Why poles of two-point function corresponds to bound states?

Question

In this article Two-time Green function method in quantum electrodynamics of high-Z few-electron atoms the author has:

Let $\mathcal{G}$ be fourier transform of the green function
$$ \begin{array}{r} \mathcal{G}\left(E ; \mathbf{x}_1^{\prime}, \ldots \mathbf{x}_N^{\prime} ; \mathbf{x}_1, \ldots \mathbf{x}_N\right) \delta\left(E-E^{\prime}\right)=\frac{1}{2 \pi i} \frac{1}{N !} \int_{-\infty}^{\infty} d x^0 d x^{\prime 0} \exp \left(i E^{\prime} x^0-i E x^0\right) \\ \times\left\langle 0\left|T \psi\left(x^{\prime 0}, \mathbf{x}_1^{\prime}\right) \cdots \psi\left(x^{\prime 0}, \mathbf{x}_N^{\prime}\right) \bar{\psi}\left(x^0, \mathbf{x}_N\right) \cdots \bar{\psi}\left(x^0, \mathbf{x}_1\right)\right| 0\right\rangle \end{array} $$

Inserting a complete se of states $$ H|n\rangle=E_n|n\rangle, \quad \sum_n|n\rangle\langle n|=I, $$ and Assuming, for simplicity, $E_0 = 0$ we obtain $$ \mathcal{G}(E)=\sum_n \frac{\Phi_n \bar{\Phi}_n}{E-E_n+i 0}-(-1)^N \sum_n \frac{\Xi_n \bar{\Xi}_n}{E+E_n-i 0}, $$ where the variables $\mathbf{x}_1^{\prime}, \ldots, \mathbf{x}_N^{\prime}, \mathbf{x}_1, \ldots, \mathbf{x}_N$ are implicit and $$ \begin{aligned} & \Phi_n\left(\mathbf{x}_1, \ldots \mathbf{x}_N\right)=\frac{1}{\sqrt{N !}}\left\langle 0\left|\psi\left(0, \mathbf{x}_1\right) \cdots \psi\left(0, \mathbf{x}_N\right)\right| n\right\rangle \\ & \Xi_n\left(\mathbf{x}_1, \ldots \mathbf{x}_N\right)=\frac{1}{\sqrt{N !}}\left\langle n\left|\psi\left(0, \mathbf{x}_1\right) \cdots \psi\left(0, \mathbf{x}_N\right)\right| 0\right\rangle \end{aligned} $$ Now let us introduce the functions $$ \begin{aligned} & A\left(E ; \mathbf{x}_1^{\prime}, \ldots, \mathbf{x}_N^{\prime} ; \mathbf{x}_1, \ldots, \mathbf{x}_N\right)=\sum_n \delta\left(E-E_n\right) \Phi_n\left(\mathbf{x}_1^{\prime}, \ldots, \mathbf{x}_N^{\prime}\right) \bar{\Phi}_n\left(\mathbf{x}_1, \ldots, \mathbf{x}_N\right), \\ & B\left(E ; \mathbf{x}_1^{\prime}, \ldots, \mathbf{x}_N^{\prime} ; \mathbf{x}_1, \ldots, \mathbf{x}_N\right)=\sum_n \delta\left(E-E_n\right) \Xi_n\left(\mathbf{x}_1^{\prime}, \ldots, \mathbf{x}_N^{\prime}\right) \bar{\Xi}_n\left(\mathbf{x}_1, \ldots, \mathbf{x}_N\right) . \end{aligned} $$ These functions satisfy the conditions $$ \begin{aligned} \int_{-\infty}^{\infty} d E A\left(E ; \mathbf{x}_1^{\prime}, \ldots, \mathbf{x}_N^{\prime} ; \mathbf{x}_1, \ldots, \mathbf{x}_N\right)= & \frac{1}{N !}\langle 0| \psi\left(0, \mathbf{x}_1^{\prime}\right) \cdots \psi\left(0, \mathbf{x}_N^{\prime}\right) \\ & \times \bar{\psi}\left(0, \mathbf{x}_N\right) \cdots \bar{\psi}\left(0, \mathbf{x}_1\right)|0\rangle \\ \int_{-\infty}^{\infty} d E B\left(E ; \mathbf{x}_1^{\prime}, \ldots, \mathbf{x}_N^{\prime} ; \mathbf{x}_1, \ldots, \mathbf{x}_N\right)= & \frac{1}{N !}\langle 0| \bar{\psi}\left(0, \mathbf{x}_N\right) \cdots \bar{\psi}\left(0, \mathbf{x}_1\right) \\ & \times \psi\left(0, \mathbf{x}_1^{\prime}\right) \cdots \psi\left(0, \mathbf{x}_N^{\prime}\right)|0\rangle \end{aligned} $$ In terms of these functions, the two point function is $$ \mathcal{G}(E)=\int_0^{\infty} d E^{\prime} \frac{A\left(E^{\prime}\right)}{E-E^{\prime}+i 0}-(-1)^N \int_0^{\infty} d E^{\prime} \frac{B\left(E^{\prime}\right)}{E+E^{\prime}-i 0}, $$ using the following commutation relations we can show $$ [Q, \psi(x)]=-e \psi(x), \quad[Q, \bar{\psi}(x)]=e \bar{\psi}(x), $$ where $Q$ is the charge operator in the Heisenberg representation. Therefore, we have $$ \mathcal{G}(E)=\int_{E_{\min }^{(+)}}^{\infty} d E^{\prime} \frac{A\left(E^{\prime}\right)}{E-E^{\prime}+i 0}-(-1)^N \int_{E_{\min }^{(-)}}^{\infty} d E^{\prime} \frac{B\left(E^{\prime}\right)}{E+E^{\prime}-i 0} $$ where $E_{\min }^{(+)}$is the minimal energy of states with electric charge $e N$ and $E_{\mathrm{min}}^{(-)}$is the minimal energy of states with electric charge $-e N$.

