Classical capacity

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: [Winter] 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 ${\displaystyle \rho _{x}}$:

${\displaystyle \mathbb {E} _{X}\left\{\left\Vert {\sqrt {\Lambda }}\rho _{X}{\sqrt {\Lambda }}-\rho _{X}\right\Vert _{1}\right\}\leq 2{\sqrt {\epsilon }}.}$

(Note that ${\displaystyle \left\Vert A\right\Vert _{1}}$ is the nuclear norm of the operator ${\displaystyle A}$ so that ${\displaystyle \left\Vert A\right\Vert _{1}\equiv }$Tr${\displaystyle \left\{{\sqrt {A^{\dagger }A}}\right\}}$.)

The following inequality is useful for us as well. It holds for any operators ${\displaystyle \rho }$, ${\displaystyle \sigma }$, ${\displaystyle \Lambda }$ such that ${\displaystyle 0\leq \rho ,\sigma ,\Lambda \leq I}$:

${\displaystyle {\text{Tr}}\left\{\Lambda \rho \right\}\leq {\text{Tr}}\left\{\Lambda \sigma \right\}+\left\Vert \rho -\sigma \right\Vert _{1}.}$

(1)

The quantum information-theoretic interpretation of the above inequality is that the probability of obtaining outcome ${\displaystyle \Lambda }$ from a quantum measurement acting on the state ${\displaystyle \rho }$ is upper bounded by the probability of obtaining outcome ${\displaystyle \Lambda }$ on the state ${\displaystyle \sigma }$ summed with the distinguishability of the two states ${\displaystyle \rho }$ and ${\displaystyle \sigma }$.

Non-commutative union bound

Lemma: [Sen's bound] The following bound holds for a subnormalized state ${\displaystyle \sigma }$ such that ${\displaystyle 0\leq \sigma }$ and ${\displaystyle Tr\left\{\sigma \right\}\leq 1}$ with ${\displaystyle \Pi _{1}}$, ... , ${\displaystyle \Pi _{N}}$ being projectors: ${\displaystyle {\text{Tr}}\left\{\sigma \right\}-{\text{Tr}}\left\{\Pi _{N}\cdots \Pi _{1}\ \sigma \ \Pi _{1}\cdots \Pi _{N}\right\}\leq 2{\sqrt {\sum _{i=1}^{N}{\text{Tr}}\left\{\left(I-\Pi _{i}\right)\sigma \right\}}},}$

We can think of Sen's bound as a "non-commutative union bound" because it is analogous to the following union bound from probability theory:

${\displaystyle \Pr \left\{\left(A_{1}\cap \cdots \cap A_{N}\right)^{c}\right\}=\Pr \left\{A_{1}^{c}\cup \cdots \cup A_{N}^{c}\right\}\leq \sum _{i=1}^{N}\Pr \left\{A_{i}^{c}\right\},}$

where ${\displaystyle A_{1},\ldots ,A_{N}}$ are events. The analogous bound for projector logic would be

${\displaystyle {\text{Tr}}\left\{\left(I-\Pi _{1}\cdots \Pi _{N}\cdots \Pi _{1}\right)\rho \right\}\leq \sum _{i=1}^{N}{\text{Tr}}\left\{\left(I-\Pi _{i}\right)\rho \right\},}$

if we think of ${\displaystyle \Pi _{1}\cdots \Pi _{N}}$ as a projector onto the intersection of subspaces. Though, the above bound only holds if the projectors ${\displaystyle \Pi _{1}}$, ..., ${\displaystyle \Pi _{N}}$ are commuting (choosing ${\displaystyle \Pi _{1}=\left\vert +\right\rangle \left\langle +\right\vert }$, ${\displaystyle \Pi _{2}=\left\vert 0\right\rangle \left\langle 0\right\vert }$, and ${\displaystyle \rho =\left\vert 0\right\rangle \left\langle 0\right\vert }$ gives a counterexample). If the projectors are non-commuting, then Sen's bound is the next best thing and suffices for our purposes here.

HSW theorem with the non-commutative union bound

We now prove the HSW theorem with Sen's non-commutative union bound. We divide up the proof into a few parts: codebook generation, POVM construction, and error analysis.

