QC Seminar

7/27/2019 QC Seminar

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Seminar: Quantum ComputingProf. Dr. Immanuel Bloch Prof. Dr. Stefan Mller-Stach

it

(t) = H

(t)

AB 12

[A,B]Hn|0= 1

2n

z{0,1}n|z

i

|E

aEb|j=

Cab

ij

yj =n1k=0

jkn xk

F =

i

(yi1yi2

yi3)

Johannes Gutenberg-Universitt MainzFachbereich Physik, Mathematik und InformatikSommersemester 2006


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Participants:Prof. Dr. Immanuel Bloch, Daniel Blinger, Egor Dranischnikow, Stephan Holzer,Carmen Kintscher, Jan Klemmer, Stefan Krause, Joschka Kupilas, Tim Langen,Lezsek Lupa, Jens Mandavid, Alexander Menk, Pavel Metelitsyn,Prof. Dr. Stefan Mller-Stach, Matthias Reinhardt, Philipp Roos, Joachim Schfer,Albrecht Seelmann, Anke Sperber, Susanne Tenhaeff, Cordula Zeller


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Preface

This manuscript contains lecture notes of an undergraduate seminar on quantum computingduring the summer term 2006 held at the Johannes-Gutenberg university in Mainz. The ideaof organizing a joint seminar about quantum information theory came up a while ago duringsome discussions between the two of us and a visit of the quantum optics laboratories. The

goal was to attract undergraduates to this interdisciplinary research field at the interface ofphysics, mathematics, and information science and to expose the students to the differentviewpoints when approaching this field of modern science.

Assuming only some basic knowledge about quantum mechanics and the mathematics offinite dimensional Hilbert spaces, the lectures gave an overview of known quantum algorithmsincluding those by Shor and Grover as well as some examples of experimental setups of workingq-bit devices. The talks also cover quantum cryptography and the promising approach oftopological quantum computing.

The lecture notes were prepared by the students participating in the seminar on differentsubjects relevant to quantum information science and are collected here. They will hopefullyprovide a fruitful reference for future work in this field.

Mainz, April 27th 2007

Immanuel BlochStefan Mller-Stach

3


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Contents

1 An Introduction to the Mathematical Basics of QC C. Zeller 7

1 Notational basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72 The qubit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3 Dynamics of a quantum system . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Classical Logic and Quantum Logic A. Sperber 17

1 Quantum gates and circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 Correspondence between classical and quantum computation . . . . . . . . . . . 193 Bases for quantum circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3 Quantum Measurement and Decoherence, . . . D. Blinger 27

1 Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 Entanglement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313 Decoherence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

4 Nonlocality in quantum mechanics . . . . . . . . . . . . . . . . . . . . . . . . . 36

4 Quantum Teleportation L. Lupa, J. Schfer 41

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 412 Concept of quantum teleportation . . . . . . . . . . . . . . . . . . . . . . . . . . 423 Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

5 Quantum Cryptography with Single Photons J. Klemmer 53

1 Classical cryptography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 532 Quantum cryptography with single photons . . . . . . . . . . . . . . . . . . . . 543 Quantum cryptography with entangled photons . . . . . . . . . . . . . . . . . . 58

6 The Algorithms of Deutsch, Jozsa, and Simon M. Reinhardt 63

1 The Deutsch algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 632 The Deutsch-Jozsa algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 653 Simons algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

7 Grovers Search Algorithm J. Kupilas 71

1 Problem and idea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 712 Implementation in quantum gates . . . . . . . . . . . . . . . . . . . . . . . . . . 733 But why O(

N/t )? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

4 Improvements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

5


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6 CONTENTS

8 Elementary Number Theory (a Crash Course) A. Menk, P. Roos 811 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 812

/n

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

3 Discrete Fourier transform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 914 Continued fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

9 Shors Factorization Algorithm C. Kintscher, S. Tenhaeff 971 Overview of Shors algorithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . 972 The quantum Fourier transformation . . . . . . . . . . . . . . . . . . . . . . . . 983 The quantum part of Shors algorithm . . . . . . . . . . . . . . . . . . . . . . . 100

10 Quantum Error Correction A. Seelmann 1051 Classical and quantum coding theory . . . . . . . . . . . . . . . . . . . . . . . . 1052 The nine-qubit code, part I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

3 Error models and properties of error-correcting codes . . . . . . . . . . . . . . . 1094 Bounds on quantum error-correcting codes . . . . . . . . . . . . . . . . . . . . . 1115 Stabilizer codes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1126 The nine-qubit code, part II, and symplectic notation . . . . . . . . . . . . . . . 116

11 Topological Quantum Computation P. Metelitsyn 121

1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1212 Flux tube model of anyons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1223 Quantum computation with anyons . . . . . . . . . . . . . . . . . . . . . . . . . 1274 Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

12 Complexity Theory and Computability S. Holzer 1371 Standard complexity classes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1372 Probabilistic complexity classes . . . . . . . . . . . . . . . . . . . . . . . . . . . 1423 Quantum computability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1454 The quantum complexity class BQP . . . . . . . . . . . . . . . . . . . . . . . . 146

13 Quantum Information in Ion Traps E. Dranischnikow 1511 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1512 Pauli trap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1533 Universal set of gates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1544 Entangled states; summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157


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Talk 1

An Introductionto the Mathematical Basics

of Quantum Computing

Cordula Zeller(April 24th, 2006)

1 Notational basics

First I will give an introduction to mathematical basics and notational conventions frequentlyused in quantum computing. As the state of a quantum system of qubits is described by avector in the tensor product of Hilbert spaces, we will now have a close look at Hilbert spaces,the tensor product, and at operators and observables.

1.1 Hilbert space

A Hilbert space is a complex vector space with a complex valued inner product, which iscomplete respect to the induced norm v =

(v, v). A complex valued inner product is a

transformation (, ) : H H

such that

(u, u) = 0 u = 0 , (u, v) = (v, u) , (u, v + w) = (u, v) + (u, w) , (u,v) = (u, v) .1.2 Dirac notation

An element a of a Hilbert space H is called ket and is denoted as |a. Similarly an elementb H := hom

(H,

) (being the space of linear transformations from the Hilbert spaceH into the complex numbers

) is called bra and is denoted as b|. The complex numberb|(|a) =: b|a is named bra-(c)-ket-product. In the case of a Hilbert space of finite dimensionthere is an isomorphism

: H H , |a |a, where the bra (

|a,) is denoted by

a|. Ifdim

H=

, is only a monomorphism.

7


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8 TALK 1. AN INTRODUCTION TO THE MATHEMATICAL BASICS OF QC

1.3 Operators

A linear operator O on a ket-space H is an element ofhom

(H, H). Likewise a linear operatorO on a bra-space H is an element of hom (H, H). Remember: There is a unique linearoperator O such that O|a, |b = |a, O|b |a, |b Hcalled the adjoint operator. Each operator O on H can be identified with the followingoperator, also denoted by O, on H:

a| a|O with a|O|b := a|O|bso the notation a|O|b is unambiguous.

1.4 Observables

In quantum mechanics an observable is a Hermitian (self-adjoint) operator on a Hilbert spaceH. Repetition:

An eigenvalue a of an operator A is a complex number such that |b H: A|b = a|b. The ket |b is called eigenket of A corresponding to the eigenvalue a. All eigenvalues of a Hermitian operator are real numbers and their corresponding eigen-

kets may be chosen orthogonal. In the case an eigenvalue has two or more linearlyindependent corresponding eigenkets, there are more possibilities of choosing eigenkets.

Definition: An eigenvalue a is called degenerate if there are two linearly independent eigen-

kets corresponding to a, otherwise it is called non-degenerate.Notation: If all eigenvalues of an observable A are non-degenerate we label the eigenkets |ai

corresponding to their eigenvalues ai such that A|ai = ai|ai.Definition: An observable is called complete, if its corresponding eigenkets form an (orthog-

onal) basis of the Hilbert space H.If the observable A is non-degenerate, we may express completeness by

i

|aiai| = 1 .

