# Highest theoretical threshold to fight single-qubit depolarizing noise for noiseless error-correction

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.

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

What we do know is the following:

• 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. Commented Jul 2 at 20:59
• 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. Commented Jul 3 at 0:14
• 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. Commented Jul 3 at 0:30
• 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 Commented Jul 3 at 3:50

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 $$p>0$$.

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

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