Quantum Computing and Information

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QUANTUM COMPUTING AND QUANTUM INFORMATION

Pavithran S Iyer, 3rd yr BSc Physics, Chennai Mathematical InstituteH-1 SIPCOT IT-Park, Siruseri, Padur Post, Chennai - 603103

Email: [email protected]& [email protected]

Typeset Using LATEX

LAST UPDATED: December 26, 2010
mailto:[email protected]:[email protected]:[email protected]:[email protected]


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Contents

I INTRODUCTION 7

1 Brief overview 91.1 Linear Alzebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2 Basic Quantum mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.3 Qubit - basic unit of quantum information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.4 Multiple Qubits. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.5 Quantum Gates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.5.1 Other single qubit gates:. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.6 Bloch Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.6.1 Generalizing Quantum Gates - Universal Gates . . . . . . . . . . . . . . . . . . . . . . 151.7 Important conventions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.8 Measurement Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.8.1 Quantum Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.8.2 Quantum Copying or Cloning circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.9 Quantum Teleportaion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.9.1 Bell States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.9.2 EPR paradox and Bells inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.9.3 Application of Bell States: Quantum Teleportation . . . . . . . . . . . . . . . . . . . . 221.9.4 Resolving some ambiguities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

1.10 Quantum Algorithms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251.10.1 Simulating classical circuits using Quantum circuits . . . . . . . . . . . . . . . . . . . 251.11 Quantum Parallelism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

1.11.1 Example of Quantum Parallelism - Deutsch Jozsa Algorithm . . . . . . . . . . . . . . 28

II PREREQUISITES - MATHEMATICS 33

2 Linear Algebra 352.1 Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.2 Linear dependence and independence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.3 Dual Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.4 Diracs Bra Ket notation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

2.5 Inner and outer products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372.6 Orthonormal Basis and Completeness Relations . . . . . . . . . . . . . . . . . . . . . . . . . . 382.7 Projection operator. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.8 Gram-Schmidt Orthonormalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 392.9 Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412.10 Hermitian Matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422.11 Spectral Theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432.12 Operator functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

2.12.1 Trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452.13 Simultaneous Diagonalizable Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

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2.14 Polar Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 462.15 Singular value decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

2.15.1 Proving the theorem for the special case of square matrices: . . . . . . . . . . . . . . . 482.15.2 Proving the theorem for the general case: . . . . . . . . . . . . . . . . . . . . . . . . . 48

3 Elementary Group Theory 51

3.1 Structure of a Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513.1.1 Cayley Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513.1.2 Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 563.1.3 Quotient Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 573.1.4 Normalizers and centralizers. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

3.2 Group Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593.2.1 Direct product of groups. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593.2.2 Homomorphism. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 593.2.3 Conjugation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

3.3 Group Actions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613.3.1 Generating set of a group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613.3.2 Symmetric group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613.3.3 Action of Group on a set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

3.3.4 Orbits and Stabilizers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 623.3.5 Orbit Stabilizer theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

III PREREQUISITES - QUANTUM MECHANICS 65

4 Identical Particles 674.0.6 Describing a two state system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674.0.7 Permutation operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674.0.8 Symmetry and Asymmetry in the wave functions . . . . . . . . . . . . . . . . . . . . . 684.0.9 Extending to many state systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 694.0.10 Bosons and Fermions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

5 Angular Momentum 71

IV PREREQUISITES - COMPUTATION 77

6 Introduction to Turing Machines 796.0.11 Informal description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 796.0.12 Elements of a turing machine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 796.0.13 Configurations and Acceptance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 816.0.14 Classes of languages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

6.1 Examples: Turing machines for some languages . . . . . . . . . . . . . . . . . . . . . . . . . . 836.1.1 L(M) = {|{a, b}} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 836.1.2 {ap| p is a prime} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

6.2 Variations of the Turing Machine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 866.2.1 Multi-Track Turing Machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 866.2.2 Multi-Tape Turing Machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 866.2.3 Multi-Dimensional Turing Machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 886.2.4 Non-Deterministic Turing Machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 896.2.5 Enumeration Machines. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 896.2.6 Equivalence of the Turing machines and Enumeration machines . . . . . . . . . . . . . 89

6.3 Universal Turing machines. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 906.3.1 Encoding Turing machines over{0, 1} . . . . . . . . . . . . . . . . . . . . . . . . . . . 916.3.2 Working of a Universal Turing machines . . . . . . . . . . . . . . . . . . . . . . . . . . 91

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6.4 Set operations on Turing machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 916.4.1 Union of two turing machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 926.4.2 Intersection of two turing machines. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 926.4.3 Complement of a Turing machine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 926.4.4 Concatenation of two Turing machine . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

6.5 Halting Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

6.5.1 Membership Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 946.6 Decidability and Undecidability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 946.7 Quantum Turing Machines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

7 Computational Complexity 95

V INFORMATION THEORY 97

8 Fundamentals of Information Theory 998.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 998.2 Axiomatic Definition of the Shannon Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . 998.3 Interpretations of the Uncertainty Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

VI CODING 109

9 Classical Coding Theory 1119.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

9.1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1119.1.2 Notations from graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1129.1.3 Unique Decipherability. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

9.2 Classifying instantaneous and uniquely decipherable codes . . . . . . . . . . . . . . . . . . . . 1149.2.1 Part 1: Krafts Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1149.2.2 Part 2: Macmillans Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1159.2.3 Part 3: Converse of Krafts inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

9.2.4 Bound on codeword length - Shannons Noiseless Coding Theorem[26] . . . . . . . . . 1189.3 Error Correcting Codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1229.3.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1229.3.2 Code parameters for a good error correcting code . . . . . . . . . . . . . . . . . . . . . 1239.3.3 Bound on the code distance - The Distance Bound . . . . . . . . . . . . . . . . . . . . 1239.3.4 Bound on the number of codewords . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1259.3.5 Parity Check codes[25] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1329.3.6 Linear Codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

9.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1339.4.1 Repetition Code[24] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

10 Quantum Codes 13510.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

10.2 Errors in Quantum codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13510.2.1 Operator sum representation[3] [22] . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13610.2.2 Lindblad form using the Master Equation[24]. . . . . . . . . . . . . . . . . . . . . . . 13710.2.3 Error Correction Condition for Quantum Codes[24] . . . . . . . . . . . . . . . . . . . 139

10.3 Distance Bound [1] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14210.4 The Quantum Singleton Bound [1] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14210.5 Quantum Hamming Bound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14310.6 The Quantum Gilbert Varshamov Bound[2]. . . . . . . . . . . . . . . . . . . . . . . . . . . . 14410.7 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

10.7.1 Bit Flip code [24] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

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10.7.2 Phase flip Errors [24]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14710.7.3 Bit-Phase flip errors - Shor Code[24][23]. . . . . . . . . . . . . . . . . . . . . . . . . . 148

11 Stabilizer Codes 15111.1 Pauli Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15111.2 Motivation for Stabilizer codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152

11.3 Conditions on stabilizer subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15311.3.1 Generating set of the stabilizer subgroup . . . . . . . . . . . . . . . . . . . . . . . . . 15311.3.2 Structure of the stabilizer subgroup . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

11.4 Error Correction for Stabilizer codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15411.4.1 Notion of an Error in a Stabilizer code . . . . . . . . . . . . . . . . . . . . . . . . . . . 15411.4.2 Measurement on the stabilizer code. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15411.4.3 Error Correction condition for Stabilizer codes . . . . . . . . . . . . . . . . . . . . . . 156

11.5 Fault tolerance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15711.5.1 Unitary gates in the stabilizer formalism. . . . . . . . . . . . . . . . . . . . . . . . . . 158

VII References 165

A Solutions to Exercises 171

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INTRODUCTION

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Chapter 1

Brief overview

1.1 Linear Alzebra

1. Tensor Product: A Tensor product is a method of multiplying two tensors (matrices), given by thegeneral (pneumonic) form: (It is represented as)

a11 a12 a13 a14 . . .a21 a22 a23 a24 . . .a31 a32 a33 a34 . . .a41 a42 a43 a44 . . .

