It is often stated that the property of spin is purely quantum mechanical and that there is no classical analog. To my mind, I would assume that this means that the classical $\hbar\rightarrow 0$ limit vanishes for any spin-observable.

However, I have been learning about spin coherent states recently (quantum states with minimum uncertainty), which do have a classical limit for the spin. Schematically, you can write down an $SU(2)$ coherent state, use it to take the expectation value of some spin-operator $\mathcal{O}$ to find

$$ \langle \mathcal{\hat{O}}\rangle = s\hbar*\mathcal{O}, $$
which has a well defined classical limit provided you take $s\rightarrow \infty$ as you take $\hbar\rightarrow 0$, keeping $s\hbar$ fixed. This has many physical applications, the result usually being some classical angular momentum value.

Why then do people say that spin has no classical analog?

It is often stated that the property of spin is purely quantum mechanical and that there is no classical analog. To my mind, I would assume that this means that the classical $\hbar\rightarrow 0$ limit vanishes for any spin-observable.

However, I have been learning about spin coherent states recently (quantum states with minimum uncertainty), which do have a classical limit for the spin. Schematically, you can write down an $SU(2)$ coherent state, use it to take the expectation value of some spin-operator $\mathcal{O}$ to find

$$ \langle \mathcal{\hat{O}}\rangle = s\hbar*\mathcal{O}, $$
which has a well defined classical limit provided you take $s\rightarrow \infty$ as you take $\hbar\rightarrow 0$, keeping $s\hbar$ fixed. This has many physical applications, the result usually being some classical angular momentum value. For example, one can consider a black hole as a particle with quantum spin $s$ whose classical limit is a Kerr black hole with angular momentum $s\hbar*\mathcal{O}$.

Why then do people say that spin has no classical analog?