_n. Any comment?
Sep 21, 2018 at 2:24 comment added Ruben Verresen @Ian sure, I don't see what would prevent one from repeating the above argument for $\hat G (z) = (z-\hat H)^{-1}$.
Sep 21, 2018 at 2:01 comment added Ian This makes a lot of sense. Let me push the question a step further. In the finite n limit, the poles are located at the eigenvalues of the Hamiltonian, such that the eigenvalues of the resolvent are 1/(z-E_n). In your example, is the pole 1/(z +/- i*Gamma) an eigenvalue of the resolvent?
Sep 18, 2018 at 7:34 history edited Ruben Verresen CC BY-SA 4.0
added 68 characters in body
Sep 17, 2018 at 22:26 history answered Ruben Verresen CC BY-SA 4.0