Codebook Generation. We first describe how Alice and Bob agree on a random choice of code. They have the channel ${\displaystyle x\rightarrow \rho _{x}}$ and a distribution ${\displaystyle p_{X}\left(x\right)}$. They choose ${\displaystyle M}$ classical sequences ${\displaystyle x^{n}}$ according to the IID\ distribution ${\displaystyle p_{X^{n}}\left(x^{n}\right)}$. After selecting them, they label them with indices as ${\displaystyle \left\{x^{n}\left(m\right)\right\}_{m\in \left[M\right]}}$. This leads to the following quantum codewords:

${\displaystyle \rho _{x^{n}\left(m\right)}=\rho _{x_{1}\left(m\right)}\otimes \cdots \otimes \rho _{x_{n}\left(m\right)}.}$

The quantum codebook is then ${\displaystyle \left\{\rho _{x^{n}\left(m\right)}\right\}}$. The average state of the codebook is then

${\displaystyle \mathbb {E} _{X^{n}}\left\{\rho _{X^{n}}\right\}=\sum _{x^{n}}p_{X^{n}}\left(x^{n}\right)\rho _{x^{n}}=\rho ^{\otimes n},}$

(2)

where ${\displaystyle \rho =\sum _{x}p_{X}\left(x\right)\rho _{x}}$.

POVM Construction . Sens' bound from the above lemma suggests a method for Bob to decode a state that Alice transmits. Bob should first ask "Is the received state in the average typical subspace?" He can do this operationally by performing a typical subspace measurement corresponding to ${\displaystyle \left\{\Pi _{\rho ,\delta }^{n},I-\Pi _{\rho ,\delta }^{n}\right\}}$. Next, he asks in sequential order, "Is the received codeword in the ${\displaystyle m^{\text{th}}}$ conditionally typical subspace?" This is in some sense equivalent to the question, "Is the received codeword the ${\displaystyle m^{\text{th}}}$ transmitted codeword?" He can ask these questions operationally by performing the measurements corresponding to the conditionally typical projectors ${\displaystyle \left\{\Pi _{\rho _{x^{n}\left(m\right)},\delta },I-\Pi _{\rho _{x^{n}\left(m\right)},\delta }\right\}}$.

Why should this sequential decoding scheme work well? The reason is that the transmitted codeword lies in the typical subspace on average:

${\displaystyle \mathbb {E} _{X^{n}}\left\{{\text{Tr}}\left\{\Pi _{\rho ,\delta }\ \rho _{X^{n}}\right\}\right\}={\text{Tr}}\left\{\Pi _{\rho ,\delta }\ \mathbb {E} _{X^{n}}\left\{\rho _{X^{n}}\right\}\right\}}$

${\displaystyle ={\text{Tr}}\left\{\Pi _{\rho ,\delta }\ \rho ^{\otimes n}\right\}}$

${\displaystyle \geq 1-\epsilon ,}$

where the inequality follows from (\ref{eq:1st-typ-prop}). Also, the projectors ${\displaystyle \Pi _{\rho _{x^{n}\left(m\right)},\delta }}$ are "good detectors" for the states ${\displaystyle \rho _{x^{n}\left(m\right)}}$ (on average) because the following condition holds from conditional quantum typicality:

${\displaystyle \mathbb {E} _{X^{n}}\left\{{\text{Tr}}\left\{\Pi _{\rho _{X^{n}},\delta }\ \rho _{X^{n}}\right\}\right\}\geq 1-\epsilon .}$

Error Analysis. The probability of detecting the ${\displaystyle m^{\text{th}}}$ codeword correctly under our sequential decoding scheme is equal to

