# Time-reversal transformation for two-component bosonic models

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Consider a two-component bosonic model $\mathcal{H}=-t\sum_{i\sigma}{b_{i\sigma}b_{i+1\sigma}^\dagger}+h.c. +\sum_{i\sigma\sigma^\prime}U_{\sigma\sigma^\prime}n_{i\sigma}n_{i\sigma^\prime}$. Here $\sigma,\sigma^\prime$ aretwo the pseudo spin, they can take $\uparrow,\downarrow$. I want to know how to define the time-reversal transformation for such two-component bosonic models. Is the definition same as fermion models?

This post imported from StackExchange Physics at 2015-10-14 17:01 (UTC), posted by SE-user fangniuwawa

edited Oct 14, 2015
It is probably not the same as for fermion models (where time reversal flips the spin) – why should time reversal change the kind of boson (though if the kinds of bosons are some effective particles like right movers and left movers it probably will). I guess the correct time reversal will be independent time reversal for both boson components (but I can be wrong).

This post imported from StackExchange Physics at 2015-10-14 17:01 (UTC), posted by SE-user Sebastian Riese

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The analysis of one-particle states spanning a representation of the Lorentz group yields up to a phase convention
$$\mathcal{T} b(E_p,\vec{p},s) \mathcal{T}^\dagger = (-1)^{1/2-s}b(E_p,-\vec{p},-s)$$

The change in the sign of spin is due to the fundamental relation $\mathcal{T} J \mathcal{T}^\dagger = -J$, so this conclusion holds for any kind of collective angular momentum. If $\sigma$ is some collective angular momentum, then the transformation above will apply, if not, you will have to look into the foundations of your model and derive the time-reversal from the transformation of the constituents.

The other question is, of course, what does $i$ stand for. If it is a position index, then it is untransformed, if some kind of fourier-transform/momentum index, then it has to be reverted (without touching the energy spectrum I assume would be somehow in matrix $U$).

answered Oct 18, 2015 by (1,645 points)

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