# A (apparently?) non-local quantum field theory

by rubbergnome
Tags: apparently, field, nonlocal, quantum, theory
 P: 12 I need to derive the euler-lagrange equations for the following non-local lagrangian density for a complex scalar field ψ $\mathcal{L} = \partial_{\mu}\psi^* \partial_{\mu}\psi - \lambda \int dy\, f(x,y) \psi^*(y) \psi(y)$ where λ is the coupling constant, f is a certain real-positive valued function linear in the first argument that satisfies f(x,y)=1/f(y,x) (which also implies f(x,x)=1). The integral is over all spacetime. Applying the usual euler-lagrange equations shouldn't be correct here. I tried taking the functional derivative of the action S=∫dx L with respect to ψ*and set it equal to zero, and I get $\partial_{\mu}\partial^{\mu}\psi(x) = -\lambda \int dy\, f(y,x) \psi(x)$ where indeed we have a non-constant mass term. On the other hand, I used the methods in this paper http://www.astro.columbia.edu/~lhui/...2012/HowTo.pdf to derive the feynman rules for the only possible vertex in the theory (this already made me think about a correction to the propagator); I get -iλ∫dxdy f(x,y) which purely depends on f. This result can also be quickly derived with eq. (136) here http://www2.ph.ed.ac.uk/~egardi/MQFT...cture_9_10.pdf The full propagator is therefore one of a free complex scalar field with mass m²= λ∫dxdy f(x,y). At least this is the result I got, and I'd like to confirm it deriving this mass term in the equations of motion. I also calculated the leading order correction to the transition amplitude between single-particle states in the canonical formalism, and the result agrees with the above procedure. The final doubt that arises is this: even if the equations of motions lead to the same result, why would the non-locality in the lagrangian be completely gone, turning into a mass term? I hope at least part of my post makes sense. Thanks in advance for helping. :)