It is stated in Griffiths in a hint to a question that $\vec L$ and $\vec S$ commute with each other but no proof is given. $\vec L$ is given in the differential form and $\vec S$ is given in matrix form. How can I prove that they commute with each other?


I think of it this way:

$\vec{L}$ and $\vec{S}$ live in two completely different Hilbert spaces, that's why you know they commute; they do not "talk" with eachother.

Usually, $\vec{L}$ lives in $L^2(\mathbb{R}^3)$ and $\vec{S}$ in $\mathbb{C}^n$.

You can visualize this by thinking that $\vec{L}$ acts as a differential operator on a 3-dimensional function and $\vec{S}$ acts on a n-dimensional vector.

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  • $\begingroup$ 1. Is your first statement valid for any two operators in different Hilbert spaces? 2. Suppose I apply first $\vec S$ on a vector then the result will be another vector. How can I now apply $\vec L$ on the vector? I might use the matrix form of $\vec L$ but there is no differential form of $\vec S$ if the case is reversed. $\endgroup$ – Asit Srivastava Aug 18 '18 at 12:10
  • $\begingroup$ 1. I never saw a proof but it seems kind of intuitive so I'd say yes (I always use this personally). 2.You can't that way, precisely because they act on different Hilbert spaces. The vector is not in $L^2(\mathbb{R}^3)$ and as such it doesn't care about you acting with $\vec{L}$ on it. $\endgroup$ – Bonsay Aug 18 '18 at 13:03

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