. . . . . . .

. . . . . . .

. . . . . . .

M=

a11M a12M a13M a14M . . .a21M a22M a23M a24M . . .a31M a32M a33M a34M . . .a41M a42M a43M a44M . . .

. . . . . . .

. . . . . . .

. . . . . . .

where,M is any matrix. If the matrices on the LHS have the

dimensions (D1r D1c) and (D2r D2c)respectively,then the result matrix, on the RHS will have the dimensions: [(D1r+ D2r) (D1c+ D2c)]

It is really important to note that the above is NOT a formal definition for a tensor product, but itis just a pneumonic form. The result in this case seems to have the same dimensions as the originalmatrix in the LHS (first matrix). Only when we expand the RHS by putting various values of M, wewill get the matrix of the dimensions (as given above).

1.2 Basic Quantum mechanics

We shall look at fundamentals of quantum mechanics in greater detail in the next chapter.

Some important definitions are:1. State of a system:1 The state of a quantum system is a vector in the infinite dimensional complex

vector space known as the Hilbert Space.The state of a classical system is represented by its position and momentum in a phase space. But thisis not possible in Quantum Mechanics because of the built in concept of the Heisenbergs Uncertaintyprinciple. So, if we close in into one definite value of position for a position, the possible valuesthat its momentum can take is infinite. So, we can only close on a given area, and say , with someprobability that the particle lies within that area. So, the (classical) state of the particle lies anywhere

1A deeper picture of this is given in the quantum mechanics section. For now, this description would do.

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1.3. QUBIT - BASIC UNIT OF QUANTUM INFORMATION CHAPTER 1. BRIEF OVERVIEW

in that continuous area that we define. So, the state of the particle has infinite position and momentumcoordinates. Hence, it is represented as a vector in an infinite dimensional complex vector space (HilbertSpace). The state of a N-independent particle system where the individual particle wave functions are|1,|2,|3 ... |N, the combined state of the N particle system is given by:

|= |1 |2 |3...... |N

In general, if |1, |2, |3, ..... are vectors inN1, N2, N2, ..... dimensional complex vector spacesrespectively, then| is a vector in a (N1 + N2 + N2 + .....) dimensional complex vector space.

2. Dirac Bra-c-Ket Notation: The dirac ket notation is a well known and frequently used here. In this

notation, every column vector is represented as: | which is also called a Ket vector. Similarly, a

row vector is represented as| which is called a Bra vector. Hence the name: BracKet notation.3. Local and non Local processes: By local process between two particles , it is meant that influ-

ences between the particles must travel in such a way that they pass through space continuously; i.e.the simultaneous disappearance of some quantity in one place cannot be balanced by its appearancesomewhere else if that quantity didnt travel, in some sense, across the space in between. In particular,this influence cannot travel faster than light, in order to preserve relativity theory.

4. Canonical Commutation Relations:The some observables in Quantum mechanics do not commute.

That is, they have a non zero commutator. The commutator for a pair of operators is defined as:

[A, B] = AB BA CommutatorThe basic commutator relations in quantum mechanics are:

[xi, pj ] = i

[xi, xj ] = 0

[pi, pj ] = 0

1.3 Qubit - basic unit of quantum information

A bit, a classical two state system, represents the smallest unit of information. A classical bit is representedby 1 or 0. It can be thought of as true and false or any two complementary quantities whose union is theuniverse, and intersection is the null set. There is a profound reason to why the smallest unit of informationis a 1 or 0, or true or false. This is because any logical querry can be split into a series of yes or noquestions. That is, with a series of yes or no answers to questions, we can perform any logical query. Thisis why a bit (which can be thought of as a most general strcture of storing information) is a 1 or a 0.A quantum bit, is just an example for a two state quantum system. A qubit can can also be an electron (withspin up and down), an ammonia molecule, etc. In quantum mechanics, a two state quantum system doesnot mean that it has only two states. This is what distinguishes a qubit from a classical bit. The differencecomes due to a very important quantum mechanical phenomena known as interferrence. Just like how aclassical bit can take a value 0 or 1, a quantum bit can take values |0 or |1, and any value produced by theinterferrence between the states |0 and |1 (like |0 +|1). Since, there are infinite such superpositions(where each of the states have 0

and 1

given by some probability amplitude and respectively), a qubit

can exist in infinite states. If each state can store a unit of information, then the qubit can hold infinite unitsof information.

A Qubit, like any other two level quantum system, is (conventionally) represented by its state:

| = | +|| = |+ +|| = |0 +|1

where||2 + ||2 = 1.

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CHAPTER 1. BRIEF OVERVIEW 1.4. MULTIPLE QUBITS

This can be misleading, since it gives us the feeling that a classical bit can at most carry 2 units of informationwhereas a quantum bit can carry infinitely many units. But, if a measurement is done on the state of theQubit, it collapses into one of the eigen states of the measurement.So, if measurement of| gives a, then after this measurement, the state of the Qubit will remain |a (theeigen state of the measurement corresponding to the eigen value a). This new state |anow, will got give thesame measurement results as |a. In fact it will not respond to any other measurement.Why this happens Postulates of Quantum Mechanics.Hence, only a single unit of information can be retrived from a Qubit.

1.4 Multiple Qubits

Any two state system (like the electron which has a spin) can be represented by a Qubit. But what aboutrepresenting the state of two electrons (which are independent of each other), using a qubit? Such a repre-sentation, we saw in the beginning of the chapter, was possible. If the state of one electron is|1 and stateof the other electron is |2, then the system of two independent electrons can be collectively represented bythe state|, where:

|

=|1

|2

Since, we take a direct product of the two states, we may represent the now state, which is the two qubitstate, as: (using the convention: |i |j |ij)

|= 00|00 +01|01 +10|10 +11|11where: |00|

2 + |01|2 + |10|

2 + |11|2 = 1

and |ij |2 is the probability of the first qubit being in state |i, and the second Qubit being in state |j. If

we want only one of them, we must sum over the other. If we want the probability of system being in state|i only, we must sum |ij|2 over all js.So, the probability of measuring the first qubit to be 0 is = |00|

2 + |01|2, and obviously, the measurement

will collapse the state of the qubit into |

=

|00|2 + |01|2|00|2 + |01|2 .Therefore, we can say that in ths 2 qubit system, we can retrieve 2 units of information. It certainlycan deliver more information than a single qubit system, but there are some difficulties too.In general, we need to carry the measurement process twice to determine the information stored in both thequbits. So, earlier we were carrying out the measurement only once, and now we need to do it twice. Canit be better? Can we get away in one measurement itself? In other words, can we store some amount ofinformation about one qubit in another, such that we can guess both the qubits, by measuring only one ofthem? The answer is yes. We can do such a trick that can, with certainty, retrieve the information stored inone qubit, by measuring the other. Such a two qubit state is called a Bell State or the EPR pair.

The Bell State is given by : | = |00 + |112

Here, the first Qubit being is measured to be 0 with the Probability 12 (changing the state to : | = |00)

and 1 with Probability 1

2(Changing the state to : | = |11)

Hence, P(measuring the first qubit to be 0) = P(measuring the second qubit to be 0) , also state afterthe measurement of the first qubit to be 0 = state after the measurement of the second qubit to be 0.Similarly, P(measuring the first qubit to be 1) = P(measuring the second qubit to be 1) , also sate after themeasurement of the first qubit to be 1 = state after the measurement of the second qubit to be 1.Therefore the first qubit always gives the same result as the measurement of the second qubit. Hence, wecan say that: By knowing the result of measurement to the first qubit, we can tell by certainity,

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the result of measurement of the second qubit.Also we can say The two states are perfectly correlated. In the language of quantum mechanics, the twostates are Entangled.

It is also easy to note that this property cannot be satisfied by any arbitrary state. Hence, we need touniquely classify these states.2 Its allpications and significance become more prominent as we procees in the

later sections.

