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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.

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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.

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  • $\begingroup$ Got it! Thank you so much! $\endgroup$ Dec 26, 2019 at 16:51
  • $\begingroup$ "an operator X that acts only on states in the subspace spanned by |ξx⟩ will act like the identity operator on all states in the orthogonal subspace spanned by |ξy⟩" Is that true? Doesn't it like a zero operator? Acting it like the identity operator means that the state is an eigen state. $\endgroup$
    – dor00012
    Oct 24, 2021 at 11:29

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