Abstract

The scheme for teleporting an unknown state without using any secredt keythrough th use of entanglement pair as a quantum channel is used.

Contents

1 Review of quantum Mechanics 31.1 Historical Note . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 Postulates of Quantum Mechanics . . . . . . . . . . . . . . . . 61.3 Vector space And The Hibert Space . . . . . . . . . . . . . . . 7

1.3.1 The linear Vector Space . . . . . . . . . . . . . . . . . 71.3.2 The Hilbert Space . . . . . . . . . . . . . . . . . . . . 8

1.4 Linear Dependence and Independence of Vectors . . . . . . . . 91.5 Dirac Notation . . . . . . . . . . . . . . . . . . . . . . . . . . 91.6 Inner Product . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.7 Outer Product . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.8 Norm of a Vector . . . . . . . . . . . . . . . . . . . . . . . . . 121.9 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.10 Orthonormality . . . . . . . . . . . . . . . . . . . . . . . . . . 121.11 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.12 Hermitian Adjoint . . . . . . . . . . . . . . . . . . . . . . . . 14

1.12.1 Hermitian and Skew-hermitian Operators . . . . . . . . 151.13 Projection Operators . . . . . . . . . . . . . . . . . . . . . . . 15

1.13.1 Properties of Projection Operator . . . . . . . . . . . . 161.14 Inverse of an Operator . . . . . . . . . . . . . . . . . . . . . . 161.15 Unitary Operators . . . . . . . . . . . . . . . . . . . . . . . . 171.16 Eigne Values and Eigen Vectors of an Operator . . . . . . . . 17

2 Concepts of Quantum Computing 182.1 Qubit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.2 Pauli Operators . . . . . . . . . . . . . . . . . . . . . . . . . . 192.3 Trace of an Operator . . . . . . . . . . . . . . . . . . . . . . . 202.4 Unitary Transformation . . . . . . . . . . . . . . . . . . . . . 212.5 Pure and Mixed States . . . . . . . . . . . . . . . . . . . . . . 22

1

2.6 Density Matrix Formulism . . . . . . . . . . . . . . . . . . . . 222.6.1 Density Operator . . . . . . . . . . . . . . . . . . . . . 222.6.2 Expectation Values of an Operator using Density Op-

erator . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.6.3 Property of Density Operator . . . . . . . . . . . . . . 23

2.7 Composite System . . . . . . . . . . . . . . . . . . . . . . . . 242.8 Tensor Product . . . . . . . . . . . . . . . . . . . . . . . . . . 242.9 Bloch Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.10 Entanglement . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.11 Measurement of Entanglement . . . . . . . . . . . . . . . . . . 26

2.11.1 Negativity . . . . . . . . . . . . . . . . . . . . . . . . . 262.12 EPR Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.12.1 Locality Principle . . . . . . . . . . . . . . . . . . . . . 272.12.2 Reality Principle . . . . . . . . . . . . . . . . . . . . . 27

2.13 Quantum Gates . . . . . . . . . . . . . . . . . . . . . . . . . . 282.13.1 Pauli Gates . . . . . . . . . . . . . . . . . . . . . . . . 282.13.2 Hamard Gate . . . . . . . . . . . . . . . . . . . . . . . 292.13.3 Controled Not Gate . . . . . . . . . . . . . . . . . . . . 292.13.4 Representation of Gate in Circuit Diagram . . . . . . . 30

2.14 Teleportation . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3 Scheme For Secure Direct Comunication 353.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363.2 Preparing EPR pairs . . . . . . . . . . . . . . . . . . . . . . . 373.3 Secure Direct Communication Using Teleportation . . . . . . . 393.4 Security of the Scheme . . . . . . . . . . . . . . . . . . . . . . 453.5 Result And Conclusion . . . . . . . . . . . . . . . . . . . . . . 49

2

Chapter 1

Review of quantum Mechanics

3

1.1 Historical Note

At the end of the nineteenth century, physics mainly consisted of Classicalmechanics, Electrodynamics and Thermodynamics. Classical mechanics wasused to study dynamics of material bodies, and electrodynamics providedthe proper frame work to study radiation, electric and magnetic phenom-ena. Interactions between matter and radiation were well explained by theLorentz force or by thermodynamics. The extraordinary success of classicalphysics made people believe that the ultimate description of nature had beenachieved. It seemed that all known physical phenomenas could be explainedwithin the framework of the classical mechanics. However, as the twenti-eth century started,classical physics, which was considered as complete, wasseriously challenged on two major fronts:Relativistic domain: Einstein�s theory of relativity (1905) showed

that Newtonian mechanics does not seem good at very high speeds (speedscomparable to that of light).Microscopic domain: It turned out that classical physics fails in

providing proper explanation for several newly discovered phenomenas atmicroscopic level (e.g. structure of atoms).It all in 1900 when Max Planck introduced the concept of the quantum

of energy. In his attempt to explain phenomenon of blackbody radiation, hesucceeded in reproducing the experimental results only after postulating thatthe energy exchange between radiation and its surroundings takes place indiscrete or quantized amounts. He argued that the energy exchange betweenan electromagnetic wave of frequency v and matter occurs only in integermultiples of hv, called the energy of a quantum. Planck�s idea of quantum ofenergy turned out to be an idea with far reaching consequences.In 1905 Einstein supported Planck�s quantum concept. In his attempt

to understand the photoelectric e¤ect, he showed that Planck�s idea of thequantization of the electromagnetic waves also applies to light as well. Fol-lowing Planck�s approach, he recognized that light itself is made of discretebits of energy (or tiny particles), called photons, each of energy hv. Thephoton concept enabled Einstein to give an elegantly accurate explanationto the photoelectric problem.Another breakthrough was due to Niels Bohr. 1911, Rutherford came

up with his atomic model. In order to address shortcomings of Rutherford�smodel�s, Bohr in 1913 presented his model of the hydrogen atom. He arguedthat atoms can be found only in discrete states of energy and that the in-

4

teraction of atoms with radiation, takes place only in discrete amounts of hvas it results from transitions of the atom between its various discrete energystates. Bohr�s model provided a satisfactory explanation to several problemssuch as atomic stability and atomic spectroscopy.In 1923 de Broglie proposed that not only does radiation exhibit particle-

like behavior but, material particles also display wave-like behavior. In 1927,Davisson and Germer experimentally con�red de Broglie idea.They obtainedinterference patterns, a property of waves, with material particles such aselectrons.Although Planck�s and Bohr ideas were outstanding but they were criti-

sized for lacking the ingreadients of a theory.This motivated Heisenberg andSchrödinger to build theoretical foundation consited of these new ideas. Theapproach succedded.By 1925, they skillfully combined various experimental�ndings and Bohr�s postulates into a new theory called quantum mechan-ics. Theory provided an accurate reproduction of the existing experimentaldata.This theory enabled physicists to explore many uncharted areas of themicrophysical world. This new theory put an end to years of patchwork(1900�1925), dominated by the ideas of Planck and Bohr, which later be-came known as the old quantum theory.Historically, there were two independent formulations of quantum me-

chanics. The �rst formulation, called matrix mechanics, was developed byHeisenberg (1925) to describe atomic structure starting from the observedspectral lines. Inspired by Planck�s quantization of waves and by Bohr�smodel of the hydrogen atom, Heisenberg founded his theory on the notionthat the only allowed values of energy exchange between microphysical sys-tems are those that are discrete: quanta. Expressing dynamical quantitiessuch as energy, position, momentum and angular momentum in terms ofmatrices, he obtained an eigenvalue problem that describes the dynamics ofmicroscopic systems; the diagonalization of the Hamiltonian matrix yieldsthe energy spectrum and the state vectors of the system. Matrix mechanicswas very successful in accounting for the discrete quanta of light emitted andabsorbed by atoms.The second formulation, called wave mechanics, was due to Schrödinger

(1926); it is a generalization of the de Broglie postulate. This method, moreintuitive than matrix mechanics, describes the dynamics of microscopic mat-ter by means of a wave equation, called the Schrödinger equation; instead ofthe matrix eigenvalue problem of Heisenberg, Schrödinger obtained a di¤er-ential equation. The solutions of this equation yield the energy spectrum and

5

the wave function of the system under consideration. In 1927 Max Born pro-posed his probabilistic interpretation of wave mechanics: he took the squaremoduli of the wave functions that are solutions to the Schrodinger equationand he interpreted them as probability densities. These two di¤erent for-mulations Schrodinger�s wave formulation and Heisenberg�s matrix approachwere shown to be equivalent. Dirac then suggested a more general formu-lation of quantum mechanics which deals with abstract objects such as kets(state vectors), bras, and operators.

