Asymptotical Information Bound of Consecutive Qubit Binary Testing

Abstract

The problem of estimating quantity of information, which can be obtained in the process of binary consecutive measuring a qubit state, is considered in the paper. The studied quantity is expressed as Shannon mutual information between the initial qubit state and the outcomes of a consecutive measurement. It is demonstrated that the maximum of information is reached by projective measurements. The maximum quantity of accessible information is calculated. It is proved that in the case of arbitrary binary test the maximum is reached asymptotically for consecutive measurements.

Full text

Asymptotical Information Bound of Consecutive
Qubit Binary Testing
Anastasiia Varava and Grygoriy Zholtkevych
V.N. Karazin Kharkiv National University
School of Mathematics and Mechanics, 4, Svobody Sqr., 61022, Kharkiv, Ukraine
{nastia.varava,g.zholtkevych}@gmail.com
Abstract. The problem of estimating quantity of information, which
can be obtained in the process of binary consecutive measuring a qubit
state, is considered in the paper. The studied quantity is expressed as
Shannon mutual information between the initial qubit state and the
outcomes of a consecutive measurement. It is demonstrated that
the maximum of information is reached by projective measurements. The
maximum quantity of accessible information is calculated. It is proved
that in the case of arbitrary binary test the maximum is reached asymp-
totically for consecutive measurements.
Keywords: Quantum information theory, quantum communication
channels, accessible classical information, Shannon mutual information,
pure qubit state, consecutive qubit binary testing.
1 Introduction
The theory of quantum information is a generalization of the classical infor-
mation theory to the quantum world. Quantum information theory aimes to
investigate what happens if information is stored in a state of a quantum sys-
tem. By state of a physical system one means the mathematical description that
provides a complete information on the system.
Talking about information from the point of view of quantum mechanics, we
take into account some specific consequences of fundamental quantum postu-
lates. The most important of them are the following: firstly, information, that
is held in a state of a quantum system, cannot be read without changing the
initial state. This property makes it largely inaccessible. Secondly, an arbitrary
quantum state can not be cloned because of the no-cloning theorem. In other
words, one can not create a perfect copy of the quantum system without already
knowing its state in advance. And finally, quantum system can embody more
information than classical, because a qubit can be in a superposition of basis
state at the same time.
Industrial applications of quantum communication channels require develop-
ing sound methods for constructing them. It is well known that Shannon infor-
mation theory is a mathematical background for such methods and the notion of
Shannon mutual information [1,10] is a basic instrument for the classical theory
V. Ermolayev et al. (Eds.): ICTERI 2013, CCIS 412, pp. 93–111, 2013.
c
Springer International Publishing Switzerland 2013

94 A. Varava and G. Zholtkevych
of communication. Thus, studying informational quantities for quantum commu-
nication channels is a very important problem for building quantum information
theory.
One of the well-known problems in this context is the problem of obtaining
information on a quantum state. Suppose that one experementalist prepares an
individual particle in a particular physical state and then sends it to another
experementalist, who is not aware of the preparation procedure, but wishes to
determine the state of the particle. By making observations on the received
particle, the second experementalist can obtain information about the physical
state by observing how it interacts with other wellcharacterized systems, such
as a measuring instrument. The amount of this information depends strongly on
whether the particle is macroscopic or microscopic. In the macroscopic, classical
case, one can observe the particle without disturbing it, and determine its state.
In quantum mechanics, on the contrary, it is impossible to learn the quantum
state of any individual physical system, because each observation will disturb
its state. The amount of the achievable information on the quantum system will
never be sufficient to determine the given state accurately. This is, in particu-
lar, the basis of quantum key distribution for cryptography. However, effective
techniques of obtaining information on the quantum state are quite important
for different applications. Such techniques can be used for characterizing optical
signals, as well as in quantum computing and quantum information theory to
estimate the actual states of the qubits.
Thus, the amount of accessible information on the quantum system is an
important information-theoretic quantity. To study the possibility of obtaining
information it is convenient to express it as Shannon mutual information between
the qubit state and the outcomes of a measurement.
One of the first investigated problems with respect to the discussed quantity is
the so-called problem of distinguishing quantum states. It consists of the follow-
ing: suppose that quantum system is prepared in a state described by one of the
density operators ρi(i=1, ..., n) with probability pi. The goal is to determine
which state is given using a single measurement of the considered quantum sys-
tem. Thus, the task is to find measurements providing the maximum of Shannon
mutual information. This problem was investigated by Holevo [5], Davies [3] and
many others. Particularly, Holevo proved in [5] that obtainable information Iis
less or equal than the so-called Holevo bound
I≤S
i
piρi−
i
piS(ρi),
where S(ρ)=Tr(ρlog ρ) is the von Neumann entropy. It follows that the amount
of information obtainable by a single measurement never exceeds log(dim H),
where His a state space of a quantum system. This result is fundamental for
the above mentioned problem. The work on this question was then continued
by many reserchers. In particular, in [2] its extension to sequential measure-
ments based on the properties of quantum conditional and mutual entropies was
presented.