In this case every ket | can be expressed as| =

i

|aiai |

and thusA =

i

ai|aiai| .

Examples for observables frequently used are the Pauli spin matrices

1 = 0 11 0

, 2 = 0 ii 0

, 3 = 1 00

1

.


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2. THE QUBIT 9

1.5 The tensor product

The tensor product of two vector spaces V and W is a vector space V W together with thebilinear application : V W V W , (v, w) v wthat satisfy the following universal property: For any space X and any bilinear transformation : V W X there is a unique linear function : V W X such that

V W //

&&v

vvvv

vvvv

vvV W

X

commutes, i. e. = . Both the space V W and the bilinear application exist and

are unique.For further reading please refer to [3] or any other book on homological algebra. The mostimportant points we have to remember about the tensor product are the following:

If e1, . . . , en is a basis of V and f1, . . . , f m is a basis of W, then {ei fj | 1 i n ,1 j m} forms a basis of V W such that

dim(V W) = dim(V)dim(W) .

The tensor product of V and W (V and W having a countable basis) is nothing but

V

W = iI

vi

wi vi V , wi W . Because is bilinear we can use

v w = (v, w) = i

viei,j

wjfj

=i,j

(viwj)(ei fj)

for calculations where v =i viei and w =

j wjfj.

Please note that not every element u ofV W can be expressed as u = v w for somev V, w W, but it may be expressed by a (finite) sum of those elements.

Notation: |v |w can be alternatively written as |v|w or as |v, w.

2 The qubit

2.1 Introducing the qubit

In classical computation the elementary carrier of information is a Shannon bit. It is either inthe state 0 or in the state 1, but never in both states at the same time. Furthermore, and inour understanding quite natural, it is possible to copy Shannon bits. Quantum computationis also based on an elementary carrier of information: the qubit. In correspondence to theShannon bit there are also two possible measurable states that it can take, denoted by

|0


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and |1. But unlike its classical counterpart, the state of a qubit can take the value of anysuperposition of those basic values, i. e. the state of a qubit can be described by

| = a|0 + b|1where (a, b) 2 \ {(0, 0)} and |0 and |1 denote an orthonormal basis of a two-dimensionalHilbert space. Thus as stated earlier the qubit is a quantum system whose state is describedby a ket-vector in a two-dimensional Hilbert space.

Two kets |a and |b describe the same state if|a = |b ,

so we can also say that a state of a quantum system is a ray in a Hilbert space or an elementof the complex projective space. Therefore we will describe a state of a qubit, unless statedotherwise, by kets of unit length

| = 1or using the above notation

| = a|0 + b|1 , a, b

, |a|2 + |b|2 = 1 .Another important difference to the Shannon bit is the fact that it is not possible to copy aqubit (see no-cloning theorem 3.3).

2.2 The Hilbert space ofm qubits

The state of one qubit is described in a two dimensional Hilbert space H, for example in 2. Ifwe have m different qubits, their state will be described by an element of the 2m-dimensional

spaceH H H

m times

= Hm e.g. = ( 2)m .

A state may be denoted as |01 . . . 1 (m qubits) or any (finite) sum of those basic kets (nor-malized to the length of one).

Please note: As mentioned in the paragraph about the tensor product, not every state | Hm can be written as | = nj=1|j, j H. This fact will be crucial later on forthe definition of entangled states.

2.3 Example: polarized light

There are a lot of possibilities of realizing a qubit. One of them is polarized light: Fromthe classical perspective light can be perceived as an electromagnetical transversal wave, itselectric field vector is perpendicular to the direction of propagation. If the direction of theelectric field vector is always parallel to a fixed line, the wave is called linearly polarized, incase it describes a circle (ellipse), it is circularly (elliptically) polarized.

In the same way in quantum mechanics photons are described to be polarized. An or-thonormal basis of the states of polarized photons is e. g. {|, |}, the linearly horizontallyor vertically polarized photons. From those two vectors we can deduce any other linear po-larization state, e. g.

|=

1

2| + |


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3. DYNAMICS OF A QUANTUM SYSTEM 11

but also circular polarization

| = 12

| + i|

.

Elliptical polarization is of course a superposition of linear and circular polarization.We will write kets as

| =

10

, | = 1

2

1

1

, | = 12

1i

,

| =

01

, | = 1

2

11

, | = 1

2

1i

.

The basis of the bra-space is

| =

1 0

, | =

0 1

.

Therefore bras are calculated as follows:

| = |, = 12

1i

,

=1

2

1 i

where in that last equality we have to note that we have a Hermitian inner product such thatav |w = av |w where a is the complex conjugate of a.

So the bras are denoted by

| = 12

1 1 , | = 1

2

1 i ,

|=

1

2 1 1 , | =1

2 1 i .The tensor product is calculated as follows

| =

12

12

12

i2

= 12

1i1i

where in the right vector we put the coefficients of the basis {|, |, |, |}.An operator is constructed by e. g.

|| = 121i 12 1 i = 12 1 ii 1 .

(This would be a projection to the eigenspace of |.)

3 Dynamics of a quantum system

3.1 Unitary transformations

The transformation of an isolated quantum system is described by the Schrdinger equation

i

t (t) = H(t)


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where denotes the Plancks constant divided by 2 and H denotes the Hamiltonian. TheHamiltonian is an observable (Hermitian transformation), in classical physics it describes thetotal energy of a system.

If we put up a transformation U such that(t) = U(0)we get

tU(t) = i

H(t)U(t) .

If the Hamiltonian does not depend explicitly on the time t, the solution is given by

U(t) = exp

i

Ht

.

U(t) is unitary because the Hamiltonian H is Hermitian and complete.

Proof Both H itself and its diagonal representation H have an orthonormal basis (H hasan orthonormal basis, because all eigenkets of an observable are orthogonal, H is representedby the standard orthogonal basis), and because H and H are also complete, the basis changefrom H to H may be represented by a unitary matrix V : H = V1HV (proof of thisstatement in [2, p. 37]). Thus

U(t) = exp

i

Ht

= exp

i

V1HtV

= V1 exp

i

Ht

V

and

U(t)U(t) = Vexpi

Ht(V1)V1 exp

i

HtV= V

exp

i

Ht

exp

i

Ht

V = VV = 1 .

The first and the last equality follow from the unitary properties of V. The middle equalityfollows from H being diagonal and

H = diag(a1, . . . , an)

= (H)k = diag(ak1 , . . . , akn)

= exp

i

Ht

= diag

exp

i

ta1

, . . . , exp

i

tan

= exp i

Ht

= diag

exp i

ta1

, . . . , exp i

tan

=

exp

i

Ht

exp

i

Ht

= diag(1, . . . , 1) .

Given that U(t) is linear, we can also proof that U(t) has to be unitary in the followingway: As we are dealing with rays we agreed to always choose the kets of unit length, i. e. thetransformation only acts on kets of unit lengths. Therefore

1 =

(t)(t)

=

(0)UU

(0)

(0)

H = UU = 1 .So a quantum system only transforms via unitary transformations.


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3. DYNAMICS OF A QUANTUM SYSTEM 13

3.2 Quantum gates

For implementing a complicated unitary transformation into a quantum computer, we will

naturally break it down into simple elementary transformations called quantum gates actingon a small set of qubits. I will now introduce some of the most important gates used inquantum computation.

First we will have a close look at the controlled NOT (CNOT) gate. It is actually anembedding from classical computation, as classical inputs (no superposition) lead to classicaloutputs. Certainly CNOT can also be applied to qubits in a superposition state, it will acton all basic states separately according to the CNOT description. In a wiring diagram we willdescribe it as follows:

b a + b|

a awhere denotes a control bit, denotes a target bit, and + stands for addition modulo 2,working here like a logic XOR (exclusive or). The CNOT gate works as follows:

|00 |00 , |01 |11 , |10 |10 , |11 |01 .It leaves the second entry unaltered and changes the first entry depending on the second entry.We can identify the basic ket |an . . . a2a1, ai {1, 0}, with the integer

i ai2

i. In this casethe CNOT gate on two qubits is also described by the permutation = (13).

