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The picture below shows the two protocols to distill EPR pairs and enhance the fidelity.

from 10.1103/PhysRevA.94.042303

EDIT:

I am trying to understand under what conditions such a protocol achieve the task, when repetead a number of times.

Since all the EPR pairs are noisy, a bit (phase) error purification correct from X (Z) noise, but in the mean time, it may propagate Z (X) noise.

My intuition says to me that, since pairs may be affected by any noise (X and Z i.i.d.), the protocol ends up in a loop where whenever you try to correct an error, you create another one.

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  • $\begingroup$ It would be helpful if you clarified what you found unsatisfying about the diagrams. One circuit fixes X errors, one fixes Z, fixing those fixes everything, and the circuits can be concatenated, therefore you can distill. Where is the step that you want more information on? $\endgroup$ Jan 8, 2023 at 7:15
  • $\begingroup$ I added some detail. $\endgroup$ Jan 8, 2023 at 11:21

1 Answer 1

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My intuition says to me that, since pairs may be affected by any noise (X and Z i.i.d.), the protocol ends up in a loop where whenever you try to correct an error, you create another one.

This only happens if the starting error rate is too high.

Concatenating the two circuits in your question is equivalent to protecting the entanglement using the 2 qubit Shor code. That code encodes a logical qubit into four physical qubits and detects any single X, Y, or Z error.

Suppose that each physical EPR pair shows up correctly 99% of the time and shows up wrong 1% of the time. Since the 2 qubit Shor code detects any single error, at least two of the EPR pairs must fail to cause the protocol to fail. The distribution of results is:

  • 0 errors (kept, succeeds): 0.99 * 0.99 * 0.99 * 0.99 ~= 96%
  • 1 error: discarded
  • 2 errors (kept, fails): ~0.01 * 0.01 * (4 choose 2) ~= 0.06%
  • more errors: negligible

The chance of rejection is roughly 4%, but given that you didn't reject the output is correct with probability 96% / (96% + 0.06%) ~= 99.9%.

So the chance of correct output increased from 99% to 99.9%. The entanglement was distilled.

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  • $\begingroup$ Thank you. I can notice that your reasoning does not care of the kind of errors. This makes me think that one may easily find different protocols, more adapted to specific scenarios. $\endgroup$ Jan 8, 2023 at 17:51
  • $\begingroup$ May you please expand the origin of your products? In particular, I find cases 0 and 2 a bit ambiguous. For example why is 0 error $0.99^4$ and not $0.99^2$? $\endgroup$ Jan 10, 2023 at 2:30
  • $\begingroup$ @DanieleCuomo it's 4 because there are 4 qubits in the 4 qubit Shor code. Four separate chances for things to go wrong. When you concatenate the X check and Z check you grow the number of things by 2x twice, requiring four EPR pairs total. $\endgroup$ Jan 10, 2023 at 4:50

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