1.5 Quantum Gates

Just as we have Classical Gates that operate on classical bit(s), we also have their quantum analogues.To Start with, consider a simple classical gate, the NOT gate:

We can now think of its quantum mechanical analogue:Consider the Gate G that is a quantum mechanical NOT gate. Gflips the state of a qubit.

G: (|0 +|1)(|1 +|0)

We can now try to see how this process of flipping the qubit is carrier out. We know that, in the case oftwo level systems, (or Spin Half) the state|0can be changed to|1and vice-versa using the ladder operatorsgiven by S and S+. The action of these operators is given by:

S+ = Sx+ iSyS= Sx iSy

S+|0= |1 Raising operatorS+|1= 0 Raising operator

Simillarly:

S|1= |0 Lowering operatorS|0= 0 Lowering operator

Therefore we can say: textbfG(S++ S)(S++ S)(|0 +|1) = (S|0 + S|0) +(S|1 + S|1)

(|1 + 0) +(|0 + 0)

|1

+|0

Hence, G(|0 +|1) = |1 +|0

A Very important aspect to take note of is that: the state |0 is very different from 0. The latter means anull vector. It represents void. While the former represents some state in which a particle is present in. Thestate|0 does not mean void.

2I still cannot get what is so special, it almost seems like they are two identical quantum states. It is like taking two classicalbits 0 and 0 and saying that the result to measurement of 0 is = to the result of measurement of the other 0 state. Whatproperty of QM is being used?

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Since the state of the qubit is a ket vector, we can also have a column vector representation for it:

| =

now the Gate G can be defined as:

G = By looking at this property, we can guess the matrix form ofG, to be:

G=

0 11 0

. We can verify that this is the x+ iy or Sx+ iSy operator, as computed above.

1.5.1 Other single qubit gates:

There are many single qubit gates. A major requirement for a quantum gate operator is that it must beUnitary. This has 2 consequences:

1. The conservation of probability.

2. Since the inverse of an unitary matrix is also an unitary matrix, each single bit quantum gate can beundone by some other single bit quantum gate. So, the input can be obtained by performing someoperation on the output. Therefore there is no loss of information, unlike the classical case wherethe gates are notinvertible.

Let us consider aZ Gate that leaves|0 unchanged, and flips the state of|1 to|1.

Z(|0 +|1) = (|0 |1)We can also guess that: Z = 1 00 1. This is similar to the z or Sz operator (pauli spin matrix).Let us now consider yet another important single qubit gate, the Hadamard Gate. This gate trans-forms the states |0 to a states between |0 and |1. This gate can be equated to an operation which isreflection of the qubit vector about the line =

8. It changes |0 to |0 + |1

2and changes |1 to |0 |1

2.

H(|0 +|1) = ( |0 + |12

+|0 |1

2)

The Matrix form ofH can be guessed as: H = 1

2 1 11

1

1.6 Bloch Sphere

The state of any two state system is represented by a point in a two dimensionalcomplexvector space. Since,this is a two dimensional complex vector space, each dimension (like we have x, y, and z dimensions in thecartesian frame) is complex. Each complex number needs two real numbers to represent itself. So, the state

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can now be described by four real quantities, rather that two complex quantities. Hence, we have:

The state in the two dimensional complex vector space is: |= |0 +|1Since, and are complex, each of them can be described by two

real quantities: = (r, i) and = (r , i)

The normalization condition: ||2 + ||2 = 1 , now translates to the condition on the

four real quantities as: |r|2 + |i|

2 + |r|2 + |i|

2 = 1. (1.1)

Just like how the equation x2 + y2 + z2 = 1 represents the surface of a sphere, placed in a three dimensionalspace, the above equation(1.1) represents the surface of the sphere kept in a four dimensional space. Now,this is the motivation for us to try to describe the state of a two state system as a point on the sphere. Afour dimensional space is still a bizarre object for us to imagine. So, we need to try and remove one degreeof freedom here so that we get our usual two dimensional sphere in a three dimensional space. The twodimensional sphere hence obtained is known as the Bloch Sphere. For this purpose, we need to work outthe above process in the polar form. So, we shall have:

The complex numbers: = rei and = re

i

Therefore, the state of the system,|= rei + rei

So, till this point, we have been working with a sphere kept in a four dimensional space, as there are fourreal quantities in the equation of the state. Now, we need toremoveone degree of freedom, that is eliminateone real quantity from the equation of the state.For doing so, we need to recollect a very important feature of quantum mechanics that any quantum mechan-ical system is invariant under rotation by a overall phase. The following sentence can be realized if we goback to our basic state of the quantum mechanical system. The state is represented by | which is actuallya probability amplitude. But what we can measure is the probability density, denoted by ||2. So, even ifthere is an overall phase factor, like , in the probability amplitude, it will not affect our measurement, andfor all values of, the system will be identical3. So, let us multiply the system byanyan overall phase angle.The choice can be decided by us, to suit our requirements.

Since the choice ofcan be artitrary, let = ei , so that, we can eliminate one real variable.

Therefore, the state of the system, ei(|) = re

i(ei) + rei(ei)

Now, the LHS remains as | since the system is invariant under overall phase change.Let: = . Then, on simplification we have: |= r|0 + rei

Now, in the above expression, the second term of the RHS has a complex coefficient.

We can write this in the complex form (x + iy):

|= r|0 + (x + iy)|1 where x = r cos and y = r sinNow, let us apply the normalization condition: |r|

2 + |x|2 + |y|2 = 1

Therefore, we now get the equation of the Bloch sphere.

Hence, we got the equation of the Bloch sphere. We now need to find how the state of a system can berepresented on the bloch sphere. For this, let us go to spherical coridinates. Let us map x, y andr on to asphere with unit radius.

Let us now make the transformations:

x = cossin

y = sinsin

r = cos

Therefore, the state now becomes: |= cos|0 + sin(cos+ sin)|1We can write this as: |= cos|0 + (ei)sin|1

3In fact, this probability amplitude is the reason for interference.

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Now, by convention, in spherical polar coridinates, goes from 0 to 2. But here, if we put = 0, we get

| = |0 and on putting = 2

, we get| = (ei)|1.So, we see that = 0 to =

2covers the entire sphere. So, we modify our by changing it to

2. Now

= 0 to =

2 covers the entire sphere. So, the final equation of the state of the system in a Bloch sphere

is:

|= cos

2

|0 + (ei)sin

2

|1 (1.2)

So, the above equation represents the single qubit in a Bloch Sphere. Now, what about multiple two statesystems, or multiple qubits ? Let us take a two qubit system represented by its state |= |1 |2. Thestate space of | is now a vector in the four dimensional complex vector space because each of the statesare vectors in a two dimensional Hilbert space.Following the exact similar argument as above, we can see that the state | can be represented as a vector ona seven dimensional sphere, kept in a eight dimensional space. This seems confusing because in the above case,we claimed that a two state system (represented as a vector in H2) can be represented on a two dimensionalsurface. Extending the same argument, one should say that a composition of two state system (representedas a vector inH2H2) should be represented as a vector on a four dimensional surface (need not be a sphere,but still it must be some four dimensional surface). Hence, we see that the surface used to represent |has more dimensions than expected. We know that a seven dimensional surface has more points than a fourdimensional one. We can now conclude that the seven dimensional surface can represent more states than afour dimensional one. Therefore,there is some missing informationthat cannot be represented on the fourdimensional surface. The seven dimensional surface carries more information than a four dimensional one.This means that when we take a tensor product of two 2-state systems, the new system formed containsinformation about each of the individual systems as well as some excess information that cannot be attributedto any single one of them. We arrived at a four dimensional surface from the assumption that all compositionsof two 2-state systems can be represented as a tensor product of some two 2-state systems. From the factthat the sphere which represents all the compositions has more states, we can say that our old assumptionthat all compositions of two 2-state systems can be represented as a tensor product, fails.So, there are some states that are composed of two 2-state systems, but cannot be expresses as a tensorproduct. These states have information (properties) that are not related to any one of the 2-state systems.