Asymptotical Information Bound of Consecutive Qubit Binary Testing 95
A more general question, the problem of estimating the amount of accessible
information about quantum states, was also studied by several researchers. In
particular, in [11] the brief review of obtained results is given. In the chapter
we continue this investigation by considering a particular case. Imagine that we
do not have any preliminary information on the possible states of a quantum
system. In this case all possible pure states are equiprobable, so, we can not work
with a finite set of initial states. Assume also that we have in our disposal a single
measuring instrument. How much classical information can we now obtain?
In the paper we consider the simplest case of this problem. Suppose we have
an arbitrary pure qubit state and an instrument doing binary test of it. Assume
that all possible initial states are equiprobable. Our aim is to obtain classical
information on the qubit state using only the given measuring instrument. Cer-
tainly, we want to get as much classical information as possible. So, this time we
investigate the amount of information accessible in this case.
It is well known that in the case of performing a single measurement the
maximum of classical information is obtainable by a projective measurement.
We present a simple proof of this fact. In addition, we calculate the maximum
value of information and then we show that in the case of arbitrary measuring
instrument this value can be attained asymptotically using consecutive testing.
This paper is organized as follows: in Section 2 the problem is set formally
and the formula for Shannon mutual information of two random variables is
obtained. The first one describes the density of the parameter of the initial
state, and the second one corresponds to the measurement result. In Section 3 a
special kind of measuring instrument which performs a pro jective measurement
is considered. It is demonstrated that in this case we obtain the maximum of
accessible information. Further, in Section 4, it is proved that in general case
consecutive measurements provide to attain this value asymptotically.
2 Qubits and Their Binary Testing
In quantum computing, a qubit is the quantum analogue of the classical bit.
Like a bit, a qubit has two basis states: |0and |1. The difference is that a qubit
can be in a superposition of the basis states at the same time. This means that
a pure qubit state can be represented as a linear combination of |0and |1:
|ψ=a|0+b|1,
where aand bare probability amplitudes and can in general both be complex
numbers such that |a|2+|b|2=1.
More precisely, a pure state of a qubit is a unit vector |ψin 2-dimensional
Hilbert space H2over the field of complex number. As usual, a density matrix
equalled the ortho-projector on the subspace C·|ψof H2is used to represent
the pure state described by the vector |ψ. This ortho-projector is denoted by
|ψψ|. Hence, one can get

96 A. Varava and G. Zholtkevych
|ψψ|=(a|0+b|1)(a0|+b1|)=|a|2|00|+ab |01|+ab |10|+|b|2|11|.
If we denote |a|2and |b|2by pand qrespectively then we can rewrite the previous
formula in the following form
|ψψ|=p|00|+eiα√pq |01|+e−iα √pq |10|+q|11|,
where p+q=1,0≤α<2π.
Now using the representation p=1
2−θ,q=1
2+θwhere −1
2≤θ≤1
2one can
identify |ψψ|with the matrix
ρ(θ, α)=⎛
⎜
⎜
⎝
1
2−θe
iα1
4−θ2
e−iα1
4−θ21
2+θ
⎞
⎟
⎟
⎠
.(1)
This matrix is called a density operator of the pure state.
In the general case a density operator for a qubit is a nonnegative definite
operator on the space H2such that its trace equals unity.
A binary test of a qubit is determined by a pair of operators {K(0),K(1)}
such that the equality K(0)†K(0) + K(1)†K(1) = 1in the following meaning:
Pr(0 |ψ)= ψ|K(0)†K(0)|ψand Pr(1 |ψ)= ψ|K(1)†K(1)|ψ;
|Eff(ψ|0)=K(0)|ψ
ψ|K(0)†K(0)|ψand |Eff(ψ|1)=K(1)|ψ
ψ|K(1)†K(1)|ψ,
where |ψdescribes the qubit state immediately before testing, Pr(s|ψ)isequal
to obtain the outcome sas testing result, |Eff(ψ|s)describes the qubit state
immediately after testing if the testing outcome equals s.
As known from theorem about polar decomposition [9, p. 28, Theorem 2.3]
operators K(0) and K(1) can be represented as
K(0) = U(0)M(0)1/2and K(1) = U(1)M(1)1/2,(2)
where M(0) = K(0)†K(0) and M(1) = K(1)†K(1) , U(0) and U(1) are unitary
operators .
In this chapter we consider a binary tests subclass only. Its members are
characterised by the condition U(0) = U(1) = 1. Such a qubit binary test has
studied in [6].
Thus, let us introduce the following definition.
Definition 1. A pair of nonnegative definite operators T={M0,M
1}on H2
such that M0+M1=1is cal led a qu bit binary test.
Probabilities of test outcomes are given by the next formulae
Pr(s|ψ)=ψ|M(s)|ψ,where s=0,1and |ψdescribes the state
of a qubit immediately before testing. (3)

