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We all know classical mechanics deal with bigger objects and quantum mechanics deal with very tiny particles.

I hear spin number in quantum mechanics, but I don't see anything like that in classic mechanics.

Why don't we calculate spin number in classical mechanics?

Qmechanic
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Pedro Collins
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Loosely speaking: Classical mechanics is the approximation of quantum mechanics in the limit of $\hbar \to 0$ (where $\hbar$ is Planck's constant).

In quantum mechanics the electron's spin is given by $$\vec{S}=\frac{1}{2}\hbar\vec{\sigma}.$$ where $\sigma_x$, $\sigma_y$, $\sigma_z$ are the Pauli matrices.

The classical limit (for $\hbar \to 0$) of this is obviously: $\vec{S}=\vec{0}$.
That means: There is no spin in classical mechanics. Spin is an entirely quantum mechanical phenomenon.

Thomas Fritsch
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  • But you could say the same thing for momentum $P=-i\hbar D$ as $\hbar \rightarrow 0$. I don't think this argument applies. – Ryder Rude Feb 06 '23 at 11:23
  • @RyderRude That is a different situation. You have $P=-i\hbar D=\hbar k=h/\lambda \to 0 / 0$ as $h\to 0$ and $\lambda\to 0$ in the classical limit. So you cannot conclude $P\to 0$. – Thomas Fritsch Feb 06 '23 at 18:58
  • Still it is possible to replace the Spin commutators with Poisson brackets to get a classical theory, classical in the sense that all the spin projections are simultaneously well defined. See :https://physics.stackexchange.com/a/285314/50583 . This classical theory describes the expected value evolution of quantum spin, just like every other classical theory describes the expected value evolution of a quantum theory. – Ryder Rude Feb 06 '23 at 19:45