In quantum information theory, the classical capacity of a quantum channel is the maximum rate at which classical data can be sent over it error-free in the limit of many uses of the channel. Holevo, Schumacher, and Westmoreland proved the following least upper bound on the classical capacity of any quantum channel ${\displaystyle {\mathcal {N}}}$:

${\displaystyle \chi ({\mathcal {N}})=\max _{\rho ^{XA}}I(X;B)_{{\mathcal {N}}(\rho )}}$

where ${\displaystyle \rho ^{XA}}$ is a classical-quantum state of the following form:

${\displaystyle \rho ^{XA}=\sum _{x}p_{X}(x)\vert x\rangle \langle x\vert ^{X}\otimes \rho _{x}^{A},}$

${\displaystyle p_{X}(x)}$ is a probability distribution, and each ${\displaystyle \rho _{x}^{A}}$ is a density operator that can be input to the channel ${\displaystyle {\mathcal {N}}}$.

Achievability using sequential decoding

We briefly review the HSW coding theorem (the statement of the achievability of the Holevo information rate ${\displaystyle I(X;B)}$ for communicating classical data over a quantum channel). We first review the minimal amount of quantum mechanics needed for the theorem. We then cover quantum typicality, and finally we prove the theorem using a recent sequential decoding technique.

Review of quantum mechanics

In order to prove the HSW coding theorem, we really just need a few basic things from quantum mechanics. First, a quantum state is a unit trace, positive operator known as a density operator. Usually, we denote it by ${\displaystyle \rho }$, ${\displaystyle \sigma }$, ${\displaystyle \omega }$, etc. The simplest model for a quantum channel is known as a classical-quantum channel:

${\displaystyle x\mapsto \rho _{x}.}$

The meaning of the above notation is that inputting the classical letter ${\displaystyle x}$ at the transmitting end leads to a quantum state ${\displaystyle \rho _{x}}$ at the receiving end. It is the task of the receiver to perform a measurement to determine the input of the sender. If it is true that the states ${\displaystyle \rho _{x}}$ are perfectly distinguishable from one another (i.e., if they have orthogonal supports such that ${\displaystyle \mathrm {Tr} \,\left\{\rho _{x}\rho _{x^{\prime }}\right\}=0}$ for ${\displaystyle x\neq x^{\prime }}$), then the channel is a noiseless channel. We are interested in situations for which this is not the case. If it is true that the states ${\displaystyle \rho _{x}}$ all commute with one another, then this is effectively identical to the situation for a classical channel, so we are also not interested in these situations. So, the situation in which we are interested is that in which the states ${\displaystyle \rho _{x}}$ have overlapping support and are non-commutative.

The most general way to describe a quantum measurement is with a positive operator-valued measure (POVM). We usually denote the elements of a POVM as ${\displaystyle \left\{\Lambda _{m}\right\}_{m}}$. These operators should satisfy positivity and completeness in order to form a valid POVM:

${\displaystyle \Lambda _{m}\geq 0\ \ \ \ \forall m}$

${\displaystyle \sum _{m}\Lambda _{m}=I.}$

The probabilistic interpretation of quantum mechanics states that if someone measures a quantum state ${\displaystyle \rho }$ using a measurement device corresponding to the POVM ${\displaystyle \left\{\Lambda _{m}\right\}}$, then the probability ${\displaystyle p\left(m\right)}$ for obtaining outcome ${\displaystyle m}$ is equal to

${\displaystyle p\left(m\right)={\text{Tr}}\left\{\Lambda _{m}\rho \right\},}$

and the post-measurement state is

${\displaystyle \rho _{m}^{\prime }={\frac {1}{p\left(m\right)}}{\sqrt {\Lambda _{m}}}\rho {\sqrt {\Lambda _{m}}},}$

if the person measuring obtains outcome ${\displaystyle m}$. These rules are sufficient for us to consider classical communication schemes over cq channels.

Quantum typicality

The reader can find a good review of this topic in the article about the typical subspace.

Gentle operator lemma

The following lemma is important for our proofs. It demonstrates that a measurement that succeeds with high probability on average does not disturb the state too much on average:

Lemma: Given an ensemble ${\displaystyle \left\{p_{X}\left(x\right),\rho _{x}\right\}}$ with expected density operator ${\displaystyle \rho \equiv \sum _{x}p_{X}\left(x\right)\rho _{x}}$, suppose that an operator ${\displaystyle \Lambda }$ such that ${\displaystyle I\geq \Lambda \geq 0}$ succeeds with high probability on the state ${\displaystyle \rho }$:

${\displaystyle {\text{Tr}}\left\{\Lambda \rho \right\}\geq 1-\epsilon .}$

Then the subnormalized state ${\displaystyle {\sqrt {\Lambda }}\rho _{x}{\sqrt {\Lambda }}}$ is close in expected trace distance to the original state $$Zdroj:https://en.wikipedia.org?pojem=Classical_capacity
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