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  How should we deal with diagrams which do not conserve particle number in a non-relativistic field theory?

+ 3 like - 0 dislike

In the last 10 years there has been more and more crossover of techniques from high energy physics being used in AMO and condensed matter scenarios, in particular diagrammatic techniques and related perturbative calculations. Unlike the relativistic case, a simple Schrodinger field theory will conserve the number of particles; however, not all diagrams that one might draw respect this conserved quantity. My question is, what should be done with such diagrams? Will the theory generically result in these diagrams not contributing to scattering processes or is this something that has to be input 'by hand?'

As an illustrative example, consider the interaction of two non-relativistic scalar fields with a local quartic interaction 
$$\mathcal{L} = \sum_{i=1}^2{(\psi^\dagger_i \dot{\psi}_i - \frac{1}{2m} \nabla \psi^\dagger_i \nabla \psi_i)} + \frac{g}{m} \psi^\dagger_1 \psi^\dagger_2 \psi_2 \psi_1,$$ which represents a system whose excitations are fermions with 2 orthogonal spin states with a contact interaction. Such a theory is considered in Braaten and Platter's [Exact Relations for a Strongly Interacting Fermi Gas from the Operator Product Expansion](http://journals.aps.org/prl/pdf/10.1103/PhysRevLett.100.205301).

Note in particular that they find that after imposing a momentum cutoff for the theory, the coupling constant must satisfy $$g(\Lambda) = \frac{4 \pi a}{1- 2 a \Lambda/\pi}$$ to recover the scattering amplitude we expect from the zero-range model with a scattering length $a$ (and $\Lambda$ is the momentum cutoff). Importantly, the coupling constant $g \sim \mathcal{O}(1/\Lambda)$ for large cutoff.

However, if one computes the corrections to the propagator (of either particle) you find that the first order correction has only one diagram and its value is $$\Delta_1(k) \sim -\frac{i g}{8 m \pi^2} \frac{\Lambda^3}{3}.$$ On the contrary, I expected this diagram to not contribute anything (similarly to how it does not in the relativistic $\phi^4$ theory)  1) because it does not conserve particle number and 2) because the parameter 'm' of the bare theory is the SAME 'm' that appears in the equivalent Schrodinger equation, meaning that the physical mass of the particle is not changed by the perturbative corrections. And then, not only is the contribution non-zero but it blows up as the cutoff becomes large. Are we required to postulate that $$m = m_0 + \mathcal{O}(\Lambda^3)$$ so that these contributions do go to zero as the cutoff increases?

Identical question for reference: http://physics.stackexchange.com/questions/200237/how-should-we-deal-with-diagrams-which-do-not-conserve-particle-number-in-a-non

asked Aug 20, 2015 in Theoretical Physics by Kevin Driscoll [ no revision ]

If this diagram is a correction to the propagator how does it not conserve particle number? The initial and final particle number should still both be 1, right? In relativistic \(\phi^4\) theory there is most definitely a self-energy contribution that blows up with the cutoff.

1 Answer

+ 0 like - 0 dislike

I can't imagine which self-energy diagram you are considering. An important occurring in the non-relativistic theory (and in the absence of chemical potentials) is that there is no correction to the propagator to any order in perturbation theory. 

answered Oct 28, 2015 by anonymous [ no revision ]

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