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Bosonization of Spin 1/2s to fields $\phi(x)$, $\theta(x)$ is defined as (Ref: 'Quantum Physics in 1-D' by Giamarchi):

  • $S^z(x)=\frac{-1}{\pi}\nabla\phi(x)+\frac{(-1)^x}{\pi a}\cos 2\phi(x)$,
  • $S^x(x)=\cos \theta(x)[(-1)^x+\cos 2\phi(x)]$,
  • $S^y(x)=\sin \theta(x)[(-1)^x+\cos 2\phi(x)]$.

Under time reversal symmetry we know that $\vec{S}_i\rightarrow -\vec{S}_i$ so to find how fields $\phi(x)$ and $\theta(x)$ will transfrom under time revrsal, if we focus on $S^z(x)$, we get

$\phi(x)\rightarrow\frac{\pi}{2}-\phi(x)$,

but now transformation of $S^{x,y}$ gets disturbed. How should field $\theta(x)$ transform such that we get $S^{x,y}\rightarrow -S^{x,y}$?

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I think your expressions for $S^x$ and $S^y$ are incorrect, I also couldn't find them in Giamarchi's book. I guess you might have deduced them from $$S^+ = e^{i\theta}((-1)^x+\cos(2\phi)\,,$$ see Giamarchi (D.10). If so, this would only hold under the assumption that $S^x$ and $S^y$ are real which is in general not the case. To my knowledge, the right expressions read $$S^x = (-1)^x\cos(2\theta) - i\sin(\theta)\cos(2\phi)$$ and $$S^y = -(-1)^x\sin(2\theta) - i\cos(\theta)\cos(2\phi).$$ Notice that there is no issue with time-reversal anymore.

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  • $\begingroup$ Should there be $\cos2\Phi$ in expression for $S^y$ because only then $S^y\rightarrow -S^y$ under T.R.S. Can you provide some reference for these. Thanks $\endgroup$
    – Barry
    Feb 1 at 13:55
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    $\begingroup$ @Barry You are right, there was a mistake. Fixed now. These expressions are used for example in <arxiv.org/abs/1903.05646>, albeit with a different convention for $\phi$ and $\theta$. $\endgroup$
    – DavidHfm
    Feb 1 at 14:28

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