4
$\begingroup$

Is there a 5-partite state such that if two parties each hold 2 shares of the state, without knowing what shares the other party holds, they can create a Bell pair between them and know which qubit in their possession is their half of the Bell pair? The two parties may not communicate.

$\endgroup$
  • 1
    $\begingroup$ Do you require this to be a deterministic protocol for producing the Bell pair, or are we allowed to, for example, use measurements and post-select on the result, so that we have a protocol that only works with some (known) probability? $\endgroup$ – DaftWullie May 9 at 7:35
  • $\begingroup$ The protocol should succeed with probability arbitrarily close to 1 (as you allow the share sizes to increase) $\endgroup$ – cplusplusguru May 9 at 15:23
  • $\begingroup$ So, when you talk about share sizes, you're allowing each party to have an arbitrarily large local Hilbert space dimension? $\endgroup$ – DaftWullie May 9 at 15:44
  • $\begingroup$ That's right. We can say that the total state lives in $\mathcal{H}^{\otimes 5}$ with $\mathcal{H}=(\mathbb{C}^2)^{\otimes n}$, i.e. each share consists of $n$ qubits and I require the protocol succeeds with probability 1 as $n$ goes to infinity. That's not to say there isn't a protocol that deterministically succeeds with finite sized shares, that would be nice, but it's not what I require. $\endgroup$ – cplusplusguru May 9 at 16:37
2
$\begingroup$

Yes, it does exist. Here's something that you could do (I'm sure there are better things):

Consider the symmetric state $$ |\psi\rangle=\frac{1}{\sqrt{\binom{5}{2}}}\sum_{x\in\{0,1\}^5:\ x\cdot x=2}|x\rangle. $$ The benefit of selecting the symmetric state is that it is entirely irrelevant which pair of qubit is assigned to Alice, and which pair is assigned to Bob.

Now I'm going to use a trick based on the fact that I know the total state has 2 excitations. Both Alice and Bob apply a controlled-not between their two qubits, and measure the control qubit in the Z-basis. If they both get answer 0, the protocol has succeeded and they share a Bell pair between the two control qubits, and the extra qubit is in the state $|0\rangle$. How did this work? the controlled-not+measurement simply worked to calculate the parity of the two qubits. Post-selecting on the 0 answer means that we forced the two qubits in the original state to be the same, either 00 or 11. If both pairs of qubits were of even parity, then one of them was 00 and the other was 11. Moreover, since the overall state was of even weight, that means the extra qubit must be in the 0 state.

I haven't carefully calculated it, but I believe the success probability of this protocol is 1/5. So, if we have $n$ copies, the probability of failure is $1-(4/5)^n$, although we expect it to take about 5 copies to produce a Bell pair on average.

There are definitely improvements that can be made to this protocol. I just used the part of the state $$ |00\rangle_A|11\rangle_B|0\rangle+|11\rangle_A|00\rangle_B|0\rangle. $$ There's another component that would do just as well, but with higher probability amplitude: $$ |00\rangle(|01\rangle+|10\rangle)|1\rangle+(|01\rangle+|10\rangle)|00\rangle|1\rangle, $$ which I could have used instead (but the measurement is a bit more complicated to describe - instead of mapping $|00\rangle$ to 0 and $|11\rangle$ to 1, we need to map $|00\rangle$ to 0 and $(|01\rangle+|10\rangle/\sqrt{2}$ to 1), and that's before one gets on to trying to think about whether there are better states... Or, for example, interacting between multiple copies....

$\endgroup$
  • $\begingroup$ This does not quite work. Even if Alice measures $0$ she doesn't know if Bob also got a $0$. Thus she doesn't know if that control qubit is actually a half of a bell pair. $\endgroup$ – cplusplusguru May 10 at 15:34
  • $\begingroup$ For clarity, imagine that Bob and Alice are each allowed to send a single qubit to Charlie, and they win if the two qubits Charlie gets from them form a bell pair $\endgroup$ – cplusplusguru May 10 at 16:05
  • $\begingroup$ So Alice and bob aren’t allowed to communicate? That’s a very strong stipulation that needs to be a clear part of the question. $\endgroup$ – DaftWullie May 10 at 18:17
  • $\begingroup$ My apologies. I added it to the question. Although if they could communicate it would be trivial, just share 5 choose 2 pairs (maybe less), and verify which one they have. $\endgroup$ – cplusplusguru May 10 at 21:31

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.