# Why can transform the $\rm SU(2)$ spin to $S^2$ space?

Spin lies in $$\rm SU(2)$$ space, i.e. $$S^3$$ space, but when we write the spin coherent state: $$|\Omega(\theta, \phi)\rangle=e^{i S \chi} \sqrt{(2 S) !} \sum_{m} \frac{e^{i m \phi}\left(\cos \frac{\theta}{2}\right)^{S+m}\left(\sin \frac{\theta}{2}\right)^{S-m}}{\sqrt{(S+m) !} \sqrt{(S-m) !}}|S, m\rangle$$ where $$\Omega$$ is the unit vector in the $$S^2$$ space: $$\boldsymbol{\Omega}(\theta,\phi)=\{\sin \theta \cos \phi,\sin \theta \sin \phi,\cos \theta\}.$$ Also, we need to choose appropriate gauge $$\chi$$, e.g. $$\chi=\pm \phi$$ to make spin coherent state $$|\Omega(\theta, \phi)\rangle$$ periodic: $$|\Omega(\theta, \phi+2\pi)\rangle=|\Omega(\theta, \phi)\rangle.$$ As we know, one property for spin is "rotate back after $$4\pi$$" ($$\rm SU(2)$$ property), but why we now can change it to $$S^2$$ space (both unit vector $$\boldsymbol{\Omega}$$ and $$|\Omega\rangle)$$ i.e. rotate back after $$2\pi$$.

Note: Someone says it can be understood as Hopf map: $$\boldsymbol{n}=\left(\begin{array}{ll}u^{*} & v^{*}\end{array}\right) \boldsymbol{\sigma}\left(\begin{array}{l}\boldsymbol{u} \\ \boldsymbol{v}\end{array}\right)$$ which maps $$S^3 \to S^2$$, but I still cannot understand its picture since I think it must lose some information?

• – Cosmas Zachos Jun 19 '20 at 13:44
• – Cosmas Zachos Jun 19 '20 at 15:21
• @CosmasZachos Thanks so much for your comments! I think combining your link and physics.stackexchange.com/questions/204090/… may be more complete. – Merlin Zhang Jun 20 '20 at 8:57
• @CosmasZachos But I am not familiar with the mathematics about fibration, I still confused that does the physical effect of "loss of phase" related with the Berry phase term obtained in spin path integral? – Merlin Zhang Jun 20 '20 at 8:58