${\displaystyle {\text{Tr}}\left\{\Pi _{\rho _{X^{n}\left(m\right)},\delta }{\hat {\Pi }}_{\rho _{X^{n}\left(m-1\right)},\delta }\cdots {\hat {\Pi }}_{\rho _{X^{n}\left(1\right)},\delta }\ \Pi _{\rho ,\delta }^{n}\ \rho _{x^{n}\left(m\right)}\ \Pi _{\rho ,\delta }^{n}\ {\hat {\Pi }}_{\rho _{X^{n}\left(1\right)},\delta }\cdots {\hat {\Pi }}_{\rho _{X^{n}\left(m-1\right)},\delta }\Pi _{\rho _{X^{n}\left(m\right)},\delta }\right\},}$

where we make the abbreviation ${\displaystyle {\hat {\Pi }}\equiv I-\Pi }$. (Observe that we project into the average typical subspace just once.) Thus, the probability of an incorrect detection for the ${\displaystyle m^{\text{th}}}$ codeword is given by

${\displaystyle 1-{\text{Tr}}\left\{\Pi _{\rho _{X^{n}\left(m\right)},\delta }{\hat {\Pi }}_{\rho _{X^{n}\left(m-1\right)},\delta }\cdots {\hat {\Pi }}_{\rho _{X^{n}\left(1\right)},\delta }\ \Pi _{\rho ,\delta }^{n}\ \rho _{x^{n}\left(m\right)}\ \Pi _{\rho ,\delta }^{n}\ {\hat {\Pi }}_{\rho _{X^{n}\left(1\right)},\delta }\cdots {\hat {\Pi }}_{\rho _{X^{n}\left(m-1\right)},\delta }\Pi _{\rho _{X^{n}\left(m\right)},\delta }\right\},}$

and the average error probability of this scheme is equal to

${\displaystyle 1-{\frac {1}{M}}\sum _{m}{\text{Tr}}\left\{\Pi _{\rho _{X^{n}\left(m\right)},\delta }{\hat {\Pi }}_{\rho _{X^{n}\left(m-1\right)},\delta }\cdots {\hat {\Pi }}_{\rho _{X^{n}\left(1\right)},\delta }\ \Pi _{\rho ,\delta }^{n}\ \rho _{x^{n}\left(m\right)}\ \Pi _{\rho ,\delta }^{n}\ {\hat {\Pi }}_{\rho _{X^{n}\left(1\right)},\delta }\cdots {\hat {\Pi }}_{\rho _{X^{n}\left(m-1\right)},\delta }\Pi _{\rho _{X^{n}\left(m\right)},\delta }\right\}.}$

Instead of analyzing the average error probability, we analyze the expectation of the average error probability, where the expectation is with respect to the random choice of code:

${\displaystyle 1-\mathbb {E} _{X^{n}}\left\{{\frac {1}{M}}\sum _{m}{\text{Tr}}\left\{\Pi _{\rho _{X^{n}\left(m\right)},\delta }{\hat {\Pi }}_{\rho _{X^{n}\left(m-1\right)},\delta }\cdots {\hat {\Pi }}_{\rho _{X^{n}\left(1\right)},\delta }\ \Pi _{\rho ,\delta }^{n}\ \rho _{X^{n}\left(m\right)}\ \Pi _{\rho ,\delta }^{n}\ {\hat {\Pi }}_{\rho _{X^{n}\left(1\right)},\delta }\cdots {\hat {\Pi }}_{\rho _{X^{n}\left(m-1\right)},\delta }\Pi _{\rho _{X^{n}\left(m\right)},\delta }\right\}\right\}.}$

(3)

Our first step is to apply Sen's bound to the above quantity. But before doing so, we should rewrite the above expression just slightly, by observing that

${\displaystyle 1=\mathbb {E} _{X^{n}}\left\{{\frac {1}{M}}\sum _{m}{\text{Tr}}\left\{\rho _{X^{n}\left(m\right)}\right\}\right\}}$

${\displaystyle =\mathbb {E} _{X^{n}}\left\{{\frac {1}{M}}\sum _{m}{\text{Tr}}\left\{\Pi _{\rho ,\delta }^{n}\rho _{X^{n}\left(m\right)}\right\}+{\text{Tr}}\left\{{\hat {\Pi }}_{\rho ,\delta }^{n}\rho _{X^{n}\left(m\right)}\right\}\right\}}$

