classical capacity

We briefly review the HSW coding theorem (the

statement of the achievability of the Holevo information rate $I(X;B)$ 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.

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 $\backslash rho$, $\backslash sigma$, $\backslash omega$, etc. The simplest model for a quantum channel

is known as a classical-quantum channel:

The meaning of the above notation is that inputting the classical letter $x$

at the transmitting end leads to a quantum state $\backslash rho\_\{x\}$ 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 $\backslash rho\_\{x\}$ are perfectly

distinguishable from one another (i.e., if they have orthogonal supports such

that $\backslash mathrm\{Tr\}\backslash ,\backslash left\backslash \{\; \backslash rho\_\{x\}\backslash rho\_\{x^\{\backslash prime\}\}\backslash right\backslash \}\; =0$ for $x\backslash neq\; x^\{\backslash prime\}$

), 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 $\backslash rho\_\{x\}$ 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

$\backslash rho\_\{x\}$ 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

$\backslash left\backslash \{\; \backslash Lambda\_\{m\}\backslash right\backslash \}\; \_\{m\}$. These operators should satisfy

positivity and completeness in order to form a valid POVM:

:$$

\Lambda_{m} \geq0\ \ \ \ \forall m

:$\backslash sum\_\{m\}\backslash Lambda\_\{m\}\; =I.$

The probabilistic interpretation of quantum mechanics states that if someone

measures a quantum state $\backslash rho$ using a measurement device corresponding to

the POVM $\backslash left\backslash \{\; \backslash Lambda\_\{m\}\backslash right\backslash \}$, then the probability $p\backslash left($

m\right) for obtaining outcome $m$ is equal to

:$$

p\left( m\right) =\text{Tr}\left\{ \Lambda_{m}\rho\right\} ,

and the post-measurement state is

:$$

\rho_{m}^{\prime}=\frac{1}{p\left( m\right) }\sqrt{\Lambda_{m}}\rho

\sqrt{\Lambda_{m}},

if the person measuring obtains outcome $m$. These rules are sufficient for us

to consider classical communication schemes over cq channels.

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

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 $\backslash left\backslash \{\; p\_\{X\}\backslash left(\; x\backslash right)\; ,\backslash rho\_\{x\}\backslash right\backslash \}$ with expected

density operator $\backslash rho\backslash equiv\backslash sum\_\{x\}p\_\{X\}\backslash left(\; x\backslash right)\; \backslash rho\_\{x\}$, suppose

that an operator $\backslash Lambda$ such that $I\backslash geq\backslash Lambda\backslash geq0$ succeeds with high

probability on the state $\backslash rho$:

Then the subnormalized state $\backslash sqrt\{\backslash Lambda\}\backslash rho\_\{x\}\backslash sqrt\{\backslash Lambda\}$ is close

in expected trace distance to the original state $\backslash rho\_\{x\}$:

(Note that $\backslash left\backslash Vert\; A\backslash right\backslash Vert\; \_\{1\}$ is the nuclear norm of the operator

$A$ so that $\backslash left\backslash Vert\; A\backslash right\backslash Vert\; \_\{1\}\backslash equiv$Tr$\backslash left\backslash \{\; \backslash sqrt\{A^\{\backslash dagger\}$

A}\right\} .)

The following inequality is useful for us as well. It holds for any operators

$\backslash rho$, $\backslash sigma$, $\backslash Lambda$ such that $0\backslash leq\backslash rho,\backslash sigma,\backslash Lambda\backslash leq\; I$:

The quantum information-theoretic interpretation of the above inequality is

that the probability of obtaining outcome $\backslash Lambda$ from a quantum measurement

acting on the state $\backslash rho$ is upper bounded by the probability of obtaining

outcome $\backslash Lambda$ on the state $\backslash sigma$ summed with the distinguishability of

the two states $\backslash rho$ and $\backslash sigma$.

