Is there a physical interpretation of the existence of poles for a Green function? In particular how can we interpret the fact that a pole is purely real or purely imaginary? It's a general question but I would be interested in the interpretation in quantum systems.

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    $\begingroup$ To fully understand this topic you need to study bound states and resonances from the point of view of scattering theory (see e.g. textbook by Taylor) and also study Kallen-Lehmann representation and LSZ formalism in quantum field theory (confront for example Peskin, Schroeder for that) $\endgroup$
    – Blazej
    Jul 24, 2017 at 20:01
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    $\begingroup$ for the relevant discussion in the case of quantum field theory, see Positivity of residues and unitarity in scattering amplitudes. $\endgroup$ Jul 24, 2017 at 21:16

3 Answers 3


Let me expand a little more on what Craig Thone just said :

Consider the energy/frequency-dependent Green function : $$ \tilde{G}(\omega)=\frac{1}{\omega-(a-\mathrm{i}b)} $$ with one single pole in $\omega=a-\mathrm{i}b$ (with $b>0$), which is Fourier transform of the time-dependent $G(t)$ Green function such as : $$ G(t)=\int\frac{\mathrm{d}\omega}{2\pi}\frac{e^{-\mathrm{i}\omega t}}{\omega-(a-\mathrm{i}b)} $$ One can show, using complex analysis (I can eventually show some details about that if needed), that it computes to : $$ G(t)=\mathrm{i}\,e^{-\mathrm{i}at-bt}\Theta(t) $$ where $\Theta$ is an Heaviside step-function.

It means that :

  1. The real part $a$ of the pole gives an oscillatory behavior to the solution. In the context of quantum systems, $a$ is often referred to as an eigen-energy.
  2. The role the imaginary part $b$ in twofold :
    • It gives a damped behavior to the solution. In the context of quantum systems, $b$ will describe how one state, which is not an eigen-state of the system, will be depleted as a function of time in a superposition of eigen-states. In the limit of a perturbation theory, $b$ is the same rate given by the Fermi Golden Rule.
    • The fact that $b>0$ ensures that the function $\tilde{G}(\omega)$ has no pole in the upper complex plane (i.e. $\forall\omega\in\mathbb{C}, \Im({\omega})>0$). This analycity of $\tilde{G}$ implies by the Cauchy's integral theorem that : $$ \forall t<0,\,G(t)=0 $$ which is needed to ensure the causality of the dynamics. To ensure such property, an Heaviside step-function $\Theta$ can be added to $G(t)$.

EDIT : Analyticity of $G$ and causality. In order to evaluate $G(t)$ for $t<0$, one can consider the path $\Gamma_R$ which is defined as : $$ \Gamma_R=\left\{\omega\in\mathbb{C},\,\omega\in[-R,R]\,\bigcup\,\mathcal{C}_R^0\right\} $$ where $\mathcal{C}_R^0$ is half of the circle centered on $\omega=0$ with a $R$ radius.

$\forall t<0,\, G(t)$ is given by the integral : $$ \forall t<0,\, G(t)=\lim_{R\rightarrow+\infty}\int_R^{-R}\frac{\mathrm{d}\omega}{2\pi}\frac{e^{-\mathrm{i}\omega t}}{\omega-(a-\mathrm{i}b)} $$ Additivity of integrals gives you then : $$ \forall t<0,\,\int_{\Gamma_R}\frac{\mathrm{d}\omega}{2\pi}\frac{e^{-\mathrm{i}\omega t}}{\omega-(a-\mathrm{i}b)}=G(t)+\lim_{R\rightarrow+\infty}\int_{\mathcal{C}_R^0}\frac{\mathrm{d}\omega}{2\pi}\frac{e^{-\mathrm{i}\omega t}}{\omega-(a-\mathrm{i}b)} $$ Since there is no pole in the complex area enclosed by $\Gamma_R$ ($\tilde{G}$ is analytic in the upper complex plan), Cauchy's integral theorem gives you that : $$ \int_{\Gamma_R}\frac{\mathrm{d}\omega}{2\pi}\frac{e^{-\mathrm{i}\omega t}}{\omega-(a-\mathrm{i}b)}=0 $$ Moreover, Jordan's lemma ensures that : $$ \lim_{R\rightarrow+\infty}\int_{\mathcal{C}_R^0}\frac{\mathrm{d}\omega}{2\pi}\frac{e^{-\mathrm{i}\omega t}}{\omega-(a-\mathrm{i}b)}=0 $$ letting you with : $$ \forall t<0,\,G(t)=0 $$ You have to realize this is the reason why sometimes you see people introduce some infinitesimal $\pm\mathrm{i}\epsilon$ which pushes the pole of a propagator out of the real axis : it ensures causality/anti-causality of the solution.

