Consider the many-body (zero temperature) fermion Green function $$ G(a,b;t)=-i\theta(t)\langle\psi_a(t)\psi_b^\dagger\rangle $$ Where I'm restricting $t>0$ for causality and that the free Hamiltonian is the usual non-interacting Hamiltonian with dispersion $\varepsilon_k=k^2/2$. Let me consider the example explained in Coleman (Intro to Many-Body Physics, Section 7.4.3) in which we consider a scattering potential of the form $$ U_q c_{k+q}^\dagger c_k $$ Where I'm leaving summation implicit. Then it's clear that we can obtain the self-energy (or corresponding the interacting Green function) via the usual Feynman diagrams. More specifically, there exists a $T$-matrix such that $$ G=G_0 +G_0TG_0, \quad T=U+UG_0T $$ Let us consider the case where we are in 2-dim (so that the density of states is constant) and the scattering potential is local, i.e., $U(x)=U\delta (x)$, so that we can actually write down the $T$ matrix as $$ T(\omega)=\frac{U}{1-UF(\omega)}, \quad F(\omega) =\sum_k G_0(k,\omega) $$ Now if $U <0$, we see that $T(\omega)$ has a pole as some real $\omega$ and thus Coleman claims that this implies a non-perturbative result in the sense that there must be a bound state since corresponding interacting Green function $G$ also has a pole.

Question. I am have trouble understanding the rigor behind this claim, and in particular, I have two question.

First of all, the Feynman diagrams is an asymptotic series and in reality should only be considered as a formal Taylor series expansion, so I'm having trouble understanding why the self-energy (or in this case, the $T$-matrix) can actually be written as the full summation of the Feynman diagrams, that is, why does $T=U+UG_0 T$ hold true on a non-perturbative level?

Second, why should a pole in the Green function $G$ correspond to a bound state? Indeed, for a non-interacting Hamiltonian, the Green function is of the form $1/(\omega -\varepsilon)$, and thus you can see that the poles corresponds to the energy levels. In fact, it is exactly equal to the resolvent $1/(z-H)$ of the single-particle Hamiltonian. However, this doesn't seem to be true in the interacting case, and thus I'm having difficulty making the connection.

Now, it could be possible that this perturbative result just happens to be non-perturbative, in the sense that there exists some rigorous proof that shows that there exists a bound state for arbitrarily weak attractive scattering interactions in 2-dims, and the logic is that the perturbative result coinciding with the rigorous result gives us confidence in our calculations. Is this the case?

Follow-up. $\DeclareMathOperator{\ad}{ad}$ I want to first thank the comments and answers referring to the spectral represenation, which indeed helped with my understanding. However, I guess I'm still having a hard time understand exactly why a pole in interacting $G$ would represent a bound state. As far as I can tell, the spectrum representation basically states that $$ G(a,b;\omega+i\delta)=\langle \psi_a \frac{1}{\omega+i\delta -H} \psi_b^\dagger \rangle $$ Where $H$ is the interacting Hamiltonian, and thus I understand the "motivation/rationale" of why a pole corresponds to a bound state, but the rigorous logic is rather tricky

  • $\begingroup$ I think the poles of 2 point function correspond to 1 particle eigenstates of the system, in general and not just the bound states. The "rationale" behind it comes from the Källén–Lehmann spectral representation, though I am not sure how much the argument can be considered "rigorous". $\endgroup$
    – Hossein
    Commented Apr 28, 2022 at 7:02
  • $\begingroup$ @Hosein I think at least for a non-interacting system, the poles of Green functions are indeed exactly the bound states. Indeed, the Green function is just the resolvent $1/(z-H)$, and so you would think that the poles correspond to all eigenstates. However, in practice, you always take the thermodynamic limit, and thus $H$ will have a continuous and discrete spectrum. The resolvent will be undefined at the continuous spectrum, but technically it doesn't correspond to a pole, since the singularity isn't isolated. $\endgroup$ Commented Apr 28, 2022 at 20:28
  • $\begingroup$ Are you considering an interacting many body system or a free system? Because in the former case there are usually vertex corrections to the scattering potential. In general, isolated poles are considered as stable particles, these can be single particles or multiparticle bound states. You can check the "Polology" section in the Vol. 1 of Weinberg's QFT book, for a more careful discussion ( in the context of relativistic field theories, though). $\endgroup$
    – Hossein
    Commented Apr 29, 2022 at 5:54
  • $\begingroup$ @Hosein I'm mainly considering interacting systems, though I have used examples of non-interacting systems to see why poles correspond to bound states. As far as I understand, these vertex corrections are perturbative Feynman diagrams (which is at best asymptotic) so I was wondering if the claim that poles (calculated via perturbation) are bound state, is indeed rigorous. $\endgroup$ Commented Apr 30, 2022 at 21:58

1 Answer 1


It can be shown that the one-particle Green's function defined in the second quantization representation, $$ G(\mathbf{x},t; \mathbf{x}', t')=-\frac{i}{\hbar}\left\langle T\left[\psi(\mathbf{x},t)\psi^\dagger(\mathbf{x},t)\right]\right\rangle, $$ is the Green's function for the corresponding Schrödinger equation (or the appropriate relativistic equation), i.e., it satisfies $$ \left[i\hbar\frac{\partial }{\partial t} - \hat{H}\right]G(\mathbf{x},t; \mathbf{x}', t')=\delta^{(3)}(\mathbf{x}-\mathbf{x})\delta(t-t') $$ (Remarks: 1) I might be missing a coefficient in the right-hand-side; 2) whether the green's function is retarded, advanced or causal depends on the boundary condition at $t=t'$.)

However, the Green's function of the latter equation can be expressed in terms of the eigenfunctions of the Hamiltonian as (for simplicity I take the retarded Green's function) $$ H\phi_n(\mathbf{x})=\epsilon_n \phi_n(\mathbf{x}),\\ G(\mathbf{x},t; \mathbf{x}', t') = -\frac{i}{\hbar}\sum_n\phi_n(\mathbf{x})\phi_n^*(\mathbf{x}')e^{-\frac{i\epsilon_n (t-t')}{\hbar}}\theta(t-t')-\theta(t'-t), $$ which has Fourier representation $$ H\phi_n(\mathbf{x})=\epsilon_n \phi_n(\mathbf{x}),\\ G(\mathbf{x},\mathbf{x}', \omega) = \sum_n\frac{\phi_n(\mathbf{x})\phi_n^*(\mathbf{x}')}{\omega - \frac{\epsilon_n}{\hbar}+i0^+}, $$ that is the poles of this Green's function correspond to the eigenenergies of the Hamiltonian.

Lehmann representation, mentioned in the comments, generalizes this property to many-particle Green's functions with interactions - a detailed discussion can be found in any standard many-body text, such as Mahan, Fetter&Walecka, AGD.


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