1.2 Postulates of Quantum Mechanics

State of a Classical System is speci�ed by two fundamental dynamical vari-ables: the position and the the momentum at any time t. Any other physicalquantity of system can be calculated in term of these fundamental variables.The complete physics of a classical system is detemined by following hamil-tons equations

dx

dt=@H

@p;

dp

dt= �@H

@t:

The Quantum mechanics formulism similar to these classic mechanics coun-terparts is the Postulates of Quantum Mechanics. These postulate are a roadmap for studying a Quantum system and enables us to understand.(i) How to write quantum state of a system mathematicaly at some time

�t�.(ii) How to calculate various physical quantities form quantum state.(iii) How to determince state of system at some later time �t0�, knowing

state at time �t�.Following are the �ve postulates of quantum mechanics that enables us

to answer the above question.Postulate 1: State of a SystemState of a system is speci�ed, a any time �t�, by a state vector j(t)i in

Hilbert space H State j(t)i contains all the necessary information aboutthe system.Postulate 2: Observables and OperatorsTo every dynamical variable A (obserable) there corresponds an oper-

ator A. The operator A is Hermitian and its eigenvectors form a completeorthonormal basis of the vector space.Postulate 3:Measurements and eigenvalues of operators

6

Measurement of an observable A is given by action of the operator  onstate j(t)i. Result of such a measurement is one of the eignevalue an ofA.State of the system after the measurement changes instantly to jni

Aj(t)i = anjni:

where an = hnj(t)i:Postulate 4: Probalistic outcome of MeasurementThe probability of obtaining one of the nondegenerated eigenvalue an of

operator A is given by

Pn(an) =jhnj(t)ij2h(t) j(t)i =

janj2h(t) j(t)i :

If the eigenvalue an is m degenerated, then Pn is expressed as

Pn(an) =

mPj=1

jhjnj(t)ij2

h(t) j(t)i =

mPj=1

jajnj2

h(t) j(t)i :

If the system is already in state jni of A: Then measurements of A giveswith certainty an : Ajni = anjni:Postulate 5: Time Evolution of SystemThe time evolution of the state vector j(t)i of a system is given by the

time-dependent Schrödinger equation .

i~@j(t)i@t

= Hj(t)i:

here H is Hamilton Operator corresponding to the total energy of the system.

1.3 Vector space And The Hibert Space

1.3.1 The linear Vector Space

A linear vector space consists of two sets of elements and two algebraic rules:

1. A set of vectors ; �; �,. and a set of scalars a, b, c,.

2. A rule for vector addition and for scalar multiplication.

7

(a) Addition ruleThe addition rule has the properties and structure of an abelian group:

� If and � are vectors (elements) of a space, their sum, + �, is alsoa vector of the same space

� Commutativity: + � = �+

� Associativity: ( + �) + � = + (�+ �)

� Existence of a zero or neutral vector: for each vector , there mustexist a zero vector O such that: O + = +O =

� Existence of an inverse vector: each vector must have a inverse vector(� ) such that: + (� ) = (� ) + = O

(b) Multiplicative RuleThe multiplication of vectors by scalars has these properties:

� The product of a scalar with a vector gives another vector.

� and � are two vectors of the space, any linear combination a + b�is also a vector of the space. (a & b being scalars)

� Distributivity: a ( + ) = a + a�; (a+ b) = a + b

� Associativity: a(b ) + (ab)

� For each element there must exist a unitary scalar I and a scaler "o"such that I = I = & o = o = 0

1.3.2 The Hilbert Space

A Hilbert space symH consists of a set of vectors , �, �,. and a set of scalarsa, b, c,which satisfy the following four properties:

� H is a linear space

� H has a de�ned scalar product that is strictly positive

� H is seperable

� H is complete

8

1.4 Linear Dependence and Independence ofVectors

Suppose we have linear combination of vector jvii with complex number �ias

�ijv1i+ �2jv2i++�njvni =nXi=1

�ijvii;

Then ifnXi=1

�ijvii = 0:

Then linear combination for a set of vectors is "linearly Independent". If oneof � = 0, we say that the set is linearly dependent.

1.5 Dirac Notation

The physical state of a system in quantum mechanics is represented by statevector in Hilbert space. These state vectors are basically elements of a Hilbertspace. The state vectors can be represented in di¤erent bases by means offunction expansion. This is analogous to specifying an ordinary (euclidean)vector by its components in various coordinate system. The components ofa vector can be represented in any coordinate system, similarly, the state ofa microscopic system has a meaning independent of the basis in which it isexpressed.Dirac introduced an invalueable notation in quantum mechanics to free

state vectors from coordinate meaning. It allows one to manipulate theformalism of quantum mechanics with ease. He introduced the concepts ofkets, bras, and bra-kets.KetsKets are the elements of a vector space. Dirac denoted the state vector

by a symbol ji which he called a Ket vector or simply a ket. Kets belongto Hilbert space H.BrasBras are the elements of a dual space. Dirac denoted the element of a

dual space by the symbol "h j" , which we call a bra vector or simply a bra.e.g. hjBra-Ket

9

This is the Dirac notation for the scalar product. Dirrac denoted thescalar (inner) product by the symbol "h j i" , which he called a Bra-ket. Forexample, the scalar product (1;2) is denoted by the bra-ket h12iProperties of Kets, Bras and Bra-Kets

� To every ket, there is a correspondding bra

ji �! hj:

� There is one-to-one correspondence between bras and kets

aji+ bj�i () a�hj+ b�h�j;

where a and b are complex numbers

jai = aji;haj = a�h:j

� We must be carefull to distinguish a scalar product from its complexconjugate; hj�i is not the same thing as h�ji

hj�i = h�ji�:

�hja11 + a22i = a1hj1i+ a2hj2i:

�ha1�1 + a2�2ji = a�1h�1ji+ a�2h�2ji:

�

ha1�1 + a2�2jb11 + b22i = a�1b1h�1j1i+ a1b�2h�1j2i

+a�2b1h�2j1i+ a�2b2h�2j2i:

� For any two states ji and j�i of HIlbert space, we can show that

jhj�ij2 � jih�j�i;

It is called Schwarz inequiliy.

10

� For any state vecctor ji and j�i, we can show thatph+ �j+ �i �

phji+

ph�j�i:

It is called triangle inequility.

� Two kets ji and j�i are said to be orthogonal if

hj�i = 0:

� Two kets ji and j�i are said to be orthonormal if they are orthogonaland if each of them has unit norm

hj�i = 0; hji = 1; h�j�i = 1:

� For ji and j�i belonging to Hilbert space, the products of the typejij�i and hjh�j are forbidden.