${\displaystyle =\mathbb {E} _{X^{n}}\left\{{\frac {1}{M}}\sum _{m}{\text{Tr}}\left\{\Pi _{\rho ,\delta }^{n}\rho _{X^{n}\left(m\right)}\Pi _{\rho ,\delta }^{n}\right\}\right\}+{\frac {1}{M}}\sum _{m}{\text{Tr}}\left\{{\hat {\Pi }}_{\rho ,\delta }^{n}\mathbb {E} _{X^{n}}\left\{\rho _{X^{n}\left(m\right)}\right\}\right\}}$

${\displaystyle =\mathbb {E} _{X^{n}}\left\{{\frac {1}{M}}\sum _{m}{\text{Tr}}\left\{\Pi _{\rho ,\delta }^{n}\rho _{X^{n}\left(m\right)}\Pi _{\rho ,\delta }^{n}\right\}\right\}+{\text{Tr}}\left\{{\hat {\Pi }}_{\rho ,\delta }^{n}\rho ^{\otimes n}\right\}}$

${\displaystyle \leq \mathbb {E} _{X^{n}}\left\{{\frac {1}{M}}\sum _{m}{\text{Tr}}\left\{\Pi _{\rho ,\delta }^{n}\rho _{X^{n}\left(m\right)}\Pi _{\rho ,\delta }^{n}\right\}\right\}+\epsilon }$

Substituting into ( 3 ) (and forgetting about the small ${\displaystyle \epsilon }$ term for now) gives an upper bound of

${\displaystyle \mathbb {E} _{X^{n}}\left\{{\frac {1}{M}}\sum _{m}{\text{Tr}}\left\{\Pi _{\rho ,\delta }^{n}\rho _{X^{n}\left(m\right)}\Pi _{\rho ,\delta }^{n}\right\}\right\}}$

${\displaystyle -\mathbb {E} _{X^{n}}\left\{{\frac {1}{M}}\sum _{m}{\text{Tr}}\left\{\Pi _{\rho _{X^{n}\left(m\right)},\delta }{\hat {\Pi }}_{\rho _{X^{n}\left(m-1\right)},\delta }\cdots {\hat {\Pi }}_{\rho _{X^{n}\left(1\right)},\delta }\ \Pi _{\rho ,\delta }^{n}\ \rho _{X^{n}\left(m\right)}\ \Pi _{\rho ,\delta }^{n}\ {\hat {\Pi }}_{\rho _{X^{n}\left(1\right)},\delta }\cdots {\hat {\Pi }}_{\rho _{X^{n}\left(m-1\right)},\delta }\Pi _{\rho _{X^{n}\left(m\right)},\delta }\right\}\right\}.}$

We then apply Sen's bound to this expression with ${\displaystyle \sigma =\Pi _{\rho ,\delta }^{n}\rho _{X^{n}\left(m\right)}\Pi _{\rho ,\delta }^{n}}$ and the sequential projectors as ${\displaystyle \Pi _{\rho _{X^{n}\left(m\right)},\delta }}$, ${\displaystyle {\hat {\Pi }}_{\rho _{X^{n}\left(m-1\right)},\delta }}$, ..., ${\displaystyle {\hat {\Pi }}_{\rho _{X^{n}\left(1\right)},\delta }}$. This gives the upper bound ${\displaystyle \mathbb {E} _{X^{n}}\left\{{\frac {1}{M}}\sum _{m}2\left[{\text{Tr}}\left\{\left(I-\Pi _{\rho _{X^{n}\left(m\right)},\delta }\right)\Pi _{\rho ,\delta }^{n}\rho _{X^{n}\left(m\right)}\Pi _{\rho ,\delta }^{n}\right\}+\sum _{i=1}^{m-1}{\text{Tr}}\left\{\Pi _{\rho _{X^{n}\left(i\right)},\delta }\Pi _{\rho ,\delta }^{n}\rho _{X^{n}\left(m\right)}\Pi _{\rho ,\delta }^{n}\right\}\right]^{1/2}\right\}.}$ Due to concavity of the square root, we can bound this expression from above by

