# What are spin-coherent states?

Trying to understand the paper; https://arxiv.org/pdf/1702.02577.pdf and ran into "spin-coherent" states. I wonder those are.

In a spin system, with states labeled by the eigenvalue of the total angular momentum operator $$\mathbf{J}^2$$ and the z-projection of the angular momentum operator $$J_z$$, $$\mathbf{J}^2||J,m\rangle=J(J+1)||J,m\rangle\quad J_z ||J,m\rangle=m||J,m\rangle,$$ the spin-coherent states can be written as $$|\theta,\phi\rangle=e^{i\theta J_z}e^{i\phi J_y}||J,J\rangle=\sum_{m=-J}^J \sqrt{\binom{2J}{J+m}}\sin^{S-m}\frac{\theta}{2}\cos^{S+m}\frac{\theta}{2}e^{-i\phi(S+m)}||J,m\rangle.$$ There are many other equivalent ways to write these states, and one can define the angles differently by choosing to start from $$||J,-J\rangle$$ instead of $$||J,J\rangle$$ but this is a great place to start. The angles $$\theta$$ and $$\phi$$ correspond to angular coordinates on the surface of a sphere and so these spin-coherent states generalize pure single-qubit states on the Bloch sphere.
We can also consider these states to be two-mode bosonic states using the Schwinger mapping. This lets us write the basis elements as $$||J,m\rangle=|n_1\rangle\otimes|n_2\rangle,$$ where we have two bosonic modes $$|n_1\rangle\otimes |n_2\rangle=\frac{a^{\dagger\,n_1}}{\sqrt{n_1!}} \frac{b^{\dagger\,n_2}}{\sqrt{n_2!}}|\mathrm{vacuum}\rangle,$$ $$n_1=J+m$$, and $$n_2=J-m$$. This means that they can be physically generated from sending a state of the form $$|N\rangle\otimes|0\rangle$$ through a beam splitter.
Edit: you might also see these states in the context of symmetric superpositions of $$N$$ qubits. Then, the state $$|n_1\rangle\otimes |n_2\rangle$$ implies that, for example, $$n_1$$ of the qubits are in the spin-up state and the other $$n_2=N-n_1$$ qubits are in the spin-down state. The symmetric superposition, in this Fock basis, implies that we cannot ascertain which of the qubits are actually in which state. For example, we could write the second-quantized state $$|1\rangle\otimes|2\rangle$$ in the first-quantized form: $$|1\rangle_{\uparrow}\otimes|2\rangle_\downarrow=\frac{|\uparrow\rangle\otimes|\downarrow\rangle\otimes|\downarrow\rangle+|\downarrow\rangle\otimes|\uparrow\rangle\otimes|\downarrow\rangle+|\downarrow\rangle\otimes|\downarrow\rangle\otimes|\uparrow\rangle}{\sqrt{3}}.$$