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My question is two-part. First, imagine a bipartite quantum state $|\Phi \rangle_{AB}$, made of $2n$-qubits, shared between Alice and Bob (with $n$-qubits each). Alice performs some unitary operation $U$ on her part of the state and then performs $Z$-basis measurements. As a result, Bob's state collapses to a mixed superposition of states. Now, if Alice measures her state to be $|0\rangle^{\otimes n}$, how do I write the state that Bob's share has collapsed to, in bra-ket notation? At first, I thought it would be $\langle 0 |^{\otimes n} (U \otimes I_n) | \Phi \rangle_{AB}$ but that is, of course, incorrect (dimensional mismatch tells me that). I should probably be using some projection operators instead of simply $\langle 0 |^{\otimes n}$ but I can't figure out exactly what.

Second, assume that $| \Phi \rangle_{AB} = \left ( \frac{|00\rangle_{AB} + |11 \rangle_{AB}}{\sqrt{2}} \right )^{\otimes n}$ so that Alice owns the first qubit from every term and Bob owns the second (essentially, they share $n$ copies of the $|\Phi^+\rangle$ Bell state between them). Now what I want to prove is $$U^{\dagger} | 0 \rangle^{\otimes n} = \color{red}{\langle 0 |^{\otimes n} (U \otimes I_n) | \Phi \rangle_{AB}} $$ where I've colored the RHS red to emphasize that I know it is wrong, but it should be replaced by the properly notated answer to my first question. How do I go about proving this? I'm only asking for a hint, not a full proof. Thanks.

(This is by no means homework; my QM skills have grown somewhat rusty but I need to use this proof in a paper that I'm working on)

Cross-posted on quantumcomputing.SE

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The notation is fine. I would probably put a subscript $A$ on the $\langle 0|_A$ state, such as to make clear that this is a state on $A$, and thus, you are left with a state on B.

The formula you want to prove is wrong, there should be a transpose, not a dagger.

Other than that, the more general statement is that for the maximally entangled state $$|\Omega \rangle = \sum_i |i\rangle_A|i\rangle_B\ , $$ it holds that $$(I\otimes M^T)|\Omega\rangle = (M\otimes I)|\Omega\rangle $$ for any matrix $M$, which immediately implies your question. You can verify the above formula by writing $M$ out in components.

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  • $\begingroup$ Yes, I realised it should be transpose a bit too late. Is this a basic enough lemma to not require proof? $\endgroup$
    – Aritra Das
    Commented Jun 29, 2020 at 3:21
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    $\begingroup$ @AritraDas That really depends on the context. In a research paper, I probably wouldn't prove it. If you want to make the line of thought more clear, you could write $$\langle 0\vert_A U\otimes I \vert\Psi\rangle_{AB} = \langle 0\vert_A I\otimes U^T \vert\Psi\rangle_{AB} = U^T\vert0\rangle_B\ .$$ $\endgroup$
    – Norbert Schuch
    Commented Jun 29, 2020 at 5:42
  • $\begingroup$ Great, many thanks! $\endgroup$
    – Aritra Das
    Commented Jun 29, 2020 at 5:58

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