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This post is a followup question to: How to get an imaginary self energy?

In the cited post, the two following representations for the one-particle Green's function are shown:

$$G(k,\omega) = \sum_n \frac{|c_k|^2}{\omega - E_n + i\eta}$$

$$G(k,\omega) = \frac{1}{\omega - \varepsilon(k) - \Sigma(k,\omega) + i\eta}$$

How does one see that these two representations have poles at the same energies? It seems to me that this is not the case, since the first formula is constrained to have poles at the real energies $E_n$, while it looks like the second formula can have poles at complex energy due to the possibly complex self-energy $\Sigma(k, \omega)$. Can anyone reconcile this apparent discrepancy?

Edit: One might also rephrase my question as an inquiry into the relation between the Lehmann representation and the Dyson equation.

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It's a good question, and it has a beautiful answer.

It is true that for any finite sum (referring to your first expression), one cannot have a complex pole. So the question is: how can the complex pole appear in the $\boldsymbol{n \to \infty}$ limit? (Spoiler: it can.)

The short answer is that in the $n\to \infty$ limit, the Green's function can develop a branch cut. What this means is that the function is more naturally defined on a larger Riemann surface, which means one can use this branch cut as a 'portal' to a new sheet. The complex pole lives on this new sheet! Not even M. Night Shyamalan could've thought of this one.

Let me illustrate this with an example. Suppose we have $$G(z) = \sum_n \frac{c_n}{z-\varepsilon_n}\qquad \textrm{with } c_n = \frac{1}{\pi} \frac{\Gamma}{\varepsilon_n^2 + \Gamma^2},$$ where we can think of the $\varepsilon_n$ being some equally spaced list of energies along the real axis. This function clearly has no complex poles. However, in the continuum limit, we can easily calculate (e.g. by using the residue theorem) that we obtain $$ G(z) = \int_{-\infty}^\infty \frac{1}{\pi}\; \frac{\Gamma}{\varepsilon^2 + \Gamma^2} \; \frac{1}{z-\varepsilon} \mathrm d\varepsilon = \left\{ \begin{array}{ccl} \frac{1}{z+i\Gamma} && \textrm{if Im } z > 0, \\ \frac{1}{z-i\Gamma} && \textrm{if Im } z < 0. \end{array} \right.$$ Clearly, this function still has no complex poles on the complex plane! However, we now have a branch cut on the real axis. In such a case, the function $G(z)$ is more naturally defined on a more general Riemann surface. In this very simple case, the situation is a bit funny: there are two complex continuations, depending on whether we come from the bottom or the top. In other words, $G(z)$ can be more naturally thought of as being defined on two separate complex planes: on one we have $\frac{1}{z+i\Gamma}$, and on the other $\frac{1}{z-i\Gamma}$ on the other. Clearly, on this more complete domain, the functions are analytic (i.e. no more branch cut) and we have complex poles at $z = \pm i \Gamma$! (Remark: these two separate functions correspond to the retarded and advanced Green's functions.)

Ending up with a Riemann surface that is the union of two disjoint spaces is a bit unusual and is an artifact of $G(z)$ having a branch cut that cut the plane in two. In more general situations (where the branch cut doesn't extend over the whole real axis), the Riemann surface will be connected. It will still be true that to get to the complex pole, you'd need to cross the branch cut, into the new branch. (Since this new branch cannot be 'accessed' without a branch cut, it makes sense that there is no complex pole for any finite sum.)

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  • $\begingroup$ 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? $\endgroup$ – Ian Sep 21 '18 at 2:01
  • $\begingroup$ @Ian sure, I don't see what would prevent one from repeating the above argument for $\hat G (z) = (z-\hat H)^{-1}$. $\endgroup$ – Ruben Verresen Sep 21 '18 at 2:24
  • $\begingroup$ If you define the resolvent in that way, it's not clear to me how the poles appear without the c_n. Any comment? $\endgroup$ – Ian Sep 21 '18 at 3:06
  • $\begingroup$ @Ian Good point! I'm stumped; a fun question to think about. $\endgroup$ – Ruben Verresen Sep 21 '18 at 3:27
  • $\begingroup$ 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. $\endgroup$ – Ian Sep 21 '18 at 4:21

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