1.6 Inner Product

Suppose, we have two vector jui and jvi. then inner product is de�ned as

hujvi:

1.7 Outer Product

Outer product is written as the product of a ket with a bra represented asj�ihj. The quantity j�ihj is basically an operator.

ji =�gh

�; j�i =

�ij

�;

h�j = (j�i)y =�i� j�

�;

jih�j =�gh

��i� j�

�;

jih�j =�gi� gj�

hi� hj�

�:

11

1.8 Norm of a Vector

Inner product can be used to �nd norm (or length) of a vector. U =phU jUi:1.9 Orthogonality

Two vector are said to be orthogonal, if their inner product is zero, i.e.

hujvi = 0:

1.10 Orthonormality

Two or more vectors or states vectors are said to be orthonormal, if

1. It�s each vector is normalized.

2. All vectors are orthogonal to each other.

Let�s check orthonomalit of computational basis.

j0i =�10

�; j1i =

�01

�Normalization

h0j0i =�1 0

��10

�= 1� 1 + 0� 0 = 1: (1.1)

h1j1i =�0 1

��01

�= 0� 0 + 1� 1 = 1: (2)

From Eq. (1.1) and (2), it�s shown that vectors are normalized.Orthogonality

h0j1i =�1 0

��01

�= 1� 0 + 0� 1 = 0: (3)

h1j0i =�0 1

��10

�= 0� 1 + 1� 0 = 0: (4)

12

From (3) and (4), Vectors are orthogonal to each other. As this set of vectorsatis�es both the normality and orthogonality condition, they are orthonor-mal.

1.11 Operators

An operator A is a mathematical rule that transforms a ket ji into anotherket j0i of the same space. In other words, An operator  is a �mathematicalobject� that maps one state vector ji into another state ji in the samespace. Same is the rule for a Bra

Aji = j0i;h�jA = h�0j:

Product of OperatorsProduct of operators obeys following rules

� Product of two operators does not obey commutative property

AB 6= BA

� Product of operators, however obeys associative property

ABC = A(BC) = (AB)C

� We can writeAnAm = An+m

� Product of operators �B applies on a ket ji in the following way

ABji = A(Bji

� When an operator  is sandwiched between a bra h�j and a ket ji, ityields a complex number

h�jAji = C

Where C is complex number.

13

� An operator is said to be linear if it obeys the distributive law and likeall operators, it commutes with constants

A(a1j1i+ a2j2i) = a1Aj1i+ a2Aj2i;

and(h1j+ h2j)A = a1h1jA+ a2h2jA:

Expectation value of an Operator

The expectation or mean value hAi of an operator A is de�ned as

hAi = hjAjihji :

1.12 Hermitian Adjoint

The hermitian adjoint or simply adjoint Ay of an operator A is de�ned as

hjAyj�i = hjAj�i�:

Properties of hermitian Conjugate

�(Ay)y = A:

� �aAy

�y= a�Ay:

� �An�y=�Ay�n:

(A+ B + C + D) = Ay + By + Cy + Dy:

� �ABCD

�y= DyCyByAy:

� �ABCDji

�y= hjDyCyByAy:

14

� The hermitian adjoint of the operator jih�j is given by

(jih�j)y = j�ihj:

� Operators act inside kets and bras,respectively as follows

j�Ai = �Aji;haAj = a�hjAy:

hjAj�i = hAyj�i = hjAy�i:

1.12.1 Hermitian and Skew-hermitian Operators

An operator A is said to be Hermitian if it is equal to its adjoint Ay

A = Ay

or

hjAj�i = h�jAji�

An operator B is said to be Skew-Hermitian or anti Hermitian if

By = �B

or

hjAj�i = �h�jAji�

1.13 Projection Operators

It is de�ned as outer product of single ket. Mathematically

P = jihj:

15

1.13.1 Properties of Projection Operator

1. A projection operator is Herniation

P = P y:

2. A projection operator is equal to its own square as

P 2 = P:

if the state ji is normalized.

3. The product of two projection operator P1 and P2 is also a projectionoperator if they commute i.e.

P1P2 = P2P1:

4. Projection operators from j0i and j1i states are

P0 = j0ih0j:P1 = j1ih1j:

P0 + P1 = j0ih0j+ j1ih1j =�1

0

��1 0

�+

�0

1

��0 1

�:

P0 + P1 =

�1 00 0

�+

�0 00 1

�=

�1 00 1

�= I:

5. Projection operators from j+i and j�i state are

P+ = j+ih+j:P� = j�ih�j:

1.14 Inverse of an Operator

Inverse of an operator is de�ned as

A�1A = AA�1 = I

where I is unit operator

16

1.15 Unitary Operators

A linear operator U is said to be unitary if it�s inverse U�1 is equall to itsadjoint U y

U y = U�1

or

U U y = U yU = I

1.16 Eigne Values and Eigen Vectors of anOperator

A state vector ji is said to be an eigne vector of an operator A if thea;;lication of A to ji gives

Aji = aji

This is called the eigen-value equation of operator AExample:For unitary operator I, we have

Iji = ji

here, eigne value is 1.

17

Chapter 2

Concepts of QuantumComputing

18

2.1 Qubit

In quantum computing, Qubit is the basic unit of information. It is de�nedas the superposition of two state vectors in a Hilbert space. Mathematicallyit�s represented as

ji = �j0i+ �j1i:Where j0i and j1i are state vectors. � and � are complex numbers. Theprobability of �nding the particle in state j0i is de�ned by j�j2, and j�j2 ofstate j1iNormalizationThe sum of probabilities must be equal to 1. i.e.

j�j2 + j�j2 = 1:

when this condition is satis�ed, we say that qubit is normalized.Matrix Form of State VectorThe state vectors j0i and j1i can be written in matrix form as

j0i =�10

�; j1i =

�01

�;

They are also called basis. The basis for a given quantum systems are notunique, for example, a qubit instead of the standard basis can also be ex-pressed in other bais such as j+i and j�i basis. In matrix form these aregiven as follows

j+i = 1p2

�11

�; j�i = 1p

2

�1�1

�:

2.2 Pauli Operators

A set of operators that turns out to be of fundamental importance in quantumcomputation is known as Pauli operators. Including the identity operator,there are four Pauli operators.Identity OperatorIt�s represented by either �0 or I. Identity operator acts on computational

basis sates as�0j0i = j0i; �0j1i = j1i:

19

In other words, an identity gate is the one that does nothing to the state ofthe system.Not OperatorThis operator is denoted by �1 or X. It acts as

�1j0i = j1i; �1j1i = j0i:

This gate corresponds to classical Not gate, because it ��ips�the basis state.Spin Flip OperatorIt�s denoted by �2 and Y . It acts on computational basis as

�2j0i = �ij1i; �2j1i = �ij0i:

Phase Flip OperatorIt�s denoted by �3 and Z. It acts on computational basis as

�3j0i = j0i; �3j1i = �j1i:

Matrix form of Pauli OperatorsFollowing are the matrix representation of Pauli operators

X =

�0 11 0

�; Y =

�0 ii 0

�; X =

�1 01 �1

�:

Z =

�1 00 1

�:

2.3 Trace of an Operator

Trace of an element is de�ned as sum of it�s diagonal elements. For example

if A =�a bc d

�then

Tr(A) = a+ d

For outer product representation of an Operator A, the trace can be writtenas

Tr(A) =

nXi=1

huijAjvii

Properties of Trace

20

1. The trace is linear.

Tr(A+B) = Tr (A) + Tr (B) :

T r (�A) = �Tr (A) :

2. Trace is cyclic

Tr(ABC) = Tr(BCA) = Tr(CAB):

