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
    $\begingroup$ 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? $\endgroup$ – dmckee --- ex-moderator kitten Feb 5 '12 at 19:19
  • $\begingroup$ I suppose it has to do with how we measure the eigenvalues for spin (perhaps multiples of bohr's magneton) $\endgroup$ – Sebgr Feb 5 '12 at 20:04
  • $\begingroup$ What I was getting at is that it's $\pm 1/2$, because the base of our units for angular momentum is the minimum non-zero rotational angular momentum. The value for $h$ was set before anyone knew about spin. That's also why the quarks carry charges in multiples of $1/3$: because we set the base before anyone knew there were quarks. So is the question about $h$ or about the difference between rotational angular momentum and spin? $\endgroup$ – dmckee --- ex-moderator kitten Feb 5 '12 at 20:10
  • $\begingroup$ I think it should be on the lines of your second option. Does it only have to do with the fact that $m_s$ is quantized on integer steps and only two states are seen in a Stern-Gerlach experiment (thus eliminating the possibility of $m_s=0$)? $\endgroup$ – Sebgr Feb 5 '12 at 20:25
  • $\begingroup$ Related: physics.stackexchange.com/q/29655/2451 $\endgroup$ – Qmechanic Apr 11 '13 at 20:22

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 $V$,


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

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 $\hat{S}^2$.

| cite | improve this answer | |
  • $\begingroup$ In a sense, you've already chosen the constant ħ in step 2, so when you get the eigenvalues ħm in step 3 the values of m are predetermined. When you use h/π instead of ħ, you still get the same eigenvalues hm'/π. $\endgroup$ – MSalters Feb 7 '12 at 15:39
  • $\begingroup$ I tried to formulate the logical reasoning better in an update. $\endgroup$ – Qmechanic Feb 7 '12 at 17:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.