This information is lost when the qubits are separated. Hence, this information is due to the tie, or (inmore sophisticated words) entanglement of the two 2-state systems. It is quite clear that all the statesdo not have this property. States that have this property are called entangled states. The Bell State

B00 = |00 + |11

2.

1.6.1 Generalizing Quantum Gates - Universal Gates

Now, since gates are unitary operators, that act on the state of the qubit, we can think of them as rotationoperators on the Bloch Sphere. Now each gate performs some rotation. But any rotation, in general can bebroken into a sequence of standard rotations, about the x-y plane, y-z plane and the x-z plane. Since, anyrotation can be broken into a sequence of standard rotations, we can drawn the same analogy and say that

any Quantum gate can be represented by a sequence of standard gates that act on the state of the qubit |,and produce the same answer as the original gate.Consider a rotation operator U. We can decompose U into several basic rotation operators:

So, U = ei

e

i

2 0

0 ei

2

cos2 sin2

sin2

cos

2

e

i

2 0

0 ei

2

In other words, these are universal operators. Similarly, we can consider universal quantum gates, that,

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when manipulated appropriately, can mimic any other quantum gate.Let us consider a simple classical universal gate, the NAND gate. The XOR Gate is not a universal gatebecause it fails to change the parity of the bits.4:

Figure 1.1: A classical univaersal gate. NAND gate.

Let us now consider a Universal Quantum Gate: A slight modification of the NOT gate, The controlled NOTgate or the CNOTGate.

Figure 1.2: A CNOT Gate

A CNOTGate is a two input NOT gate. It takes two inputs: a DATA and CONTROL,flips DATA if CONTROL = 1. Classically, this is analogous to the XOR Gate.

leaves DATA unchanged if CONTROL = 0.

The Gate operation is represented ast: CNOT:(|A + |B) |A, B AThe action of this Gate can be described explicitly:(|DATA,CONTROL |DATA,CONTROL)|00 |00|01 |11|10 |10|11 |01

4This is not clear to me

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In a Quantum circuit, the CNOT gate is represented as:

Figure 1.3: Circuit representation of a CNOT gate. The top wire has the control bit and the bottom has theTarget or the Data bit.

We can now look at the matrix representation of this Gate.

CNOT =

1 0 0 00 1 0 00 0 0 10 0 1 0

1.7 Important conventions

|0 + |1 X N OT gate |1 + |0

|0 + |1 Z Z gate |0 - |1

|0 + |1 H Hadamard gate |0 + |12

+ |0 |1

2

A most remarkable or unique feature that we notice about a quantum gate, is that it operates on single bits,which are like a superposition of 2 probabilistic classical bits. The unitary(existence of inverse) nature ofthese operators enables the input Qubit to be easily retrieved.

1.8 Measurement Basis

Measurement is an operation that is performed on system, to determine with certainty, the state of the

system. Measurement also has a meaning in classical bits. Each classical bit had a even probability of itbeing 0 or 1. When a measurement is done, we know by certainty what value that bit has. The classical bitis 0 with probability one halfor1 with probability one half. Both these choices forever exclude each other.Whereas in the quantum case, the difference comes here. We have superposition states. For example, a qubitcan be|0 with probability one halfand |1 with probability one half. Note the usage of andand or.The meaning of measuring some property (operator) of a state | is nothing but finding the eigen valuesof that operator (corresponding to the property). The allowed values of any property (or, the result ofmeasurement of any property) are limited to the eigen values of the operator representing this property. Butit is not so straight forward: what if we are trying to measure the property corresponding to the operatorA,and the given state | is not an eigen state of A?

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For example, till now we took| = |0+ |1. This means we took the ket vectors |0and |1 as our basisstates. These vectors are nothing but the eigen vectors of the z or the Sz operator. So, we were in theSzeigen basis. That is, the operator Sz is diagonal in this basis. Now, if we measure a property correspondingto theSz operator, then the measurements results can have only two allowed values: 1 and -1, because theseare the eigen values ofSz. But what if we want to measure a property corresponding to the Sx operator ?Here, |0 and |1 are certainly not the eigen vectors of the Sx operator (they better not be because if theyare the eigen vectors ofSx also, that would mean Sx andSz commute). Here, we make use of a key propertyof the eigen basis. They form a complete basis. So, any state | can be expressed as a superposition of theSx eigen states, and the corresponding eigen values are the result of the measurement. So, to start with,we have a state | in the Sz basis. Now when we want to measure a property corresponding to Sx, wefind the eigen states ofSx, and write | as a superposition of those eigen states. Now the coefficients ofthe eigen states are the measurement results, and the eigen states are those to which | shall collapse aftermeasurement.

So, whichever property (operator) we are measuring, we must expand | in the eigen basisof that operator(correspond to the property), and that find the coefficients of those states in the expansion of|.

1.8.1 Quantum Circuits

A quantum circuit, like any other, is read from left to right. A line in the circuit represents a wire. Awire here may just denote the path of the photon (or passage of time, etc. There is no physical path orchannel.) The input to a circuit is the qubit (represented by |, which by convention, is assumed to be inthe computational basis state).An important feature not allowed in quantum circuits is loops. There is no feedback from one part of thecircuit to another. Thsee feedback forms are of 2 general types: several wires joined together FANIN , and a

copy of the qubit in one wire, going to multiple wires FANOUT. These are not allowed. Refer to figure. Asan interesting consequence of this, the copying of a qubit is an impossible task.

Figure 1.4: FANIN and FANOUT characteristics:

Also, we can generalize the representation of a controlled not gate which takes N qubits. The generalizedrepresentation is shown in the figure:

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Figure 1.5: Generalized CNOT gate

Another important aspect of quantum circuits are meters used to measure the quantum bits.5 As we havealready discussed, the measurement will collapse the qubit into a probabilistic classical bit (distinguished by

drawing a double arrow line)6. |(= |0+ |1), upon measurement, will change to a classical bit thatgives result 0 with probability ||2 and 1 with probability ||2. A representation of the quantum meter isshown below:

Figure 1.6: Representation of a Quantum Bit measuring device:

1.8.2 Quantum Copying or Cloning circuits

One of the key features of quantum computation is that it disallows the replication or copying of Qubits.We can see why this happens. Let us take up the issue of replicating or the copying of a classical bit. Toaccomplish this classically, one could do the following:

5measurement here means to identify the qubit. measuring a state|0 + |1 means to extract out and .6The subtle difference here is that a probabilistic classical bit is fundamentally different from a quantum bit. This is because

a classical has some finite probability of carrying only one piece of information, and it gives that upon measurement. Qubiton the other hand carries infinite information, but when measured, it has some probability of collapsing on to one piece

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Figure 1.7: Classical method of copying a bit. Here stands for the XOR operation. Here, we take anarbitrary bit x and perform an XOR operation with that bit and y, where y is another bit that is constantly= 0.

Let us now try the following with the qubit also: Here, we blindly replace the classical XOR with thequantum XOR or

. A valid question would be : have we copied all the information stored in |

. The

answer obviously is NO. It is obvious because, we agreed that an infinite amount of information is stored in|. So, all of it cannot possibly be copied. This theorem is called the no cloning theorem. There is ashort proof of this theorem.Suppose we have two states |x and |y, such that there exists a machine U which copies the state |x into|y (|y is called the target state), then the system of these two qubits is in the state: |x |y. So, wehave:

U(|x |y) = (|x |x) (1.3)Here U is copying the value of a state into another. We claim that U is some unitary, universal operatorthat can clone any quantum state. Let us not take two arbitrary quantum states | and |. Let the targetstate be represented as |s. According to the above claim, we have:

U(| |s) = (| |) (1.4)similarly,U(| |s) = (| |) (1.5)

We can take an inner product of equations (1.4) and (1.5). To take an inner product of the equations meansto take the inner products on the two RHS to form a new RHS and inner product of two LHS to form a newLHS. The inner product of any two matrices (applies to even column vectors - since they too are matrices)is nothing but the product of the hermitian conjugate of the first with the second matrix.The LHS of equation (1.4) is of the form A|x. This is nothing but a productAB. The hermitian conjugateof (AB) isBA. The LHS will then be:

[U(| |s)][U(| |s)](| |s)U[U(| |s)]

(|

s|)UU(|

|s

)

In the last step, we see that (A B) = (A B). Also since U is unitary, UU = I. Hence the LHSbecomes:

(| s|)(| |s)(|) (s|s)In deriving the above step, we have used the formula: (A B)(C D) = (A C) (B D). Also, sincethe target state |s is normalized,s|s =1.