${\displaystyle 2\left[\mathbb {E} _{X^{n}}\left\{{\frac {1}{M}}\sum _{m}{\text{Tr}}\left\{\left(I-\Pi _{\rho _{X^{n}\left(m\right)},\delta }\right)\Pi _{\rho ,\delta }^{n}\rho _{X^{n}\left(m\right)}\Pi _{\rho ,\delta }^{n}\right\}+\sum _{i=1}^{m-1}{\text{Tr}}\left\{\Pi _{\rho _{X^{n}\left(i\right)},\delta }\Pi _{\rho ,\delta }^{n}\rho _{X^{n}\left(m\right)}\Pi _{\rho ,\delta }^{n}\right\}\right\}\right]^{1/2}}$

${\displaystyle \leq 2\left[\mathbb {E} _{X^{n}}\left\{{\frac {1}{M}}\sum _{m}{\text{Tr}}\left\{\left(I-\Pi _{\rho _{X^{n}\left(m\right)},\delta }\right)\Pi _{\rho ,\delta }^{n}\rho _{X^{n}\left(m\right)}\Pi _{\rho ,\delta }^{n}\right\}+\sum _{i\neq m}{\text{Tr}}\left\{\Pi _{\rho _{X^{n}\left(i\right)},\delta }\Pi _{\rho ,\delta }^{n}\rho _{X^{n}\left(m\right)}\Pi _{\rho ,\delta }^{n}\right\}\right\}\right]^{1/2},}$

where the second bound follows by summing over all of the codewords not equal to the ${\displaystyle m^{\text{th}}}$ codeword (this sum can only be larger).

We now focus exclusively on showing that the term inside the square root can be made small. Consider the first term:

${\displaystyle \mathbb {E} _{X^{n}}\left\{{\frac {1}{M}}\sum _{m}{\text{Tr}}\left\{\left(I-\Pi _{\rho _{X^{n}\left(m\right)},\delta }\right)\Pi _{\rho ,\delta }^{n}\rho _{X^{n}\left(m\right)}\Pi _{\rho ,\delta }^{n}\right\}\right\}}$

${\displaystyle \leq \mathbb {E} _{X^{n}}\left\{{\frac {1}{M}}\sum _{m}{\text{Tr}}\left\{\left(I-\Pi _{\rho _{X^{n}\left(m\right)},\delta }\right)\rho _{X^{n}\left(m\right)}\right\}+\left\Vert \rho _{X^{n}\left(m\right)}-\Pi _{\rho ,\delta }^{n}\rho _{X^{n}\left(m\right)}\Pi _{\rho ,\delta }^{n}\right\Vert _{1}\right\}}$

${\displaystyle \leq \epsilon +2{\sqrt {\epsilon }}.}$

where the first inequality follows from ( 1 ) and the second inequality follows from the gentle operator lemma and the properties of unconditional and conditional typicality. Consider now the second term and the following chain of inequalities:

${\displaystyle \sum _{i\neq m}\mathbb {E} _{X^{n}}\left\{{\text{Tr}}\left\{\Pi _{\rho _{X^{n}\left(i\right)},\delta }\ \Pi _{\rho ,\delta }^{n}\ \rho _{X^{n}\left(m\right)}\ \Pi _{\rho ,\delta }^{n}\right\}\right\}}$

${\displaystyle =\sum _{i\neq m}{\text{Tr}}\left\{\mathbb {E} _{X^{n}}\left\{\Pi _{\rho _{X^{n}\left(i\right)},\delta }\right\}\ \Pi _{\rho ,\delta }^{n}\ \mathbb {E} _{X^{n}}\left\{\rho _{X^{n}\left(m\right)}\right\}\ \Pi _{\rho ,\delta }^{n}\right\}}$

${\displaystyle =\sum _{i\neq m}{\text{Tr}}\left\{\mathbb {E} _{X^{n}}\left\{\Pi _{\rho _{X^{n}\left(i\right)},\delta }\right\}\ \Pi _{\rho ,\delta }^{n}\ \rho ^{\otimes n}\ \Pi _{\rho ,\delta }^{n}\right\}}$

${\displaystyle \leq \sum _{i\neq m}2^{-n\left[H\left(B\right)-\delta \right]}\ {\text{Tr}}\left\{\mathbb {E} _{X^{n}}\left\{\Pi _{\rho _{X^{n}\left(i\right)},\delta }\right\}\ \Pi _{\rho ,\delta }^{n}\right\}}$