3. Trace of an Operator is equal to sum of it�s eigenvalues

Tr (A) =

nXi=1

�i;

4. Trace of an outer product is inner product

Tr(jih�j) = hj�i:

2.4 Unitary Transformation

It is a methodology of transforming the matrix form of an operator from onebasis to another basis. For simplicity, let consider two dimensional vectorspace C2. The change of basis matrix from a basis juii to jvii is given by

U =

�hv1ju1i hv1ju2ihv2ju1i hv2ju2i

�:

A state vector ji given in the juii basis in terms of the new jvii basis as

j0i = U ji;

here j0i is the same vector but with new basis jvii and the operator in newbasis will be given as

A0 = UAU y:

Do remember that U = U y:

U =

�h+j0i h+j1ih�j0i h�j1i

�=

0@h� 1p2h0j+ h1j

�j0i h

�1p2h0j+ h1j

�j1i

h�1p2h0j � h1j

�j0i h

�1p2h0j � h1j

�j1i

1A :

21

U =1p2

�1 11 �1

�:

Now!

ji =

�1

�2

�; j0i = U ji;

j0i =1p2

�1 11 �1

��1

�2

�=

1p2

��13

�:

2.5 Pure and Mixed States

In most of the cases, we need to study a collection of systems rather than asingle quantum system called an ensemble. Elements of ensemble are usuallyin di¤erent quantum states. There a particular probability for each quantumstate in ensemble. At ensemble level the use of probability is acting inclassical way, so ensemble is a statistical mixture of quantum states.If a quantum system is in jii states with probability probability pi for

each respective state. In simple words, if a system contains more than onequantum state, such a system is said to be in Mixed state. While when asystem is in a de�nite state, it�s said to be in Pure state. Pure state is aspecial case of mixed state in which pi = 1 for a particular state.

2.6 Density Matrix Formulism

2.6.1 Density Operator

When the state vector of a quantum mechanical system is not completelyknown, an alternate approach known as the density operator (matrix) for-malism is used. For an ensemble of pure states jii that are available for thesystem, the density operator is written as

� =X

pijiihij;

where pi is the probability of the system to be in the state jii. if the systemis known to be in a particular state jki, then pk = 1 and the density matrixreduces to

� = jiihij:

22

The density operator represents a pure state if and only if

Tr��2�= 1;

While for a mixed state, the following inequality holds

Tr(�2) < 1:

2.6.2 Expectation Values of an Operator using DensityOperator

We can calculate the expectation value of an operator A by using Denistyoperator

hAi =nX

k;l=1

huljAjukihukjAjuli;

=nXl=1

hulj�

nXk=1

jukihukj!Ajuli;

=nXl=1

hulj�Ajuli;

hAi = Tr(pA):

2.6.3 Property of Density Operator

An operator is a density operator if and only if it satis�es the followingrequirements

1. Denisty operator is Hermition

� = �y:

2. Trace of denisty operator is

Tr (�) = 1:

3. � is positive i.ehujpjui � 0:

Representation of Denisty Matrix in Computational Basis

23

Density Matrix in computation basis {j0i; j1i} can be written as

� =

�h0j�j0i h0j�j1ih1j�j0i h1j�j1i

�:

If a measurement is made, the probability of �nding a member of ensemblein state j0i is given by h0j�j0 /i and that of in state j1i is given by h1j�j1i

2.7 Composite System

More than one particle in a system forms a composite system. Compositesystem can be further cateforized into two categoriesSeparable StatesThese states can be seperated and making measurement on one particle

does not e¤ects other particle state.Product states

2.8 Tensor Product

Tensor product puts vector spaces togetherr into a single vector space. Formaking the hilbert space of composite systtem, we take tensor product ofindiviual hilbert spaces. The dimensions of hilbert space oof composite sys-tem is the product of dimensions of H1 and H2. If N1 be the dimension ofH1 and N2 be the dimension of H, then dimension of the composite Hilbertspace H will be

Dim(H) = N1N2:

Then, tensor product is given by

j�i ji:

Let ji =�ab

�, and j�i =

�cd

�, then tensor product of these state will be

ji j�i =

�a

b

��c

d

�

=

0BB@acadbcbd

1CCA :

24

Properties of Tensor Product

1. The tensor product of two vectors is linear i.e

ji (j�1i+ j�2i) = ji j�1i+ ji j�2i:

2. If A and B are positive, then AB will also be positive.

3. If A and B are projection operators, then A B is also a projectionoperator.

4. If A and B are hermition, then AB is hermition

2.9 Bloch Sphere

Bloch Sphere gives the geometrical visualization of "Qubit". In Bloch Sphererepresentation, a qubit is written as

ji = cos

��

2

�j0i+ ei� sin(

�

2)j1i;

ji = cos

��

2

�j0i+ (cos�+ i sin�) sin(

�

2)j1i

�ei� = cos�+ sin�

:

Figure.3.1.The Bloch sphere Representation of a Qubit

Where � and � are parameters used to de�ne a point on a unit three dimen-sional sphere known as Bloch Sphere. � and � ranges from 0 to 2�: � = 0corresponds to basis j0i while � = � to basis j1i. j0i and j1i lies on the northand south poles of the sphere.

25

2.10 Entanglement

It is a very fascinating and unusual phenomena of Quantum Mechanics thatgives priority to quantum over classical mechanics. It states

"Two particles are linked together in such a way that measurement of oneparticle�s state determines the possible state of other linked particle"

2.11 Measurement of Entanglement

Entanglement is an unusual phenomena and it can be used to teleport in-formtion. There�s a number of method for measuring Entanglement. one ofwhich is

2.11.1 Negativity

It is a quanti�er as well as an identi�er. Besides identifying, it also tells howmuch a state is entangled. Suppos we have a Bipartite system and want tocheck it�s entanglement.Following are the steps for calculating negativity.

1. Find the denisty matrix of the state under consideration. We assumehere that the denisty matrix of the state is

� =

0BB@a b c d0 0 0 00 0 0 0w x y z

1CCA :

2. Take partial transpose of denisty matrix. Procedure for �nding PartialTranspose is given in section(3.10). Partial transpose of � will be

�pt =

0BB@a 0 c 0b 0 d 00 w 0 y0 x 0 z

1CCA :

3. Find eigenvalues of �pt by making charesteristic equaiton i.e.

det(�pt � �I) = 0:

26

We have calculated �i eigenvalues If one or more eigenvalues are negativethat means state is a product state (engtangled state). If non of the eignevalue is negative then state is a separable state. Quantity of entanglement isgive by

N(�) =

Pi �i � 12

;

N(�) tells how much a state is entangled. Value of N(�) resides betweeninterval 0 � N(�) � 1

2: N(�) has a value of 1/2 for maximally entangled

states.

2.12 EPR Paradox

The Phenomena of Entanglement has no classical analogy. The unusualproperties of Entanglement were challenged in a paper published in 1935 byEinstein, Podolsky and Rosen. The authors of the paper, on the basis of"locality" and "reality" principles, showed that Quantum Mechanics leads toa contradiction.

2.12.1 Locality Principle

It states that

If two system are causally disconnected the result of any measurementperformed on one system cannot in�uence the state of other system

Following the theory of Relativity, we say that two measurement eventsare disconnected if

(�x)2 > c2 (�t)2 :

Where, �x = x1 � x2 is spatial separation between system and �t = t1 � t2is temporal separation.

2.12.2 Reality Principle

It states that

Physical quantities are in reality already determined before themeasurement.

27

The system already has these physical quantities determined. They areelements of reality. This concept of classical reality is known as EinsteinReality.

