The Exercise

Alice sends classical messages to Bob through a noisy quantum channel $\mathcal{E}$. Over a block of $n$ uses she encodes each message as a product of $n$ input states, one for each use, and Bob identifies the message by a single measurement on the $n$ outputs together. The product state capacity $C^{(1)}(\mathcal{E})$ is the highest rate at which messages can be sent this way with a probability of wrong decoding that vanishes as $n$ grows.^✦ Theorem 12.8 of the book is the theorem of Holevo, Schumacher and Westmoreland.

Systems and States

A quantum system $Q$ is a physical system whose state space is a finite dimensional complex Hilbert space.^✦ A state $\rho$ of $Q$ is a positive operator of unit trace on the state space of $Q$.^✦ The mixture $\sum_i p_i \rho_i$ of states $\rho_i$ of $Q$, with $i$ running over a finite set and weights $p_i \ge 0$ summing to one, is the state of $Q$ whose density operator is that convex combination.^✦

The composite system $\bigotimes_i Q_i$ of a finite family of systems $Q_i$ is the system whose state space is the tensor product of the state spaces of the $Q_i$^✦; the system $Q^{\otimes n}$ of $n$ copies of $Q$ is the composite of $n$ copies of $Q$. For states $\rho_i$ of the $Q_i$, the product state $\bigotimes_i \rho_i$ is the state of $\bigotimes_i Q_i$ whose density operator is the tensor product of the $\rho_i$.^✦

Measurements

A measurement on a system $Q$ with a finite set of outcomes is a family of measurement operators $M_i$ on the state space of $Q$, one for each outcome $i$, that satisfies the completeness relation $$\sum_i M_i^\dagger M_i = I.$$(1)^✦ The positive operator valued measure (POVM) of a measurement $\{M_i\}$ is the family of its effects $E_i = M_i^\dagger M_i$, one for each outcome. When the measurement is performed on a state $\rho$ of $Q$, the outcome $i$ occurs with the probability given by the Born rule, $$p(i) = \operatorname{tr}(M_i \rho M_i^\dagger) = \operatorname{tr}(E_i \rho).$$(2)^✦

Product State Codes

A channel $\mathcal{E}$ on $Q$ is a map that takes each state of $Q$ to a state of $Q$. A product state code for $\mathcal{E}$ with $K$ messages and $n$ uses of the channel has two parts. Its encoder assigns to each message $M \in \{1, \dots, K\}$ input states $\rho_{M,1}, \dots, \rho_{M,n}$ of $Q$, whose product is Alice’s codeword for $M$. Its decoder is a measurement by Bob on $Q^{\otimes n}$, with the $K$ messages as outcomes and one outcome more, $0$, read as inconclusive; its effects are $E_1, \dots, E_K$ and $E_0$.

The output codeword $\sigma_M$ of the message $M$ is the product state of $Q^{\otimes n}$ that Bob receives, the channel acting on each use separately: $$\sigma_M = \mathcal{E}(\rho_{M,1}) \otimes \cdots \otimes \mathcal{E}(\rho_{M,n}).$$(3)^✦ Bob decodes $M$ with probability $\operatorname{tr}(E_M \sigma_M)$, the probability (2) of the outcome $M$ when his measurement is performed on $\sigma_M$^✦, and the error probability of $M$ is $$p^e_M = 1 - \operatorname{tr}(E_M \sigma_M).$$(4)^✦ The maximum error probability of the code is the largest $p^e_M$ over its messages.^✦

A rate $R$ is achievable for $\mathcal{E}$ when there are product state codes $\mathcal{C}_n$, one for each $n$, with $n$ uses of the channel and $\lceil e^{nR} \rceil$ messages, whose maximum error probabilities tend to zero: $$\lim_{n \to \infty} \max_M p^e_M(\mathcal{C}_n) = 0.$$(5)^✦ The product state capacity of $\mathcal{E}$ is the least upper bound of the rates achievable in the sense of (5), $$C^{(1)}(\mathcal{E}) = \sup\, \{ R : R \text{ is achievable for } \mathcal{E} \}.$$(6)^✦

The Holevo Quantity

The von Neumann entropy of a state $\rho$ of $Q$, measured in nats, is $$S(\rho) = -\operatorname{tr}(\rho \log \rho).$$(7)^✦ The Holevo quantity of an ensemble that prepares the state $\rho_i$ of $Q$ with probability $p_i$, for $i$ in a finite set, is, with the entropy (7), $$\chi = S\Bigl(\sum_i p_i \rho_i\Bigr) - \sum_i p_i\, S(\rho_i).$$(8)^✦

For a channel, the ensembles that count are those of its outputs: a finite ensemble of inputs that prepares $\rho_i$ with probability $p_i$ gives the output ensemble that prepares $\mathcal{E}(\rho_i)$ with the same probability. The Holevo quantity $\chi(\mathcal{E})$ of the channel is the least upper bound of the set of values that (8) takes on output ensembles, $$\chi(\mathcal{E}) = \sup_{\{p_i, \rho_i\}} \Bigl[ S\Bigl(\sum_i p_i\, \mathcal{E}(\rho_i)\Bigr) - \sum_i p_i\, S\bigl(\mathcal{E}(\rho_i)\bigr) \Bigr].$$(9)^✦

The Claim

What Is Checked

The claim (10) is stated over the Holevo quantity (9) of a channel and its product state capacity (6), the latter built from the output codewords (3) of product state codes and the error probabilities (4) of Bob’s measurement.

The library proves that every output codeword $\sigma_M$ is a state of $Q^{\otimes n}$: a product state is positive with unit trace because each of its factors is, and the state space of a composite system is finite dimensional, with the products of basis vectors of the factors as a basis. The effects of a measurement are positive operators.