LHS: (|)

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Coming to the RHS and applying similar simplifications, we have:

(| |)(| |)(| |)(| |)

(|)2

Therefore, we can write the new equation formed as:

(|) = (|)2

This equation is of the form x = x2. The solutions to this equation are x = 1 and x = 0. The first case(|) = 1 means that both| and | are identical states. But then, this would mean that U only copiesstates that are identical. That is, U can only copy some specific state. But this contradicts our assumptionthat U is an universal operator. In the other case where (|) = 0, we see that | and | are bothorthogonal states. So, the cloning operator U will only clone orthogonal states.Therefore, we can conclude that there is no universal operator that can copy any arbitrary state. If it cancopy a given quatum state, then the only other state that it can copy is one which is orthogonal to the givenstate.In the figure given below:

Figure 1.8: Repeating the same above process for a quantum case

The process can be concisely described as:

[|0 +|1]|0 CNOT |00 +|11

The output however, is not equal to || because when we multiply [|0+ |1] by [|0 + |1], weget: 2|00 +|01 + |10 +2|11. But this certainly isnt equal to our result obtained from the CNOTgate since the cross terms |01and |10 are not present. But one can now say that we can copy the quantumbit if= 0. But this must mean that at least one of them must be 0. If this is so, then the bit will nolonger be a qubit.Henceit is impossible to copy the state of a Qubit.

1.9 Quantum Teleportaion

1.9.1 Bell States

We have encountered these Bell States earlier. They are known to be the most correlated pair. The result ofthe measurement on the first qubit is the same as the result of the same measurement on the second qubit.In this case, on measuring the first qubit of the Bell State, one obtains 2 possible answers: 0 with probability1

2and 1 with probability

1

2. On measurement on the second qubit one obtains 2 possible answers: 0 with

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probability 1

2and 1 with probability

1

2. The measurement on the always yields the same result. These Bell

states are formed by taking the two qubit system (in the computational basis state), and pass them into aHadamard gate followed by a CNOT gate. A simple table for this would be:

Table 1.1: Table Showing the input and output states of a Hadamard Transform, used to produce Bell States:

In Process and Out

|00 H |00 + |102

CNOT |00 + |11

2|B00

|01 H |01 + |112

CNOT |01 + |10

2|B01

|10 H |00 |102

CNOT |00 |11

2|B10

|11 H |01 |112

CNOT |01 |10

2|B11

The generalized bell state is given by:

|Bx,y |0, y + (1)x|1, y

2

1.9.2 EPR paradox and Bells inequality

We shall look at these in detail in the next chapter: Fundamentals of Quantum mechanics.We just saw that if we have the result of measurement of one qubit in the Bell state, the result of measurementon the second qubit is determined. Not only this, but even if the two qubits are (practically) infinitely faraway, the result of measurement os determined instantaneously, after measuring one of them. It is surprisinghow this is possible because we know that information cannot be transmitted at a speed greater than thevelocity of light. Therefore, following this difficulty, EPR suggested that the two halfs of the Bell statesalready had some more information, that describes the state of the two partice system at any given time,and that we have not accounted for that information in our formulation of Quantum mechanics. Also, EPRcalled this information, hidden variables. Later on ,John Bell suggested a mechanism to test for the presenceof local hidden variables. He devised an inequality which and he claimed that, if a quantum system does notsatisfy, it is impossible for it to have a local hidden variable.

1.9.3 Application of Bell States: Quantum Teleportation

Consider the following problem:Bob and Alice are two friends living far apart. They together generated an EPR pair or a Bell State. Aliceand Bob have the 2 halfs of the state. Now Bob is hiding, and Alices mission is that she must deliver a state| to Bob.From looking at the problem, we can see that things are bad for Alice. She can only send classical information.She cannot copy the state|. Alice does not even know the state |, and even if she knew it, it would takeinfinite amount of information (and time) to describe| since it takes values in a continuous space.So, the only thing left for Alice is to utilize the EPR pair (take advantage of the property of coherence ofmeasurement). So, briefly, what Alice does is the following:

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1. Alice interacts | with her half of the EPR pair, and gets one of the four possible results: 00, 01, 10,11.

2. She sends the obtained result as classical information to Bob.

3. Bob, knows that his measurement7, i.e; when he interacts | with his half of the EPR pair, the resultwill be the same. So since Bob knows the result (sent by Alice), he decides an appropriate operation8

on his half of the EPR pair, and he gets |.Note that here the | has been communicated from Alice to Bob, without being actually transmitted. Theinformation was conveyed without any transport. So, it can be called Teleported, and hence the name :Quantum Teleportation.

Now let us look into the process more closely:

let |00 + |11

2be the Bell state or the EPR state that Alice and Bob together created. Let the state to be

conveyed to Bob be| (the state to be teleported) = |0 +|1, where and are unknown amplitudes.Now, on acting the EPR pair, with the state |, we get:

|0=||B00

|0= 12 [|0(|00 + |11) +|1(|00 + |11)]

Now, the first qubit represents the message to be teleported and the second

is Alices half of the Bell state. The last is Bobs half of the Bell state.

Alices now will get her qubits into a CNOT gate and obtain |1, where:|1= 1

2[|0(|00 + |11) +|1(|10 + |01)]

Now, on sending|1 through a Hadamard gate, we get|2 where:

|2= 12

|0 + |12

(|00 + |11) +|0 |12

(|10 + |01)

12

[(|0 + |1)(|00 + |11) +(|0 |1)(|10 + |01)]On expanding and rearranging the terms, we can obtain the following expressions:

|= 12

[|00(|0 +|1) + |01(|1 +|0) + |10(|0 |1) + |11(|1 |0)]

So, as per the convention,

Figure 1.9: The first two bits are Alices and the next is Bobs as shown.

7measurement means any interaction with the system8kind of inverse operation of what Alice did

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if Alice performs a measurement and obtains 00, she will send it to Bob, then Bob will see which Qubit ofhis in the expression will also give the same result. So, he will come to the conclusion that the state he wantsis with the Qubit |00. So, we have:

Table 1.2: Alices measurement and the corresponding state |:Alices measurement result State|

3that Bob shall recover and get as the final message from Alice:

00 |3(00) [|0 +|1]

01 |3(01) [|1 +|0]

10 |3(10) [|0 |1]

11 |3(11) [|1 |0]

From the measurement results, Bob will apply a certain operation and retrieve the corresponding Qubit. For

example, if the measurement result is1. 00, then Bob will let | as it is.2. 01, then he will act the| state with an X gate9.3. 10, he will apply Z gate to|.4. 11 then he will first apply X gate to| and then Z gate on the rest.

Therefore, the resultant operation summarizes to:

ZM1XM2 |2=|

whereM1and M2 are the two bits of information send by Alice. These are the results of Alices measurement.

Summarizing all the above operations, we can not draw a circuit diagram that Bob must follow, to retrievethe state|:

Figure 1.10: The top 2 lines of input are for Alice (the first 2 qubits belong to Alice). The bottom qubitbelongs to Bob. The measurement operation that Bob must follow is described in the generalized fashion asZM1XM2 .

9NOT gate, these conventions will be used in most of the places.