2.13 Quantum Gates

As any classical computation can be broken down into a sequence of classicallogic gates that act on only a few classical bits at a time, so too any quantumcomputation can be broken down into a sequence of quantum logic gatesthat act on only a few qubits at a time. Gates are active components thattransform the state of a bit in a desired fashion. A gate can be thought ofas an abstraction that represents information processing. In order to processinformation and perform di¤erents measurements we need quantum gate justlike classical gate. Gates in quantum computing are unitary operation. Aunitary operator is the one for which we have U = U y: Just like as operatorscan be represented in matrix form, gates can also be represented in matrixform. A quantum gate with n inputs and outputs can be represented by amatrix of degree 2n. For a single qubit a gate must be a matrix of degree 2(21 = 2): So a quantum gate acting on a single qubit will be a unitary matrixof order 2 X2. Gates can be categorized as singleand multiple quibt gates.Single quantum gate are the one that are represented by a 2 � 2 unitarymatrix. Most basic and well know single qubit gates are Pauli gates.

2.13.1 Pauli Gates

For single qubits, the �Pauli matrices�(I,X,Y,Z), which happen to be bothhermitian and unitary, are of special interest. We have discussed pauli oper-ators in section (3.11) in detail.

X =

�0 11 0

�; Y =

�0 ii 0

�; X =

�1 00 �1

�;

I =

�1 00 1

�:

The Pauli X gate is synonymous with the classical NOT gate and alter thestate on whict it acts i.e

Xj0i =�0 11 0

��1

0

�=

�0

1

�= j1i:

28

The pauli Y gates is a bit �ip gate. it acts on a qubit as

Y j0i =�0 ii 0

��1

0

�=

�0

i

�= ij1i:

The Z operator is sometimes called the phase �ip gate because it takes aqubit ji = �j0i+ �j1i into a state ji = �j0i � �j1i. In matrix form

Zji =�1 00 �1

���

�

�=

��

��

�:

2.13.2 Hamard Gate

One of the most useful single qubit gates, in fact perhaps the most usefulone, is the Hadamard gate, H. The Hadamard gate is de�ned by the matrix

H =1p2

�1 11 �1

�:

The action of the Hadamard gate on the standard or computational basisstates is to map the fj0i; j1ig states into the superposition states respectively�

j0i+ j1ip2

;j0i+ j1ip

2

�:

2.13.3 Controled Not Gate

Controlled NOT Gate (CNOT) is a two qubit gate and is represented by amatrix of 4� 4. This gate helps us to implement an if-else type of constructwith quantum gate in analogy to controlled classical gate. CNOT gate usesa control qubit to determine whether NOT gate should be applied to targetqubit or not.

Figure. 3.2. Circuit Diagram Representation of CNOT gate

29

As in the �gure (2.13.3), CNOT gate has two inputs, a controlled input (j�i)and a target input (j�i). CNOT gate does nothing to target qubit if controlqubit is j0i but applies NOT gate on target qubit if control qubit is j1i:The possible input states to the CNOT gate are j00i; j01i; j10i and j11i

and the action of the CNOT gate on these states is

j00i 7�! j00i:j01i 7�! j01i:j10i 7�! j11i:j11i 7�! j10i:

The matrix representation of this gate is

CN =

0BB@1 0 0 00 1 0 00 0 0 10 0 1 0

1CCA ;

And the outer product form is

CN = j00ih00j+ j01ih01j+ j10ih11j+ j11ih10j:

2.13.4 Representation of Gate in Circuit Diagram

We can represent the action of a quantum gate by drawing a circuit diagram.Each gate is represented by a block with lines (wires) used to represent inputand output. Circuit diagram of some of the single qubit gate are given infollowing �gure

30

Figure. 3.3. Circuit Diagram Of X, Y and Z gate (up to down respectivily)

Representation of measurement by circuit diagram

Figure. 3.4. Measurement

2.14 Teleportation

Now we are ready to use our understanding of qubits and quantum gates foran application that is Teleportation. Teleportation is a procedure that allowsone party (Alice) to send a quantum state to other party (Bob) withoutthat state being transmitted in the usual sense. The purpose of quantumteleportation is to transmit an unknown quantum state of a qubit such thatthe recipient reproduces exactly the same state as the original qubit state.Note that in teleportation a qubit actually is not sent but the informationrequired to reproduce the quantum state is transmitted. The original state isdestroyed during teleportation so it does not violate the no-cloning theorem.

31

The task here is that Alice wants to send an unknown state to Bob. Letdonate the state that Alice want to transmit to Bob by j�i. The state is aqubit

j�i = �j0i+ �j1i: (2.1)

Unknown state mean that we don�t know � and � but we assume that j�j2+j�j2 = 1: Teleportation takes place in series of steps. Following is a simpli�ed�gure for whole process

Figure. 2.3 Teleportation Process

1. Alice and Bob share an Entangled pair of particles

Alice and Bob prepares the entangled state

j�00i =j0iAj0iB + j1iAj1iBp

2=j00i+ j11ip

2; (2.2)

here we assume that �rst member of pair is of Alice and second of Bob.Alice and Bob separates physically.

2. Encoding

Now Alice wants to send state (2.1). To do so, she needs to link un-known qubit with her qubit (encoding). She do so by taking tensorproduct of j�i with EPR pair. We represent overall state after encod-ing unknown state as ji.

ji = j�i j�00i = (�j0i+ �j1i)�j00i+ j11ip

2

�;

=�(j000i+ j011i) + � (j100i+ j111i)p

2:

The �rst two qubit belongs to Alice while third to Bob.

32

3. Processing The State

Alice applies CNOT gate to qubits under her possession. She usesunknown state as control qubit and her member of EPR pair as targetqubit. State of system after applying CNOT gate changes to j0i

j0i = UCNOT ji;

=1p2

��(UCNOT j000i+ UCNOT j011i)+� (UCNOT j100i+ jUCNOT111i)

�;

=1p2[�(j000i+ j011i) + � (j110i+ j101i)] :

Next, she applies hadamard gate to �rst qubit and state of systemchanges to j00i

j00i = Hji;

=1p2[�Hj0i (j00i+ j11i) + �Hj1i (j10i+ j01i)] ;

=1p2

24 ��j0i+j1ip

2

�(j00i+ j11i)

+��j0i�j1ip

2

�(j10i+ j01i)

35 ;=

1p2

��j000i+ �j011i+ �j100i+ �j111i+�j010i+ �j001i � �j110i � �j101i

�;

=1p2

�j00i (�j0i+ �j1i) + j01i (�j1i+ �j0i)+j10i (�j0i � �j1i) + j11i (�j1i � �j0i)

�: (2.3)

4. Alice Measures her Pair

Alice makes a measurement on both qubits in her possession. She canchose one of these four measurements fj00i; j01i; j10i; j11ig. Possibleoutcome of each measurement is given in following table

Measurement Outcome Relation to state j�ij00i 1p

2(�j0i+ �j1i) j�i

j01i 1p2(�j1i+ �j0i) Xj�i

j10i 1p2(�j0i � �j1i) Zj�i

j11i 1p2(�j1i � �j0i) ZXj�i

(2.4)

33

5. Alice Tells Her Measurement Outcome To Bob

Alice contact Bob via public channel (Call, mail,fax etc.) and tell�sbob his measurement outcome or she sent two classical bit either ofone (00; 01; 10; 11) in public channel to Bob.

