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

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

  • 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).

  • $\begingroup$ If $m$ is the physical mass the branch cut would stay at $(2m)^2$. Perturbation theory is around the physical scattering state, not at the bare quantities. This is why we have counter terms in perturbation theory given finite answers, because we are expand around the right hamiltonian $H_0$, that have physical mass and charges for the particles. $\endgroup$ – Nogueira Oct 30 '16 at 16:48
  • $\begingroup$ $2m$ is pricesely the minimum energy of two far away particles that don't interact each other. This is the physical mass, so all loops corrections (self-interaction) are tanking account. Energies that is bellow the $2m$ are necessarily bound states, by conservation of energy can give rise to a scattering state, than having discrete spectra. So, what the loops corrections would do is move this discrete spectra that are bellow $2m$. $\endgroup$ – Nogueira Oct 30 '16 at 16:55

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 $(2m)^2$.


The branch cut anticipates the formation of scattering states. The branch point is at the square of the minimum energy of a scattering state, namely $E=2mc^2$, the energy of two far away particles not interacting each other. So, the branch cut are determined only by the self interactions of each particle, and this is already taken account by the physical mass of the particles. So, seeking for a correction of the location of the branch point is the same as seeking for corrections to the mass of the one-particle scattering state.

What than the interaction, and consequently the loop corrections, can do bellow the branch cut? They can create and move poles between the branch point and the $m^2$ pole. They are the bound states of the system. This states should be bound states because there is not sufficient energy to make particles flying apart (scattering state of two or more particles), and there is to much energy to put in only one particle.

They should be poles because of the boundary conditions that this states should obeys, namely the "impossibility" to have two particles in bound state at very far distances (some exponential fall off of a sort).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.