Bra-Ket Notation and Proof of a Ket Equation in Two-Party Shared-Entanglement Setting

Question

Disclaimer: I had posted this question previously on the physics StackExchange, but received no response there.

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)

Answer 1

Let's consider the following $4$ qubit state (taking $n=2$ from the quesion):

$$|\psi_{in} \rangle = \frac{1}{2} \big( |0 0\rangle \otimes |00\rangle + |1 1\rangle \otimes |1 1\rangle + |01\rangle \otimes |01\rangle + |10\rangle \otimes |10\rangle\big)$$

The first two qubits are Alice's qubits and the last two qubits are Bob's qubits. We can describe this operation by projective measurements (for definition: M. Nielsen and I. Chuang textbook's page 87) for observable $M$:

$$M = m_{00} P_{00} + m_{01} P_{01} + m_{10} P_{10} + m_{11} P_{11}$$

where $P$s are the corresponding projectors onto eigenspaces of $M$ with their eigenvalues $m$:

$$ P_{00} = |0 0\rangle \langle 0 0| \otimes II \qquad m_{00} = 1 \\ P_{01} = |0 1\rangle \langle 0 1| \otimes II \qquad m_{01} = 2 \\ P_{10} = |1 0\rangle \langle 1 0| \otimes II \qquad m_{10} = 3 \\ P_{11} = |1 1\rangle \langle 1 1| \otimes II \qquad m_{11} = 4 $$

Here it can be proved that $M$ is a Hermitian operator. The one projector whose action is described in the question (obtaining the $|00\rangle$ state after the measurement) is the $P_{00}$ projector. The resulting state after applying $P_{00}$ projector (the formula can be found from the same textbook's page 88):

$$|\psi_{out}\rangle = \frac{P_{00} |\psi_{in}\rangle}{\sqrt{\langle \psi_{in}| P_{00} |\psi_{in} \rangle}} = |0 0\rangle \otimes |00\rangle $$

If we apply some $U$ to Alice's qubit's before the measurement, then:

$$|\psi_{out}\rangle = \frac{P_{00} \big( U \otimes I \big)|\psi_{in}\rangle}{\sqrt{\langle \psi_{in}| \big( U^\dagger \otimes I \big) P_{00} \big( U \otimes I \big)|\psi_{in} \rangle}} $$

If we disregard Alice's qubits, then Bob's state will be as follows:

$$|\psi_{B}\rangle = \frac{\big( \langle 0 0| \otimes I \big) \big( U \otimes I \big)|\psi_{in}\rangle}{\sqrt{\langle \psi_{in}| \big( U^\dagger \otimes I \big) P_{00} \big( U \otimes I \big)|\psi_{in} \rangle}} $$

Here $I$ operators are 4x4 identity matrices.