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)

  • $\begingroup$ Have a look at the partial trace. $\endgroup$ – M. Stern Jun 28 at 18:51
  • $\begingroup$ Also, since you already know the resulting state is a mixed state: use density matrices. $\endgroup$ – M. Stern Jun 28 at 18:52
  • 1
    $\begingroup$ 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!) $\endgroup$ – Norbert Schuch Jun 28 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.

| improve this answer | |
  • $\begingroup$ 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.? $\endgroup$ – Aritra Das Jun 27 at 13:59
  • 1
    $\begingroup$ @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. $\endgroup$ – Davit Khachatryan Jun 27 at 14:11
  • 1
    $\begingroup$ @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. $\endgroup$ – Davit Khachatryan Jun 28 at 14:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.