# What do people mean when they say orbital and spin angular momentum operators act on different spaces?

I was looking for an explanation of why the orbital and spin angular momentum operators commute, and I found many sources saying they act on different vector spaces.

I am confused about the use of different vector spaces in quantum mechanics. Spin states are in a finite dimensional space, whereas energy, position, momentum etc. are described in an infinite dimensional space. I think these spaces are separate and cannot be converted between (i.e., you can't find a position wavefunction from the spin state) since one has finite and one has infinite dimensions. However, I am confused about their interaction. Operators like $$\hat J$$, which is $$\hat S + \hat L$$ make me think that you can act on both at the same time.

Does any particle need to be described by a combination of a spin state vector and a state vector in the infinite dimensional space? Would that mean that $$\hat L_n$$ and $$\hat S_n$$ just act on different parts of the total combined description of the state? I think I am misunderstanding this, because that would mean that the total description of the state would never be an eigenstate of $$\hat L_n$$ or $$\hat S_n$$. What is the correct way to think about this (in basic, non-relativistic QM)?

• Consider e.g. a single electron moving on the real line. The corresponding Hilbert space can be chosen as $H=L^2(\mathbb R)\otimes \mathbb C^2$. The angular momentum operators act non-trivially only on the first part, whereas the spin operators only on the second. I.e., you consider operators of the form $L^2\otimes I_{\mathbb C^2}$ and $I_{L^2(\mathbb R)}\otimes S^2$ etc. Commented Apr 9 at 13:20
• Do I need to learn about tensor products to understand the answer? (I am only in QM1 and have not learned about that yet) Commented Apr 9 at 13:22
• Your basic QM text basically teaches you about tensor products in instructing you on how $\hat S$ and $\hat L$ act , "leaving each other's space alone". Think of the latter acting on a surface, and the former on columns coming off that surface... Commented Apr 9 at 13:27

Concretely, if you choose a basis of the spin states (for example, according to the projection $$\sigma$$ along the $$z$$ axis), this means a state is given by a finite collection of wave-functions $$\psi_\sigma(x)$$. The spin operator $$S^i$$ will act by multiplying by some matrix $$S^i_{\sigma,\sigma'}$$, sending the state $$\psi_\sigma(x)$$ to the state $$\sum_{\sigma'} S^i_{\sigma,\sigma'}\psi_{\sigma'}(x)$$.
On the other hand, the orbital angular momentum operators $$L^j$$ are differential operators acting on the $$x$$ variable (and acting the same on the different wave functions corresponding to the different values of $$\sigma$$).
Just to clariy: $$L + S$$ indeed doesn't make sense, precisely because they act on different spaces. But this is a commonly used shorthand notation for $$L\otimes \mathbb{1} + \mathbb{1}\otimes S$$, where the identity acts on the respective other space.