Asymptotical Information Bound of Consecutive Qubit Binary Testing 97
The vector described the qubit state immediately after testing are given by the
next formulae
|Eff(ψ|s)=M(s)1/2|ψ
Pr(s|ψ),where s=0,1and |ψdescribes the state
of a qubit immediately before testing.
(4)
The mathematical model of a qubit binary test is described and studied in [12].
3 Shannon Mutual Information between the Qubit State
and the Outcomes of a Consecutive Measurement
To set the problem formally we firstly need some basic definitions. We use a
density operator as a mathematical model of a qubit state and describe the
considering pure state of a qubit by the corresponding density operator ρ(θ, α)
represented by the matrix ρ(θ, α) (see formula (1)).
Let Θbe the equiprobability distribution on the segment −1
2,1
2.Itisused
to model uncertainty of our knowledge about parameter θfor an initial state of
the qubit.
Let Mbe a set of all possible qubit binary tests. Consider T∈M,T=
{M0,M
1}.Let{|0,|1} be the ortho-normal basis in H2such that |0,|1are
eigenvectors of the operator M0corresponding to eigenvalues 0 ≤m1≤m2≤1
respectively. In this case the operators M0and M1are represented in a such
way:
M0=m10
0m2,M
1=1−m10
01−m2.
Denote by Σ={0,1}the set of outcomes for the given qubit binary test. The
probability distribution on this set is defined by the following formulae:
Pr(0 |ρ)=Tr(ρ·M0),Pr(1 |ρ)=Tr(ρ·M1).
As we already mentioned, our aim is to obtain classical information value of
the initial state as much as possible, using only the given instrument to measure.
For this purpose we perform consecutive qubit binary testing. In other words, we
repeat our measurement ntimes to observe the result of each iteration. After n
iterations we obtain a sequence x(n)=(x1,...,x
n), where xi∈Σ(i=1, ..., n).
As it is shown in [12],
Pr(x(n)|ρ(θ, α)) = mk
1(1 −m1)n−k1
2−θ+mk
2(1 −m2)n−k1
2+θ,
where kis a number of ’1’ in the outcomes sequence x(n).
In further computations it is suitable to use some properties of Bernstein basis
polynomials of degree n[8]:
Bn,k(z)=n
kzk(1 −z)n−k,

98 A. Varava and G. Zholtkevych
so, we rewrite the last equation as follows:
Pr(x(n)|ρ(θ, α)) = n
k−11
2−θBn,k(m1)+ 1
2+θBn,k(m2).
Let X(n)∈{0,1}nbe a random variable describing an outcomes sequence.
The density of X(n)with respect to Θis defined by the following formula
p(x(n)|θ)=Pr(x(n)|ρ(θ, α)).
The main objective of this section is to compute Shannon mutual information
[1,10] of random variables X(n)and Θ.Itisequalto
I(X(n);Θ)=H(X(n))−H(X(n)|Θ),(5)
where H(X(n)) is Shannon entropy of X(n),andH(X(n)|Θ) is conditional en-
tropy [1,10]:
H(X(n)|Θ)=1
2
−1
2
H(X(n)|Θ=θ)dθ .
Let us compute the marginal distribution of the random variable X(n),which
corresponds to the joint density function p(x(n),θ). The latter is described in a
such way:
p(x(n),θ)=n
k−11
2−θBn,k(m1)+1
2+θBn,k(m2).
The marginal distribution p(x(n)) is defined as follows:
p(x(n))=1
2
−1
2
p(x(n),θ)dθ =n
k−1Bn,k(m1)+Bn,k (m2)
21
2
−1
2
dθ−
(Bn,k(m1)−Bn,k (m2)) 1
2
−1
2
θdθ=n
k−1Bn,k(m1)+Bn,k (m2)
2.
Firstly, let us compute entropy of X(n):
H(X(n))=−
x(n)∈{0,1}n
p(x(n))logp(x(n))=−
n

k=0 n
kp(x(n))logp(x(n)).
Substituting the expression of marginal distribution, we obtain
H(X(n))=
n

k=0 Bn,k(m1)+Bn,k (m2)
2log n
k−
n

k=0 Bn,k(m1)+Bn,k (m2)
2·log Bn,k(m1)+Bn,k(m2)
2.(6)

