Question

What happens to a Luttinger liquid under time reversal?

Answer 1

Suppose you a have an ordinary Luttinger liquid with

H=∫dx∑η=±1,σ=↑,↓ψ†η,σ(x)(−ivη∂x)ψη,σ(x).H=∫dx∑η=±1,σ=↑,↓ψη,σ†(x)(−ivη∂x)ψη,σ(x).

You then bosonize it using

ψη,σ=12πα−−−√Fη,σe−iϕη,σψη,σ=12παFη,σe−iϕη,σ

where F is the Klein factor and ϕϕ a bosonic field.

My question is, what happens to F and ϕϕ under time reversal?

The problem here is that the usual result for fermions,

Tci,↑T−1=ci,↓Tci,↑T−1=ci,↓

Tci,↓T−1=−ci,↑Tci,↓T−1=−ci,↑

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 ϕ↑ϕ↑ turn into ϕ↓ϕ↓

Answer 2

Hey mate ^_^

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Answer:
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A Luttinger liquid under time result for fermions,

Tci, (up) T−1=ci,

Tci, (down) T−1=−ci,

Where 'T' is the time reversal operator, is not so obvious when there is separation between the Klein factor and the phase.

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