In quantum computation we are always dealing with unitary transformations and certainlywe also express them by matrices. The CNOT gate can therefore also be characterized by thematrix corresponding to the lexicographically ordered basis [|00, |01, |10, |11]:

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

.Another quantum gate that also works in classical computation is the Toffoli (CCNOT) gate:

c c|

b b|

a

a + bc

where multiplication works like a logic AND and addition once again like logic XOR. Usingthe notation as before this gate corresponds to the permutation = (67).

Other examples for embeddings of classical gates are NOT and SWAP:

NOT: a NOT a + 1 or

0 11 0

,

SWAP:b a

| | |a b

or

1 0 0 00 0 1 00 1 0 00 0 0 1

.


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Note: Embeddings of classical gates into quantum computation there suddenly gain morecomputational possibilities. Their inputs and outputs may be superpositions of manystates and a gate may also produce an entangled state (see third talk).

We can easily see that in quantum computation we have a lot more possibilities of transfor-mations such as:

The square root of NOT (NOT) 1 i2

i 11 i

.

The square root of SWAP (SWAP)

1 0 0 0

0 1+i21i

2 0

0 1i21+i

2 0

0 0 0 1

.

More important gates are:

The Hadamard gate 12

1 11 1

. The Hadamard gate produces superpositions of the

basis such that

H|0 = 1

2

|0 + |1 , H|1 = 12

|0 |1 .In the example of polarized light this gate corresponds to a rotation of 45 degrees.

Rotation gates such as

cos isin icos cos = ei1 , cos sin sin cos = ei2 , ei 00 ei = ei3where 1, 2, 3 are the Pauli spin matrices (see observables).

3.3 No-cloning theorem

Now we will prove the quite disconcerting fact that we cant copy a qubit: We will use theearlier result that quantum systems only transform via linear unitary transformations whereascopying is inherently quadratic:

Supposition: There is a unitary transformation U that can copy

U: HA H H HA H H , U|0|a| = |1|a|awhere HA is an auxiliary Hilbert space and | ist the blank where we want to copy ato.

As usual we assume that |0, |a, and | are of unit length, and as U is unitary, |a is alsoof unit length. Let |b denote a second ket in H such that

0 n) and satisfies the

condition W(| |0Nn) = (U|) |0Nn for all | Bn. This means that we add 0bits for the computing at the beginning and give the same number of 0 bits back at the end.Or to be more precise after the computation we get a state that can be represented as follows:| | where | is the information we are interested in and | is arbitrary.

2 Correspondence between classical

and quantum computation

2.1 Problems in combing both theories

Boolean circuits are not included in quantum circuits as a special case!What we already know is that in the classical permutation G : k k corresponds to

a unitary operator G acting on Bk. The condition for this correspondence is G| := |G.The following definition of reversible classical circuits and their realization of permutationsare analog to those of the operators.


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20 TALK 2. CLASSICAL LOGIC AND QUANTUM LOGIC

Definition 2.1Let A be a set of permutations. A reversible classical circuit over the basis A has alreadybeen defined. A permutation realized by reversible circuit using ancillas is a permutation such

that product of permutations W = GL[AL] G1[A1] fulfills the condition W(x, 0Nn) =(Gx, 0Nn) for arbitrary x

n.

2.2 Embedding the Boolean logic into the quantum logic

The solution to combine both theories is to use an extended function instead of the generalBoolean function.

The Boolean function F:

n

m is not computed, but the function F : n+m n+mcan be computed and is given by the formula F(x, y) = (x, y F(x)) where means bitwiseaddition modulo 2. If we set y := 0, we get the desired F(x, 0) = (x, F(x)). Unfortunatelytwo-bit permutations are not sufficient to realize all functions F

, but permutations on three

bits do so.

Lemma 2.2 (controlled NOT

)A Boolean circuit (size L, depth d, basisA) with fan-in and fan-out bounded by a constant isthe realization of a function F:

n m. Then we can realize a map (x, 0) (F(x), G(x))by a reversible circuit (size O(L), depth O(d)) over the basis A consisting of the functionsf (f A) and the function

: (x, y) (x, x y).

Proof Construction of the Boolean circuit that computes F:Input variables: x1, . . . , xn;auxiliary variables: xn+1, . . . , xn+L started by 0;Boolean: xn+k := fk(xjk , . . . , xk), fk

A, jk, . . . , k < n + k;

corresponding reversible circuit is action of the permutation (fk) that is xn+k := xn+k fk(xjk , . . . , xk).

At the beginning, the auxiliary variables have the value 0 and they have to be 0 at theend, so the positions of the bits change.

We have already recognized that several assignments can occur simultaneously. But theconcurrent ones should not share their input variables. Otherwise, we would have to insertcopy gates between the layers, each would be a

. The result would in depth be increased

by the fan-out-constant.

F(x) G(x)

x xn+1, . . . , xLm F(x)

x 0 0

n L m m

Remark 2.3The result F(x) is needed, but G(x) is produced, too. Since () : (a, b) (b, a) can berepresented in the form ()[j, k] =

[j, k]

[k, j]

[j, k], the CNOT allows to interchange

bits in memory.


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3. BASES FOR QUANTUM CIRCUITS 21

Lemma 2.4 (Garbage removal Mllbeseitigung)A Boolean circuit (size L, depth d, basis A) with fan-in and fan-out bounded by a constantis the realization of a function F:

n

m. Then F

can be realized by a reversible circuit

with size O(L + n + m) and depth O(d) with use of ancillas.

Proof

F(x) G(x) F(x)

y

F(x) G(x) y

x 0 F(x)

y

x 0 y

m L + n m m

n L m

Theorem 2.5Let F and F1 be computed by Boolean circuits (size L, depth d). Then F can berealized by a reversible circuit of the size O(L + n) and depth O(d) using ancillas.

Proof

F(x) 0

F(x) x

x F(x)

x 0

n n

Corollary 2.6A complete basis for reversible circuits is formed by negation (note ) and the Toffoli gate(note

: (x,y ,z)

(x,y ,z

xy)). (Realization with ancillas is meant!)

3 Bases for quantum circuits

Because of the fact that there exists an uncountable number of unitary operators, we have toeither choose a complete basis with an infinite uncountable number of gates or approximate.

3.1 Exact realization

Theorem 3.1The basis consisting of all one-qubit and two-qubit unitary operators allows the realization ofan arbitrary unitary operator.


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3.1.1 Operators with quantum control

Definition 3.2

The relations(U)|0 | = |0 | and(U)|1 | = |1U| define for every operatorU (U: Bn Bn) an operator (U) : B Bn B B n. Analogously for operators withseveral controlling qubits we get

k(U)|x1, . . . , xk | =

|x1, . . . , xk | if x1 xk = 0 ,|x1, . . . , xk U| if x1 xk = 1 .

Example 3.3Let x =

0 11 0

and

(x)|0 | = |0 | , ()(x)

|1 |

=|1

x

|

. (

)

We want to calculate (x):

(x) :=

u1 u2u3 u4

and | = c0|0 + c1|1 =

c0c1

.

Look at (): u1 u2u3 u4

|0 | = |0 | .

Left hand side:

u1 u2u3 u410 c0c1 = u1u3c0c1 = u1c0|00 + u1c1|01 + u3c0|10 + u3c1|11 .Right hand side:

10

c0c1

= c0|00 + c1|01 .

Compare coefficients:

u1 = 1 , u3 = 0 = L(x) =

1 u20 u4

.

Look at (): Left:1 u20 u4

01

c0c1

=

u2u4

c0c1

= u2c0|00 + u2c1|01 + u4c0|10 + u4c1|11 .

Right:01

1 u20 u4

c0c1

=

01

c0 + u2c1u4c1

= 0|00 + 0|01 + u4c1|11 + (c0 + u2c1)|10 .

Compare coefficients:

u2 = 0 , u4 = 1 = (x) = 1 00 1

=

.


