Just above Eq (20.7) in Mark Wilde's book, while discussing the classical capacity of a quantum channel, he says:

These results then suggest that the ultimate classical capacity of the channel is the regularization of the accessible information of the channel, $I_{\mathrm{reg}}(\mathcal{N}) \equiv \lim _{k \rightarrow \infty} \frac{1}{k} I_{\mathrm{acc}}\left(\mathcal{N}^{\otimes k}\right)$ where $I_{\mathrm{acc}}(\mathcal{N}) \equiv \max _{\left\{p_X(x), \rho_x, \Lambda\right\}} I(X ; Y)$.

The Holevo quantity of a channel $\chi(\mathcal{N}) \equiv \max _\rho I(X ; B)$ is an upper bound to $I_{\rm acc}(\mathcal{N})$.

The Holevo-Schumacher-Westmoreland theorem states that the classical capacity of a quantum channel is $C(\mathcal{N}) = \chi_{\mathrm{reg}}(\mathcal{N}) \equiv \lim _{k \rightarrow \infty} \frac{1}{k} \chi\left(\mathcal{N}^{\otimes k}\right).$

Doesn't this mean that the Holevo quantity of a channel equals the accessible information of a channel as opposed to being an upper bound? I don't think that's true so maybe there is some other misunderstanding here?


1 Answer 1


No they are not the same. Given some quantum channel $\mathcal{N}$ we can consider an encoding map $\mathcal{E}$ and a decoding map $\mathcal{D}$ such that $\mathcal{C}_1= \mathcal{D}\circ \mathcal{N} \circ \mathcal{E}$ is a classical channel. Then if we apply Shannon's theorem we get a capacity which is the accessible information $I_{\mathrm{acc}}(\mathcal{N})$ which is effectively optimizing over encoding and decoding maps.

Now if we are a bit crafty, we can notice that applying Shannon's theorem directly to the classical channel $\mathcal{C}_1$ is maybe not optimal. This is because Shannon's theorem asks about the rate of $$ \mathcal{C}_1^{\otimes n} = \mathcal{D}^{\otimes n} \circ \mathcal{N}^{\otimes n} \circ \mathcal{E}^{\otimes n} $$ and so we are restricting ourselves to encoding and decoding maps that are in tensor product form. Instead, we could use $m$ copies of our original channel by constructing a classical channel $$ \mathcal{C}_m = \mathcal{D}_m \circ \mathcal{N}^{\otimes m} \circ \mathcal{E}_m $$ where the encoding maps $\mathcal{D}_m$ and $\mathcal{E}_m$ can now act on all systems simultaneously and are not restricted to be in a tensor product form. We can now apply Shannon's theorem again to this new channel we consider many copies of the channel $\mathcal{C}_m^{\otimes n}$ and Shannon's theorem tells us the capacity is $I_{\mathrm{acc}}(\mathcal{N}^{\otimes m})$. Because we use $\mathcal{N}$ $m$-times inside $\mathcal{C}_m$ we divide by $m$ to find the capacity of $\frac1m I_{\mathrm{acc}}(\mathcal{N}^{\otimes m})$. Because the accessible information is superadditive we can take the limit to see that a capacity of $\lim_{m \to \infty} \frac1m I_{\mathrm{acc}}(\mathcal{N}^{\otimes m})$ is achievable.

This is still however not the ultimate limit. The problem is that when you invoke Shannon's theorem you are still restricting to some form of product structure on the encoding and decoding maps. That is for any $m \in \mathbb{N}$, Shannon's theorem considers the asymptotic rate of information transfer for the maps $$ \mathcal{C}_m^{\otimes n} = \mathcal{D}_m^{\otimes n} \circ \mathcal{N}^{\otimes nm} \circ \mathcal{E}_m^{\otimes n} $$ so the encoding and decoding maps can only act on $m$ copies of $\mathcal{N}$ simultaneously. Instead, to get the ultimate classical capacity it makes sense to NOT invoke Shannon's theorem and instead directly consider copies of our quantum channel $\mathcal{N}^{\otimes n}$ together with decoding/encoding channels $\mathcal{D}_n$/ $\mathcal{E}_n$ that act on ALL copies of $\mathcal{N}$. This is the setting of the Holevo-Schumacher-Westmoreland theorem and this is why you can achieve something more than the regularized accessible information.

  • $\begingroup$ Thanks for the answer! Trying to still process exactly what happens here: When one looks at $I_{\mathrm{acc}}\left(\mathcal{N}^{\otimes m}\right)$, we are looking at a sequence of encoder/decoder pairs for each $m$. In the limit $m \rightarrow \infty$, this does NOT necessarily give us an encoder/decoder pair that uses all copies of the channel $\mathcal{N}$. Is that correct? $\endgroup$ Commented Nov 18, 2023 at 16:37
  • 1
    $\begingroup$ No because for every $m$ you are considering copies of the classical channel $\mathcal{C}_m$ which fixes your encoding and decodings to act only on blocks of size $m$. $\endgroup$
    – Rammus
    Commented Nov 18, 2023 at 19:06

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.