When the commutator of two operators vanishes then we can measure one without affecting the other. I'm not sure how this translates in the case of density matrices.

If the density matrices are representing pure states then the density matrices would represent projection operators onto the subspace spanned by the given state. So I think a vanishing commutator on two density matrices could mean that either the subspace spanned by the two states are orthogonal or that they are the same. Is this correct?


If two operators commute, they have the same eigenvectors. For the density matrix of a qubit, the eigenvectors can be visualised as being along a particular axis of the Bloch sphere, corresponding to the direction of the Bloch vector. So, two density matrices commute if their Bloch vectors ($\vec{n}$ and $\vec{m}$) are parallel, i.e. there exists a real number $\alpha$ such that $\vec{n}=\alpha\vec{m}$ (note the $\alpha$ may be negative. I'm also ignoring the trivial case of $\vec{m}=0$).

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  • $\begingroup$ As the OP mentions there is also a third case where the two density matrices have their supports on orthogonal subspaces. $\endgroup$ – Rammus May 21 at 7:52
  • $\begingroup$ @Rammus That is already taken care of. For qubits, this can only happen for pure states where $\vec{n}=-\vec{m}$. $\endgroup$ – DaftWullie May 21 at 7:53
  • $\begingroup$ Sorry, I didn't read the title properly (didn't see the restriction to the Bloch sphere). $\endgroup$ – Rammus May 21 at 7:57
  • $\begingroup$ @DaftWullie Thank you. In the case that our density matrices were describing mixed state, does this change at all? $\endgroup$ – user12101 May 21 at 10:19
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    $\begingroup$ @Kay45 No, my answer covers that (pure is just the special case $\alpha=\pm 1$). $\endgroup$ – DaftWullie May 21 at 11:08

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