quantum mechanics - Why does the branch cut of the self-energy begin at $2m$?

Consider a scalar field $\phi(x)$, and let its two-point function be $$ \frac{1}{p^2-m^2-\Pi(p^2)}=\int_0^\infty\mathrm d\mu^2\ \rho(\mu^2)\ \frac{1}{p^2-\mu^2} $$

We usually have $\Pi(m^2)=0$, which implies that there is a one-particle state with mass $m^2$. In terms of the spectral density, this reads $$ \rho(\mu^2)=\delta(\mu^2-m^2)+\sigma(\mu^2) $$ where the support of $\sigma(\mu^2)$ is disconnected from $m^2$.

On the other hand, the continuum contribution usually begins at $(2m)^2$, meaning that $\Pi(k^2)$ has a branch cut from $(2m)^2$ to $\infty$. Again, in terms of the spectral function, this is written as $\text{supp}(\sigma)=[(2m)^2,\infty)$.

My question

I would expect that the first bound state to lie close to $(2m)^2$, but slightly below, something around $(2m)^2-\mathcal O(\alpha^n)$, where $\alpha$ is the coupling constant and $n\in\mathbb N$. For example, in a simplified QED model, the bound state of an electron and a positron has binding energy $\sim 7\mathrm{eV}\sim m\alpha^2$, and so I would expect the branch point to be somewhere around $(2m)^2-m^2\alpha^4$.

I know the value of the self-energy to one loop, and the branch point is indeed exactly at $(2m)^2$. I don't know where to find higher loop corrections, so I cannot conclude whether

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  1) at higher loops, the branch point moves a bit closer to $m^2$, or

  
  


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  2) it stays at $(2m)^2$.

  
  

I would expect that the correct behaviour is 1), but this contradicts the fact that all books I've read always depict the threshold of pair production as an hyperboloid with its bottom sitting exactly at $(2m)^2$. On the other hand, if it turns out that the correct option is 2), I am lead to ask why is perturbation theory unable to correctly predict the position of the branch point (I know this is a subtle issue because bound states are, to some extent, non-perturbative entities; but AFAIK, perturbative calculations do contain a lot of information about bound states).

Answer

First, remember that the spectral density is written for states with zero total momentum.

That branch cut comes not from bound states but simply from unbound two-particle states that form continuous spectrum (you can always take particles with arbitrary opposite momenta which will give you arbitrary total energy). In contrast two-particle bound states usually form discrete spectrum. That mean that they will contribute as poles below $(2m)^2$.