# Why do operators belonging to different Hilbert spaces commute?

Every time I run into a commutator of two observables such as $$[\hat{X},\hat{Y}]$$, with $$\hat{X}$$ and $$\hat{Y}$$ being two operators from different state spaces: $$\hat{X}\in\xi_{x}\land\hat{Y}\in\xi_{y}$$; it is said that their commutator equals zero because each operator acts on its own subspace.

Could someone give the proof for that statement? Thank you.

• See my answers here : Total spin of two spin-1/2 particles, especially equations from (47) to (57) in my T H I R D___ A N S W E R. Dec 26, 2019 at 13:45

If our Hilbert space can be separated to two (or more!) orthogonal subspaces $$|\xi_x\rangle$$ and $$|\xi_y\rangle$$, then every state in it can be written as a a direct product of states from this two orthogonal subspaces $$|\xi_x, \xi_y \rangle = |\xi_x\rangle \otimes |\xi_y\rangle$$ In this case, an operator $$\hat{X}$$ that acts only on states in the subspace spanned by $$|\xi_x\rangle$$ will act like the identity operator on all states in the orthogonal subspace spanned by $$|\xi_y\rangle$$, and likewise for an operator $$\hat{Y}$$ that acts only on states in the subspace spanned by $$|\xi_y\rangle$$.
From here, it is clear to see that for every state $$\hat{X} \hat{Y} |\xi_x, \xi_y\rangle = \hat{Y} \hat{X} |\xi_x, \xi_y\rangle$$ As this is true for every state, they commute.