Asymptotical Information Bound of Consecutive Qubit Binary Testing 99
Secondly, we can compute H(X(n)|Θ):
H(X(n)|Θ)=−1
2
−1
2
H(X(n)|Θ=θ)dθ =−1
2
−1
2
n

k=0 n
kp(x(n)|θ)logp(x(n)|θ)dθ .
By direct calculations it is easy to check that
H(X(n)|Θ)=
n

k=0 Bn,k(m1)+Bn,k (m2)
2log n
k−
n

k=0 1
2
−1
21
2−θBn,k(m1)+1
2+θBn,k(m2)·
log 1
2−θBn,k(m1)+1
2+θBn,k(m2)dθ . (7)
Now, using (5), (6) and (7), one can obtain the expression for Shannon mutual
information
I(X(n);Θ)=
n

k=01
2
−1
21
2−θBn,k(m1)+1
2+θBn,k(m2)·
log 1
2−θBn,k(m1)+1
2+θBn,k(m2)dθ−
Bn,k(m1)+Bn,k (m2)
2·log Bn,k(m1)+Bn,k(m2)
2.
It is easy to see that if kis such that Bn,k(m1)=Bn,k (m2), then the corre-
sponding summand is equal to zero:
1
2
−1
2
(Bn,k(m1)·log Bn,k(m1)) dθ −Bn,k (m1)·log Bn,k (m1)=0.
Note that in the case of equality m1=m2the polynomials Bn,k (m1)and
Bn,k(m2) are equal for each k, so, we can not obtain any information about
the initial state. Thus we futher consider the case m1<m
2.
Let Anbe a set of non-negative integers defined as follows:
An={k∈{0, ..., n}|Bn,k(m1)=Bn,k (m2)}.
Denote by Anthe complement of the set An, i.e. An={0, ..., n}\An.Now
we can consider only values of kfrom this set, but we keep on working with all
summands. We demonstrate further that it is more suitable. Instead of omitting
zero summands, let us present them in the following way:

100 A. Varava and G. Zholtkevych
0=Bn,k(m1)−Bn,k(m1)
2ln(2) −2ln2−1
2ln2 ·Bn,k (m1).
Thus, we have
I(X(n);Θ)= 
k∈An1
2
−1
21
2−θBn,k(m1)+1
2+θBn,k(m2)·
log 1
2−θBn,k(m1)+1
2+θBn,k(m2)dθ−
Bn,k(m1)+Bn,k (m2)
2·log Bn,k(m1)+Bn,k(m2)
2+

k∈AnBn,k(m1)−Bn,k (m1)
2ln(2) −2ln2−1
2ln2 ·Bn,k (m1).
After integration we obtain
I(X(n);Θ)= 
k∈An−Bn,k(m1)+Bn,k (m2)
4ln2 +
Bn,k(m2)2log Bn,k(m2)−Bn,k (m1)2log Bn,k (m1)
2(Bn,k(m2)−Bn,k (m1)) −
Bn,k(m1)+Bn,k (m2)
2·log Bn,k(m1)+Bn,k(m2)
2+

k∈AnBn,k(m1)−Bn,k (m1)
2ln(2) −2ln2−1
2ln2 ·Bn,k (m1).
Taking into account that for kfrom the set Anthe equality Bn,k (m1)=
Bn,k(m2) is held, we can rewrite this formula as follows:
I(X(n);Θ)=
n

k=0
2ln2−1
4ln2 (Bn,k (m1)+Bn,k (m2))−
k∈An
2ln2−1
2ln2 Bn,k (m1)+

k∈AnBn,k(m2)2log Bn,k (m2)−Bn,k (m1)2log Bn,k (m1)
2(Bn,k(m2)−Bn,k (m1)) −
(Bn,k(m1)+Bn,k(m2)) ·log (Bn,k (m1)+Bn,k (m2))
2.

