# What happens to a Luttinger liquid under time reversal?

Suppose you a have an ordinary Luttinger liquid with
$$H = \int dx \sum _{\eta= \pm 1 , \sigma =\uparrow,\downarrow } \psi^\dagger_{\eta, \sigma} (x) (-i v \eta \partial _x) \psi _{\eta,\sigma} (x).$$

You then bosonize it using $$\psi_{\eta,\sigma}=\frac{1}{\sqrt{2 \pi \alpha}} F_{\eta ,\sigma} e^{-i \phi_{\eta, \sigma}}$$ where F is the Klein factor and $\phi$ a bosonic field.

My question is, what happens to F and $\phi$ under time reversal?

The problem here is that the usual result for fermions, $$T c_{i,\uparrow} T^{-1}= c_{i,\downarrow}$$ $$T c_{i,\downarrow} T^{-1}=- c_{i,\uparrow}$$

where T is the time reversal operator, is not so obvious when there is separation between the Klein factor and the phase. I mean, why would $\phi_\uparrow$ turn into $\phi_\downarrow$ ?

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