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1.9.4 Resolving some ambiguities

The whole process is slightly surprising because it causes several doubts that imply that the teleportationviolates the laws of Quantum Computation, that we earlier agreed upon. The following are the ambigui-ties:

1. This seems to imply that| is being copied by Bob from Alice. This is against the law of no cloningtheorem that we discussed. The subtlety that is hidden here is that, both copies of | never coexistbecause as Bob gets the result of Alices measurement (in odder for him to create his |, Alice hadalready destroy his copy due to the measurement.

2. The process of Teleportation seems to say that the information | is being conveyed instantly, sinceit does not explicitly involve passage of any data. This is misleading because the fact that Bob mustget Alices measurement result, which is in the form of classical information, is a very very key conceptwithout which the teleportation is impossible. So, the speed of conveying the information is limited bythe speed of light10, and does not violate the theory of relativity.

1.10 Quantum Algorithms

Now, we can look at a few like why Quantum Computation is preferred to Classical computation, can aquantum computer do all that a classical computer is capable11, etc. However, Quantum gates cannotbe used directly as classical logic gates because the former are inherently reversible, while the later are

irreversible. But still we can build classical reversible gates.

1.10.1 Simulating classical circuits using Quantum circuits

Any classical circuit or gate can be replaced by an equivalent reversible gate known as the Toffoli gate. ATafolli gate is has three input bits. The last input bit, the target bitis flipped if all the other bits are 1. Itleaves the first two bits unchanged and performs an XOR operation of (AND(all other bits)) with the last

bit. The Taffoli gate can now be used to simulate NAND12

. This gate is now reversible and has an inversewhich is the Taffoli gate itself. The gate can be used as a quantum as well as a classical gate. In the quantumcase, the Taffoli gate takes a state |110 and gives |111. It simply permutes the computational basis states.

10speed of classical information cannot cross the velocity of light - Theory of relativity11It would be surprising if this was not possible because we know that all classical phenomena can be explained through

Quantum mechanics.12If it can simulate a NAND gate, which is a classical universal gate, then it can simulate all classical gates.

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1.11. QUANTUM PARALLELISM CHAPTER 1. BRIEF OVERVIEW

Figure 1.11: A Toffoli gate - Reversible classical gate: Truth table and circuit representations. The last bitis flipped if the first two bits are 1.

1.11 Quantum Parallelism

Parallelism has a different meaning in the quantum case, as compared to the classical case. In the classicalcase, at any given unit of time, only one unit of task is in execution. While, in the quantum case, parallelismactually takes its real meaning. At any given unit or instance in time, all the tasks run in parallel. Thisfeature in Quantum Mechanics (for the multiple qubit case) is achieved due to presence of the superpositionof bits like: [ij|ij]i,j . One of the most simple operations that can be performed on two qubits is an XORoperation (denoted by). For all practical purposes, let us assume XOR gate to be a universal gate. So, ifwe can do in parallel two XOR operations, then we have showed quantum parallelism in action. Let us nowclaim that corresponding to any function f where f:xf(x) we can always define a unitary transformationUf, such that

Uf|x, y

=|x, y

f(x)

We also impose a condition that for the quantum transformation or circuit given by Uf, the inputs shouldnot be in the computational basis.A rough circuit diagram of our set up is given below: (Here, since y is permanently set to |0 the value of|x, 0 f(x) is equal to f(x) itself). In other words,

Uf|x, 0=|x, 0 f(x) f(x)

Figure 1.12: Quantum circuit that computes f(x) (since |y = |0) simultaneously for 0 and 1

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CHAPTER 1. BRIEF OVERVIEW 1.11. QUANTUM PARALLELISM

The output, in this case is:|0, f(0) + |1, f(1)

2. Hence, in one run of the function Uf, we have computed

both f(0) and f(1). A single execution of Ufwas able to compute f(x) for multiple values of x. Hence thename parallelism.This idea of parallelism can be extended to multiple qubit systems also. We can have a quick look at howthis is done. So, our main objective (when generalized to n qubit case) is that, given n qubit states, we must

compute the values for f(x) for all the n states in parallel, that is, in a single evaluation of Uf. To do this,just like the previous case, let us take a (n+1) qubit system with the last qubit equal to|0 (It is similar tothe|ythat we set to|0in the two qubit case). We now need all permutations of the first n qubits (with eachqubit in the computation basis) in the system. To produce all the permutations of n qubit system (whereinitially, all qubits are set to |0) , we require n hadamard gates. We first set all the qubits to |0 initially.Each of the hadamard gate, acting on a qubit produces

H|0= |0 + |12

therefore: (H|0)(H|0) =

|0 + |12

|0 + |1

2

(H|0)(H|0) = |0 + |01 + |10 + |11

22 This can now be generalized to:

(H|0)(H|0)(H|0).......(H|0) =

|0 + |12

|0 + |1

2

|0 + |1

2

....

|0 + |1

2

(H|0)(H|0)(H|0).......(H|0) =

i{all permutations of n qubits} |i2n

The above statement is in short represented as:

Hn|0n = i{all permutations of n qubits}

|i

2n (1.6)whereHn denotes the parallel action of n hadamard gates. This is also called a hadamard transform to thefirst n qubits.Now this (n+1) qubit system can be sent to the function Uf, and for each permutation, the function Ufwill produce f(x) where x{ permutations of n qubits}. The final answer that will come as output from thefunctionUf is:

Uf(Hn|0n|0) = 1

2n

x{all permutations of n qubits}|x, f(x)

Therefore, we can compute 2n values of f(x) for different values of x, in one single evaluation of the function

f.It is clear that this parallelism could not be possible if the qubits were not in the superposition of states.In other words, we have shown that the parallelism works only in the quantum case and not in the case ofprobabilistic classical bits because the classical bits cannot be in superposition of states.But the story does not end here. There is a catch. The problem is that even if Uf computes all thevalues of f(x) in one single evaluation, and returns the result, as shown above, we are still restricted (by thenature of measurement that can be made in quantum mechanics) to only obtain one value of f(x). After thismeasurement, the entire output will collapse to an eigen state. So, though we have shown to compute f(x)for 2n values of x, we are allowed to look at f(x) for only one value of x. But there is a way of getting out ofhere, and we shall see how, in the later sections.

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1.11.1 Example of Quantum Parallelism - Deutsch Jozsa Algorithm

In the previous few sections, we have been talking about features that are present only on quantum computersand not in classical ones. In this section, we shall see how these features can come together and outperformthe classical computer (for this problem which is going to be presented). This problem is given by Deutschand Jozsa. The problem formulation is given as the following situation:

Alice in Amsterdam selects a number between 0 to2n 1 and mails it in a letter to Bob in Boston. Bob takesthe number x(sent by Alice) and computes f(x) and replies with the result, which is either 0 or 1. Now Bobhas promised to use a function which is of one of the two kinds; either f(x) is constant for all values of x, or

f(x) is balanced, i.e; equal to 1 for exactly half of all possible values and 0 for the other half. Alices goal is todetermine with certainty weather Bob has chosen a constant or a balanced function, corresponding with himas little as possible. How far can she succeed ?