Asymptotical Information Bound of Consecutive Qubit Binary Testing 101
Simplifying this expression and using the evident equality,
n

k=0
Bn,k(z)=1,
we get the following expression for mutual information:
I(X(n);Θ)=2ln2−1
2ln2 −
k∈An
2ln2−1
2ln2 Bn,k (m1)+
1
2·
k∈AnBn,k(m2)2log Bn,k (m2)−Bn,k (m1)2log Bn,k (m1)
(Bn,k(m2)−Bn,k (m1)) −
(Bn,k(m2)2−Bn,k (m1)2)·log (Bn,k (m1)+Bn,k (m2))
(Bn,k(m2)−Bn,k (m1)) .(8)
4 Extremal Property of Projective Measurements
In this section we consider a special kind of measurement. Let m1=0and
m2= 1. In this case the qubit binary test T={M0,M
1}looks as follows:
M0=00
01
,M
1=10
00
.
The first interesting property of this measurement is its repeatability. Actu-
ally, note that Mi·Mj=δi,j ·Mi.So,M0and M1are orthoprojectors, and T
is a projective measurement. Thus, in this case the repeated measurements give
the same result. We demonstrate further that due to this property consecutive
testing do not provide any extra information. The second and the most remark-
able property is that we can obtain the maximum of the accessible information
if and only if we use a projective measurement.
First of all let us compute the amount of information obtainable by the con-
sidering measurement. In the considered case we have only two values of ksuch
that the corresponding polymomials Bn,k(m1)andBn,k (m2) are not equal. So,
An={0,n},and
∀k∈An:Bn,k(m1)=Bn,k (m2)=0.
Now using (8) it is easy to see that considering m1=0andm2=1we obtain
I(X(n);Θ)=2ln2−1
2ln2 .
Our next goal is to show that the obtained amount of information can not be
reached using any other qubit binary test. For this purpose we investigate the
function describing the accessible information (8).
At first let us rewrite this function in a more suitable way. Consider the general
case, in which 0 <m
1<m
2<1. We will demonstrate futher that the exceptional
cases, when 0 = m1<m
2<1or0<m
1<m
2= 1, are similar to the latter.

102 A. Varava and G. Zholtkevych
Taking into account that in the general case ∀k∈{0, ..., n}Bn,k(m1)>0, let
us denote the ratio between Bn,k(m2)andBn,k (m1) by a new function:
tn,k(m1,m
2)=Bn,k (m2)
Bn,k(m1).
Thus, by direct calculations we obtain the following formula for mutual
information:
I(X(n);Θ)=2ln2−1
2ln2 −2ln2−1
2ln2 
k∈An
Bn,k(m1)−1
2
k∈An
Bn,k(m1)·
⎛
⎜
⎜
⎝
tn,k(m1,m
2)2·log tn,k(m1,m
2)
tn,k(m1,m
2)+1+log(tn,k(m1,m
2)+1)
1−tn,k(m1,m
2)⎞
⎟
⎟
⎠
.(9)
Let f(t) be a function defined as follows:
f(t)=
t2·log t
t+1+ log(t+1)
1−t.
Now it is easy to see that the formula (9) can be written as
I(X(n);Θ)=2ln2−1
2ln2 −⎛
⎝
2ln2−1
2ln2 
k∈An
Bn,k(m1)+
1
2
k∈An
Bn,k(m1)·f(tn,k (m1,m
2)).(10)
We claim that in the considered case (0 <m
1<m
2<1) mutual information
is less than 2ln2−1
2ln2 .
To prove this fact it is enough to notice that the variable part of expression (10)
is always negative. Actually, we know that Bn,k (m1)>0andtn,k (m1,m
2)isa
ratio of two positive values, so, it is also positive. In addition, it is easy to show
in the classical way that for all t∈(0,1) ∪(1,∞) function f(t) is greater than
zero.
Now we need to consider the special case, in which 0 <m
1<1,m
2=1.Itis
evident that the case when m1=0,0<m
2<1 is similar to the latter.
Suppose that 0 <m
1<1,m
2= 1. Thus, on the one hand, for all kwe have
Bn,k(m1)>0. On the other hand, for all k<nB
n,k(m2) = 0, and only for
k=nB
n,k(m2) = 1. It is easy to see that in this case we obtain