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3.1.2 The realization of the Toffoli gate

First, we have to find a pair of operators X, Y for which the following relation holds:

XY X1

Y1

= ix

.

X =1

2

i 11 i

, Y =

0 1

1 0

,

det(X) =

1

2

2(i2 + 1) = 1 , det(Y) = 0 (1) = 1 ,

A1 =1

det(A)ACr =

1

det(A)

a bc d

Cr=

1

det(A)

d b

c a

= X1 = 12

i 1

1 i

, Y1 =

0 11 0

.

Testing:

XX1 =1

2

i 11 i

i 1

1 i

=1

2

i2 + (1)2 i + ii i 12 i2

=

1 00 1

,

XY X1Y1 =1

2

i 11 i

0 1

1 0

i 11 i

0 11 0

=

1

2

1 ii 1

1 ii 1

=1

2

1 + i2 i ii i i2 + 1

=

0 ii 0

= ix .

3.1.3 Implementation of the Toffoli gate

X Y Y1X1

i

3.2 The realization ofk for U U(B)

3.2.1 Implementation of the operator k for U using ancillas

P1 P

0

x2

x1 x1 x2 xk

...G(x)

... x1...

xk

... 0

U


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3.2.2 Implementation of the operator k(ix) without ancillas

X Y Y1X1

x1

x2...xk/2xk/2+1

...xk

...

...

...

...

3.2.3 Ancilla-free realization ofk(U), U U(2)

Uk

Zk1

Z2

Z1

Z0

. . .

. . .

. . .

. . .

. . .

......

......

x1

x2

xk1

xk

3.2.4 The realization on an arbitrary operator

Lemma 3.4An arbitrary unitary operator U on the space

M can be represented as a product of M(M

1)/2 matrices of the form:

1 0 ...

. . . 0 0 1 0 0

a bc d

0

0 1 0 . . . 00

1

, where

a bc d

U(2) .


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3.3 Approximate realization

Now I will try to give you the approximate alternative. So a norm on the operator space is

needed. As I hope you already know the Euclidean norm, which is defined by | = |.The Euclidean norm satisfies the norm conditions.Definition 3.5The norm of an operator X is named operator-norm:

X = sup|=0

X|| .

The operator norm has several properties:

XY X Y , X = X ,

X

Y

=

X

Y

,

U

||= 1 if U is unitary.

Definition 3.6The approximation of an operator U is U. The operator is approximated with a precision .So

U U .Corollary 3.7IfU U = U1 U1 .Proof

U U , to show: U1 U1 :

U U U1(U U) U1

I U1U U1(I U1U)U1 U1U1

U1 U1

Corollary 3.8For U = UL U1 andUk Uk k:

UL

U2U1

UL

U2U1

j j .Proof

U2U1 U2U1 = U2(U1 U1) + (U2 U2)U1 U2(U1 U1) + (U2 U2)U1 U2 U1 U1 + U2 U2) U1= U1 U1 + U2 U2) .

To be able to use ancillas and we have to generalize definition 3.6:


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Definition 3.9The operatorU: Bn Bn is approximated by the operatorU: BN BN with precision using ancillas, if for arbitrary

|

in

Bn the inequalityU| |0Nn U| |0Nn |

is satisfied.

Another formulation: If we introduce a map V : Bn Bn in the way V : | ||0Nn,then V is isometric. Then we see:

U V V U .

Definition 3.10

{H,K,K1

, (x

), 2

(x

)}, whereH =

12

1 11 1

, K =

1 00 i

is called standard basis.

Remark 3.11The standard basis given in definition 3.10 is complete.

References

[1] Skript Physikalisches Praktikum fr Naturwissenschaftler Teil II PMC, 2004

[2] G. Fischer: Lineare Algebra, Vieweg

[3] I. N. Bronstein: Taschenbuch der Mathematik, Thun und Frankfurt am Main, VerlagHarri Deutsch, 2001

[4] A. J. Kitaev: Classical and quantum computation, American Mathematical Society,United States of America, 2002


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Talk 3

Quantum Measurement andDecoherence, Entanglement and

Nonlocality

Daniel Blinger(May 15th, 2006)

This text is written for the seminar on quantum computing at the universityMainz. First, the basics of quantum mechanical measurement will be described.In the next chapter, entangled states are formally introduced. At the end, a briefintroduction of decoherence is given and EPRs paradox is discussed.

1 Measurement

1.1 General principles

Unlike in classical mechanics, in quantum mechanics the measurement process is a fundamentalpart of the theory. In this section its general principles will be described.

Let us start with two quantum mechanical axioms:

Axiom: Observables are characterized by complete hermitian operators A hom

(H, H).

Axiom: Possible results of measurement of an observable are eigenvalues ai of the corre-sponding observable A.

Now let A denote a complete nondegenerate observable with eigenvalues ai and orthonormaleigenkets |ai in a Hilbert space H. Let Q be the corresponding quantum system.Remember: Complete means that the {|ai} form an (orthonormal) basis of H and that

i|aiai| = .Due to completeness, we can write every state | as

| = ici|ai , ci = ai | . (1)

27


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28 TALK 3. QUANTUM MEASUREMENT AND DECOHERENCE, . . .

From | = ai |ai = 1 follows

i|ci|2 = 1 (2)

which makes it reasonable to interpret the |ci|2s as probabilities. This leads us to anotherquantum mechanical axiom:

Axiom: The measurement of an observable A of a quantum system Q in the state | pro-duces the eigenvalue ai as the measured result with the probability1 |ci|2 = |ai ||2.

The last axiom I would like to point out is at the same time the most special. It dividesquantum mechanics from classical mechanics containing the fact that every measurementdirectly affects the measured quantum system Q:Axiom: A measurement (of our observable A) always causes (the quantum system Q) to

jump into an eigenstate of the dynamical variable that is being measured, i. e. if ai is

being measured, the Q will jump into the corresponding eigenket2 |ai.Example: Let again Q be represented by the state |bm1 =

i|aiai | and let us measure

our observable A. What will be the result? Due to quantum mechanics, we are not ableto predict the exact result but to give probabilities for the possible outcome. With theprobability |aj |bm1|2 we will measure the (eigen-)value aj and the system jumps intothe state |am1 = |aj. The indices bm1 and am1 are for before and after measurement#1. Now let us repeat the measurement of A at the system Q, thus renaming |aj =|am1 |bm2. What is going to happen now? Again, the probability is given by|ak |bm2|2 for measuring ak. Because ofaj |ak = jk and |aj = |bm2 we know thatall probabilities are zero except in the case k = j; that means we are going to measure

aj again for sure.Summarizing:

| =i

|aiai | aj meas.=prob.=|aj ||2

| = |aj aj meas.=prob.=1

| = |aj . (3)

1.2 Expected values

In this section we derive the expected value A we obtain when measuring the observable Aat (infinitely) many copies from our quantum system Q. We start from the standard formula

A

= i ai Prob(observing ai) . (4)

If Q is in the state |, then Prob(observing ai) is given by |ai ||2. Using this, we canrewrite (4) as

A =i

ai ai |2 =

i

|aiaiai | =i,j

|aiai|A|ajaj | = |A|

= A = |A| . (5)1in case of degeneracy, i. e. A|ai,k = ai|ai,k k K the probability is given by

k |ci,k|

2, ci,k =ai,k |.

2In case of degeneracy (see above), the system will jump into the state | = N

kK ci,k|ai,k where N

is for normalization.


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1. MEASUREMENT 29

1.3 Measurement of different observables

In this section we want to work out how the measurement of an observable A affects the result

of the measurement of a complete nondegenerate observable B.First we assume that there exists a complete set of simultaneous eigenkets of A and B

labeled as |ai, bi, i.e.

