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?

  • $\begingroup$ Fibration (WP). $\endgroup$ – Cosmas Zachos Jun 19 '20 at 13:44
  • $\begingroup$ Linked. $\endgroup$ – Cosmas Zachos Jun 19 '20 at 15:21
  • $\begingroup$ @CosmasZachos Thanks so much for your comments! I think combining your link and physics.stackexchange.com/questions/204090/… may be more complete. $\endgroup$ – Merlin Zhang Jun 20 '20 at 8:57
  • $\begingroup$ @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? $\endgroup$ – Merlin Zhang Jun 20 '20 at 8:58

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