The classical channel capacity ($C_{ea}$) and the quantum channel capacity ($Q$) as defined here (eqs. 1 and 2) are given by \begin{equation} C_{ea} = \text{sup}_{\rho} \Big[S(\rho) + S(\Phi_t \rho) - S(\rho,t)\Big], \end{equation} and \begin{equation} Q(\Phi_t) = \text{sup}_{\rho} \Big[ S(\Phi_t \rho) - S(\rho,t)\Big]. \end{equation} Here, $S(\cdot)$ is the von-Neumann entropy. Also, $\Phi_t \rho$ is the state evolved under the action of the map $\Phi$, and $S(\rho,t)$ is the entropy at the output of the complementary channel. The two definitions look exactly the same apart from $S(\rho)$ in the first case. How does the former correspond to classical and the late to quantum case? Further, for a pure state $\rho$, $S(\rho) = 0$, so the two definitions seem to coincide. Everything right here?


These are not really the definitions of classical and quantum capacity, as I will explain. Before doing that, let me adjust the notation being used slightly: let $\Phi:\text{L}(\mathcal{X}) \rightarrow \text{L}(\mathcal{Y})$ be the channel whose capacities we are interested in and let $\Psi:\text{L}(\mathcal{X}) \rightarrow \text{L}(\mathcal{Z})$ be a channel that is complementary to $\Phi$.

First, we define the coherent information of $\Phi$ with respect to a state $\rho \in \text{D}(\mathcal{X})$ like this: $$ \text{I}_\text{C}(\rho;\Phi) = \text{S}(\Phi(\rho)) - \text{S}(\Psi(\rho)). $$ (This value is the same for all choices of a complementary channel $\Psi$.) We then define the maximum coherent information of $\Phi$ by maximizing over all $\rho$: $$ \text{I}_\text{C}(\Phi) = \max_{\rho\in\text{D}(\mathcal{X})} \text{I}_\text{C}(\rho;\Phi). $$ This is equivalent to the expression written in the second equation of the question. (In particular, the supremum is taken over a compact set and the coherent information is continuous, so we can safely replace it by a maximum.) It is not, however, equal to the quantum capacity in general; the quantum capacity is defined in a completely different, operationally motivated way that I won't explain in this answer.

There is, however, a theorem that connects the two quantities, known alternatively as the quantum capacity theorem or the LSD theorem (named after Lloyd, Shor, and Devetak). The theorem states that the quantum capacity is given by this expression: $$ \text{Q}(\Phi) = \lim_{n\rightarrow\infty} \frac{\text{I}_{\text{C}}(\Phi^{\otimes n})}{n}. $$ This is a regularization of the maximum coherent information, and in general the regularization is requied. There are examples of channels, however, including all so-called degradable channels, for which the regularization is not required and we have $\text{Q}(\Phi) = \text{I}_{\text{C}}(\Phi)$.

The other quantity under consideration is the entanglement assisted classical capacity. Again this quantity is defined in a different, operationally motivated way, but this time there is a theorem that characterizes this quantity perfectly: $$ \text{C}_{\text{E}}(\Phi) = \max_{\rho\in\text{D}(\mathcal{X})} \Bigl( \text{I}_\text{C}(\rho;\Phi) + \text{S}(\rho) \Bigr). $$ This theorem is sometimes called the entanglement assisted classical capacity theorem, and it was proved by Bennett, Shor, Smolin, and Thapliyal. It is a rare example in quantum Shannon theory where the characterization of a channel capacity does not require a regularization.

In both cases, it is helpful to see the theorems as being quantum analogues of Shannon's noisy channel coding theorem. The states $\rho$ that achieve the maximum values should really not be thought of so much as actual states that are sent through the channels, but "suggestions" for codes that stretch across multiple independent uses of $\Phi$. It is true, as the question suggests, that the values of these expressions are both zero for pure states, but the codes that these states suggest are effectively trivial and useless for transmitting information.

The proofs of both theorems are rather difficult. There are intuitive ways of thinking about the different expressions, but I don't think it is healthy to think about the term $\text{S}(\rho)$ as arising from something classical and disappearing in the quantum case. It is immediate from the definitions that the entanglement assisted classical capacity should be at least as large as the quantum capacity, but because one of these capacities is an entanglement assisted notion and the other is not, we are comparing apples with oranges to some extent. The entanglement assisted quantum capacity, for instance, is always exactly half the entanglement assisted classical capacity (and that is probably why people don't bother talking about it).


Your Answer