The first equality follows because the codewords ${\displaystyle X^{n}\left(m\right)}$ and ${\displaystyle X^{n}\left(i\right)}$ are independent since they are different. The second equality follows from ( 2 ). The first inequality follows from (\ref{eq:3rd-typ-prop}). Continuing, we have

${\displaystyle \leq \sum _{i\neq m}2^{-n\left[H\left(B\right)-\delta \right]}\ \mathbb {E} _{X^{n}}\left\{{\text{Tr}}\left\{\Pi _{\rho _{X^{n}\left(i\right)},\delta }\right\}\right\}}$

${\displaystyle \leq \sum _{i\neq m}2^{-n\left[H\left(B\right)-\delta \right]}\ 2^{n\left[H\left(B|X\right)+\delta \right]}}$

${\displaystyle =\sum _{i\neq m}2^{-n\left[I\left(X;B\right)-2\delta \right]}}$

${\displaystyle \leq M\ 2^{-n\left[I\left(X;B\right)-2\delta \right]}.}$

The first inequality follows from ${\displaystyle \Pi _{\rho ,\delta }^{n}\leq I}$ and exchanging the trace with the expectation. The second inequality follows from (\ref{eq:2nd-cond-typ}). The next two are straightforward.

Putting everything together, we get our final bound on the expectation of the average error probability:

${\displaystyle 1-\mathbb {E} _{X^{n}}\left\{{\frac {1}{M}}\sum _{m}{\text{Tr}}\left\{\Pi _{\rho _{X^{n}\left(m\right)},\delta }{\hat {\Pi }}_{\rho _{X^{n}\left(m-1\right)},\delta }\cdots {\hat {\Pi }}_{\rho _{X^{n}\left(1\right)},\delta }\ \Pi _{\rho ,\delta }^{n}\ \rho _{X^{n}\left(m\right)}\ \Pi _{\rho ,\delta }^{n}\ {\hat {\Pi }}_{\rho _{X^{n}\left(1\right)},\delta }\cdots {\hat {\Pi }}_{\rho _{X^{n}\left(m-1\right)},\delta }\Pi _{\rho _{X^{n}\left(m\right)},\delta }\right\}\right\}}$

${\displaystyle \leq \epsilon +2\left[\left(\epsilon +2{\sqrt {\epsilon }}\right)+M\ 2^{-n\left[I\left(X;B\right)-2\delta \right]}\right]^{1/2}.}$

Thus, as long as we choose ${\displaystyle M=2^{n\left[I\left(X;B\right)-3\delta \right]}}$, there exists a code with vanishing error probability.

See also

• Entanglement-assisted classical capacity
• Quantum capacity
• Quantum information theory
• Typical subspace

References

• Holevo, Alexander S. (1998), "The Capacity of Quantum Channel with General Signal States", IEEE Transactions on Information Theory, 44 (1): 269–273, arXiv: quant-ph/9611023 , doi:10.1109/18.651037 .
• Schumacher, Benjamin; Westmoreland, Michael (1997), "Sending classical information via noisy quantum channels", Phys. Rev. A, 56 (1): 131–138, Bibcode:1997PhRvA..56..131S, doi:10.1103/PhysRevA.56.131 .
• Wilde, Mark M. (2017), Quantum Information Theory, Cambridge University Press, arXiv: 1106.1445 , Bibcode:2011arXiv1106.1445W, doi:10.1017/9781316809976.001
• Sen, Pranab (2012), "Achieving the Han-Kobayashi inner bound for the quantum interference channel by sequential decoding", IEEE International Symposium on Information Theory Proceedings (ISIT 2012), pp. 736–740, arXiv: 1109.0802 , doi:10.1109/ISIT.2012.6284656, S2CID   15119225 .
• Guha, Saikat; Tan, Si-Hui; Wilde, Mark M. (2012), "Explicit capacity-achieving receivers for optical communication and quantum reading", IEEE International Symposium on Information Theory Proceedings (ISIT 2012), pp. 551–555, arXiv: 1202.0518 , doi:10.1109/ISIT.2012.6284251, S2CID   8786400 .