5
$\begingroup$

I read about some QEC codes such as Shor code. It encodes a logical qubit to 9 physical qubits, to correct the bit-flip and phase-flip error. To do this, it needs multiple $\mathrm{CNOT}$ and $\mathrm{H}$ gates. As we know, the increase of the number of operations brings errors. How can this procedure then decrease the overall error rate?

$\endgroup$
5
$\begingroup$

Quantum error correction concerns errors that happen on qubits; it does not provide any protection against errors in operations on those qubits.

Note however, that an error on an operation can be seen as the perfect operation plus some error 'on the qubit'. It is, however, the case that, without any precaution, the added operations introduced by error correction invoke many more extra errors than that the code might be able to correct. This is indeed a very though issue.

The answer is, in short, what is collectively known as fault-tolerance. This is essentially taking much, much effort to implement all operations included in error correction in such a way that the extra errors that might arise during this operation do not pile up, thereby exceeding the error correction capabilities of the code.

It has been proven that, taking enough precaution, one can correct errors quicker than they arise: The threshold theorem. Note that fault-tolerance requires very much resources (qubits and operations). Fault-tolerance is quite tough, but for an idea behind it see the second half of the introduction by Gottesman.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for your answer. For example, if we create a circuit using qiskit. Since the extra errors are caused by the error correction, how can we use the error correction code to decrease the error rate on the real device? Does it mean that the quantum error correction is just a thought experiment and it cannot apply to any real device? $\endgroup$ – peachnuts Jan 6 at 14:51
  • 3
    $\begingroup$ No, QEC is very much needed for a 'full-scale universal' quantum computer. However, only using QECC's will never be enough, as I explained in my answer it is only one side of the coin; the other side being fault-tolerance. Hardware needs to be good enough so that errors can be corrected faster than that they arise. The extra errors caused by the error correction itself can just be regarded as any other error (if the implementation is fault-tolerant). Such a full scale universal quantum computer is also often referred to as a fault-tolerant quantum computer. $\endgroup$ – JSdJ Jan 6 at 16:40
-5
$\begingroup$

It was my understanding that error-correction does not really apply to quantum computing. The elements in the set are not numbers, they are probabilities. Probability elements are never in one state or another. They are always in all states. That is the beauty of quantum computing.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for your answer. But I am still very confused. If error-correction does not apply to quantum computing, what is it used for? $\endgroup$ – peachnuts Jan 6 at 15:42
  • $\begingroup$ EC applies to computation of sets where the elements are deterministic. $\endgroup$ – Daro Gross Jan 6 at 16:40
  • 3
    $\begingroup$ This is incorrect, there is a whole subfield in quantum computing concerning (quantum) error correction and fault-tolerance. A universal quantum computer will very, very likely need some sort of error correction. Please elaborate more on what you mean by 'the elements in the set are not numbers, they are probabilities.' Thanks in advance. $\endgroup$ – JSdJ Jan 6 at 16:43
  • $\begingroup$ Sure. A qbit exist in all states at different probabilities depending on how one manipulates it. For some algorithms, this is a plus. It is possible to look at a problem from many different angles at the same time. $\endgroup$ – Daro Gross Jan 6 at 17:35
  • $\begingroup$ @DaroGross: But eventually you have to measure qubits and you get some answer. This answer is one partiuclar state the qubits collapsed to after measurement. It is true that qubits are described by probability distribution. However, because of noise the distribution is distorted and this cause errors. You have to have some mechanisms how to correct them. $\endgroup$ – Martin Vesely Jan 6 at 17:43

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.