# Confusion regarding bra-ket notation and proof of a ket equation

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

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.
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.
• @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 $$\langle 0\vert_A U\otimes I \vert\Psi\rangle_{AB} = \langle 0\vert_A I\otimes U^T \vert\Psi\rangle_{AB} = U^T\vert0\rangle_B\ .$$ Jun 29 '20 at 5:42