In condensed matter physics (like in QFT) we can use Feynman's diagrams to compute the self-energy. From here we can obtain the spectral function as: $$ A_{\mathbf k} (\omega) = \frac{-\frac{1}{\pi}\mathrm{Im}\Sigma_{\mathbf k}(\omega)} {\left[\omega - (\epsilon_{\mathbf k }- \mu) - \mathrm{Re}\Sigma_{\mathrm{k}}(\omega)\right]^2 + \left[ \mathrm{Im}\Sigma_{\mathbf k}(\omega) \right]^2} $$

At low temperatures this spectral function typically has some satellites structures and a quasi-particle peak. From here we can read-off quasi-particle properties... However I want to know if (and how!) can we see if there are any bound states in the system? Actually, I do know that it can be done (in BCS theory for example), but I do not know how?


The poles of the spectral density give you the particles of the system, whether they are fundamental or bound states. There is no fundamental difference between these two types: they behave exactly the same, phenomenologically, and they are indistinguishable theoretically.

This becomes even clearer in light of the concept of duality: the same theory often has several equivalent descriptions, and the fundamental particles in one description look composite in the other, and vice-versa. There is no way to tell whether a given state is fundamental or composite, because this depends on which variables you use in defining the theory.

When one says that e.g. a meson has substructure, it doesn't mean it behaves differently from a fundamental particle. It just means we haven't find an appropriate description of QCD where the meson is fundamental. Finding this description would be essentially the same as solving QCD, so it is a very hard problem. But, as a matter of principle, if we could find such description, the field that describes the meson would be just a regular field in your lagrangian, akin to the electron field in the standard model. In supersymmetric models we can in fact find such dual description thanks to the wonderful work of Seiberg, cf. Seiberg duality.


Bound vs. extended states
In condensed matter physics bound states are the states with the wave functions decaying towards infinity, as opposed to the extended states, as, e.g., plane waves. In this sense the definitive answer to the question can be given only by the exact diagonalization of the Hamiltonian and studying the behavior of the wave functions. Without that the question is necessarily vague, and the discussions of the bound states usually carry qualitative character.

Gaps between the states
Another way to define bound states is as isolated states, i.e., they are not a part of a continuous spectrum. In non-interacting case (or in the spectrum of a diagonalized Hamiltonian), these are immediately seen from the dependence of $\epsilon_k$ on $k$. In other words, they are separated by a gap from the adjacent bound states or the continuum. In this picture, including the interactions broadens the states, and whether they remain (quasi-)bound depends on how the broadening compares to the gap. When teh broadening is very strong we say that the bound states wash out, becoming continuous.

Gap opening
Finally, there is a class of problems where we are interested in appearance of bound states. In some cases one can re-express the problem in terms of an effective Schrödinger equation (e.g., by summing the ladder diagrams - see Fetter&Walecka for a clear presentation). In this case one distinguishes bound and extended states by the propreties of the solutions of thsi effective Schrödinger equation (i.e., decaying to infinity or not).

Another case is gap opening, which can be studied by a number of techniques, e.f., renormalization group.

Clarification about the spectral function
The textbooks on QFT in condensed matter physics teach us that the partcle (and multiparticle) excitations appear as singularities in the Green's function. It is necessary to stress here that Green's function is not the same thing as the spectral function (like the one given in the question). In fact, the spectral function is analytical, i.e., it does not have singularities. It is convenient to use mathematically, but it is less suitable for the question that interests us. In fact, although it gives a good intuition when discussing broadening of states or their delocalization, as I discussed above, it can be very misleading in terms of finding the true excitations, since a lot of $\omega$-dependence, and all the new physics due to the interaction, is hidden in the self-energy.

If we work with the Green's function, then its singularities will be either isolated poles or branch cuts - corresponding respectively to isolated (likely bound) states and the continuous spectrum.

Another point - the spectral function given in the question is that for a one-particle Green's function. If we are interested in two-particle bound states, operating with a two-particle Green's function and the corresponding spectral representation.

  • $\begingroup$ Everything you said is of course correct. I completely agree. However, I wanted to know if we could see the existence of bound states from the spectral function? From what you wrote (if I understood correctly) we cannot determine the existence of bound states from the spectral function alone? Right? $\endgroup$
    – RedGiant
    Jan 6 at 10:13
  • $\begingroup$ You seem to be talking about the $\omega$-dependence of the spectral function, when $k$ is fixed, i.e. you are talking about a single state, whereas the existence of gaps between states is seen only by comparing the position of different states. Another thing to keep in mind: the spectral function is analytical, whereas the Green's functions have poles for the bound states and branch cuts for continuum states. $\endgroup$ Jan 6 at 10:28
  • $\begingroup$ Yes, I was talking about $\omega$ dependence. What you said makes sense. So, I would search for the bound states by comparing the spectral functions for different $\mathbf{k}$ and see if there is any "intersection" or not? $\endgroup$
    – RedGiant
    Jan 6 at 10:42
  • $\begingroup$ Formally, the method is to look for the poles/branch cuts in the Greens function (not the spectral function). In practice, it is one of the approaches in my answer, but it depends on the problem you are dealing with - there is no general method. $\endgroup$ Jan 6 at 10:46
  • $\begingroup$ I expanded the answer to incorporate these points. $\endgroup$ Jan 6 at 11:28

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