Asymptotical Information Bound of Consecutive Qubit Binary Testing 103
I(X(n);Θ)=2ln2−1
2ln2 +
1
2·Bn,n(m2)2log Bn,n (m2)−(Bn,n(m2)2−1) ·log (1 + Bn,n (m2))
Bn,n(m2)−1=
2ln2−1
2ln2 −1
2·f(Bn,n(m2)) <2ln2−1
2ln2 .
Thus, now we know that
I(X(n);Θ)≤2ln2−1
2ln2 ,
and, in particular, the equality is held if and only if the considered binary test
is projective.
5 Asymptotic Properties of Consecutive Measurements
Intheprevioussectionwehaveconsidered the case when the given qubit binary
test is a projective measurement. We have proved that only this type of measure-
ment allows to achieve the maximum of information about the initial state. As
far as the measurement is projective, repeating of the measuring procedure does
not provide any extra information. In addition, we have found the maximum
value of the accessible information:
max
T∈M,n∈N{I(X(n);Θ)}=2ln2−1
2ln2 .
In this section we return to considering of the general view of a qubit test,
and we work with consecutive qubit testing. So, this time we investigate the
dependence of the amount of information on n– the number of iterations. The
objective of this section is to prove that the maximum of accessible information
can be reached asymptotically by performing consecutive measurements using
an arbitrary qubit binary test.
More strictly, our aim is to prove the next theorem:
Theorem 1. Suppose we have a pure qubit state and we perform consecutive
qubit binary testing using the given test T={M1,M
2}. Then for arbitrary
ε>0there exists a corresponding number of iterations n(ε)such that for all
subsequent iterations (n>n(ε)) the following inequality is held:
max
T∈M,m∈N{I(X(m);Θ)}−I(X(n);Θ)<ε.
In other words, as far as the mutual information can be written as
I(X(n);Θ)=2ln2−1
2ln2 −⎛
⎝
2ln2−1
2ln2 
k∈An
Bn,k(m1)+
1
2
k∈An
Bn,k(m1)·f(tn,k (m1,m
2)),

104 A. Varava and G. Zholtkevych
we need to find n1(ε) such that for all n>n
1(ε)

k∈An
Bn,k(m1)·f(tn,k (m1,m
2)) <ε
2,(11)
and n2(ε) such that for all n>n
2(ε)

k∈An
Bn,k(m1)<ε
2.(12)
Therefore, for n>n(ε)=max{n1(ε),n
2(ε)}both of these inequalities are held.
Let us fix a certain positive value of ε. At first we consider the left side of
inequality (11). Let us divide the set Aninto two non-intersecting subsets:
Γn(m1)={k∈An}:
k
n−m1<˜
δ(m1,m
2),
Δn(m1)={k∈An}:
k
n−m1≥˜
δ(m1,m
2),
where ˜
δ(m1,m
2) is a certain positive function.
It was demonstrated in [8], that for 0 <m
1<1andδ>0:

k∈Δn(m1)
Bn,k(m1)≤1
4nδ2.(13)
On the one hand, it is easy to see that f(t) is a bounded function. Suppose
that for all t∈(0,1) ∪(1,∞)f(t)<C,whereCis a certain positive constant.
Thus we have

k∈Δn(m1)
Bn,k(m1)·f(tn,k (m1,m
2)) ≤C
4n˜
δ2(m1,m
2).
So, we can choose a value n1,1(ε) such that for all n>n
1,1(ε)

k∈Δn(m1)
Bn,k(m1)·f(tn,k (m1,m
2)) <ε
4.
On the other hand, we can see that when kis close to n·m1, the value of
tn,k(m1,m
2) goes to zero as ngoes to infinity. As far as lim
t→0f(t)=0,there exists
avaluen1,2(ε) such that for all n>n
1,2(ε)

k∈Γn(m1)
Bn,k(m1)·f(tn,k (m1,m
2)) <ε
4.
Now let n1(ε)beamaximumofvaluesn1,1(ε)andn1,2(ε). Thus, for all n>n
1(ε)
inequality (11) is held.
Finally, let us consider inequality (12). Note that in the case of inequality
m1<m
2the set Ancontains at most one element. Actually, by construction

Asymptotical Information Bound of Consecutive Qubit Binary Testing 105
k∈Anif and only if Bn,k (m1)=Bn,k (m2). Solving this equation for the variable
k,wehave
k0=n·
ln 1−m1
1−m2
ln (1 −m1)·m2
(1 −m2)·m1
If n, m1and m2are such that k0is an integer then An={k0}.Ifnot,theset
Anis empty. It is easy to show that lim
n→∞ Bn,k(m1)=0,so, we can easily find
n2(ε) such that for all n>n
2(ε)inequality(12)isheld.
Now let us build a rigorous proof of the considering statement using this
heuristic consideration. To do it, we firstly need several trivial propositions.
Proposition 1. The following statements are correct:
1. for all x∈(0,1) the inequality
x>ln (x+1)
is true;
2. for all x>0the inequality
ln x
x+1<−1
x+1
is true too.
Proof. Consider the first statement. One can directly obtain it using the Tailor
series expansion of ln (x+1):
ln (x+1)=
∞