6. Decoding

On receiving information from Alice, Bob applies the required gate torestore the state to the original state j�i:

Let Alice sent Bob two classical bit 11 in public channel. That mean sheperformed measurement of j11i, So state on Bob particle must be

=1p2(�j1i � �j0i) :

And he knows from table (2.4) which gate he must apply to restore the stateto j�i. He applies X-gate �rst

1p2(�Xj1i � �Xj0i) = 1p

2(�j0i � �j1i) ;

Now he applies Z-gate

1p2(�Zj0i � �Zj1i) = 1p

2(�j0i+ �j1i) = j�i:

34

Chapter 3

Scheme For Secure DirectComunication

35

In this chapter we are reviewing a paper "A secure direct communicationusing EPR pairs and teleportation" published in "The European PhysicalJournal B" in 2004. In this paper a novel scheme for secure direct communi-cation is proposed, where there is no need for establishing a shared secret key.Communication is based on EPR pair (quantum Channel) and teleportationbetween Alice and Bob. After checking the security of quantum channel,Bob encodes the secret message directly on a sequence of particles statesand teleport them to Alice. In this scheme their no chance for a potentialeavesdropper to get information. Since there no need for shared secret key,this scheme is completely secure while quantum channel is perfect.

3.1 Introduction

Cryptography is an art to transmit information so that it is unintelligibleand therefore useless to those who are not meant to have access to it. Cryp-tography schemes are only completely secure when the two communicatingparties, Alice and Bob, establish a shared secret key before the transmissionof a message. This means they �rst should create a secret key which is com-posed of a random bit sequence, not known to anyone else, and of the samelength as the message.It is di¢ cult to share secret key security in a classical channel. Likely one

can use quantum channel for transmission of secret key which is more secureand di¢ cult to interrupt.First quantum key distribution protocol was proposed in 1984 by Bennett

and Brassard, which is know as standard BB84 QKD protocol[7]. At presentthere�s a lot number of QKD protocols.In 2002, Beige et al. [8] proposed a quantum secure direct communica-

tion (QSDC) scheme, where the message is deterministically sent through thequantum channel, but can only be deduced after a �nal transmission of clas-sical information. Same year, Bostrom and Felbinger presented a Ping-PongQSDC scheme[9] Wojcik in 2003 [11] proved that Beige�s and Bostrom�sschemes are insecure in operated in noisy quantum channel. Then Deng etal [10] presented two step quantum direct communication protocol and it ap-peared fairly secured. All of the above mention schemes have a basic security�aw. In all these schemes, a qubit carrying secret message must be sent in apublic channel (classical channel). So it is quite easy for an eavesdropper toattack qubit during transmission in public channel.

36

We present here a scheme for secure direct communication using EPRpair and teleportation where there is no need of establishing a shared secretkey. Since there is no need for transmission of qubit carrying secret key inpublic channel so this scheme is completely secure for direct communicationif quantum channel is perfect. This protocol can be divided into three steps,�rst preparing EPR pair, second checking security of the quantum channeland the last step is teleportation of information.

3.2 Preparing EPR pairs

We need EPR pairs for Teleportation. So our �rst step will be to prepareEPR pairs. We use Hadamard gate and Cnot gate to get EPR pairs. Diagramfor the quantum circuit to prepare EPR pairs is given in �g. 3.2.

Fig. 4.1

Quantum circuit has two inputs jai and jbi. Time move from left to right.First Hadamard gate is applied to qubit jai to generate a superposition state.Then this superposition state is take as input for CNOT gate as control qubit.Another qubit jbi (either j0i; j1i) is applied to circuit as target qubit. At theend the result turn out to be a bell state jabi. Bell state in equation formcan be represented by following equation

j�abi =j0; bi+ (�1)a j1; �aip

2;

where, a represents NOT aNow we will explain the whole process by setting qubit jai as j0i and jbi

as j1i:

37

Hadamard gate will act on j0i to generate superposition state

Hj0i = j0i+ j1ip2

:

Output of Hadamard gate will be passed to CNOT gate as control qubit.Superposition state consists of two qubits j0i and j1i.Both these qubit willbe passed as a control qubit one by one. So in �rst step j0i is control qubit.We know that if control qubit is j0i CNOT gate does nothing to target qubitthat is also j0i in this case. So we get j00i: In second step control qubit isj1i so the state of target gubit will be reversed i.e. j0i ! j1i: Hence outputwill be j11i. By combining these two output we get a bell state

j�00i =j00i+ j11ip

2: (3.1a)

By taking qubits jaiand jbi as j0i, we get bell state j�00i;

j�01i =j01i+ j10ip

2: (3.1b)

If jai is taken as j1i and jbi as j0i, then we get j�10i:

j�10i =j00i � j11ip

2: (3.1c)

And at the end, when both input to circuit in �g.3.2 are j1i, we get.

j�11i =j01i � j10ip

2: (3.1d)

These four states are represented by general form of Bell states that is j�abi,but we will use a di¤erent notation in this chapter for EPR states. we willdonate j�ooi as j�+i, j�10i ! j��i; j�01i ! j+i; j�11i ! j�i .Suppose that Alice and Bob share a set of entangled pairs of qubits,

considering �rst particle is of Alice and second is of Bob. So we can writeEPR pair j�00i as

j�+iAB =j00iAB + j11iABp

2: (3.2)

here A stands for Alice and B for Bob.

38

We know that j+i = j0i+j1ip2and j�i = j0i�j1ip

2. Adding j+i and j�i and

simplifying we get j0i = j+i+j�ip2.

Subtracting j+i and j�i simplifying we get j1i = j+i�j�ip2

: Now puttingvalues of j+i and j�i in eq. (3.2) and simplifying, we get

j�+iAB =j+iAj�iB + j�iAj+iBp

2:

By doing same maths, we can get other EPR pairs in j+i and j�i. Here weare summarizing these four state in table given below

EPR pair In j0i and j1i basis In j0i and j1i basisj�+iAB j00iAB+j11iABp

2

j+iAj+iB+j�iAj�iBp2

j��iAB j00i�j11ip2

j+iAj�iB+j�iAj+iBp2

j+iAB j01i+j10ip2

j+iAj+iB�j�iAj�iBp2

j�iAB j01i�j10ip2

j�iAj+iB�j+iAj�iBp2

There are many ways to obtain these states. As an example, Bob couldprepare the pairs and then send half of each to Alice. Another way is, athird party could prepare and then send half of each Alice and Bob. Inorder to keep it general, All EPR pairs used in our scheme are the Bell statej�+iAB.

3.3 Secure Direct Communication Using Tele-portation

After insuring the security of thee quantum channel (EPR pair), we beginsecure direct communication. Suppose Bob wishes to communicate infor-mation to Alice, and his actual message is 01001. He makes his particle insequence according to the message sequence, j+i corresponding to the 0 andj�i corresponding to 1. So message and sequence of particles will be

Message Particle sequence in states

01001 j+ij�ij+ij+ij�i

39

Alice and Bob will use quantum channel of EPR pairs as a channel fortransmission of messages encoded in the sequence of particle states. This isthe so called quantum teleportation process, which we now describe in detailsWe will use subscript A and B for the particles of Alice and Bob respec-

tively and subscript C for BoB�s particle with messages. We will be utilizingj�+iAB EPR pair as quantum channel. Particles B and C are in Bob�s pos-session while particle A in Alice possession. And the qubit state carryingmessage is written as

jic =1p2(j0i+ bj1i)

Here b = 1 and b = �1 corresponds to j+i and j�i states respectively.Now, bob will encode the state jiC with EPR pair j�+iAB by taking

tensor product jiC of state with j�+iAB. The overall state of the systemsABC after taking tensor product will be

j�+iABjiC =1

2(j00iAB + j11iAB)(j0iC + bj1iC) (3.3)

=1

2(j000iABC + bj001iABC + j110iABC + bj111iABC)(3.4)