Let us start by taking the naive classical method, which is when Alice shall query Bob at most 2n1 + 1times (until she gets a different number from Bob). If she gets a different answer at any instant, she canconclude that f(x) is balanced. If she gets a particular 2 n1 + 1, then it means that the function is constant(this is because, if it hadnt been constant, it must be balanced and hence it must have 2n1 values of theother kind. But our function in this case has 2n1 + 1 values of one kind.). So, in this solution, Alice and

Bob need to communicate for at most for 2n1 + 1 values of x.Let us try to do better by taking the quantum analogue of this problem. Let Alice and Bob be allowed toexchange Qubits instead of classical bits, and let f(x) now be a function that operates on classical bits. Theoutput of the function, however the type of output remains the same. Let us now work through the processfor the simple n = 2 case. So, we need Alice to choose a number between 0 and 3. Alice has 4 choices, and weneed two qubits to distinctly represent 4 different quantities. So, let us suppose Alice chooses some numberthat is represented by the state |0 = |01. Now, let us send each qubit through a Hadamard gate, andobtain the new state |1. The parallel action of two Hadamard gates on the state |0 produces |1. So,we have:

input state (Alices number): |0=|01now:H2|0= |1

|1= |0 + |12 |0 |12 Now, let us apply the transformation Uf to the two qubits obtained. (Uf is the quantum circuit that wediscussed in the last subsection). Just for recap,

Uf|x, y=|x, y f(x)We can now send the two qubits of |1 into the quantum circuit Uf, as |x and |y respectively. Let usdefine the new qubit|2, which is obtained after acting|1 with Uf. So, we get:

|2= Uf

|0 + |12

|0 |1

2

On expanding the product, we get: |2

= Uf |0(|0 |1) + |1(|0 |1)2

On separating the terms: |2=

Uf|0(|0 |1)

2 + Uf

|1(|0 |1)2

(1.7)

Here, both the terms which are separated out and being added, are of the

same form: Uf|x

|0 |12

(1.8)

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Therefore, let us first try to compute a general form for the expression: Uf|x

|0 |12

Hence, we get: Uf|x

|0 |12

Uf

|x, 0 |x, 1

2

On the action of Uf the expression becomes: |x, 0 f(x) |x, 1 f(x)

2 Which can be further simplified to: |x

|0 f(x) |1 f(x)

2

Assuming that we know the property of the(XOR) operation, let us consider two cases: f(x) = 0 and f(x)= 1.

Taking the first case: f(x) = 0:

|x

|0 0 |1 02

|x

|0 |1

2

Similarly, taking the second case: f(x) = 1:

|x |0 1 |1 12 |x |1 |02 Therefore, we can summarize by giving a general form for |x

|0 |1

2

by inspecting the above two cases,

as the following:

|x

|0 |12

= (1)f(x)|x

|0 |1

2

(1.9)

Now, since we have obtained a general form for the expression, we can substitute values of x to get the termsfor the expressions for |2, that we obtained earlier in (1.7). The first and second terms of the RHS (ofexpression in (1.7)) can be evaluated directly by substituting values of x (of expression in (1.9)) to be 0 and1 respectively.Therefore, we obtain:

Putting x = 0 in the first term and x = 1 in the second term, of the RHS:

|2= (1)f(0)|0

|0 |12

+ (1)f(1)|1

|0 |1

2

The above expression gives the same value if f(0) = f(1), and a different value otherwise. So, we have twodifferent possibilities of evaluation: one is if f(0) = f(1), the other is when f(0)= f(1).

Let us consider the case f(0) = f(1). So, let f(0) = f(1) = c. Hence, we get:

|2= (1)c

|0 + |12

|0 |1

2

. Here, we can write (1)c as

Similarly, for the other case, f(0) = f(1), we have the following argument.If f(0)

= f(1), then (

1)f(0)

(

1)f(1). So, we have: let (

1)f(0) =

(

1)f(1) = c

Hence, the expression becomes: |2= c

|0 |12

|0 |1

2

. Here, we can write c as

So, we obtain two expressions for |2. Now we can write them as:

{|2=

|0 + |1

2

|0 |1

2

f(0) = f(1)

|0 |1

2

|0 |1

2

f(0)= f(1)

} (1.10)

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Now, we have obtained|2. So, let us now act the first qubit with |2, to produce |3. So, we get: H|2=|3. Let us see the action ofHon the first qubit, separately for the two cases which are (1.10) and (1.10).Let us expand the product in (1.10) and see the action of H on the first qubit.

H|2=H

|00 + |01 + |10 + |112

We can act H with the first qubit and leave the second one unchanged. So, we get:

H|2= 12

|0 + |1

2

|0

|0 + |1

2

|1 +

|0 |1

2

|0

|0 |1

2

|1

Hence, on simplifying: H|2= 12

2[|00 |01 + |10 |11 + |00 |10 |01 + |11]

H|2=|0

|0 |12

Similarly, on carrying out the calculations, for the other case of|2, we get:

H|2=|1

|0 |12

So, we obtain two expressions for |3 just like how we got for |2. Now we can write them as:

|3=

|0

|0 + |12

f(0) = f(1)

|1

|0 |12

f(0)= f(1)

(1.11)

The above expression (1.11) can be writtern in a more shorter and concise manner, in the following way.:

|3= |f(0) f(1)

|0 + |12

(1.12)

Hence, by measuring the first qubit, Alice can, by certainty can say weather f is balancedorconstant. Hence,by a single run of the function, Alice was able to determine a global property of the function. Earlier we sawthat though we could compute the function for lots of inputs, but we could measure only one result. Here,with the same constraint, we have accomplished our task. So, on summarizing we have done the followingsteps:

1. The input state was prepared.

2. Hadamard Transform on the input state was performed.

3. The qubit state was now passed to theUftransformation.

4. The first qubit of resultant state was passed through the Hadamard gate.

We can now generalize the whole procedure for the multiple qubit case, that is Alice can choose any numberbetween 0 to 2n where n can be as large. So, in the multiple qubit case, Alices query register can berepresented by a n qubit state, all of which are initially set to zero. Alices input state is a (n+1) qubit statewith the last qubit set to |1.

Alices input state: |0=|0n|1

Now, let us perform a Hadamard Transform on the query register and pass the last qubit (answer register,which Bob will modify and send |1 into a Hadamard gate, to get |1 . (this is similar to (1.8)). In theprevious case, instead of the transform, we had only one gate. From (1.6), we know the result of Hadamard

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Transform on N qubits, where all are initially set to |0, gives

x{0,1}n

|x2n

, and the hadamard gate on the

single qubit|1 gives

|0 |12

. So, on their parallel action, we get:

|1= x{0,1}n

|x

2n |0

|1

2 (1.13)where (x{0, 1}n) means all permutations of 0 and 1 which are of length n.Bob now takes the input state sent by Alice and computes the function f using the transformation Uf. Bobnow sends the answer in |2 (the form of|2 resembles the equation in (1.9)):

|2=

x{0,1}n

(1)f(x)|x2n

|0 |1

2

(1.14)

Alice has now a set of Qubits in which the result of Bobs function is stored in the amplitude of the superpo-

sition state:

x{0,1}n(1)f(x)|x

2n

|0 |12

. She now interferes the terms in superposition with a Hadamard

transform, to get|3. To calculate the effect of the Hadamard transform, we can first see how the transformin general works with the n-qubit state |x1, x2, x3, ....xn:

Hence, we need to calculate: Hn|x1, x2, x3, ....xn

Let us now go one more level down, and first try to calculate the result of Hadamard transform on a singlequbit: Hn|x

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Part II

PREREQUISITES - MATHEMATICS

33


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Chapter 2

Linear Algebra

2.1 Vector Spaces

A vector space over a field is a set that us closed under the finite addition and scalar multiplication, wherethe scalar belongs to the field. In this context, we shall deal with vector spaces over the field of complexnumbers, denoted by C. A vector space has the following properties:

1. Closure under vector addition:(u + v)V (u,v)V

2. Closure under scalar multiplication:cuV uV , cF

3. Vector addition and scalar multiplication are commutativeu + v = v + u u,vV(cd)u = c(du) = d(cu)uV , (c,d)F

4. Vector addition distributivec(u + v) = cu + cv (u,v) V , cF

5. Existance of an additive inverse u V, vV such that u + v = 0The vector v is denoted as -u and called the inverse of u, under addition.

6. Existance of a scalar identity under multiplication u V, cF such that cu = uThe identity is denoted by 1.

7. Existance of a zero vector u V, vV such that u + v = uv is then called the zero vector and denoted by 0.

In this context, we shall deal with vector spaces over the field of complex numbers, denoted by C.A vector can be represented by a matrix. The entries of the matrix are called the components of the vector.

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For example:

v =

v1v2v3.

..vn

, u =

u1u2u3.

