<p>For your question (1), the answer is yes. The 'hashing construction' for quantum encoding is independent of the quantum channel you use, so if you use this construction for encoding, you can send information over an ensemble of Pauli channels at the right rate. (Actually, this is slightly incorrect ... you do need that the input state you use in the formula for coherent information is the same for all the channels, but in the construction discussed by Preskill it is.)</p>

<p>The original paper Shor and Smolin does not mention an ensemble of channels. If you take a depolarizing channel near the quantum-capacity threshold you can achieve a larger quantum capacity by considering a new superchannel where one signal of the superchannel is five consecutive uses of the original channel. When you apply the 'hashing bound' to this superchannel (acting on a 32-dimensional quantum space rather than a 2-dimensional one), you find that the formula for quantum capacity is larger than five times the formula for the original channel. There is only one channel being considered here: the superchannel composed of five uses of the original channel, and not an ensemble of channels. The ensemble of channels comes in when Preskill gives the intuition for what is happening in Shor-Smolin.</p>

<p>(And of course you should replace five in the above paragraph with the appropriate value of $m$, which is only five for some of these constructions.)</p>

CC BY-SA 3.0
2012-02-21T13:34:18.170Z
Oh, you're using the same type of random code for each channel anyway so there should exist a random code that is good for all the channels. I'll have to think a little bit about it but I'm pretty sure that's the answer. Thanks! As for your second paragraph I'm not sure if I'm putting words into Preskill's mouth or if he means to explain it the same way you do. But I think thanks to your answer the way I was trying to understand it does work.
Martin Leslie
CC BY-SA 3.0
2012-02-21T12:26:21.217Z
2012-02-21T12:50:18.007Z
Peter Shor 
<p>In Preskill's quantum computing notes <a href="http://www.theory.caltech.edu/people/preskill/ph229/notes/chap7.pdf" rel="nofollow">Chapter 7</a> approximate page 82, he shows that a Pauli channel has capacity $Q \geq 1-H(p_I,p_X,p_Y,p_Z)$ where $H$ is Shannon entropy and $p_I, p_X, p_Y, p_Z$ are the probabilities of the channel acting like the appropriate Pauli matrix. In particular this gives us the 'hashing bound' or 'random coding bound' for the quantum capacity of the depolarizing channel $Q(p) \geq 1-H(p,1-p)-p\log_23$.</p>

<p>He then describes work of Shor and Smolin [1]: if you take a $m$-repetition code and concatenate it with a suitable random code you can do better than the hashing bound. The argument for this is that taking $m-1$ measurements the inner repetition code thought of as a super channel is a Pauli channel with entropy $H_i$. Then averaging over the $2^{m-1}$ possible classical measurements you can find the average entropy of the superchannel $\langle H \rangle$.</p>

<p>[1] P.W. Shor and J.A. Smolin, “Quantum Error-Correcting Codes Need Not Completely Reveal the Error Syndrome” <a href="http://arxiv.org/abs/quant-ph/9604006" rel="nofollow">quant-ph/9604006</a>; D.P. DiVincen, P.W. Shor, and J.A. Smolin, “Quantum Channel Capacity of Very Noisy Channels,” <a href="http://arxiv.org/abs/quant-ph/9706061" rel="nofollow">quant-ph/9706061</a>.</p>

<p>Then by random coding on this new channel you can achieve a rate $R=\frac{1-\langle H \rangle}{m}$ (dividing by $m$ to get this rate in bits/original channel use).</p>

<p>I don't see how random coding works. You have a random code which is optimal for each particular channel but how do you decide which one to use? By the time you know the classical measurements for your channel you have already sent the codeword.</p>

<p>So two questions:</p>

<p>1) If you have an ensemble of Pauli channels with average entropy $\langle H\rangle$, can you by using random coding achieve a rate $1-\langle H \rangle$?</p>

<p>2) If you can't do this, am I misinterpreting the results of Shor and Smolin or Preskill's exposition?</p>

