# Time reversal symmetry and Bosonization

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 $$S^{x,y}\rightarrow -S^{x,y}$$?

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.
• 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 Feb 1 at 13:55
• @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 $\theta$. Feb 1 at 14:28