..un

, v + u=

v1+ u1v2+ u2v3+ u3

.

..vn+ un

Here n is called the dimension of the vector space. The vector space can also be infinite dimensinal.

2.2 Linear dependence and independence

A set of vectorsv1,v2,v3 . . . . vn are said to be linearly dependent iff some constants cis suchthat:

ni=1

civi = 0.

Exception: The trivial case where all cis are = 0.

Each vector depends on the rest of the vectors by a linear relationship. Hence the name linearly dependent.The vectors are called linearly independent iff they do not satisfy the above conditions. For linearlyindependent vectors, no vector is related to the rest by a linear relationship. The maximum number oflinearly independent vectors in a vector space is called the dimension of the vector space, denoted by n.Any vectorx in this vector space can be uniquely described as a linear combination of these n vectors of thevector space. The proof for this is quite simple. A n-dimensional vector space will have a set of n linearlyindependent vectors. If we add the vectorx to this set, then we will still have a n-dimensional space. So,this means the setv1,v2,v3 . . . .vn,x is linearly dependent. So:

ni=1

civi + cxx = 0

so:n

i=1

civi =cxx

n

i=1

civi =x

Hence, we have shown that the general vectorx can be expressed uniquely as a linear combination of then-linearly independent vectors of the vector space. The set of n linearly independent vectors is called the

basisof the vector space. The set of all the vectors represented by this basis is called the spanning setofthe basis.

2.3 Dual Spaces

A linear function can be defined over the elements of the vector space. Let f be a function which is linear.f :x f(x ), such that f(x ) C. The function maps an element in the vector space to an element inthe field over which the vector space is defined. By our convention we represent a vectorx by a coloumn

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CHAPTER 2. LINEAR ALGEBRA 2.4. DIRACS BRA KET NOTATION

vector. So, we can see by inspection thatfshould be represented using a row vector. Then on multiplyingthe row and the column vector, we get a single scalar. The function satisfies the requirement:

f(ax + by ) = af(x ) + bf(x )where: {a, b}F and {x , y }V

So, we can propose a representation for the function:

if x =

x1x2x3...

xn

then we can say that f=

f1 f2 f3 . . .f n

Hence, on matrix multiplication of f withx , we get f1x1+ f2x2+ f3x3....fnxnwhich is a dimensionless number F.

The space of such functions is called the dual space to the corresponding vector space. If the vector spaceisV, then the corresponding dual space is represented by V.

2.4 Diracs Bra Ket notation

The Dirac bra-ket notation is widely used in Quantum Mechanics. Here any vectorx (represented as acolumn matrix) is represented as |x and this vector is called a ketvector. The corresponding dual vectoris called the bra vector, denoted byx|. Hence the name bra ket notation. We would be following thismethod from now on. The reasons will become evident as we proceed. So, we have the conventions:

x||x (2.1)x|cc|x (where c is some complex number) (2.2)

2.5 Inner and outer products

Consider a vector |x and its dual vectorx|. We saw that the former was a column matrix and the latterthe row matrix.x| is called the bra dual of |x. So, the product of a ket vector and its corresponding bravector is a dimensionless scalar quantity (as we saw this is multiplying a row matrix with a column matrix).The product is represented as: x|x. This product is called inner product. The inner product of a braand a ket vector is in short defined as:

x|y=i

yi xi (2.3)

The inner product also satisfies some linearity properties:

x|c1y1+ c2y2= c1x1|y + c2x2|y (2.4)c1x1+ c2x2|y= c1x1|y + c2x2|y (2.5)

The first can be derived by pulling the constant out of the ket vector. The second can be obtained by thesame technique, but using the rule given in (2.2). Another important property satisfied by the inner productis:

x|y= (y|x) . (2.6)

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2.6. ORTHONORMAL BASIS AND COMPLETENESS RELATIONSCHAPTER 2. LINEAR ALGEBRA

This can be proved as follows:

given from equation (2.3): x|y=

i

yi xi

Consider the RHS:

iyi xi

i

(yixi )

i

yixi

coming back to the initial definition of the inner product, we get back:

(y|x)Therefore, hence we have proved that:x|y= (y|x)

The inner product of a vector and its bra dual gives the square of the normof the vector. It is representedas:

x|x=|| |x||2 (2.7)

The norm is a scalar dimensionless quantity. However there is also another possibility which we havent yetexplored. What about the product |xx|. This is a product of a column matrix with a row matrix. Thisis a matrix whose dimensions will be (nn) where n is the dimension of the vector space containing |x.This product is called an outer product. A matrix can be thought as any transformation on a vector. Inquantum mechanics, these transformations or matrices are called operators.

2.6 Orthonormal Basis and Completeness Relations

We saw that any set of n linearly independent vectors {|v1, |v2, |v3,...|vn}form a basis for a n dimensionalvector space. We impose the condition that:

vi|vj= ij for all{i , j}{1..n}

The basis that satisfies this condition is called the orthonormal basis. The orthonormal basis is conven-tionally represented by {|e1, |e2, |e3,...|en}. The basis has some useful properties. Consider a vector |xin the vector space spanned by the basis {ei}. We can use this othonormality property to get the coefficientsmultiplying the basis states:

|x=

i

ci|ei

multiply both sides byej |: ej |x= i

ciej |ei

sinceej |ei = ji , it will be 1 only when j = i

Hence, when summed over we get: ej |x= cj (2.8)

Therefore, to get the coefficient multiplying the jth basis (orthonormal) state, in some basis expansion for avector, we must take an inner product of this basis state with the vector. Now, we can see some results withthe outer product of the basis states. The outer product leads to an operator. Now, let us take the basis

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CHAPTER 2. LINEAR ALGEBRA 2.7. PROJECTION OPERATOR

expansion of the vector, and substitute the result obtained in (2.8):

|x=

i

ci|ei

from (2.8), |x=

i

ei|x|ei

The expression inside the summation is a scalar multiplication with a vector, which is commutative.

Hence: |x=

i

|eiei|x

Now, consider the RHS of the form A|x, where A is some operator:

Hence: |x=

i

|eiei|

|x

Now, looking at the form of this operator, we can easily guess that the operator is the identity operator.Therefore:

i

|eiei|= I (2.9)

This relation given above is called the completeness relation.

2.7 Projection operator

Consider the operator |ekek|. This operator when acts on any vector, gives the coefficient of |ek, in the{ek} basis expansion of that vector. So, in other words, it takes any vector and projectsthe vector alongthe|ek direction. The projection occurs due to the formation of the inner product of the bra ek| with thearbitrary ket vector. Hence this operator |ekek| is called the projection operator. It is denoted byPk.The projection operator satisfies some important properties:

(Pk)n

= (Pk) (2.10)

PiP

j= 0 (2.11)

i

Pi = I (2.12)

The first (2.10) can be proved by considering Pk =|ekek|. So,(Pk)

n=|ekek|ekek|ekek| . . . |ekek|.

Now, we can simplify this statement by taking inner product. We use the fact that |ekek| = 1. Only theterminal terms |ek andek| will remain after all the others become 1 through formation of inner product.Hence, the RHS will be Pk.The second statement (2.11) can be proved by considering: PiPj = |eiei|ejej |. Now the inner productei|ej is = 0. The entire expression reduces to 0. Hence the statement is proved.The third statement (2.12) is already proved in (2.9)when we were looking at the orthonormal basis andcompleteness relation.

2.8 Gram-Schmidt Orthonormalization

Given any linearly independent basis that span some vector space, we can construct an orthonormal basisfor the same vector space. If we could not do that, then the original set of vectors would not be a basis sincewe have a vector that cannot be uniquely described by them. Therefore, we proceed by trying to constructan orthonormal set of basis from a given basis. So, our objective is that we must construct a set {Oi},

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2.8. GRAM-SCHMIDT ORTHONORMALIZATION CHAPTER 2. LINEAR ALGEBRA

corresponding to the set {vi}. By defenition of orthonormal basis, we need tha

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