Lemma: [Sen's bound] The following bound

holds for a subnormalized state $\backslash sigma$ such that $0\backslash leq\backslash sigma$ and

$Tr\backslash left\backslash \{\; \backslash sigma\backslash right\backslash \}\; \backslash leq1$ with $\backslash Pi\_\{1\}$, ... , $\backslash Pi\_\{N\}$ being

projectors:

\text{Tr}\left\{ \sigma\right\} -\text{Tr}\left\{ \Pi_{N}\cdots\Pi

_{1}\ \sigma\ \Pi_{1}\cdots\Pi_{N}\right\} \leq2\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:

where $A\_\{1\},\; \backslash ldots,\; A\_\{N\}$ are events. The analogous bound for projector

logic would be

:$$

\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 $\backslash Pi\_\{1\}\backslash cdots\backslash Pi\_\{N\}$ as a projector onto the intersection of

subspaces. Though, the above bound only holds if the projectors $\backslash Pi\_\{1\}$,

..., $\backslash Pi\_\{N\}$ are commuting (choosing $\backslash Pi\_\{1\}=\backslash left\backslash vert\; +\backslash right\backslash rangle$

\left\langle +\right\vert , $\backslash Pi\_\{2\}=\backslash left\backslash vert\; 0\backslash right\backslash rangle\; \backslash left\backslash langle$

0\right\vert , and $\backslash rho=\backslash left\backslash vert\; 0\backslash right\backslash rangle\; \backslash left\backslash langle\; 0\backslash right\backslash vert$

gives a counterexample). If the projectors are non-commuting, then Sen's

bound is the next best thing and suffices for our purposes here.

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 $x\backslash rightarrow\backslash rho\_\{x\}$ and a

distribution $p\_\{X\}\backslash left(\; x\backslash right)$. They choose $M$ classical sequences

$x^\{n\}$ according to the IID\ distribution $p\_\{X^\{n\}\}\backslash left(\; x^\{n\}\backslash right)$.

After selecting them, they label them with indices as $\backslash left\backslash \{\; x^\{n\}\backslash left($

m\right) \right\} _{m\in\left[ M\right] }. This leads to the following

quantum codewords:

The quantum codebook is then $\backslash left\backslash \{\; \backslash rho\_\{x^\{n\}\backslash left(\; m\backslash right)\; \}\backslash right\backslash \}$

. The average state of the codebook is then

where $\backslash rho=\backslash sum\_\{x\}p\_\{X\}\backslash left(\; x\backslash right)\; \backslash rho\_\{x\}$.

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 $\backslash left\backslash \{\; \backslash Pi\_\{\backslash rho,\backslash delta\}$

^{n},I-\Pi_{\rho,\delta}^{n}\right\} . Next, he asks in sequential order,

"Is the received codeword in the $m^\{\backslash text\{th\}\}$

conditionally typical subspace?" This is in some sense

equivalent to the question, "Is the received codeword the

$m^\{\backslash text\{th\}\}$ transmitted codeword?" He can ask these

questions operationally by performing the measurements corresponding to the

conditionally typical projectors $\backslash left\backslash \{\; \backslash Pi\_\{\backslash rho\_\{x^\{n\}\backslash left(\; m\backslash right)$

},\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:

:$$

\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\}

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

:$\backslash geq1-\backslash epsilon,$

where the inequality follows from (\ref{eq:1st-typ-prop}). Also, the

projectors $\backslash Pi\_\{\backslash rho\_\{x^\{n\}\backslash left(\; m\backslash right)\; \},\backslash delta\}$

are "good detectors" for the states $\backslash rho\_\{x^\{n\}\backslash left(\; m\backslash right)$

} (on average) because the following condition holds from conditional quantum

typicality:

Error Analysis. The probability of detecting the $m^\{\backslash text\{th\}\}$

codeword correctly under our sequential decoding scheme is equal to

where we make the abbreviation $\backslash hat\{\backslash Pi\}\backslash equiv\; I-\backslash Pi$. (Observe that we

project into the average typical subspace just once.) Thus, the probability of

an incorrect detection for the $m^\{\backslash text\{th\}\}$ codeword is given by

and the average error probability of this scheme is equal to

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:

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

:$$

1 =\mathbb{E}_{X^{n}}\left\{ \frac{1}{M}\sum_{m}\text{Tr}\left\{

\rho_{X^{n}\left( m\right) }\right\} \right\}

:$=\backslash mathbb\{E\}\_\{X^\{n\}\}\backslash left\backslash \{\; \backslash frac\{1\}\{M\}\backslash sum\_\{m\}\backslash text\{Tr\}\backslash left\backslash \{\; \backslash Pi$

_{\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\}

:$=\backslash mathbb\{E\}\_\{X^\{n\}\}\backslash left\backslash \{\; \backslash frac\{1\}\{M\}\backslash sum\_\{m\}\backslash text\{Tr\}\backslash left\backslash \{\; \backslash Pi$

