View source for Classical capacity

{{short description|Term in quantum information theory}}
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. [[Alexander Holevo|Holevo]], Schumacher, and Westmoreland proved the following least upper bound on the classical capacity of any quantum channel <math>\mathcal{N}</math>:

:<math>
\chi(\mathcal{N}) = \max_{\rho^{XA}} I(X;B)_{\mathcal{N}(\rho)}
</math>

where <math>\rho^{XA}</math> is a classical-quantum state of the following form:
:<math>
\rho^{XA} = \sum_x p_X(x) \vert x \rangle \langle x \vert^X \otimes \rho_x^A ,
</math>
<math>p_X(x)</math> is a probability distribution, and each <math>\rho_x^A</math> is a density operator that can be input to the channel <math>\mathcal{N}</math>.

== Achievability using sequential decoding ==

We briefly review the HSW coding theorem (the
statement of the achievability of the [[Holevo information]] rate <math>I(X;B)</math> 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 <math>\rho</math>, <math>\sigma</math>, <math>\omega</math>, etc. The simplest model for a [[quantum channel]]
is known as a classical-quantum channel:
<div class="center"><math>
x\mapsto \rho_{x}.
</math></div>
The meaning of the above notation is that inputting the classical letter <math>x</math>
at the transmitting end leads to a quantum state <math>\rho_{x}</math> 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 <math>\rho_{x}</math> are perfectly
distinguishable from one another (i.e., if they have orthogonal supports such
that <math>\mathrm{Tr}\,\left\{  \rho_{x}\rho_{x^{\prime}}\right\}  =0</math> for <math>x\neq x^{\prime}
</math>), 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 <math>\rho_{x}</math> 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
<math>\rho_{x}</math> 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
<math>\left\{  \Lambda_{m}\right\}  _{m}</math>. These operators should satisfy
positivity and completeness in order to form a valid POVM:
:<math>
\Lambda_{m}   \geq0\ \ \ \ \forall m</math>
:<math>\sum_{m}\Lambda_{m}  =I.
</math>
The probabilistic interpretation of [[quantum mechanics]] states that if someone
measures a quantum state <math>\rho</math> using a measurement device corresponding to
the POVM <math>\left\{  \Lambda_{m}\right\}  </math>, then the probability <math>p\left(
m\right) </math> for obtaining outcome <math>m</math> is equal to
:<math>
p\left(  m\right)  =\text{Tr}\left\{  \Lambda_{m}\rho\right\}  ,
</math>
and the post-measurement state is
:<math>
\rho_{m}^{\prime}=\frac{1}{p\left(  m\right)  }\sqrt{\Lambda_{m}}\rho
\sqrt{\Lambda_{m}},
</math>
if the person measuring obtains outcome <math>m</math>. 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 <math>\left\{  p_{X}\left(  x\right)  ,\rho_{x}\right\}  </math> with expected
density operator <math>\rho\equiv\sum_{x}p_{X}\left(  x\right)  \rho_{x}</math>, suppose
that an operator <math>\Lambda</math> such that <math>I\geq\Lambda\geq0</math> succeeds with high
probability on the state <math>\rho</math>:
<div class="center"><math>
\text{Tr}\left\{  \Lambda\rho\right\}  \geq1-\epsilon.
</math></div>
Then the subnormalized state <math>\sqrt{\Lambda}\rho_{x}\sqrt{\Lambda}</math> is close
in expected trace distance to the original state <math>\rho_{x}</math>:
<div class="center"><math>
\mathbb{E}_{X}\left\{  \left\Vert \sqrt{\Lambda}\rho_{X}\sqrt{\Lambda}
-\rho_{X}\right\Vert _{1}\right\}  \leq2\sqrt{\epsilon}.
</math></div>
(Note that <math>\left\Vert A\right\Vert _{1}</math> is the nuclear norm of the operator
<math>A</math> so that <math>\left\Vert A\right\Vert _{1}\equiv</math>Tr<math>\left\{  \sqrt{A^{\dagger}
A}\right\}  </math>.)

