Hanbury Brown Twiss

# Noise correlations in an expanding 1d Bose gas

Just out in PRA is a cute result concerning the expansion of a 1D gas from a regular array.

The initial state -- a 1D Mott insulator -- acts as an array of incoherent sources. Thus one doesn’t expect to see an interference pattern in the density after the atoms of the gas are released. There are fringes, however, in the density-density correlation function. This is the famous Hanbury Brown and Twiss effect. The effect is sensitive to the statistics of the particles: an array of sources gives rise to peaks in the correlation function for bosons, dips for fermions.

The effect of interactions hasn’t been too important historically, either because one dealt with noninteracting particles (photons, say) and / or because the spatial separation of the sources (different parts of a star in the original experiment!).

That will change in the situation discussed here: cold atoms in one dimension. The reason is that the effect involves the interferences of trajectories that differ by an exchange of a pair of particles -- that’s how statistics enters -- thus the particles must necessarily pass each other and interact.

Now comes the interesting part. It’s known that when bosons have infinite short-range repulsion in one dimension, we can describe them in terms of fermions, as least as far as any observable involving density is concerned. Thus with increasing interaction, we must somehow cross over from the bosonic (peaks) to the fermonic (dips) HBT effect. This paper describes how.

The answer turns out to be quite simple to state. In general there are still a series of features in the density-density correlation function, but they are Fano resonances.

A Fano resonance has a lineshape given by

$\frac{\left[q\Gamma/2+\varepsilon\right]^{2}}{\Gamma^{2}/4+\varepsilon^{2}}$

where $$\varepsilon$$ is the distance from the resonance, $$\Gamma$$ is the width and $$q$$ is an asymmetry parameter that goes from $$\infty$$ (a peak or resonance) to $$0$$ (a dip or antiresonance). In this case $$q$$ depends on the scattering phase of the pairs of particles contributing to that peak, and so goes from $$0$$ at low relative momenta where the interaction appears impenetrable and the fermion picture applies, to $$\infty$$ at high relative momenta where the noninteracting bosonic HBT effect is recovered. Simple, eh?

Although the answer is simple, the calculation can only be done because of the integrability of the 1D Bose gas, and in particular the existence of a remarkable formula due to Craig Tracy and Harold Widom for the N-particle propagator.