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I understand how spin is defined in analogy with orbital angular momentum. But why must electron spin have magnetic quantum numbers $m_s=\pm \frac{1}{2}$? Sure, it has to have two values in accordance with the Stern-Gerlach experiment, but why precisely those values?

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    $\begingroup$ There is a definition of the units thing here, and a why does the intrinsic angular momentum have a lower limit than the rotational angular momentum question. Can you say which you care about? $\endgroup$ Commented Feb 5, 2012 at 19:19
  • $\begingroup$ Related: physics.stackexchange.com/q/29655/2451 $\endgroup$
    – Qmechanic
    Commented Apr 11, 2013 at 20:22
  • $\begingroup$ Albeit it's not good to draw analogy between electron spin and classical rotation, but here it helps to understand nature of electron spin. Imagine earth flips it's rotation direction, so that angular momentum changes as $\vec L \to -\vec L$. That's why electron has spin angular momentum of $+{\frac {\hbar }{2}},-{\frac {\hbar }{2}}$ $\endgroup$ Commented Jan 12, 2022 at 8:41

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I don't know if this is what OP is really asking(v1), but it is a remarkable fact in representation theory, that it is possible to deduce just from the assumptions that

  1. the Hilbert space $V$ of states is $2$-dimensional, and

  2. the real $so(3)$ Lie algebra $$[\hat{S}_i, \hat{S}_j ] ~=~i\hbar \epsilon_{ijk} \hat{S}_k \tag{A}\label{eq:A}$$ of spin operators $\hat{S}_i$ acts on $V$,

that

$$ {\rm the~eigenvalues}~\hbar m_s~{\rm of~the~spin~operator}~\hat{S}_z \tag{B}\label{eq:B}$$

can only be one out of the following two alternatives:

  1. $m_s=\pm \frac{1}{2} $. ($V=\underline{2}$ is a dublet representation with spin $s=\frac{1}{2}$.)

  2. $m_s=0$. ($V=\underline{1}\oplus\underline{1}$ is a sum of singlet representations with spin $s=0$.)

Of course, the second alternative is not relevant for electrons, which have spin $s=\frac{1}{2}$.

For a proof using ladder operators, see e.g. section 5 of 't Hooft's lecture notes. The pdf file is available here.

To summarize the logic, once we have adopted the scaling convention of $\eqref{eq:A}$ and $\eqref{eq:B}$, there is no ambiguity left in what we mean by the variable $m_s$. Once we agree on the meaning of $m_s$, we can have a meaningful discussion of the possible values of $m_s$. We next use representation theory to conclude that the values of $m_s$ are half integers. Similarly, the definition of the spins $s\geq 0$ are not arbitrary, but scaled in such a way that $\hbar^2s(s+1)$ become the eigenvalues for the Casimir operator $\hat{S}^2$.

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Particles with non-zero mass have spin 1/2 because space is three-dimensional. A spinor has four real components, representing probability density and three functions describing the three (independent) components of momentum. There is a preprint on this on arxiv.

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  • $\begingroup$ as it stands now, this is not helpful unless you explain how the 3d space is connected to spin-1/2, or why you could not just have any 2 values (say, 1/3 and 7/5). $\endgroup$ Commented Mar 13, 2023 at 23:13

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