  • $\begingroup$ Could you clarify how the analycity in the upper half plane implies $G(t)=0$ for $t<0$? $\endgroup$ Jul 30, 2017 at 10:17
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    $\begingroup$ See edit for more details. $\endgroup$
    – dolun
    Jul 31, 2017 at 8:43
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    $\begingroup$ This is a good answer, and both this and David's are worthy of the bounty; I'm awarding it to him to encourage new contributors. $\endgroup$ Aug 1, 2017 at 11:52
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    $\begingroup$ Thanks for your clear and helpful answer. Has it a typo in the third formal ? a minus sign is omitted before "$iat$"?: $G(t) =\mathrm{i} e^{-\mathrm{i} at - b t}$ $\endgroup$
    – Zoe Rowa
    Sep 21, 2017 at 3:00
  • $\begingroup$ Glad it helped! Yes, you are right! Thanks for pointing it out! $\endgroup$
    – dolun
    Sep 21, 2017 at 15:35

This is a somewhat broad question, because there are a number of different Green's functions in quantum physics. Perhaps the simplest one is the resolvent Green's function for a single-particle system. Its definition is $$G(\omega^{\pm})=\lim_{\delta \rightarrow 0^+}\left[ \omega \pm i\delta - H \right]^{-1}\equiv \frac{1}{\omega\pm i\delta - H},$$ where $H$ is the Hamiltonian. Ignoring the $\delta$, and replacing $\omega \rightarrow E$, you can see how this relates to the time-independent Schrödinger equation: $$H\left|\psi \right>=E\left|\psi \right> \longrightarrow \left[ E - H\right]\left|\psi \right> = 0.$$ Basically, if you apply the operator $G$ to a solution to the Schrödinger equation, $\left|\psi_n \right>$ with energy level $E_n$, you get a pole (the "denominator" vanishes) at $\omega \pm i\delta = E_n$. That is, $$G_{nn}(\omega^{\pm})\equiv\left<\psi_n \right|G(\omega^{\pm}) \left|\psi_n \right>=\frac{1}{\omega \pm i\delta - E_n},$$ has a pole. Then it is clear that the poles of (Tr = trace) $$\mathrm{Tr}\{G(\omega^{\pm})\}\equiv\sum_{n}^{\mathrm{all\,states}}G_{nn}(\omega^{\pm}),$$ give you the full spectrum. In fact, you can show that the quantity (Im = imaginary part) $$\rho(\omega)=-\frac{1}{\pi}\mathrm{Im}\{\mathrm{Tr}\{G(\omega^+)\}\},$$ gives you the density of states of the Hamiltonian $H$.

