(adsbygoogle = window.adsbygoogle || []).push({}); isn't classical "spin" an instance of spin 1?

A spin 1 field A(x) is a 3-component vector field which transforms under rotations by

[tex]A(x)\rightarrow exp[i(S+L)\cdot \hat{n}\theta]A(x) = exp(iS\cdot \hat{n}\theta)A(R^{-1}(\theta)x)[/tex]

where S are the three 3x3 matrices

[tex]S_1=\left( \begin{array}{c1 ... cn}

0 & 0 & 0 \\

0 & 0 & i \\

0 & -i & 0

\end{array}\right) [/tex]

[tex]S_2=\left( \begin{array}{c1 ... cn}

0 & 0 & -i \\

0 & 0 & 0 \\

i & 0 & 0

\end{array}\right) [/tex]

[tex] S_3=\left( \begin{array}{c1 ... cn}

0 & i & 0 \\

-i & 0 & 0 \\

0 & 0 & 0

\end{array}\right)

[/tex]

This is certainly true in quantum theory. Multiplying S by hbar (and so replacing i with i/hbar in the transformation) gives the quantum spin operators for spin 1. But, as far as I can tell, there is nothing particularly "quantum" about this. The classical electromagnetic vector potential also transforms like this. Any vector field transforms like this.

Furthermore (entering less certain ground now) a rotating rigid body with intrinsic angular momentum [tex]L_{spin}[/tex] located at position x is essentially a vector [tex]L_{spin}(x)[/tex] attached to the point x. It is similar to the field A(x), except that rather than a field it is a single vector attached to a particular x. And does it not also transform under rotations through the operation exp(iS\cdot \hat{n}\theta)[/tex]? So is it not then a spin 1 instance.

In short, if spin 1 is vector, and since classical spin is a vector, is not "spin 1" in this sense? That is, their generators of rotations are the same representation of SO(3).

The quantum case is not "quantum" because its generators of rotations do not commute. It is quantum because the structure of the theory requires eigenvectors of the spin matrices and we cannot have simultaneous eigenvectors of noncommuting matrices. The classical case has no restriction on its vectors.

Am I stating this correctly?

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# Isn't classical spin an instance of spin 1?

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