A|ai, bi = ai|ai, bi , B|ai, bi = bi ketai, bi . (6)

Nowanalogous to eq. (3)we execute two measurements on Q, but now A is measured first,then B is. Then we repeat the A-measurement:

| =i

|ai, biai, bi |

aj meas.=prob.=|aj ,bj ||2

| = |aj , bj bj meas.=prob.=1

| = |aj , bj aj meas.=prob.=1

| = |aj, bj . (7)

In words: When measuring A the first time, the wave-function collapses and jumps into thecorresponding eigenket |aj , bj to the measured value aj . Now we can measure B withoutdestroying the actual state, i. e. if we measure A again, we will certainly gain the same resultagain. That means that we can sharply measure A and B at the same time.

Now we assume that there does not exist a set of simultaneous eigenkets of A and B, i.e.

A|ai = ai|ai , B|bi = bi|bi , |ai = |bj i , j . (8)

Note that both sets are complete, which means we can write

|ai =j

|bjbj |ai , |bj =i

|aiai |bj . (9)

We now do the same thing again, first measuring A, then B, finally A again:

| =i

|aiai |

aj meas.=

prob.=|aj ||2| = |aj =

i

|bibi |aj bk meas.=prob.=|bk |aj|2

| = |bk =i

|aiai |bkanmeas.

=prob.=|an |bk|2 | = |an . (10)

In words: When measuring B, we destroy the state again, thus destroying the measured valuefor A. When measuring A again, it is possible to gain another result. That means we cannotmeasure A and B at the same time.

How can we decide if we can measure two observables A and B at the same time? In orderto do so we introduce the commutator [A, B] = AB BA and prove the following theorem inthe nondegenerate case:

Theorem: If and only if [A, B] = 0, then there exists a set of simultaneous eigenkets of theobservables A and B.


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Proof =: We prove that there does not exist a complete set of simultaneous eigenketsif [A, B] = 0. To show this let us assume the converse to be true, i. e. there exists a set ofeigenkets as in eq. (6). Then we obtain

AB|ai, bi = Abi|ai, bi = aibi|ai, bi = biA|ai, bi = BA|ai, bi . (11)

Hence, AB|ai, bi = BA|ai, bi and thus [A, B] = 0 in contradiction to the assumption.=: Now let A and B commute and {|ai}i be the complete set of eigenkets of A. We

observe thatai|[A, B]|aj = (ai aj)ai|B|aj = 0 . (12)

So ai|B|aj must vanish unless ai = aj , which means that the matrix ai|B|aj representingB is diagonal, i. e. ai|B|aj = ijai|B|ai. Using the completeness-relation we can write Bas

B = i |

ai

ai|

B|ai

ai|

.

Now suppose that this operator acts on an eigenket of A:

B|aj =i

|aiai|B|aiai |aj = B|aj . (13)

That means the eigenket |aj to the eigenvalue aj ofA is eigenket to the eigenvalue B ofB.Therefore, the set {|ai}i is a set of simultaneous eigenkets of A and B.

Note: We can measure A and B at the same time if [A, B] = 0.

We now want to quantify the uncertainty in measurement. Therefore, we define the uncertaintyof an observable as the standard deviation of the observed eigenvalues; hence it is given by

A =

(A A)2 . (14)Using the Schwarz inequality and the fact that the expectation value of a Hermitian operatoris purely real, we gain

Heisenbergs uncertainty principle:

AB 12 [A, B] (15)

where A and B are two observables3.

As an example, take A = xi the ith component of the position operator and B = pj the jthcomponent of the momentum operator. Due to [xi, pj ] = ij eq. (15) becomes

xipj 2

ij .

Thus, we can measure different components of position and momentum simultaneously withunlimited precision. This is wrong if we try to measure the same component.

3For proof, see e. g. [1] or [3].


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2. ENTANGLEMENT 31

2 Entanglement

2.1 Juxtaposition of quantum states

We now want to describe two isolated quantum systems Q1 and Q2 as one system Q. Isolatedhere does not mean that the ingredients (like electrons, photons or whatever) are isolatedspatially but that the quantum states |1 ofQ1 and |2 ofQ2 do not interact. The Hilbertspace corresponding to Q1 is H1, to Q2 it is H2. The state of the global system to Q is givenas the tensor product

| = |1 |2 H1 H2 = H .|1 |2 usually is denoted as |1, 2.

Let the Qi (i {1, 2}) be two qubits, i. e. dim H1 = dim H2 = 2. For both Hi we choosea basis {|0i, |1i}. The most general state ofQi then is |i = i|0i + i|1i, i, i ,

|i

|2 +

|i

|2 = 1. Therefore, corresponding states in

Qlook like

| = 1|01 + 1|11 2|02 + 2|12= 12|00 + 12|11 + 12|01 + 12|10 (16)

where the indices have been suppressed in the last step.

2.2 Entanglement of quantum states

Obviously, a basis of the Hilbert space H describing our 2-qubit register Q is given by{|00, |01, |10, |11}. Therefore, the most general state of Q is

| = c00|00 + c11|11 + c01|01 + c10|10 (17)with4 |c00|2 + |c11|2 + |c01|2 + |c10|2 = 1, cij . Thus, H contains more states than can beconstructed through juxtaposition. These states are called entangled.

Example: The state

| = 12

|00 + |11 H = H1 H2 (18)is an entangled state, i. e. it is not a juxtaposition of two states |1 |2 with |i Hi.

Proof Assume the converse, then (18) has to look like (16) with

12 =1

2, 12 = 0 , 12 = 0 , 12 =

12

which has no solution obviously.

Now let me give the general definition of quantum entanglement:

Definition: Let Q1, Q2, . . . , Qn be quantum systems with underlying Hilbert spaces H1, H2,. . . , Hn, respectively. Then the global quantum system Q consisting of the quantum

4Often the digits in the kets are interpreted as binary numbers, i. e. our state gets | = c00|0 + c01|1 +c10|2 + c11|3.


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systems Q1, Q2, . . . , Qn is said to be entangled if its state | H =nj=1 Hj cannot

be written in the form

| =nj=1

|j (19)where each ket |j lies in the Hilbert space Hj for j = 1, 2, . . . , n. We also say thatsuch a state | is entangled.

You should keep in mind two additional facts on entanglement:

The subsystems Qi do not have to be at the same place for the global system Q to beentangled, they might be spatially separated.

Measurement of a subsystem Qi is correlated to measurement of Qj, i = j, only if theglobal state

|

is entangled.

Example: Let us examine what happens when measuring an unentangled state, here | =(|00 + |01 + |10 + |11)/2. First we measure Q1, then Q2 afterwards:

| 11 in Q1 meas.=prob.=1/2

| = 12

|10 + |11 12 in Q2 meas.=prob.=1/2

| = |11 . (20)

That means: If we measure 1 in Q1 we still have the same probability for measuring 0or 1 in Q2. Otherwise, if we measure an entangled state, here | = (|00 + |11)/

2,

we gain another result:

|11in

Q1meas.

=prob.=1/2 | = |1112in

Q2meas.

=prob.=1 | = |11 . (21)

When measuring Q1 we instantaneously pinpoint what will be measured in Q2, bothresults are 100 % correlated.

2.3 Entangling states

We now show in an example how to produce an entangled state out of a juxtaposed one. Letthe initial state be

|

t=0 =

1

2|00 |10 =1

2|01 |11 |02=

12

10

01

10

=

12

10

10

(22)

and the Hamiltonian

H2

|1010| |1011| |1110| + |1111| = 2

0 0 0 00 0 0 00 0 1 10 0

1 1

(23)


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3. DECOHERENCE 33

which is applied to the system from t = 0 to t = 1. To derive the time evolution of |, wecompute the propagator and obtain

U(t = 0 t = 1) = e(i/)H =

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

. (24)This done as described in the footnote5. Then at t = 1, the state is

|t=1 = U(t = 0 t = 1) |t=0 = 12

100

1

= 12|00 |11 (25)

which is an entangled state obviously, compare with (18).Summarized: Using the unitary transformation6 U(t = 0 t = 1) we managed to make

an entangled state out of an unentangled one. Applying H another timestep long, we regainour initial unentangled state |t=2 = |t=0

3 Decoherence

3.1 The density operator

We now introduce the density operator7, an important tool for describing the quantum statis-tics properly. Let

|

be the state of our system

Q. The density operator is defined as

p = || . (26)

Some properties of are (without proof8)

A = tr(pA) , tr p = 1 , 2p = p , = p . (27)

The real power of the density operator shows up when you are trying to handle the measure-ment of several systems {Qi}iI. When all systems are in the same state |, we talk of apure ensemble and p (p for pure) can be applied. More often we deal with the situation that

5We have to calculate U = e(i/)H eB. First we note that B2 = iB and therefore Bn = (i)n1B.

