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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)

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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

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