n=1
(−1)n+1
nxn=x+
∞

n=1
x2nx
2n+1−1
2n<
x+
∞

n=1
x2n1
2n+1−1
2n<x
Now let us prove the second statement. By using Euler’s transform on Tailor
series expansion shown above, we have
ln ˜x
˜x−1=
∞

n=1
1
n˜xn>1
˜x
for every |˜x|>1.
Let ˜x=x+1, then |˜x|>1and
ln x+1
x>1
x+1,
so,
ln x
x+1 <−1
x+1.


106 A. Varava and G. Zholtkevych
Proposition 2. Let x, y ∈(0,1) and x=ythen the following inequality is held:
x
yy1−x
1−y(1−y)
<1.
Proof. Let g(x)=xc·(1 −x)(1−c)be a function of x∈(0,1), where c∈(0,1) is
a certain constant. It is easy to prove in a standart way that
max
x∈(0,1) g(x)=g(c)
Let c=y.Thenforx=y:
xy·(1 −x)(1−y)<y
y·(1 −y)(1−y),
so,
x
yy1−x
1−y(1−y)
<1.

Now we can prove the above formulated theorem.
Proof (of Theorem 1). As we already know, the mutual information can be
presented as
I(X(n);Θ)=2ln2−1
2ln2 −⎛
⎝
2ln2−1
2ln2 
k∈An
Bn,k(m1)+
1
2
k∈An
Bn,k(m1)·f(tn,k (m1,m
2)).
We also know that
max
T∈M,n∈N{I(X(n);Θ)}=2ln2−1
2ln2 .
To prove the theorem it is enough to show that there exists n(ε) such that for
all n>n(ε)

k∈An
Bn,k(m1)·f(tn,k (m1,m
2)) + 
k∈An
Bn,k(m1)<ε.
Consider an arbitrary ε>0. Let us divide the set Aninto subsets Γn(m1)
and Δn(m1) in the following way:
Γn(m1)={k∈An}:
k
n−m1<˜
δ(m1,m
2),
Δn(m1)={k∈An}:
k
n−m1≥˜
δ(m1,m
2),

Asymptotical Information Bound of Consecutive Qubit Binary Testing 107
where ˜
δ(m1,m
2) is a certain positive function of m1and m2defined further.
Our aim is to prove that there exists such n(ε)thatforalln>n(ε):

k∈Δn(m1)
Bn,k(m1)·f(tn,k (m1,m
2)) <ε
4,(14)

k∈Γn(m1)
Bn,k(m1)·f(tn,k (m1,m
2)) <ε
4,(15)

k∈An
Bn,k(m1)<ε
2.(16)
Firstly, let us consider inequality (14). We had already mentioned that for all
t∈(0,1) ∪(1,∞)f(t)≥0, and it is easy to see that for considered values of t
f(t)<2 . So, using the above mentioned property of Bernstein basis polynomials
(13), we have:

k∈Δn(m1)
Bn,k(m1)·f(tn,k (m1,m
2)) ≤1
2n˜
δ2(m1,m
2).
Let n1,1(ε) be sufficiently great. Then for all n>n
1,1(ε) the considering
inequality (14) is held.
Secondly, let us consider inequality (15). On the one hand, it is not hard to
find such δ(ε)that
∀t∈(0,δ(ε)) : f(t)=
t2·log t
t+1+ log(t+1)
1−t<ε
4.
On the other hand, for sufficiently great values of nfor all k∈Γn(m1)wehave
tn,k(m1,m
2)<δ(ε). It follows that for great values of nwe obtain

k∈Γn(m1)
Bn,k(m1)·f(tn,k (m1,m
2)) <ε
4·
n

k=0
Bn,k(m1)= ε
4.
Let δ(ε)=min√1+(ε/2·ln(2))2−1
ε/2·ln(2) ;1
2.Thenfort∈(0,δ(ε)) we have
t<ε·ln(2)
4(1 −t2).
As far as 0 <t≤1
2<1,we obtain
t−t2
t+1
(1 −t)·ln(2) =t
(1 −t2)·ln(2) <ε
4.

