4
$\begingroup$

Does the fact that the initial and final states : $|\phi_0\rangle$ and $S(\infty,-\infty) |\phi_0\rangle$ are not the same in a non-equilibrium setting - meaning essentially a time dependent Hamiltonian,(where $|\phi_0\rangle$ is a state of the time-independent hamiltonian part and $S(\infty,-\infty)$ being the S-matrix in the interaction picture) have something to do with the absence of a Gellman-Low like theorem for time dependent perturbations?

Gellman-Low states that both $\frac{U_{\epsilon I}(0,+\infty)|\Phi_0\rangle }{\langle \Phi_o|U_{\epsilon I}(0,+\infty)| \Phi_0 \rangle}$ and $\frac{U_{\epsilon I}(0,-\infty)|\Phi_0\rangle }{\langle \Phi_o|U_{\epsilon I}(0,-\infty)| \Phi_0 \rangle}$, in the limit $\epsilon \rightarrow 0$ are eigenstates of $H=lim_{\epsilon \rightarrow 0} (H_0+e^{-\epsilon |t|}H_1)$, $H_0$ being the non-interacting part, effectively implying their equality in the non-degenerate case.

For finite $\epsilon$, $U_{\epsilon I}(0,\pm\infty)|\Phi_0\rangle$ is an eigenstate of the full Hamiltonian with energy $E$.

$\endgroup$
0
$\begingroup$

I might be reversing your question, but the development of the Feynman-Dyson expansion in Abrikosov, Gorkov and Dzyaloshinski's book explicitly relies the Gel-Mann-Low theorem: by adiabatically turning on and the turning off the perturbation one can assure that the initial and the final states are the same (more precisely they differ by a phase factor), which enables the zero-temperature expansion formalism

| cite | improve this answer | |
$\endgroup$

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.