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Prove that if

$\text{Proj}_{M_1}\text{Proj}_{M_2}= \text{Proj}_{M_2}\text{Proj}_{M_1}$ then

$\text{Pr}(\text{span}[M_1, M_2]) = \text{Pr}(M_1) + \text{Pr}(M_2) − \text{Pr}(M_1 \cap M_2)$.

In the case where the projection operators are non-commutative, I understand how to show that the above formula is actually false. However, I am unsure of how the commutativity of the projectors implies the above equation. I have included an image from the text to provide the definitions of $M_1$ and $M_2$.

Image from text to provide definitions

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  1. It follows that $M_1$ is invariant subspace of operator $\text{Proj}_{M_2}$. Indeed, if $v \in M_1$ then $$\text{Proj}_{M_1}\text{Proj}_{M_2}v=\text{Proj}_{M_2}\text{Proj}_{M_1}v = \text{Proj}_{M_2}v,$$ so $$\text{Proj}_{M_1}(\text{Proj}_{M_2}v) = \text{Proj}_{M_2}v,$$ but this can happen only if $\text{Proj}_{M_2}v \in M_1$.

  2. Similarly, it can be proved that $M_1^{\perp}$ is invariant for $\text{Proj}_{M_2}$, and also $M_2,M_2^{\perp}$ are invariant for $\text{Proj}_{M_1}$ by symmetry.

  3. Since $M_1$ is invariant for $\text{Proj}_{M_2}$ and $\text{Proj}_{M_2}$ has two eigenspaces $M_2,M_2^{\perp}$, then $M_1$ can be split into $$ M_1 = (M_1 \cap M_2) \oplus (M_1 \cap M_2^{\perp}) $$ Similarly, $$ M_2 = (M_2 \cap M_1) \oplus (M_2 \cap M_1^{\perp}) $$

  4. Now, clearly, $(M_1 \cap M_2^{\perp}) \perp (M_2 \cap M_1^{\perp})$, so $$ M_1 + M_2 = (M_1 \cap M_2) \oplus (M_1 \cap M_2^{\perp}) \oplus (M_2 \cap M_1^{\perp}), $$ hence $$ \text{Pr}(M_1+M_2) = \text{Pr}(M_1 \cap M_2) + \text{Pr}(M_1 \cap M_2^{\perp}) + \text{Pr}(M_2 \cap M_1^{\perp}) = $$ $$ = \text{Pr}(M_1) + \text{Pr}(M_2) - \text{Pr}(M_1 \cap M_2) $$

A more general case is discussed here https://quantumcomputing.stackexchange.com/a/6469/5870

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  • $\begingroup$ Thank you! Is $M_1$ invariant under $Proj_{M_2}$ because the operator projects onto a space with basis vectors $|0>$ and $|1>$ and $M_1$ is a space that has only $|0>$ as a basis? $\endgroup$
    – imconfused
    Apr 12 '20 at 18:22
  • $\begingroup$ No, my answer is general. Also for $M_1, M_2$ from that exercise there will be no commutativity of projectors. $\endgroup$
    – Danylo Y
    Apr 12 '20 at 19:01

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