The following inequality is useful for us as well. It holds for any operators
<math>\rho</math>, <math>\sigma</math>, <math>\Lambda</math> such that <math>0\leq\rho,\sigma,\Lambda\leq I</math>:
<div class="center">{{NumBlk|:|<math>
\text{Tr}\left\{  \Lambda\rho\right\}  \leq\text{Tr}\left\{  \Lambda
\sigma\right\}  +\left\Vert \rho-\sigma\right\Vert _{1}.
</math>|{{EquationRef|1}}}}</div>
The quantum information-theoretic interpretation of the above inequality is
that the probability of obtaining outcome <math>\Lambda</math> from a quantum measurement
acting on the state <math>\rho</math> is upper bounded by the probability of obtaining
outcome <math>\Lambda</math> on the state <math>\sigma</math> summed with the distinguishability of
the two states <math>\rho</math> and <math>\sigma</math>.

==Non-commutative union bound==
Lemma: [Sen's bound] The following bound
holds for a subnormalized state <math>\sigma</math> such that <math>0\leq\sigma</math> and
<math>Tr\left\{  \sigma\right\}  \leq1</math> with <math>\Pi_{1}</math>, ... , <math>\Pi_{N}</math> being
projectors:
<math>
\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\}  },
</math>

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:
<div class="center"><math>
\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\}  ,
</math></div>
where <math>A_{1}, \ldots, A_{N}</math> are events. The analogous bound for projector
logic would be
:<math>
\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\}  ,
</math>
if we think of <math>\Pi_{1}\cdots\Pi_{N}</math> as a projector onto the intersection of
subspaces. Though, the above bound only holds if the projectors <math>\Pi_{1}</math>,
..., <math>\Pi_{N}</math> are commuting (choosing <math>\Pi_{1}=\left\vert +\right\rangle
\left\langle +\right\vert </math>, <math>\Pi_{2}=\left\vert 0\right\rangle \left\langle
0\right\vert </math>, and <math>\rho=\left\vert 0\right\rangle \left\langle 0\right\vert
</math> 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 <math>x\rightarrow\rho_{x}</math> and a
distribution <math>p_{X}\left(  x\right)  </math>. They choose <math>M</math> classical sequences
<math>x^{n}</math> according to the IID\ distribution <math>p_{X^{n}}\left(  x^{n}\right)  </math>.
After selecting them, they label them with indices as <math>\left\{  x^{n}\left(
m\right)  \right\}  _{m\in\left[  M\right]  }</math>. This leads to the following
quantum codewords:
<div class="center"><math>
\rho_{x^{n}\left(  m\right)  }=\rho_{x_{1}\left(  m\right)  }\otimes
\cdots\otimes\rho_{x_{n}\left(  m\right)  }.
</math></div>
The quantum codebook is then <math>\left\{  \rho_{x^{n}\left(  m\right)  }\right\}
</math>. The average state of the codebook is then
<div class="center">{{NumBlk|:|<math>
\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},
</math>|{{EquationRef|2}}}}</div>
where <math>\rho=\sum_{x}p_{X}\left(  x\right)  \rho_{x}</math>.

'''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 <math>\left\{  \Pi_{\rho,\delta}
^{n},I-\Pi_{\rho,\delta}^{n}\right\}  </math>. Next, he asks in sequential order,
"Is the received codeword in the <math>m^{\text{th}}</math>
conditionally typical subspace?" This is in some sense
equivalent to the question, "Is the received codeword the
<math>m^{\text{th}}</math> transmitted codeword?" He can ask these
questions operationally by performing the measurements corresponding to the
conditionally typical projectors <math>\left\{  \Pi_{\rho_{x^{n}\left(  m\right)
},\delta},I-\Pi_{\rho_{x^{n}\left(  m\right)  },\delta}\right\}  </math>.

Why should this sequential decoding scheme work well? The reason is that the
transmitted codeword lies in the typical subspace on average:
:<math>
\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\}  </math>
:<math> =\text{Tr}\left\{  \Pi_{\rho,\delta}\ \rho^{\otimes n}\right\}  </math>
:<math> \geq1-\epsilon,</math>
where the inequality follows from (\ref{eq:1st-typ-prop}). Also, the
projectors <math>\Pi_{\rho_{x^{n}\left(  m\right)  },\delta}</math>
are "good detectors" for the states <math>\rho_{x^{n}\left(  m\right)
}</math> (on average) because the following condition holds from conditional quantum
typicality:
<div class="center"><math>
\mathbb{E}_{X^{n}}\left\{  \text{Tr}\left\{  \Pi_{\rho_{X^{n}},\delta}
\ \rho_{X^{n}}\right\}  \right\}  \geq1-\epsilon.
</math></div>

