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?
2 Answers
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
the Hilbert space $V$ of states is $2$-dimensional, and
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:
$m_s=\pm \frac{1}{2} $. ($V=\underline{2}$ is a dublet representation with spin $s=\frac{1}{2}$.)
$m_s=0$. ($V=\underline{1}\oplus\underline{1}$ 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 $\hat{S}^2$.
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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