In a text (Introduction to Quantum Mechanics by Griffiths) I am using it states without motivation that spin angular momentum has the same commutations relations as orbital angular momentum (these relations with the ladder operators were used to find the eigenvalue equations of orbital angular momentum) These are the spin angular momentum commutation relations: $$[\hat{S}_x, \hat{S}_y] = i \hbar \hat{S}_z,~~~[\hat{S}_y, \hat{S}_z] = i \hbar \hat{S}_x, ~~~ [\hat{S}_z, \hat{S}_x] = i \hbar \hat{S}_y$$ it follows then that spin angular momentum has the same eigenvalue equations as orbital angular momentum:
$$\hat{S}^2 |s m \rangle = \hbar^2 s (s+1)|s m \rangle;~~~\hat{S}_z |s m \rangle = \hbar m |s m \rangle;$$
Further in the text we consider a two particle system of two spin-$\frac{1}{2}$ particles-for example the electron and proton of a hydrogen atom in ground state where we define the spin operator as $$\hat{S} := \hat{S}^{(1)} + \hat{S}^{(2)}.$$ It then states that in order to confirm eigenvectors belonging to this operator, we have to ensure that the eigenvalue equations above are satisfied. I just want to know if the commutation relations holds for any spin operator, even for multi-particle systems and is that then why the same eigenvalue equations arise? If this is true then we should also be able to apply the ladder operator in each case in about the same way as orbital angular momentum to derive the eigenvectors of a certain spin operator?
Best Answer
Mathematically, the orbital angular momentum operators and the spin angular momentum operators are really two sides of the same coin. In group-theoretic language, we say that these operators arise from two different representations of the rotation group $SO(3)$ (to be precise, in quantum mechanics we are interested in projective representations because physically, two vectors that differ by a phase are indistinguishable. This requires a representation of the double cover of $SO(3)$, which is $SU(2)$). The group encodes information about the symmetries of the system and a representation of the group on a particular space gives us a way to realize these symmetries as operators on our space of states.
The difference between the spin operators and the angular momentum operators is really just what type of vector space they operate on. However, the group has a certain structure associated with it that carries through in its representations (This is related to the structure of the Lie-algebra of $SO(3)$, $\mathfrak{so}(3) \cong \mathfrak{su}(2)$ which I can elaborate on later if you would like). Hence, any representation of the group $SO(3)$ will have the same commutation relations. This includes extensions to multiple particle states. If we denote the Hilbert space of a single particle as $\mathcal{H}_1$ and the space of a second as $\mathcal{H}_2$, then the total space describing the two particles together is denoted $\mathcal{H}_1 \otimes \mathcal{H}_2$. This is nothing but a new vector space that we may represent $SO(3)$ on!
Thus, when we want to talk about the spin of a two-particle system, we are just talking about a different representation of $SO(3)$. There are some subtleties involved in this procedure due to the fact that the representation on the space $\mathcal{H}_1 \otimes \mathcal{H}_2$ is not irreducible. However, the Clebsh-Gordon decomposition gives us a way of decomposing this representation into a sum of reducible representations. This procedure gives the Clebsch-Gordon coefficients that arise when talking about multiple particle systems.