'''Error Analysis'''. The probability of detecting the <math>m^{\text{th}}</math>
codeword correctly under our sequential decoding scheme is equal to
<div class="center"><math>
\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\}  ,
</math></div>
where we make the abbreviation <math>\hat{\Pi}\equiv I-\Pi</math>. (Observe that we
project into the average typical subspace just once.) Thus, the probability of
an incorrect detection for the <math>m^{\text{th}}</math> codeword is given by
<div class="center"><math>
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\}  ,
</math></div>
and the average error probability of this scheme is equal to
<div class="center"><math>
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\}  .
</math></div>
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:
<div class="center">{{NumBlk|:|<math>
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\}  .
</math>|{{EquationRef|3}}}}</div>

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
:<math>
1   =\mathbb{E}_{X^{n}}\left\{  \frac{1}{M}\sum_{m}\text{Tr}\left\{
\rho_{X^{n}\left(  m\right)  }\right\}  \right\}  </math>
:<math> =\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\}
</math>
:<math> =\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\}  </math>
:<math> =\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\}  </math>
:<math> \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
</math>
Substituting into ({{EquationNote|3}}) (and forgetting about the small
<math>\epsilon</math> term for now) gives an upper bound of
:<math>
\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\} </math>
:<math>
-\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\}  .
</math>
We then apply Sen's bound to this expression with <math>\sigma=\Pi_{\rho,\delta
}^{n}\rho_{X^{n}\left(  m\right)  }\Pi_{\rho,\delta}^{n}</math> and the sequential
projectors as <math>\Pi_{\rho_{X^{n}\left(  m\right)  },\delta}</math>, <math>\hat{\Pi}
_{\rho_{X^{n}\left(  m-1\right)  },\delta}</math>, ..., <math>\hat{\Pi}_{\rho
_{X^{n}\left(  1\right)  },\delta}</math>. This gives the upper bound
<math>
\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\}  .
</math>
Due to concavity of the square root, we can bound this expression from above
by
:<math>
 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}</math>
:<math> \leq2\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},
</math>
where the second bound follows by summing over all of the codewords not equal
to the <math>m^{\text{th}}</math> 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:
:<math>
 \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\}  </math>
:<math> \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\}  </math>
:<math> \leq\epsilon+2\sqrt{\epsilon}.
</math>
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:
:<math>
  \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\}  </math>
:<math>  =\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\}  </math>
:<math> =\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\}  </math>
:<math>  \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\}
</math>
The first equality follows because the codewords <math>X^{n}\left(  m\right)  </math> and
<math>X^{n}\left(  i\right)  </math> 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
:<math>
  \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\}  </math>
:<math>  \leq\sum_{i\neq m}2^{-n\left[  H\left(  B\right)  -\delta\right]
}\ 2^{n\left[  H\left(  B|X\right)  +\delta\right]  }</math>
:<math>  =\sum_{i\neq m}2^{-n\left[  I\left(  X;B\right)  -2\delta\right]  }</math>
:<math>  \leq M\ 2^{-n\left[  I\left(  X;B\right)  -2\delta\right]  }.
</math>
The first inequality follows from <math>\Pi_{\rho,\delta}^{n}\leq I</math> 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:
:<math>
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\}  </math>
:<math>\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}.
</math>
Thus, as long as we choose <math>M=2^{n\left[  I\left(  X;B\right)  -3\delta
\right]  }</math>, there exists a code with vanishing error probability.

== See also ==

* [[Entanglement-assisted classical capacity]]
* [[Quantum capacity]]
* [[Quantum information theory]]
* [[Typical subspace]]

== References ==
*{{citation|first=Alexander S.|last=Holevo|authorlink=Alexander Holevo|arxiv=quant-ph/9611023|title=The Capacity of Quantum Channel with General Signal States|year=1998|doi=10.1109/18.651037|journal=IEEE Transactions on Information Theory|volume=44|issue=1|pages=269–273}}.
*{{citation|first1=Benjamin|last1=Schumacher|first2=Michael|last2=Westmoreland|doi=10.1103/PhysRevA.56.131|title=Sending classical information via noisy quantum channels|journal=Phys. Rev. A|volume=56|pages=131–138|year=1997|issue=1|bibcode = 1997PhRvA..56..131S }}.
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{{Quantum computing}}

[[Category:Quantum information theory]]
[[Category:Limits of computation]]