<p>You have two questions here.</p>
<ol>
<li><p>Why can't Shannon's Noisy Coding Theorem be used for a quantum channel?</p>
</li>
<li><p>What gaps are there to proving a quantum noisy channel coding theorem?</p>
</li>
</ol>
<p>I will concentrate on the first question.</p>
<p>Shannon's noisy coding theorem says that for a memoryless channel,</p>
<p><span class="math-container">$$ {\displaystyle \ C=\sup _{p_{X}}I(X;Y)},$$</span></p>
<p>where <span class="math-container">$p_X$</span> is maximized over all probability distributions over the input variable <span class="math-container">$X$</span>.</p>
<p>If you try to apply this naïvely to quantum channels, you run into a bunch of problems:</p>
<ul>
<li>The input and the output of a quantum communication channel are never around simultaneously, so the classical definition of <span class="math-container">$I(X;Y)$</span> doesn't work at all.</li>
<li>The proof depends on the fact that the channel is a probabilistic map taking the input to the output. This isn't true for quantum channels; a <a href="https://en.wikipedia.org/wiki/Quantum_operation" rel="nofollow noreferrer">quantum channel</a> is a completely positive trace-preserving linear operator.</li>
<li>Classical channels essentially only have one capacity. There are many different capacities for quantum channels. Are you trying to send classical or quantum information?  Do the sender and receiver have access to quantum entanglement? Can the receiver communicate on a classical channel that sends information back to the sender? Neither of these last two abilities change the capacity of classical channels, but they do change the capacity of quantum channels.</li>
<li>There are two equivalent expressions for mutual information:
<span class="math-container">$$I(X;Y) =H(X) +H(Y) - H(X,Y) = H(Y) - H(Y|X). $$</span>
The quantum equivalents of these two expressions are not equal. In fact, each of these corresponds to a different quantum channel capacity; the first corresponds to the <a href="https://en.wikipedia.org/wiki/Entanglement-assisted_classical_capacity" rel="nofollow noreferrer">entanglement-assisted classical capacity</a>  and the second to the <a href="https://en.wikipedia.org/wiki/Classical_capacity#HSW_theorem_with_the_non-commutative_union_bound" rel="nofollow noreferrer">Holevo-Schumacher-Westmoreland capacity</a>.</li>
</ul>
<p>So to summarize, you can't apply Shannon's Noisy Channel Coding theorem directly to quantum channels because not only does the proof not work, but the standard formulation of Shannon's capacity formula isn't even defined for quantum channels, and mutual information has two plausible extensions to the quantum case.</p>
<p>For your second question, there are several different quantum noisy channel coding theorems. The main problem with most of them is that they have a regularization step in them. Instead of getting an answer that you can compute by maximizing over inputs to a single channel use, you get an answer that is difficult to compute: you need to maximize a quantity that depends on <span class="math-container">$n$</span> channel uses and divide by <span class="math-container">$n$</span>, and then take <span class="math-container">$n$</span> to infinity. There may not be a single-channel-use formula; even a lot of quantities in classical information theory don't have a single-symbol formula.</p>

CC BY-SA 4.0
2021-01-02T14:37:31.833Z
2021-01-02T14:59:03.510Z
2021-01-02T14:59:03.510Z
Peter Shor 
<p>Why Shannon's <a href="https://en.wikipedia.org/wiki/Noisy-channel_coding_theorem" rel="nofollow noreferrer">noisy channel coding theorem</a> can't be used for quantum communication applications?</p>
<p>Schumacher proved the first <a href="http://www.tandfonline.com/doi/abs/10.1080/09500349414552191?journalCode=tmop20" rel="nofollow noreferrer">Noiseless theorem</a> and there are quantum error correction mechanisms out there.</p>
<p>I guess I am not sure what is the gap in knowledge to prove the <a href="https://en.wikipedia.org/wiki/Noisy-channel_coding_theorem" rel="nofollow noreferrer">noisy channel coding theorem</a>?</p>