108 A. Varava and G. Zholtkevych
If we combine this with Proposition 1, then for all t∈(0,δ(ε)) we have
f(t)=
t2·ln t
t+1+ln(t+1)
(1 −t)·ln(2) <
t+t2·−1
t+1 
(1 −t)·ln(2) =t−t2
t+1
(1 −t)·ln(2) <ε
4.
Now we need to find such n1,2(ε)thatforallk∈Γn(m1):tn,k(m1,m
2)<δ(ε).
By definition,
tn,k(m1,m
2)=m2
m1k
·1−m2
1−m1n−k
.
As far as m2
m1
>1, tn,k(m1,m
2) strictly increases with respect to k.So,if

k
n−m1<˜
δ(m1,m
2)then
tn,k(m1,m
2)<m2
m1m1+˜
δ(m1,m2)
·1−m2
1−m11−m1−˜
δ(m1,m2)n
.
Consider the right side of this inequality. Note that
m2
m1m1+˜
δ(m1,m2)
·1−m2
1−m11−m1−˜
δ(m1,m2)
srtictly increases with respect to ˜
δ(m1,m
2). It is equal to 1 when ˜
δ(m1,m
2)=
˜
δ∗(m1,m
2), where
˜
δ∗(m1,m
2)=⎛
⎜
⎜
⎝
ln 1−m2
1−m1
ln (1 −m2)·m1
(1 −m1)·m2−m1⎞
⎟
⎟
⎠
.
AccordingtoProposition2,
m2
m1m1
·1−m2
1−m11−m1
<1,
we see that for ˜
δ(m1,m
2)∈(0,˜
δ∗(m1,m
2)) we have
m2
m1m1+˜
δ(m1,m2)
·1−m2
1−m11−m1−˜
δ(m1,m2)
<1.
Let
˜
δ(m1,m
2)= ˜
δ∗(m1,m
2)
2.
Now it is easy to put n1,2(ε) such that for all n>n
1,2(ε) inequality (15) is held.

Asymptotical Information Bound of Consecutive Qubit Binary Testing 109
Finally, let us find such n2(ε)thatforn>n
2(ε) condition (16) is satisfied.
We have already seen that |An|≤1, and the equality is held when
k0=n·
ln 1−m1
1−m2
ln (1 −m1)·m2
(1 −m2)·m1
is an integer. Let us denote the right side as n·c(m1,m
2) and write k0as
k0=n·c(m1,m
2).
Referring to the standard way of proving the Stirling’s formula, we can write
the following inequality:
√2πn ·n
en≤n!≤e·√n·n
en
.
It follows that
n
k≤e
2πn
k(n−k)·n
kk·n
n−kn−k
.
So, it is now easy to see that
Bn,k(m1)≤e
2πn
k(n−k)·n·m1
kk·n·(1 −m1)
n−kn−k
.
Substituting k0=n·c(m1,m
2)forkin the last inequality, we get
Bn,k0(m1)≤e
2π1
n·c(m1,m
2)(1 −c(m1,m
2))·
m1
c(m1,m
2)c(m1,m2)
·1−m1
1−c(m1,m
2)1−c(m1,m2)n
.(17)
It follows from Proposition 2 that
m1
c(m1,m
2)c(m1,m2)
·1−m1
1−c(m1,m
2)1−c(m1,m2)
<1.
Now using (17) it is not hard to put n2(ε) such great that for n>n
2(ε)
inequality (16) is held.
Finally, let n(ε)=max{n1,1(ε),n
1,2(ε),n
2(ε)}.Nowforalln>n(ε)
max
T∈M,m∈N{I(X(m);Θ)}−I(X(n);Θ)<ε.
The theorem is proved. 

110 A. Varava and G. Zholtkevych
6 Conclusions
In the paper the problem of obtaining classical information about the pure qubit
state using a single qubit binary test has been considered. It has been demon-
strated that the maximum of information is reached if and only if the using mea-
surement is projective. The maximum value of information has been calculated:
max
T∈M,n∈NI(X(n);Θ)=2ln2−1
2ln2 .
It follows, in particular, that to distinguish two arbitrary pure qubit states using
a single binary test it is necessary to have at least four pairs of qubits prepared
in the considered states.
It has been shown that the maximum of reachable information can be attained
asymptotically using an arbitrary consecutive qubit binary test. Thus, if we have
a certain measuring instrument performing a qubit binary test, we can obtain
an amount of information arbitrary close to the maximum.
As known [4,7], Yu. Manin and R. Feynman proposed to use quantum systems
to simulate others quantum systems. The results obtained in the paper show that
this idea should be refined: one should take into account all dependences between
an input data, a behaviour of a simulating system, and a structure of an output
data.
Our further research will deal with generalizing results of the paper for the
case of an n-level quantum system and a measurement with moutcomes.
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