Then

eB =

+ B +1

2!B2 + . . . +

1

n!Bn + . . .

= + B

1 +

1

2!(i) + . . . +

1

n!(i)n1 + . . .

=

+B

i

1 + 1 + (i) +

1

2!(i)2 + . . . +

1

n!(i)n + . . .

ei = 1

=

+2

iB = (24) .

6By the way, the used unitary transformation is the CNOT operation which inverts the state of Q2 only if|1 = |11.

7Often called density matrix.8For proofs, see e. g. [2].


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we cannot tell in which state the system Qi is, but give probabilities {pi}i1,...,n;n

for thesystem being in one of the states |1, . . . , |n,

ni=1pi = 1. If one of the pis equals 1 we

have the situation of a pure state, if all pis are smaller than 1, we talk of a mixed ensemble.

Then the density operator looks like

=ni=1

pi|ii| (28)

with the propertiesA = tr(A) , tr = 1 , 2 = , = . (29)

Note: The density operator (28) describes the situation that the system is in the state |iwith the probability pi. This situation is fundamentally different from a system whichis in a coherent superposition | =

ni=1pi|i.

The description of mixed ensembles is not possible without the density operator, there-fore its importance.

Example: We are interested in the z-spin states of a beam of silver atoms.

1. Pure ensemble: All atoms of the beam have been prepared to be in the state| = (| + |)/2. Then the density operator is

= p = || = 12

|| + || + || + || . (30)2. Mixed ensemble: The atoms are in a random state, i. e. we expect that 50 % are in

the state | and 50% in |. Then the density operator is given by

= 12

|| + || . (31)Please note the difference between (30) and (31) which can be verified in experiment:In situation (30) interference effects can be observed; there are not any in (31).

3.2 The reduced density matrix

Let again be H = H1 H2 the space describing the system Q consisting of two subsystems Q1and Q2. The state ofQ is considered to be the pure state |, which might be an entangled one.We are now interested in an observable A that pertains only to subsystem Q1, A = A1 2.Then it is easy to show that

A = tr(A1 ) = tr1

tr2()A1

(32)

where tri means tracing over the degrees of freedom of system Qi9. Therefore we define thereduced density matrix 1 = tr2() which yields exactly the same statistics generated of A as.

The same result holds in general for any pure state | H = H1 HN of a resolutionof a system into N subsystems and any observable A =

1 j1 Aj j+1 N.Then j = tr1,...,j1,j+1,...,N|| and

A = tr(A) = trj(jAj) . (33)9The partial trace is a map trI : hom

jJHj,jJHj hom jJ\IHj,jJ\IHj, I J.


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3. DECOHERENCE 35

3.3 The challenge of decoherence

The measurement process described in section 1 conforms to the Copenhagen interpretation

of quantum mechanics. There the measurement is performed by a classical apparatus whichstimulates the wave function to collapse.[6]: The key feature of the Copenhagen interpretation is the dividing line between quantum

and classical. Bohr emphasized that the border must be mobile, so that even the ultimateapparatusthe human nervous systemcan be measured and analyzed as a quantum object,provided that a suitable classical device is available to carry out the task. In the absence of acrisp criterion to distinguish between quantum and classical, an identification of the classicalwith the macroscopic has often been tentatively accepted. The inadequacy of this approachhas become apparent as a result of relatively recent developments: A cryogenic version of theWeber bara gravitywave detectormust be treated as a quantum harmonic oscillator eventhough it can weigh a ton [. . .], and many other examples.

Thus the mobility of the border of quantum mechanics has paradox consequences, e. g.Schrdingers cat. Why is it not possible for macroscopic objects to be in a superpositionof states, like (|dead + |alive)/2 in the cats case or (|here + |there)/2 regarding thelocalization of any object. All these problems disappear in the theory of decoherence. Itsbasic proposition is that macroscopic objects cannot even approximately be treated as isolatedsystems. When including the environment into our calculations about quantum statistics, weare able to achieve the correct behavior of macroscopic objects. We are even going to avoidthe strange collapse of wave functions.

3.4 General concepts of decoherence

We are now analyzing the measurement of an apparatus (referred to as system QA) on a(quantum) system QS. Now both systems are treated as quantum systems. The measurementis assumed to be ideal that means only the state of the apparatus QA will change appreciably,while the state ofQS remains unchanged. Note that this assumption is roughly in contrast tothe Copenhagen interpretation.

Let HS be the Hilbert space of the system QS with basis {|n}n . Then an appropriateHamiltonian describing the interaction between both systems has the form

Hint =n

|nn| An . (34)

The operators An acting on states in HA are rather arbitrary depending on the explicit typeof interaction. Of course they must depend on the quantum number n because the finalstates of the apparatus should macroscopically distinguish the states |nthe basic task of areal measurement apparatus. The operator acting on the system QS is the identity operator:remember S =

n|nn|. If QS is in the state |n and QA is in the initial state |0, the

time evolution of the global state | H = HA HS reads10

| = |n |0t |t = e(i/)Hintt|n |0 = |n e(i/)Ant|0 =: |n

n(t) . (35)10The process described in (35) is often called premeasurement.


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Now let the state ofQS be a superpositionn cn|n. Then analogous to (35) the linearity of

time evolution leads to

| = n

cn|n|0 t |t = n

cn|nn(t) (36)which is an entangled state in general. We determine the reduced density matrix of systemQS. Before time evolution, we have11

S= trA|| =n,m

cncm|mn| , (37)

which refers to a pure ensemble. After time evolution, we achieve

tS= n,mcncm

m(t)n(t)

|mn| . (38)

If the states |n are approximately orthogonal12, i. e. n(t)|m(t) 1 the density matrixbecomes approximately diagonal in this basis,

tSn

|cn|2|nn| . (39)

Surprisingly, tS refers to a mixed ensemble. The coherence of (37) is gone, we measure eitherthe one or the other with the probabilities |cn|2, all interference terms vanished.

But is coherence really destroyed? Here I would like to quote [5] for clarity: Certainlynot, the right-hand side of (36) still displays a superposition of different n. The coherence isonly delocalized into the larger system. That means that the circumstance that we performlocal measurements only on QS is responsible for the apparent decoherence.

Premeasurement interactions do frequently and unavoidably happen with all macroscopicobjects, like scattering: Surprisingly, even the scattering of a single photon can lead to dra-matic effects. Thus, coherence is carried away by this photon and no interference effect onmacroscopic objects are observable. We have recovered our classical world.

4 Nonlocality in quantum mechanics

As we have seen before, measurements performed on entangled states are correlated, no matterhow far the subsystems are separated. Let us consider an entangled state of polarization of

two photons which were emitted in opposite directions (ez) by an atomic transition13

| = 12

|| + || (40)11Little help in calculating this: Let {k} be a basis of HA, then

S =k,n,m

cncm|mk |00 |kn| =n,m

cncm|m 0|

k

|kk|

=

|0

= 1

n| .

12The physical argument here is the possibility to discriminate the states n from each other.13With symmetry arguments, one can show that (40) is the correct state in which the photons are emitted.


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4. NONLOCALITY IN QUANTUM MECHANICS 37

where | stands for linear polarization in ex direction and | for linear polarization in eydirection. Measurement on photon 1 instantaneously pinpoints the polarization of photon 2which might be interpreted in two ways:

1. Information between photon 1 and 2 is exchanged with superluminal velocity which isnot possible according to special relativity.

2. Quantum theory violates Einsteins principle of locality which states that local opera-tions (like the performed measurement) can only have local effects.

