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

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)

• Have a look at the partial trace. – M. Stern Jun 28 '20 at 18:51
• Also, since you already know the resulting state is a mixed state: use density matrices. – M. Stern Jun 28 '20 at 18:52
• It is a transpose, not a dagger. (You can see this even from linearity: The rhs is linear in $U$, while the lhs in antilinear!) – Norbert Schuch Jun 28 '20 at 21:39

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.

• Are you sure about this? I'm not, especially the $\langle 00 | \otimes I$ part. I think it should be something like $| 00 \rangle \langle 00 | \otimes I$. Also, maybe I have a different version of the book but I can't find anything relevant on the page numbers you've mentioned. Could you specify the subheadings/sections/chapters etc.? – Aritra Das Jun 27 '20 at 13:59
• @AritraDas, from the book: 2.2.5 Projective measurements. I have defined $P_{00} = |00\rangle \langle 00| \otimes II$ and described its action on the state... the last lines where I have written $\langle 00| \otimes II$ comes after disregarding Alice's qubits. Note that the state is denoted as $| \psi_B \rangle$ to show that it is only the state of Bob's qubits. – Davit Khachatryan Jun 27 '20 at 14:11
• @AritraDas, I have edited the answer. I have changed only the observable, the rest of the calculations haven't changed. For the initial observable there were equal $m$s and I wasn't sure if we can have equal $m$s (degeneracy) when we define projectors for the observable (from the textbook's definition this is not clear for me). So, I have chosen another observable that doesn't have equal $m$s. – Davit Khachatryan Jun 28 '20 at 14:29