Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\phi_\downarrow$ ?

share|cite|improve this question

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.