Both possibilities are strange and unsatisfying for people used to classical mechanics14. There-fore, Einstein, Podolski and Rosen proposed a theory of hidden variables: In this theories,there exists an additional set of variables which pinpoints the outcome of every performedmeasurement. This variables are set at time of the decay, where both photons were at thesame place. Because the outcome of the measurements are predefined, no nonlocality ap-pears, and no signal has to travel at superluminal speed. The apparent statistic effect inquantum measurements is simply explained by the lack of knowledge of the value of thesehidden variables.

We now want to proof that the concept of a (linear) theory of hidden variables is wrong,i. e. it does not describe nature correctly. Therefore, we first proof

4.1 Bells inequalities

Bells inequalities give a statement of the probability of measuring specific values in determin-istic theories. Let = (1, 2, . . .) denote a set of parameters15 and () the probabilitydistribution of

with

() d = 1 . (41)

Let us suppose that we have an object which has four properties Aa, Aa , Bb, Bb which areindependent of each other and which take values 1. Our instruments measure these values.We can write the expectation value of the product AaBb as

P(AaBb) =

()A(a, )B(b, ) d (42)

where A(a, ) and B(b, ) are the measured values of Aa and Bb depending on the parameters. In any case A(a, ) 1 , B(b, ) 1 . (43)Then

P(AaBb) P(AaBb) =

()

A(a, )B(b, ) A(a, )B(b, )

d

=

()

A(a, )B(b, )

1 A(a, )B(b, )d

()

A(a, )B(b, )

1 A(a, )B(b, )d .14Even if quantum mechanics is hard to accept, this is even worse . . .15which correspond to our hidden variables in case of EPR


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Using (43), it follows that 1 A(a, )B(b, ) 0. With |a b| |a| + |b| we gain

P(AaBb) P(AaBb) ()A(a, )B(b, )1 A(a, )B(b, )d

+

()A(a, )B(b, )1 A(a, )B(b, ) d

()1 d

()A(a, )B(b, ) d

+

()1 d

()A(a, )B(b, ) d

= 2 P(AaBb) + P(AaBb) (44)which leads to

P(AaBb) P(AaBb)+ P(AaBb) + P(AaBb) 2 (45)which is one of the formulations of Bells inequalities.4.2 Application on QM

We now want to compare predictions of quantum theory versus predictions of Bells inequalities(45). Consider the photons described above in the state (40). The index a (b) now denotesthe measurement of the polarization of a photon in a direction having an angle a (b) with theex axis in the xy plane. The basis states of polarization in direction a is given by

|a = cos a | + sin a | , |a = sin a | + cos a | . (46)

Photon 1 is measured by the operatorAa = 2|aa| = cos(2a)

|||| + sin(2a)|| + || (47)with eigenvalues 1 and eigenkets |a, |a , photon 2 by Bb defined analogous to (47). ThenP(AaBb) is given by

P(AaBb) Aa Bb = | Aa Bb| . (48)With (40) eq. (48) simply becomes

P(AaBb) = cos

2(a b) . (49)Now taking 2a = 0, 2b = 0, 2a = 45, and 2b = 90, we get with (45)

cos0 cos(90)+ cos(45) + cos 45 = 1 + 2 ! 2 . (50)We learn that quantum mechanical predictions are violating Bells inequality. In experiment,quantum theory is validated which means that the hidden variables approach is incorrect.

References

[1] J. J. Sakurai: Modern Quantum Mechanics

[2] Schwabl: Quantenmechanik


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REFERENCES 39

[3] P. G. J. van Dongen: Quantenmechanik I

[4] Samuel J. Lomonaco Jr: A rosetta stone for quantum mechanics with an introduc-

tion to quantum computation[5] Claus Kiefer and Erich Joos: Decoherence: Concepts and Examples

[6] Wojciech H. Zurek: Decoherence and the transition from quantum to classical

[7] A. A. Grip and W. A. Rodrigues: Nonlocality in Quantum Physics


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Talk 4

Quantum Teleportation

Leszek Lupa and Joachim Schfer(Math 22nd, 2006)

The following scheme discusses the teleportation of polarized particle statesover arbitrary distances. The process of teleportation is realized with three pho-tons. An initial photon carries a predefined polarization and is superposed withone of an entangled photon pair. The joint measurement transfers the initial po-larization state to the other photon of the entangled pair which can be verifiedby polarized dependent measurement. The first experimental teleportation wassucceeded by Anton Zeilinger in 1997 by demonstrating the transmission and re-construction of45 polarized photons.

1 Introduction

1.1 Motivation

First we explain what teleportation is. Teleportation is a term created by science fictionauthors. It is a process, which lets a person or an object disappear, and construct an exactreplica immediately at another place. From the classical point of view, every person can bedescribed by its properties. These can all be determined by a physical measurement. To createa replica now, we need to do nothing but send these information and to create a person with

the same properties.If we look at this scheme from the quantum point of view, we will see that the measurement

is a weak point. If we want to get a perfect replica of the person or object, it is inevitableto determine the states of the atoms and electrons. As a result we have to measure quantumproperties. But according to Heisenbergs uncertainty principle, these can not be determinedwith arbitrary precession. Therefore teleportation in this way is not possible. Thus we needto construct a new scheme of teleportation.

In 1993 Charles H. Bennett has suggested that it is possible to transfer the quantum statesof a particle onto another. He provided one does not get any information about the state ofhis particle. Hence Heisenbergs principle still works. The central idea of Bennett is to usethe entanglement of particles. Entanglement describes correlations between quantum systems

41


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42 TALK 4. QUANTUM TELEPORTATION

much stronger than any classical correlation does. To circumvent Heisenbergs uncertaintyprinciple one need the help of so called pair of entangled particles.

The first experimental realization of quantum teleportation succeeded in 1997 by Anton

Zeilinger. By producing pairs of entangled photons with the process of parametric down-conversion and using two-photon interferometry for analyzing entanglement, he was able totransfer a quantum state from one photon to another. An important thing we will see isthat teleportation transfers the quantum state, but does not transfer mass. We also callit teleportation and not of cloning, because the original state is destroyed in the course ofteleportation. This is due to the no-cloning theorem, which says that it is not possible withinthe quantum theory to produce a clone of a given quantum system [2]. At least we willlearn that teleporting of a quantum state has the natural speed limit of the speed of light, inaccordance with Einsteins theory of relativity.

1.2 The basic concepts of quantum mechanicsIn this section we show the theoretical basics of quantum mechanics. In general we deal withtwo-level systems. Such a system can be represented by a polarization state of a photon orthe spin of the particle. To stay general we call the two states |1 and |0. The general wavefunction is the superposition of these states

| = |0 + |1

where and are both complex numbers with ||2 + ||2 = 1.For a system enclosing entangled photons the wave function is

|12 =1

2|01|12 |11|02with a special superposition of the states for the so-called entangled states or EPR states.Entanglement is essential for quantum teleportation. The entangled state describes a singlesystem in an equal superposition of the states. But involving two of these particles they losetheir identities in a certain sense. The entangled state contains no information about thestates of the individual particle. It only indicates that two of the particles are in the oppositestates.

This means that, if you measure state |0 on the first particle, the state of the secondparticle is in the state |1. Notice the fact that |12 has no restrictions on the spatialdistance between the both particles. Therefore the influence must be instantaneous. This

effect is so unbelievable that even Einstein could not believe it. He called it a spooky actionat a distance. Nevertheless experiments have shown that entangled states can be realized.

After this introduction we are going to have a closer look on problems we have withquantum teleportation and explain the solution of Bennett.

2 Concept of quantum teleportation

2.1 The problem of quantum teleportation

Suppose that a sender, we call it Alice, has a quantum system it wants to teleport. We assumethat Alice does not know the exact wave function of its particle and there is no way to learn


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2. CONCEPT OF QUANTUM TELEPORTATION 43

Figure 1: Theoretical scheme

it. Let us consider a two-level system with the general wave function| = |0 + |1 .

