1
$\begingroup$

There are several quantum error correction techniques, such as 3-qubit bit-flip code, and Shor’s 9-qubit code. 3-qubit bit-flip code is a straightforward technique for correcting a single error (either bit-flip or phase-flip but not both). Shor’s 9-qubit code can correct both bit-flip and/or phase-flip errors.

Working with qubits is easy as we can encode and decode, but working with entangled particles has puzzled me.

Let us assume Alice and Bob share an entangled Bell state, $|00 \big \rangle _{AB}$, where $|00 \big \rangle _{AB} = \frac{1}{\sqrt{2}} (|00 \big \rangle + |11 \big \rangle) $. Assume this state is prepared by Charlie and distributed to Alice and Bob. Alice recieves the first particle ($|0 \big \rangle_A$) and Bob recieves the second one ($|0 \big \rangle_B$).

How can Alice and Bob make sure to some extent that their particle is error-free?

I am thinking about applying the principle of the repetition code (3-qubit bit-flip) to the basis states and obtains the following code:

$|00 \big \rangle _{AB}$ - > $|000000 \big \rangle _{AB}$, where $|000000 \big \rangle _{AB} = \frac{1}{\sqrt{2}} (|000000 \big \rangle + |111111 \big \rangle) $.

Alice receives the first three particles ($|000 \big \rangle_A$) and Bob recieves the other three particles($|000 \big \rangle_B$).

After that, Alice and Bob apply the same principle of the 3-qubit bit-flip; however, he/she can only fix the bit-flip error by following the majority(Ex, 101 means 111, 110 also means 111 ).

How can they fix both bit-flip and phase-flip errors?

Alice (Bob) ends up with a a GHZ entangled state ($\frac{1}{\sqrt{2}}((|0 \big \rangle^n + |1 \big \rangle^n)$) where n is the number of particles.

How can Alice(Bob) utilize the n particle to get a single but correct particle? Let say n equals 9, can they apply Shor's code? Who will do the encoding? Is it Charlie?

Any suggested solutions or references?

Implementation of @Craig Gidney answer

enter image description here

I have made random errors in the initial entangled state (0 and 10) and some random errors in the coding states. The results of measurement of the initial states (0 and 10) are always correct as in the following figure

enter image description here

Am I doing it right? Any comments?

$\endgroup$
2
$\begingroup$

Error correcting codes work the same on entangled qubits as any other qubit. All Alice and Bob have to do is separately encode their qubit into a Shor code. They each run an encoding circuit, apply noise, then run a decoding circuit and apply the corrections it inferred.

$\endgroup$
5
  • $\begingroup$ It is not clear to me how they can do that. Let us assume these two scenarios: 1- Charlie prepares an entangled state ($|00>$) and sends the first particle to Alice and the second to Bob. How Alice and Bob perform Shor code. 2- Charlie prepares an entangled state of 18 particles ($|0>^{18}$) and sends 9 particles to Alice and the other nine to Bob. In this case, is Charlie responsible for the coding, and Alice(Bob) will just do the decoding? $\endgroup$
    – m.aldarwbi
    Aug 24 at 18:58
  • $\begingroup$ I have edited my question, please have a look. The entanglement is done by one of the parties or by a third party(Charlie). In this case, who will do the encoding? and who will do the decoding of Shor's code? $\endgroup$
    – m.aldarwbi
    Aug 24 at 19:08
  • 1
    $\begingroup$ @m.aldarwbi All that matters is that the qubits are encoded before the noise you are trying to protect against is applied, and decoded after. E.g. if you were trying to protect the qubits during transit from Charlie to Alice/Bob then Charlie would have to do the encoding. The fact that the qubits are in an entangled state, instead of a separable state, has no bearing. Solve the problem with Charlie sending separable qubits to Alice/Bob, then just do exactly what you'd have done in that case. $\endgroup$ Aug 24 at 23:36
  • $\begingroup$ Thanks, @Craig, I will try to implement it in a simulator and see if it works. $\endgroup$
    – m.aldarwbi
    Aug 25 at 12:43
  • $\begingroup$ I have edited my question by including a circuit that I have made using Qiskit. Please have a look. $\endgroup$
    – m.aldarwbi
    Aug 27 at 12:56

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.