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How can we explain spin of electron, or the spin of other fundamental particles?

If we think the spin of electron is similar to the spin of a ball or planet we make a mistake.

We say it is an intrinsic property. However, in calculating magnetic momenta and other cases we consider it as spinning entity. It's too difficult to abandon the model which we see. Without which to explain abstract one becomes challenging.

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marked as duplicate by ACuriousMind, Neuneck, Qmechanic Aug 15 '14 at 14:15

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The best way to understand spin is actually to consider the Dirac Equation

$$ i\hbar \frac{\partial }{\partial t}\Psi=\left[c\sum_i{\alpha_i p_i}+mc^2\beta\right]\Psi $$

or more compactly:

$$(i\gamma^\mu\partial_{\mu}-m)\psi=0$$

The solutions to the Dirac equation are collections of complex valued fields called spinors.

The spinor solution actually encodes not only the spin of the particle but also the existence of its anti-particle and its spin as well. This means the spinor is a four valued complex vector:

$$\psi(x) = \begin{bmatrix}\psi^1(x)\\\psi^2(x)\\\psi^3(x)\\\psi^4(x)\end{bmatrix} $$

Where, for instance, a negatively charged electron with spin up would be represented as:

$$\left|e^-,\, +\tfrac{1}{2}\right\rangle = \begin{bmatrix}1\\0\\1\\0\end{bmatrix}$$

The point of explaining it this way is to convey the fact that particle spin is only manifest in quantum theory. In fact the existence of particle spin and anti-particles is prima facie proof of quantum theory as a means to explain the physical world; there simply is no classical counterpart.

This is sometimes very hard for people to understand, but basically spin is a notion of having a value in some direction in a complex vector space, which is about as close as one can really get to a classical description.

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For massive particles the intuition of thinking of spin as a rotation is correct. In the rest frame, a massive ($M^2>0$) particle has momentum

$$p_0^\mu=(M,0,0,0).$$

Remember that the quantity $p^2=p_\mu p^\mu$, for arbitrary $p_\mu$ is invariant under Lorentz transformations. In the case above the subgroup of the Lorentz group leaving $p_0^\mu$ invariant is clearly the Rotation group. So, if we transform a massive particle described by the state $|p,s\rangle$ with a unitary representation $U(\Lambda)$ of an arbitrary Lorentz transformation $\Lambda$, you will find that it will only rotate the $s$ index:

$$U(\Lambda)|p,s\rangle=\sum_{s^\prime}D_{ s^\prime s}|\Lambda p,s^\prime\rangle$$

where $D_{s^\prime s}$ are the elements of the rotation matrix $e^{i \vec{J}\cdot\vec{\theta}}$ and $J_i$ are the three rotation generators of the $SU(2)$ algebra.

So for massive particles described by $|p,s\rangle$ with $M^2>0$, spin does corresponds to $SU(2)$ rotations. For massless states $M=0$ this is no longer true, the spin index $s$ does not transform as above and thinking spin as angular momentum does not work.

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