angular momentum - Spin coherent state for general spin $S$

In Wen's book on Many-body QFT, he claimed that the coherent state for a spin-$S$ particle can be written as a tensor product of $2S$ spin-1/2 coherent states:

$$|\hat{n}\rangle=|z\rangle\otimes|z\rangle...\otimes|z\rangle$$ where each $|z\rangle$ is a spin-1/2 spinor.

If we look at the degrees of freedom of both sides of the above equation, the spin-$S$ coherent state has $2S+1$ free parameters, whereas the $2S$ spin-1/2 have $4S$ free parameters in total. How do we understand this?

Answer

By definition, a spin coherent state $\vert \theta,\phi\rangle_S$ is just a rotation of the $\vert S,S\rangle$ state.

What makes the state "coherent" is that all the individual spin states point in the same $(\theta,\phi)$ direction (are "coherently aligned"), thus

$$\begin{align} \vert \theta,\phi\rangle_S &:= \vert \theta,\phi\rangle_1\otimes\ldots\otimes \vert \theta,\phi\rangle_{2S} \tag{1} \\ &=R_z(\phi)R_y(\theta)\vert S,S\rangle \\ & \; = [R_z(\phi)R_y(\theta)\vert +\rangle_1]\otimes \ldots \otimes [R_z(\phi)R_y(\theta)\vert +\rangle _{2S}] \end{align}$$ $\vert \theta,\phi\rangle_S$ clearly depends on only 2 parameters, as it should, and indeed there is no freedom in the orientation of any of the particle spins (except the first) are they must all be aligned in the same $(\theta,\phi)$ direction as the first. Thus, the right hand side of (1) depends on only two angles also.

The state in Eq.(1) is clearly symmetric under permutation of the particle indices, so taking $\theta=\phi=0$ gives

$$\vert S,S\rangle = \vert +\rangle_1\otimes \vert +\rangle_2\otimes\ldots \otimes \vert +\rangle_{2S} \tag{2}$$ This state is an eigenstate of $L_z$ $$L_z= L_z^{(1)}+L_z^{(2)}+\ldots +L_z^{(2S)}$$ with eigenvalue $M=S$, and is killed by the raising operator $$L_+= L_+^{(1)}+L_+^{(2)}+\ldots +L_+^{(2S)}$$ where $L_+^{(k)}$ acts on the state of particle $k$ alone.