quantum mechanics - Why is the value of spin +/- 1/2?

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?

Answer

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 $$[\hat{S}_i, \hat{S}_j ] ~=~i\hbar \epsilon_{ijk} \hat{S}_k \qquad\qquad (A)$$ of spin operators $\hat{S}_i$ acts on $V$,

   
   

that

$$ {\rm the~eigenvalues}~\hbar m_s~{\rm of~the~spin~operator}~\hat{S}_z\qquad\qquad (B) $$

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 adapted the scaling convention of (A) and (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 $\hat{S}^2$.