one can show with the help of standard mathematical methods that the equation $$ \mathcal{G}(E)=\int_{E_{\text {min }}^{(+)}}^{\infty} d E^{\prime} \frac{A\left(E^{\prime}\right)}{E-E^{\prime}}-(-1)^N \int_{E_{\min }^{(-)}}^{\infty} d E^{\prime} \frac{B\left(E^{\prime}\right)}{E+E^{\prime}} $$ defines an analytical function of $E$ in the complex $E$ plane with the cuts $\left(-\infty, E_{\min }^{(-)}\right]$and $\left[E_{\min }^{(+)}, \infty\right)$. This equation provides the analytical continuation of the Green function to the complex $E$ plane.

Now the author claims the bound states correspond to the poles of the function $\mathcal{G}(E)$ on the right-hand real semiaxis. If the interaction between the electron-positron field and the electromagnetic field is switched off, the poles corresponding to bound states are isolated. Switching on the interaction between the fields transforms the isolated poles into branch points.

Why poles of $\mathcal{G}(E)$ corresponds to bound states? How that the continuous states just disappear?

Answer 1

The states $|n\rangle$ are your bound states. Your complete set is, by definition, a basis that diagonalizes $P^\mu$, namely they represent states with definite momentum and energy, an object that we often simply call a particle, most often referred to as a bound state (a notion that has no intrinsic meaning but has clear historical significance).

Hence, since $G(E)\sim 1/(E-E_n)$, the two point function has a pole at every $E=E_n$, i.e., at each eigen-energy, i.e., at each energy of a bound state. (You are implicitly looking at zero momentum so the energy is the same as mass; in general you have poles at $p^2=m^2$. As a function of $p^0$ this is not really a pole but a branch cut instead, since you have "a pole for every $\boldsymbol p$", a continuum of poles, a.k.a. a branch cut).

(Just to be extra clear, $G(E)$ has a pole at every $E=E_n$ only when the numerator does not vanish; the norm $|\Phi_n|^2$ does not usually vanish unless enforced by symmetry à la Wigner-Eckart)