<quantum-computing><quantum-information>
Quantum capacity for ensemble of Pauli channels
Martin Leslie
# Quantum capacity for ensemble of Pauli channels

(Asked by: username_0 on 2012-02-21)

In Preskill's quantum computing notes [Chapter 7](http://www.theory.caltech.edu/people/preskill/ph229/notes/chap7.pdf) approximate page 82, he shows that a Pauli channel has capacity $Q \geq 1-H(p\_I,p\_X,p\_Y,p\_Z)$ where $H$ is Shannon entropy and $p\_I, p\_X, p\_Y, p\_Z$ are the probabilities of the channel acting like the appropriate Pauli matrix. In particular this gives us the 'hashing bound' or 'random coding bound' for the quantum capacity of the depolarizing channel $Q(p) \geq 1-H(p,1-p)-p\log\_23$.

He then describes work of Shor and Smolin [1]: if you take a $m$-repetition code and concatenate it with a suitable random code you can do better than the hashing bound. The argument for this is that taking $m-1$ measurements the inner repetition code thought of as a super channel is a Pauli channel with entropy $H\_i$. Then averaging over the $2^{m-1}$ possible classical measurements you can find the average entropy of the superchannel $\langle H \rangle$.

[1] P.W. Shor and J.A. Smolin, “Quantum Error-Correcting Codes Need Not Completely Reveal the Error Syndrome” [quant-ph/9604006](http://arxiv.org/abs/quant-ph/9604006); D.P. DiVincen, P.W. Shor, and J.A. Smolin, “Quantum Channel Capacity of Very Noisy Channels,” [quant-ph/9706061](http://arxiv.org/abs/quant-ph/9706061).

Then by random coding on this new channel you can achieve a rate $R=\frac{1-\langle H \rangle}{m}$ (dividing by $m$ to get this rate in bits/original channel use).

I don't see how random coding works. You have a random code which is optimal for each particular channel but how do you decide which one to use? By the time you know the classical measurements for your channel you have already sent the codeword.

So two questions:

1) If you have an ensemble of Pauli channels with average entropy $\langle H\rangle$, can you by using random coding achieve a rate $1-\langle H \rangle$?

2) If you can't do this, am I misinterpreting the results of Shor and Smolin or Preskill's exposition?

**Tags:** quantum-computing, quantum-information


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## 1 answers

### Answer by username_1 on 2012-02-21. Score: 5 | Accepted answer

For your question (1), the answer is yes. The 'hashing construction' for quantum encoding is independent of the quantum channel you use, so if you use this construction for encoding, you can send information over an ensemble of Pauli channels at the right rate. (Actually, this is slightly incorrect ... you do need that the input state you use in the formula for coherent information is the same for all the channels, but in the construction discussed by Preskill it is.)

The original paper Shor and Smolin does not mention an ensemble of channels. If you take a depolarizing channel near the quantum-capacity threshold you can achieve a larger quantum capacity by considering a new superchannel where one signal of the superchannel is five consecutive uses of the original channel. When you apply the 'hashing bound' to this superchannel (acting on a 32-dimensional quantum space rather than a 2-dimensional one), you find that the formula for quantum capacity is larger than five times the formula for the original channel. There is only one channel being considered here: the superchannel composed of five uses of the original channel, and not an ensemble of channels. The ensemble of channels comes in when Preskill gives the intuition for what is happening in Shor-Smolin.

(And of course you should replace five in the above paragraph with the appropriate value of $m$, which is only five for some of these constructions.)

> **Comments:**
> **username_0**: Oh, you're using the same type of random code for each channel anyway so there should exist a random code that is good for all the channels. I'll have to think a little bit about it but I'm pretty sure that's the answer. Thanks! As for your second paragraph I'm not sure if I'm putting words into Preskill's mouth or if he means to explain it the same way you do. But I think thanks to your answer the way I was trying to understand it does work.  

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