The resolvent Greens function is valid for a single-particle system, but the concept caries over well to many-body physics, and thus to quantum field theory. However, the definition of the Green's function is not as direct in those cases. The Green's function is given by the probability amplitude that a particle will be added to the vacuum state at a time $t$ and state $n$, and that after time evolution it will be removed at time $t'$ and state $n'$: $$\mathcal{G}(t',n';t,n)=-i\left<\Phi \right|\psi_{n'}(t')\psi_n^\dagger (t) \left|\Phi \right>,$$ where the factor of $i$ appears just as a convention, and $\left|\Phi \right>$ represents the ground state. This definition has a very transparent interpretation, because this quantity is the answer to the question: if in the quantum vacuum, which is (to quote Lawrence Krauss' favorite phrase) "a boiling bubbling brew of particles popping in and out of existence," a particle were to pop into existence at time-state $(t,n)$, what is the probability (amplitude) that it would propagate to time-state $(t',n')$? There are a number of different versions of this type of Green's function, each useful for a different thing. One thing all of these have in common is that handling them in this form is complicated, and it is much easier to work with their Fourier transforms $$\mathcal{G}_{n'n}(\omega^{\pm})=\int\mathrm{d}(t'-t)\,\mathrm{e}^{i(\omega\pm i \delta )(t'-t)}\mathcal{G}(t',n';t,n).$$ As you can see, a complex shift $\delta$ is introduced here, which has to do with the convergence of the integral. This is precisely the origin of it in the resolvent Green's function. In fact, if you have a system where the particles don't interact with each other, and instead each particle simply follows the Hamiltonian $H$ with energies $\{E_n\}$, then you can show that $$\mathcal{G}_{n'n}(\omega^{\pm})=\frac{\delta_{n',n}}{\omega\pm i\delta - E_n},$$ and you get a full analogy with the single-particle Green's function, and the answer to the question above is "if the particle is created at an eigen-energy of the Hamiltonian, and propagates to be annihilated at the same eigen-energy, you get a peak in probability (a pole); if not, you get zero."

However, things get more interesting when you do have interactions, and your Hamiltonian becomes $H + U$. For that case, it can be shown that $$\mathcal{G}_{n'n}(\omega^\pm) = \frac{1}{\omega - E_n - \Sigma_{n'n}(\omega^\pm)},$$ where $\Sigma_{n'n}(\omega^\pm)$ is called the self energy, and the equation above is called the Dyson equation. In many cases this will have a simple form when you calculate specifically $\mathcal{G}_{nn}$: $$\mathcal{G}_{nn}(\omega) = \frac{1}{\omega - (E_n + \Lambda(\omega) ) - i \Gamma_{n}(\omega) }.$$ This has a very simple meaning, when you compare it to the non-interacting case: first, if you ignore the $\Gamma$, you have that because of the interactions the pole has moved from $E_n$ to $E_n + \Lambda$. Then, if you consider $\Gamma$, you will find that the pole is not infinitely "narrow," but instead has a width $\approx \Gamma$. In fact, this expression is nothing but a type of Lorentz distribution. Basically, because of interactions, your energy level is not longer infinitely lived, but now decays into the vacuum in a time scale given by $\hbar/\Gamma$. $\Gamma$ is sometimes called the scattering rate, or decay rate. But because there is still an analogy with the resolvent Green's function, one says that this state, which has a finite lifetime, is an energy state for a new particle---a quasi-particle---that emerges from the interacting system.

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    $\begingroup$ Nice and clear answer. Yet, I suggest adding a paragraph also on the meaning/rôle of Green's functions in linear response theory. That will give poles another (rather different) meaning, and connects them directly to some observed phenomena. $\endgroup$
    – AlQuemist
    Jul 31, 2017 at 7:20
  • $\begingroup$ That's a good point. I'll come back after work and add a few details on that, and perhaps on the connection to the path integral formalism. Thanks for the suggestion! $\endgroup$ Jul 31, 2017 at 14:31

The pole of Green's function is related to the spectrum of the particle which is propagating. One dimension for example $$\tilde{G}(\omega)= \frac{i}{\omega-(\epsilon+i\Gamma)}$$ If pure real, G(t) is some oscillation function which shows that the particle is stable. If pure imaginary, G(t) has some exponential decay behavior which shows that the particle is unstable.

  • $\begingroup$ For classical linear response the GF can be neither purely real nor purely imaginary on any finite interval because of Kramers-Kronig. I.e., a real GF violates causality except in vacuum. Is this not true in QM? $\endgroup$
    – user27777
    Aug 12, 2013 at 15:22
  • $\begingroup$ The picture is similar. For free case without interaction, GF is real. For intacting case, there is a imaginary part which is related to self energy. Linear response is some kinds of interacting. $\endgroup$
    – thone
    Aug 13, 2013 at 3:07

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