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For orbital angular momentum defined as $L= r \times p $ we can prove, in quantum mechanics, the commutation relations. Also, we could prove these relationships through the study of rotations (infinitesimal) in space. These are: $$[L_i , L_j]=i \hbar \sum_k ε_{ijk}L_k. $$

Since there isn't an analogous definition for spin angular momentum like that of the orbital angular momentum,

  1. How can we prove the commutation relations: $$[S_i , S_j]= i \hbar \sum_k ε_{ijk}S_k. $$

  2. Can we follow a path similar to that of the orbital angular momentum, that is the study of rotations in some space and if yes, in what space and what would this space represent?

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    $\begingroup$ 1. What do you mean with "proving" the CR for $S_i$? How do you define the $S_i$ if not as precisely the spin operators fulfilling that relation? 2. No, the whole point of spin is that it is absent from classical considerations. (Of course, the operators, being a rep of $\mathfrak{so}(3)$, still act as rotating the basis of some Hilbert space, but I have the feeling you don't mean that. $\endgroup$
    – ACuriousMind ♦
    May 14, 2015 at 15:19
  • $\begingroup$ You may want to read up on the general subject of representation theory, Lie algebras, and Lie groups. They deal with (among other things) taking a set of operators that obey some set of commutation relations, constructing an abstract vector space on which they act, and then "exponentiating" them (much as you exponentiate $L_z$ to get a rotation about the $z$-axis) to form a group of transformations on that vector space. The set of rotations in three dimensions is just a special case of this process. $\endgroup$
    – Michael Seifert
    May 14, 2015 at 15:22
  • $\begingroup$ @ACuriousMind Hi. Do you mean in 1), that the definition of spin operators is the commutation relations? 2)In two, that is something like what I had in mind. $\endgroup$
    – Constantine Black
    May 14, 2015 at 15:22
  • $\begingroup$ @MichaelSeifert Hi. For the orbital angular momentum, do we prove the commutation relations or do we take them as definitions? In the class we have proved them, and that's what I though. If that's the case then, having read 's answer, why doesn't a method like that apply to spin? $\endgroup$
    – Constantine Black
    May 14, 2015 at 15:25
  • $\begingroup$ From the perspective of Lie algebras, the commutation relations are fundamental; any such set of operators that satisfy these relations are a realization of that particular Lie algebra. We can then take a set of operators acting on some vector space (such as $L_z = -i \hbar (x \partial_y - y \partial_x)$ and so forth, acting on the Hilbert space of wavefunctions) and prove that they satisfy these relations. But from the perspective of Lie algebras, what you've proven is that these operators are a particular instantiation of the commutation relations, which are fundamental. $\endgroup$
    – Michael Seifert
    May 14, 2015 at 15:32

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You appear confused by how spin is introduced in ordinary QM. It is rather ad hoc:

Given a Hilbert space without spin degrees of freedom of a particle $\mathcal{H}_0$, and the spin $s$ of the particle, we take the total space of states of the particle to be $\mathcal{H}_0\otimes \mathcal{S}_s$, where $\mathcal{S}_s$ is a $2s+1$-dimensional complex Hilbert space carrying the unique irreducible representation of $\mathrm{SU}(2)$ labeled by $s$.

By construction, there are three anti-Hermitian generators $T_i\in\mathfrak{su}(2)\cong\mathfrak{so}(3)$ acting on $\mathcal{S}_s$ fulfilling the commutation relations $$ [T_i,T_j] = \sum_k\epsilon_{ijk}T_k$$ from which you get the usual Hermitian spin operators by multiplying by $\mathrm{i}$.

For $s=1$, the space $\mathcal{S}_1$ is three-dimensional, and the action of the $T_i$ is just a real-valued rotation around the $i$-axis, but, in general, the representations of $\mathrm{SU}(2)$ are not rotations, although they may be, whenever the representation map $\mathrm{SU}(2)\to\mathrm{U}(2s+1)$ hits only the real orthogonal matrices $\mathrm{O}(2s+1)\subset\mathrm{U}(2s+1)$, which happens for integer $s$.

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  • $\begingroup$ Thanks. May I ask why it is introduced like this? $\endgroup$
    – Constantine Black
    May 14, 2015 at 17:17
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    $\begingroup$ @ConstantineBlack: Initially, because we needed something to explain the Stern-Gerlach experiment and this works (for $s=1/2$). Nowadays, one would say that this comes from QM being a non-relativistic limiting case of QFT, where every field naturally has to transform in some representation of the (universal cover of the) Lorentz group, and what remains of the Lorentz group in the non-rel limit is $\mathrm{SO}(3)$, and so everything in QM should/could transform in a representation of $\mathrm{SU}(2)$ (being the universal cover of the rotation group). $\endgroup$
    – ACuriousMind ♦
    May 14, 2015 at 17:24

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