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To turn the probability of the projection over the Hilbert space $\mathcal H_A \otimes \mathcal H_B$ into the POVM probabilty over $\mathcal H_A$ we we use this equality: $$tr_Atr_B(ρΠ_i)=tr_A(ρ_Atr_B(ρ_BΠ_i))$$

My syllabus states that $\rho=\rho_A\otimes \rho_B$ but says nothing about the projector $\Pi_i$, as far as I know this is just an operator that can't necessarily be split into a tensor product and works on the full Hilbert space.

How does one obtain this equality?

What does $tr_B(\rho_B \Pi_i)$ even mean? As $\rho_B$ and $\Pi_i$ are matrices of different sizes.

What I have tried:
If I do assume that $\Pi_i=\Pi_A\otimes \Pi_B$ I get: $$ \begin{align} tr_Atr_B(ρΠ_i)&= tr_Atr_B((\rho_A \otimes \rho_B)\cdot(\Pi_A\otimes \Pi_B))\\ &=tr_Atr_B((\rho_A\cdot \Pi_A)\otimes(\rho_B\cdot \Pi_B)\\ &=tr_A((\rho_A\cdot \Pi_A) tr_B(\rho_B\cdot \Pi_B) \end{align} $$ Which is almost correct but now there is this $\Pi_A$ term. I'm not sure if I can move this $\Pi_A$ back into the $tr_B$ somehow. This is also all assuming I can even decompose the projector as a tensor product.

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1 Answer 1

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You can use the identity $$ \mathrm{tr}_{B}\left( (Y_A \otimes \mathbb{I}_{B}) X_{AB} \right) = Y_A \mathrm{tr}_B(X_{AB}) $$ where $X_{AB}$ can be any operator acting on the joint system $AB$. You can prove this quite quickly by writing the operators in the standard basis $Y_A = \sum_{i,j} y_{ij} |i \rangle \langle j |$ and $X_{AB} = \sum_{a,b,c,d} x_{abcd} |a \rangle \langle b| \otimes |c \rangle \langle d|$ and showing the two expressions are equal.

Using that expression you have $$ \begin{aligned} \mathrm{tr}_A (\mathrm{tr}_B ( \rho \Pi)) &= \mathrm{tr}_A(\mathrm{tr}_B((\rho_A \otimes \rho_B) \Pi_{AB})) \\ &= \mathrm{tr}_A(\mathrm{tr}_B((\rho_A \otimes \mathbb{I}_B)(\mathbb{I}_A \otimes \rho_B) \Pi_{AB})) \\ &= \mathrm{tr}_A(\rho_A\mathrm{tr}_B((\mathbb{I}_A\otimes \rho_B)\Pi_{AB})) \end{aligned} $$ which is what you wanted. Note that $\rho_B \Pi_{AB}$ is usually shorthand for $(\mathbb{I}_A \otimes \rho_B) \Pi_{AB}$.

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