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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)?

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    $\begingroup$ 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. $\endgroup$ Commented Apr 9 at 13:20
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    $\begingroup$ Do I need to learn about tensor products to understand the answer? (I am only in QM1 and have not learned about that yet) $\endgroup$ Commented Apr 9 at 13:22
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    $\begingroup$ 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... $\endgroup$ Commented Apr 9 at 13:27

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The answer, as explained in the comments, is that the Hilbert space is the tensor product of two spaces and the orbital and spin operators only act on one factor.

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 $\sigma$).

It is easy, then, to check that the spin and orbital angular momentum operators commute with each other.

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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.

Edit: Tobias already mentioned it in the comments.

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