CC BY-SA 3.0
2018-02-01T21:32:38.673Z
Do you know about the [Holevo](https://arxiv.org/abs/quant-ph/9611023)-[Schumacher-Westmoreland](https://journals.aps.org/pra/abstract/10.1103/PhysRevA.54.1869) theorem?
Peter Shor 
CC BY-SA 4.0
2021-01-02T14:48:23.343Z
Why do you think the quantum noisy channel coding theorem hasn't been proved already?
Peter Shor 
<quantum-information><open-problem>
Noisy channel coding theorem in quantum information
0x90
# Noisy channel coding theorem in quantum information

(Asked by: username_0 on 2018-01-28)

Why Shannon's [noisy channel coding theorem](https://en.wikipedia.org/wiki/Noisy-channel_coding_theorem) can't be used for quantum communication applications?

Schumacher proved the first [Noiseless theorem](http://www.tandfonline.com/doi/abs/10.1080/09500349414552191?journalCode=tmop20) and there are quantum error correction mechanisms out there.

I guess I am not sure what is the gap in knowledge to prove the [noisy channel coding theorem](https://en.wikipedia.org/wiki/Noisy-channel_coding_theorem)?

**Tags:** quantum-information, open-problem
> **Comments:**
> **username_1**: Do you know about the [Holevo](https://arxiv.org/abs/quant-ph/9611023)-[Schumacher-Westmoreland](https://journals.aps.org/pra/abstract/10.1103/PhysRevA.54.1869) theorem?  
> **username_1**: Why do you think the quantum noisy channel coding theorem hasn't been proved already?  

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## 1 answers

### Answer by username_1 on 2021-01-02. Score: 4 | Accepted answer

You have two questions here.

1. Why can't Shannon's Noisy Coding Theorem be used for a quantum channel?
2. What gaps are there to proving a quantum noisy channel coding theorem?

I will concentrate on the first question.

Shannon's noisy coding theorem says that for a memoryless channel,

$$ {\displaystyle \ C=\sup \_{p\_{X}}I(X;Y)},$$

where $p\_X$ is maximized over all probability distributions over the input variable $X$.

If you try to apply this naïvely to quantum channels, you run into a bunch of problems:

* The input and the output of a quantum communication channel are never around simultaneously, so the classical definition of $I(X;Y)$ doesn't work at all.
* The proof depends on the fact that the channel is a probabilistic map taking the input to the output. This isn't true for quantum channels; a [quantum channel](https://en.wikipedia.org/wiki/Quantum_operation) is a completely positive trace-preserving linear operator.
* Classical channels essentially only have one capacity. There are many different capacities for quantum channels. Are you trying to send classical or quantum information? Do the sender and receiver have access to quantum entanglement? Can the receiver communicate on a classical channel that sends information back to the sender? Neither of these last two abilities change the capacity of classical channels, but they do change the capacity of quantum channels.
* There are two equivalent expressions for mutual information:
  $$I(X;Y) =H(X) +H(Y) - H(X,Y) = H(Y) - H(Y|X). $$
  The quantum equivalents of these two expressions are not equal. In fact, each of these corresponds to a different quantum channel capacity; the first corresponds to the [entanglement-assisted classical capacity](https://en.wikipedia.org/wiki/Entanglement-assisted_classical_capacity) and the second to the [Holevo-Schumacher-Westmoreland capacity](https://en.wikipedia.org/wiki/Classical_capacity#HSW_theorem_with_the_non-commutative_union_bound).

So to summarize, you can't apply Shannon's Noisy Channel Coding theorem directly to quantum channels because not only does the proof not work, but the standard formulation of Shannon's capacity formula isn't even defined for quantum channels, and mutual information has two plausible extensions to the quantum case.

For your second question, there are several different quantum noisy channel coding theorems. The main problem with most of them is that they have a regularization step in them. Instead of getting an answer that you can compute by maximizing over inputs to a single channel use, you get an answer that is difficult to compute: you need to maximize a quantity that depends on $n$ channel uses and divide by $n$, and then take $n$ to infinity. There may not be a single-channel-use formula; even a lot of quantities in classical information theory don't have a single-symbol formula.


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