Some processing is required here to obtain desired result. We will proceedstep by step here

� Step 1: Bob applies hadamard gate on his particle. Since state threeparticle are linked together, so we apply IHI. So by applying this,Hadamard gate will act on Bob�s particle while there will be no e¤ect onAlice and message particles since identity gates will act on there states.One more thing we will omit subscripts A;B ;C during processing andlet that �rst particle is of Alice, second of Bob and third of message

=1

2

�I H Ij000i+ I H I (bj001i)+I H Ij110i+ I H I (bj111i)

�;

=1

2

24 j0i�1p2(j0i+ j1i)

�j0i+ bj0i

�1p2(j0i+ j1i)

�j1i

+j1i�1p2(j0i � j1i)

�j0i+ bj1i

�1p2(j0i � j1i)

�j1i

35 ;=

1

2p2

�j000i+ j010i+ bj001i+ bj011i+j100i � j110i+ bj101i � bj111i

�:

40

Step 2: Applying I H I again

1

2p2

2664I H Ij000i+ I H Ij010i

+I H I(bj001i) + I H I(bj011i)+I H Ij100i � I H Ij110i

+I H I(bj101i)� I H I(bj111i)

3775 ;

=1

2p2

26666664j0�1p2(j0i+ j1i)

�0i+ j0

�1p2(j0i � j1i)

�0i

+bj0�1p2(j0i+ j1i)

�1i+ bj0

�1p2(j0i � j1i)

�1i

+j1�1p2(j0i+ j1i)

�0i � j1

�1p2(j0i � j1i)

�0i

+bj1�1p2(j0i+ j1i)

�1i � bj1

�1p2(j0i+ j1i)

�1i

37777775 ;

=1

2p2p2

2664j000i+ j010i+ j000i � j010i

+bj001i+ bj011i+ bj001i � bj011i+j100i+ j110i � j100i+ j110i

+bj101i+ bj111i � bj101i+ bj111i

3775 ;After canceling terms

=1

2p2p2[2j000i+ 2bj001i+ 2j110i+ bj111i] ;

Taking common

=1

2p2p2

�j0i [2j00i+ 2bj01i]+j1i [2j10i+ bj11i]

�;

=1

2p2p2

2664 j0i�

j00i+ j00i+ j11i � j11i+bj01i+ bj01i+ bj10i+ bj10i

�+j1i

�j10i+ j10i+ j01i � j01i

+bj11i+ bj11i+ bj00i � bj00i

�3775 :

41

Step 3: Rearranging

=1

2p2

266664j0i

j00i+j11ip2

+ j00i�j11ip2

+b j01i+j10ip2

+ b j01i�j10ip2

!

+j1i

j10i+j01ip2

� j10i�j01ip2

+b j00i+j11ip2

� b j00i�j11ip2

!377775 ; (3.5)

=1

2p2

�j0i (j�+i+ j��i+ bj+i+ bj�i)+j1i (j+i � j�i+ bj�+i � bj��i)

�;

j�+iABjiC =1

2

266641p2(j0iA + bj1iA) j�+iBC

+ 1p2(j0iA � bj1iA) j��iBC

+ 1p2(bj0iA + j1iA) j+iBC

+ 1p2(j0iA � bj1iA) j�iBC

37775 : (3.6)

Encoding is done, Now Bob preforms a Bell measurement on his particleBC. He can perform one of measurement (j�+iBC , j��iBC , j+iBC ,j�iBC) on particles BC. Here, Bob can chose one of the measurementhe want to preforms randomly. Probability of performing a certainmeasurement is 1/4. So each outcome of the measurement will occurrandomly with equal probability 1/4. Hence after measurement, theresulting state of Alice�s particle will be 1p

2(j0iA+ bj1iA) if he performs

measurement of j�+iAB. So the result of these four measurements willbe respectively

1p2(j0iA + bj1iA);

1p2(j0iA � bj1iA);

1p2(bj0iA + j1iA);

1p2(bj0iA � j1iA):

Moreover, in each case state of Alice particle is related to jiC by a �xedunitary transformation Uij independent of the identity of jiC. So we will usehere unitary transformation Uij to transform Alice state to the state jiC .

42

Unitary Transformation Uij

U00 =

�1 00 1

�; U01 =

�1 00 �1

�;

U10 =

�0 11 0

�; U11 =

�0 1�1 0

�:

After applying Unitary transformation, Alice particle states will be

1p2(j0iA + bj1iA) = U00jiA:;

1p2(j0iA � bj1iA) = U01jiA;

1p2(bj0iA + j1iA) = U10jiA;

1p2(bj0iA � j1iA) = U11jiA:

Now, Bob sends his actual Bell measurement outcome to Alice in a publicchannel. He can send his actual measurement outcome to Alice via a phonecall, E-mail, fax etc. He tells Alice which inverse unitary transformation sheshould apply on her particle to restore it�s state to state jiA Alice applies theinverse transformation that Bob tell�s her by contacting in a public channel.After that she measures the basis (j+i and j�i), and read out the state thatBob has sent her. She performs measurement of j+i, and if the outcome isone that mean the state on her particle is j+i and if outcome is zero it meansstate of her particle is j�iWe want to demonstrate here how Bob will send j+i state. For this

state, we but b=1 in eq. (3.6) and proceed forward. After this bob preformsmeasurement on particle BC, a measurement of j��iBC So state of Aliceparticle after Bob�s measurement will be

43

1p2(j0iA � j1iA) = U01jiA:

1p2(j0iA � bj1iA) =

1p2

�1�1

�; U01 =

�1 00 �1

�:

j+i =1p2

�11

�:

U01j+i =

�1 00 �1

��1p2

�11

��=

1p2

�1�1

�=

1p2(j0iA � j1iA):

So state of Alice particle is related to the state that bob has sent by unitarytransformation U01. So by applying inverse unitary transformation we getthe state that Bob has sent as U01U�101 = I:Bob tells her to perform U�101 inverse unitary transformation. She applies

inverse transformation, and restores state to original state.

U�101 =

�1 00 �1

�;

And state of Alice particle is

1p2(j0iA � j1iA) =

1p2

�1�1

�;

So applying inverse transformation, we have

U�101 =

�1 00 �1

��1p2

�1�1

��=

1p2

�11

�= j+i:

So by Applying inverse transformation, she restored the state to the originalstate that Bob has sent. Still She don�t know which state she have on herparticle. So she measures the basis and gets h+j+i = 1 that means Bob hassent her state j+i:The message Bob wants to send to Alice is 01001. So he will take state

jic = j+i and teleport it to Alice then he will take jic = j�i and teleportit, and so on. One by one, In a sequence Bob will teleport all required statesto Alice. Alice already knows that j+i corresponds to 0 and j�i to 1. So atthe end she will get following states in sequence

44

j+ij�ij+ij+ij�iThat translates to

01001

Quantum Teleportation has two notable features.

1. The entanglement of EPR pairs is independent of spatial location ofAlice and Bob. So Bob can send information to Alice without evenknowing the position of Alice. He need only to broadcast his Bellmeasurement outcome.

2. State to be transmitted is independent of the processing need to be donefor teleportation. Also the Bell measurement outcomes are random. Soin this scheme, teleportation transmits Bob�s message without reveal-ing any information to a potential eavesdropper if quantum channel isperfect EPR pairs.

