If not, does that mean that when doing partial measurements on two different shares of an entangled state, the results (expressed as a proability mass function) can depend on the order (i.e who measured first between Alice and Bob)?


1 Answer 1


Let $A,B$ be any two bounded operators acting over a Hilbert space and $I$ the identity operator. Then, $A\otimes I$ always commutes with $I \otimes B$.

$$[A\otimes I,I\otimes B]=(A\otimes I) (I\otimes B)-(I\otimes B)(A\otimes I)$$

The composition related to the tensor product goes as $(A_1 \otimes A_2)(B_1 \otimes B_2) = A_1B_1\otimes A_2B_2$. Therefore,

$$(A\otimes I) (I\otimes B) = A\otimes B = (I\otimes B)(A\otimes I)$$ implying $[A\otimes I,I\otimes B]=0$.


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