_{\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\}

:$=\backslash mathbb\{E\}\_\{X^\{n\}\}\backslash left\backslash \{\; \backslash frac\{1\}\{M\}\backslash sum\_\{m\}\backslash text\{Tr\}\backslash left\backslash \{\; \backslash Pi$

_{\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\}

:$\backslash leq\backslash mathbb\{E\}\_\{X^\{n\}\}\backslash left\backslash \{\; \backslash frac\{1\}\{M\}\backslash sum\_\{m\}\backslash text\{Tr\}\backslash left\backslash \{$

\Pi_{\rho,\delta}^{n}\rho_{X^{n}\left( m\right) }\Pi_{\rho,\delta}

^{n}\right\} \right\} +\epsilon

Substituting into ({{EquationNote|3}}) (and forgetting about the small

$\backslash epsilon$ term for now) gives an upper bound of

:$$

\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\}

:$$

-\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 $\backslash sigma=\backslash Pi\_\{\backslash rho,\backslash delta$

}^{n}\rho_{X^{n}\left( m\right) }\Pi_{\rho,\delta}^{n} and the sequential

projectors as $\backslash Pi\_\{\backslash rho\_\{X^\{n\}\backslash left(\; m\backslash right)\; \},\backslash delta\}$, $\backslash hat\{\backslash Pi\}$

_{\rho_{X^{n}\left( m-1\right) },\delta}, ..., $\backslash hat\{\backslash Pi\}\_\{\backslash rho$

_{X^{n}\left( 1\right) },\delta}. This gives the upper bound

\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

:$$

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}

:$\backslash leq2\backslash left[\; \backslash mathbb\{E\}\_\{X^\{n\}\}\backslash left\backslash \{\; \backslash frac\{1\}\{M\}\backslash sum\_\{m\}\backslash text\{Tr\}\backslash left\backslash \{$

\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 $m^\{\backslash text\{th\}\}$ 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:

:$$

\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\}

:$\backslash leq\backslash mathbb\{E\}\_\{X^\{n\}\}\backslash left\backslash \{\; \backslash frac\{1\}\{M\}\backslash sum\_\{m\}\backslash text\{Tr\}\backslash left\backslash \{\; \backslash left($

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\}

:$\backslash leq\backslash epsilon+2\backslash sqrt\{\backslash epsilon\}.$

where the first inequality follows from ({{EquationNote|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:

:$$

\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\}

:$=\backslash sum\_\{i\backslash neq\; m\}\backslash text\{Tr\}\backslash left\backslash \{\; \backslash mathbb\{E\}\_\{X^\{n\}\}\backslash left\backslash \{\; \backslash Pi$

_{\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\}

:$=\backslash sum\_\{i\backslash neq\; m\}\backslash text\{Tr\}\backslash left\backslash \{\; \backslash mathbb\{E\}\_\{X^\{n\}\}\backslash left\backslash \{\; \backslash Pi$

_{\rho_{X^{n}\left( i\right) },\delta}\right\} \ \Pi_{\rho,\delta}

^{n}\ \rho^{\otimes n}\ \Pi_{\rho,\delta}^{n}\right\}

:$\backslash leq\backslash sum\_\{i\backslash neq\; m\}2^\{-n\backslash left[\; H\backslash left(\; B\backslash right)\; -\backslash delta\backslash right]$

}\ \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 $X^\{n\}\backslash left(\; m\backslash right)$ and

$X^\{n\}\backslash left(\; i\backslash right)$ are independent since they are different. The second

equality follows from ({{EquationNote|2}}). The first inequality follows from

(\ref{eq:3rd-typ-prop}). Continuing, we have

:$$

\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\}

:$\backslash leq\backslash sum\_\{i\backslash neq\; m\}2^\{-n\backslash left[\; H\backslash left(\; B\backslash right)\; -\backslash delta\backslash right]$

}\ 2^{n\left[ H\left( B|X\right) +\delta\right] }

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

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

The first inequality follows from $\backslash Pi\_\{\backslash rho,\backslash delta\}^\{n\}\backslash leq\; I$ 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:

:$$

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\}

:$\backslash leq\backslash epsilon+2\backslash left[\; \backslash left(\; \backslash epsilon+2\backslash sqrt\{\backslash epsilon\}\backslash right)$

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

Thus, as long as we choose $M=2^\{n\backslash left[\; I\backslash left(\; X;B\backslash right)\; -3\backslash delta$

\right] }, there exists a code with vanishing error probability.

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

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{{Quantum computing}}

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