3.4 Security of the Scheme

The security of this scheme only depends on the perfect quantum channel(Pure EPR pairs). So as long as the quantum channel is perfect, the schemeis secure.We should point out that it is necessary to test the security of quantum

channel, since a potential eavesdropper may obtain information as following:

1. Eve can use the entanglement pair to obtain information. Suppose thatEve has a particle pair in the state j�+iDE. When Eve obtains particleB in preparing EPR pair, she links her particle with Bob�s particle bytaking tensor product of states j�+iAB and j�+iDE. Here again we needto perform certain steps to achieve our desired form.

j�+iAB j�+iDE =1p2(j00i+ j11i) 1p

2(j00i+ j11i) (3.7)

j�+iABj�+iDE =1

2[j0000i+ j0011i+ j1100i+ j1111i]ABDE

45

Applying Hadamard gate on 2nd qubit (Bob�s particle)

1

2

24 j0�j0i+j1ip

2

�00i+ j0

�j0i+j1ip

2

�11i

+j1�j0i�j1ip

2

�00i+ j1

�j0i�j1ip

2

�11i

35ABDE

1

2p2

�j0000i+ j0100i+ j0011i+ j0111i+j1000i � j1100i+ j1011i � j1111i

�1

2p2

�j0000i+ j1000i+ j0011i+ j1011i

�j1100i+ j0111i � j1111i

�Applying Hadamard on 3rd -qubit.

=1

2p2p2

2664j0000i+ j0010i+ j1000i+ j1010i+j0001i � j0011i+ j1001i � j1011i+j0100i+ j0110i � j1100i � j1110i+j0101i � j0111i � j1101i+ j1111i

3775Taking common

=1

2p2p2

2664j00i (j00i+ j01i+ j10i � j11i)+j01i (j00i+ j01i+ j10i � j11i)+j10i (j00i+ j01i+ j10i � j11i)+j11i (�j00i � j01i � j10i+ j11i)

3775=

1

2p2

�j00i fj��i+ j+ig+ j01i fj��i+ j+ig+j10i fj��i+ j+ig � j11i fj��i+ j+ig

�=

1

2p2

��j00i+ j01i+ j10i

�j11i

��j��i+ j+i

�=

1

2

��j��i+ j+i

� �j��i+ j+i

��=

1

2

�j��ij��i+ j��ij+i+j+ij��i+ j+ij+i

�(3.8)

Taking second part of eq. (3.8) that is j��ij+i:

j��ij+i =1p2(j00i � j11i) 1p

2(j01i+ j10i)

=1

2[j0001i+ j0010i � j1101i � j1110i]

46

Applying Z-gate on 3rd qubit.

=1

2[j0001i � j0010i � j1101i+ j1100i]

=1

2[j01i (j01i � j10i)� j10i (j01i � j10i)]

= j�ij�i (3.9)

j+ij��i =1p2(j01i+ j10i) 1p

2(j00i � j11i)

=1

2[j0100i � j0111i+ j1000i � j1011i]

Applying X-gate on 2nd qubit.

=1

2[j0000i � j0011i+ j1100i � j1111i]

Applying Z-gate on 3rd qubit.

=1

2[j0000i+ j0011i+ j1100i+ j1111i]

=1

2[j00i (j00i+ j11i) + j11i (j00i+ j11i)]

= j�+ij�+i (3.10)

Using value of eq.(3.9) and eq. (3.10) in eq. (3.8), we have

j�+iABj�+iDE =1

2

�j�+iBDj�+iAE + j�iBDj�iAE+j�+iBDj�+iAE + j+iBDj+iAE

�(3.11)

Now she perform measurement of particles BD. So after the measure-ment state of particle AE will be in one of the following entangled statefj�+iAE; j��iAE; j+iAE; j�iAEg. The entangled state is determinedby the measurement on eq. (3.11). Suppose after the measurementthe state of particles BD collapses to the state j��iBD so the particlesAE will be in state j��iAE. Then Eve will transmit particle B to Bob.Both Bob and Alice do not know that there is Eve which is trying toget information. If they do not test the quantum channel, Bob willproceed as usual, Therefore a part of messages might be leaked to Eve.

47

However by testing quantum channel Eve and Bob can �nd Eve. Incasethey locate Eve, they discard EPR pair and prepare new EPR pair.Infect after the bell measurement performed by Eve of particles BD,Particles AE are in an entangled state j��iAE

j��iAE =1p2(j+iAj�iE + j�iA) j+iE;

And particle BD are also in an entangled state.

j��iBD =1p2(j+iBj�iD + j�iB) j+iD;

But there is no correlation between Alice�s Particle A and BoB�s par-ticle B. So when Alice and Bob performs measurements on fj+i; j�igindependently, the result will be random without any correlation. Ifthis is the case that means there exist an Eve who has entangled herparticle with Alice�s or Bob�s particle. Hence EPR pair must be dis-carded and New EPR pair should be used for Teleportation of unknownstate.

2. Eve can obtain information by Coupled EPR pair with her probe inpreparing EPR pair. We can check whether quantum channel is per-fect or not by following simple strategy. We select a random subset ofEPR pairs. Alice and Bob perform a measurement in basis fj0i; j1igor basis fj+i; j�ig randomly. If the measurement outcomes are com-pletely correlation in the same basis of Alice and Bob, then the quan-tum channel is completely perfect or secure, because because EPR pairstate is the simultaneous eigenvalues of the operators �Ax �

Bx and �

Az �

Bz

with the same eigenvalue 1. here �x and �z are Pauli operators. How-ever if measurement outcome of Alice and Bob are random and theyare not completely correlation in measurement perform on same basis,that means their might be a potential Eve, who have coupled his EPRpair with pair of Alice and Bob.

Here we give an example describing brie�y if there an Eve then how Aliceand Bob �nd her. Let Alice�s particle A, Bob�s particle B and Eve�s particleE be in following entangled state

j�+iABE =1p2(j000i+ j111i)ABE ; (3.12)

j0i =1p2(j+i+ j�i) ; j1i = 1p

2(j+i � j�i) ;

48

Putting value of j0i and j1i in eq. (3.12) and simplifying, we have

j�+iABE =1

2

�j+iA (j+iBj+iE + j�iBj�iE)+j�iA (j+iBj�iE + j�iBj+iE)

�: (3.13)

If Alice and Bob performs measurement on fj+i; j�ig basis, Bob will obtainj+iB and j�iB with the same probability 1

2, whether Alice�s measurement

outcome is j+iA or j�iA. This mean Alice�s and Bob�s measurements onare not in correlation. There is an Eve, so they must discard the EPR pair,prepare new one and check the security of new EPR before sending state.One advantage of this scheme over schemes mentioned in section (3.1) is

that we check security before sending actual information. Whenever a po-tential eavesdropper tries to get information, we locate him/her and discardthe e¤ected EPR pairs. Prepare new EPR and check security again. In thisscheme we only send state when we are 100% sure that quantum channel isperfect.

3.5 Result And Conclusion

We worked on a scheme for secure direct communication. The most obviousadvantage of this scheme is that there is no need for establishing a shared se-cret key. Communication is based on EPR pairs as quantum channel and thephenomena of Teleportation. In �rst step, EPR pairs are prepared. Securityof quantum channel is checked. In case of perfect quantum channel Bob en-codes secret message directly on a sequence of particle states and transmitsthem to Alice by teleportation. Teleportation sends Bob�s message withoutrevealing any information to a potential eavesdropper. Alice can read out themessage by measurement on her particles. Since there is no need of sharedsecret key, so the scheme is completely secure if quantum channel is perfect.In the schemes [7, 8, 9] a qubit carrying secret message is transmitted in

a pubic channel, so there are chances for an eavesdropper of getting informa-tion. However in our scheme there in no chance at all for an eavesdropperto get information. In other words, our protocol has high capacity to defendsignal against interference.

49

I am Awais Chattah and never give a dam care to what people usuallycalled hard work

50

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