1
$\begingroup$

Let's consider that each qubit in the lab faces a single-qubit depolarizing channel $\mathcal{N}(\rho)=(1-p) \rho + p \mathbb{I}/2$.

Is there a theoretical result indicating the largest value of $p$ such that quantum error-correction can completely suppress the effect of the noise asymptotically (by encoding the qubit in logical qubits composed of sufficiently many physical qubits).

In different terms: is the maximum theoretical threshold to fight depolarizing noise known?

I am assuming that the implementation of error-correction is noiseless here.

$\endgroup$

3 Answers 3

3
$\begingroup$

One way of defining the quantum capacity of a quantum channel $\mathcal{E}$ is to ask, asymptotically, as $n\to \infty$, how much quantum information can I send with error rate going to $0$ over $\mathcal{E}^{\otimes n}$?

For the family of depolarizing channels $$\mathcal{E}_p (\rho) = (1-p) \rho + \frac{p}{3} X\rho X + \frac{p}{3}Y \rho Y + \frac{p}{3}Z \rho Z$$

Your question is equivalent to asking for the largest $p$ such that the quantum capacity is positive. Call this $p^*$. Unfortunately, the quantum capacity of the depolarizing channel is not known in closed form since the coherent information can be superadditive over multiple uses of the channel.** Lower bounds on $p^*$ by constructing coding schemes with a lower bound on their threshold.

Figure 7.1 in Graeme Smith's thesis plots some bounds on the capacity of the depolarizing channel

enter image description here

What we do know is the following:

** For the most extreme example, there are channels that individually have zero capacity but together have finite capacity

$\endgroup$
4
  • $\begingroup$ What is the no cloning argument that limits it to 25%? I ask because my understanding is that iterative entanglement purification, alternating between X-basis and Z-basis [2,1,2] repetition codes, should work for any amount of depolarizing noise less than 50%. At 50% it becomes impossible because the Bell pair is so depolarized that it can be prepared using LOCC. $\endgroup$ Commented Jul 2 at 20:59
  • $\begingroup$ Using this channel, only one-way classical communication is possible. Two-way is a different channel and indeed it is known to have finite capacity up to 1/2. $\endgroup$ Commented Jul 3 at 0:14
  • $\begingroup$ Ah, you're assuming the classical internet doesn't exist to assist the quantum channel. I always do analysis assuming the classical internet exists, because it does. $\endgroup$ Commented Jul 3 at 0:30
  • 1
    $\begingroup$ Both channels are reasonable to analyze, but for different reasons. I agree that for communication, two-way is likely more realistic. However, this is a distinct channel, and one should be careful about which one is under consideration. For example, if I would like to store a qubit for some long time (without interacting with it**) and then return to it, depolarizing noise with one-way classical communication may be a reasonable model. In the literature, "the depolarizing channel" frequently refers to one-way communication if not stated. ** Say, by shelving the state in a nuclear spin $\endgroup$ Commented Jul 3 at 3:50
0
$\begingroup$

If you assume the error correction to be noiseless and only consider a depolarizing channel on all qubits (say $p$ is the same on all qubits), then there will always be a probability of stabilizing a logical operator, this will be $\mathcal{O}(p^d)$, where $d$ is the code distance.

So if you are looking for the correction of an arbitrary number of errors, this will not be possible for any finite $p>0$.

$\endgroup$
1
  • $\begingroup$ Thanks for your answer. I know that with a finite number of qubits it is not possible. But asymptotically it is (for $d\to +\infty$). $\endgroup$ Commented Jul 2 at 16:48
0
$\begingroup$

Realistically, probably the zero-rate Hashing bound, so like 18-19%. Technically this could be violated? But I doubt by much.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.