A measurement on this system would lead to a projection onto an eigenstate of the measuredobservable. And if | is not an eigenstate of this observable (extremely unlikely), one has noway to learn the exact wave function. Thus we conclude that measuring | in general leadsto a loss of information about the original states. A reconstruction of the state is impossible.The only possibility for Alice is to send the particle itself. This, of course, is the trivial way,thus we search for another way, where we do not have to send the original particle.

A solution of this problem was given by Bennett presented in his article Teleporting anUnknowing Quantum State via Dual Classical and Einstein-Podolsky-Rosen-Channels [1]. Aswe can see on the title he needed an entangled pair of particles. There is also another importanthint in the title. The information encoded in | can be divided into two parts, the classicaland non-classical one. To understand Bennetts scheme we are going to have a closer look athis theory yet.

2.2 Bennetts solution of the problem

Let us suppose that Alice has a particle in the state [1]

|1 = |01 + |11 .

For the non-classical channel between Alice and the receiver Bob we need an entangled pairof particles. Alice holds particle 2 and Bob particle 3 (see figure 1.1). The wave function ofthese particles can be expressed by

|23 = 12

|02|13 |12|03 .In the following we want to distinguish between the classical and the non-classical channel.

Non-classical information channel The fact that Alice and Bob share an EPR pairestablishes the possibility of an non-classical correlations between them. But note that theEPR pair does not carry any information about particle 1. Now we are going to couple


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particle 1 with Alices EPR particle. This can be done when Alice performs a measurementon the quantum system of the particles 1 and 2. This measurement is called Bell statemeasurement. It projects the system onto one of the four Bell states

|12 = 12

|01|12 |11|02 , |12 = 12

|01|02 |11|12 .These states are forming a complete orthonormal basis, called Bell basis, for particles 1 and2. The complete state of the EPR pair coupled with particle 1 is:

|123 = 2

|01|02|13 |01|12|03 + 2

|11|02|13 |11|12|03 .This equation can also be expressed in the terms of the Bell basis

|123 = 12

|12

|03 |13

+ |+12

|03 + |13

+ 12|12|13 + |03 + |+12|13 |03 .

From this equation we conclude that each measurement result of Alices Bell state measure-ment is equal likely with a probability of 1/4.

What we are also observing is that Bobs particle 3 is influenced by Alices Bell statemeasurement. It is instantaneously projected into one of the four pure states noting in thefollowing equation. When we define that

|0

10

and

|1 01we see that each possible resultant state for Bobs EPR particle.

|3

,

1 00 1

|3,

0 11 0

|3,

0 11 0

|3 .

In this equations we see that each possible resultant state for Bobs EPR particle is relatedwith simple unitary transformations to the state of |1 which Alice wanted to teleport.

Classical information channel In the first case for example, Alice measures |12 andBobs particle 3 is in the state of particle 1, except for an irrelevant phase factor. In theother three cases of Alices results of the Bell measurement, Bob only has to apply one ofthe unitary transformations to get the state of particle 1. At this point we see that we needa second channel which transmits us the information of Alices Bell measurement to Bob.And this information can only be transmitted via a classical information channel. Hence wesee that quantum teleportation is limited by the speed of light as we expected. Neverthelessteleportation theoretically could happen over arbitrary distances.

An important thing is while Bob has an exact replica of particle 1, Alice on the other handis left with the particle 1 and 2 in the states of |12 or |12 without any trace of the stateof particle 1. Therefore particle 3, in the state of |1, is not a clone! We call it the teleportedparticle 1. It is important to notice that the theoretical quantum teleportation is possibleonly because the Bell state measurement does not give any informations on the particles. Therealization of this theory we will clarify in the next chapter.


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3. EXPERIMENT 45

Figure 2: type-II parametric down-conversion

Figure 3: High speed infrared film exposed with light from type-II down conversion. The two ringsare the emitted cones. Polarized entangled photons are observed at the intersection of the two circles.

3 Experiment

3.1 Introduction

The heart of quantum teleportationthe creation of an entangled particle pair, e. g. photonsis a very difficult task in experimental realization. In 1997 Zeilinger [2] succeeded in the firstexperimental demonstration of quantum teleportation by transferring the polarization statefrom one photon onto another. The entangled pairs in Zeilingers experiment were created viatype-II parametric down conversion using a BBO-crystal. The outcoming entangled photonscould be projected onto at least two of the four Bell states using a standard 50:50 beamsplitterand coincidence measurement. The following sections focus on the realization of Zeilingersexperiment.

3.2 Type-II parametric down-conversioncreation of the entangled pair

Let us first discuss the production of an entangled photon pair and then its description inquantum mechanics. As seen in figure 2 an initial UV-pump can decay spontaneously with asmall propability ( 106) into two photons. The used BBO crystal (BaB2O4) is birefrin-gent and has a nonlinear electric susceptibility (2) which permits the decay. Each photon isemitted into a cone. The one in the top cone is vertically polarized, the one in the bottomcone, which is its exactly opposite partner, is horizontally polarized. Along the intersection ofthe two cones the polarization is undefined; the only information is that it has to be correlatedorthogonally. This leads to a polarization entanglement between the two photons in beams Aand B.


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Energy and momentum before and after the decay are conserved

p = A + B , kp

kA + kB

with p, A, B the frequencies of the pump signal, the extraordinary (top photon), and theordinary (bottom photon), respectively, and kp, kA, kB their wave vectors. The approximationfor the momentum conservation is made as the result of photon-phonon interaction in thecrystal grid.

Now we discuss the construction of the entangled polarization state quantum mechanically[3]. The initial state of the extraordinary and the ordinary photon is represented by

|0 = |0VA |0HA |0VB |0HB (1)

where A is one emitted photon e. g. the top photon, B its exactly opposite partner, e. g. thebottom photon. Both photons have no polarization, so it is necessary to use creation oper-ators for horizontal (aHA , a

HB

) and vertical polarization (aVA, aVB

). The Hamilton operator(calculation of perturbations in first order) describing this is given by

H1 =

aVA aHB

+ aHAaVB

, (2)Ep (2)

where (2) is the second-order electric susceptibility and Ep the amplitude of the classicalcoherent field.

The state vector |0 evolves according to

(t)

= e(i/)H1t|0 (3)

which we now expand in second order in time

(t) 1 i

H1t +1

2

i

H1t2|0 . (4)

By replacing |0 with |0VA |0HA |0VB |0HB we receive

(t) = 1 22

|0VA |0HA |0VB |0HB

i2

|1VA |0HA |0VB |1HB + |0VA |1HA |1VB |0HBwhere = t.

Now we define |0 := |0V|0H, |V := |1V|0H, and |H := |0V|1H and obtain(t) = 1 2

2

|0A|0B it

|VA|HB + |HA|VB .The second term, when normalized is given by

|+ = 12

|VA|HB + |HA|VB (5)and is one member out of a set of the four previously mentioned Bell states.


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3. EXPERIMENT 47

Figure 4: 50:50 standard beam splitter

Figure 5: 50:50 standard beam splitter

3.3 The Bell state analysis

After discussing the possibility of achieving the different Bell states with the BBO crystal weneed to study a way to analyze them [4].

To achieve projection of photons 1 and 2 onto a Bell state, we have to make them indis-tinguishable. This indistinguishability can be obtained with a 50:50 standard beam splitter.

As shown in figure 4 the beam splitter has two spatial input modes a and b and two outputmodes c and d. Quantum mechanically the process of the beam split can be described as

|a i2|c + 1

2|d , |b 1

2|c + i

2|d (6)

where |a is the quantum state of the input beam a and |b the quantum state of the inputbeam b. The factor i is nothing more then a phase jump as a result of reflection.

Thus one initial photon can be found in eather of the output modes |c and |d with thesame probability 1/2. Let us now assume that each initial photon 1 and 2 is in a superposedpolarization state

|1 = |H1 + |V1 , |2 = |H2 + |V2 , ||2

+ ||2

= 1 , ||2

+ ||2

= 1 .

Photon 1 and photon 2 have the same probability of being transmitted or reflected. As seenin figure 5 there are four possibilities how two photons leave the beam splitter, each with theprobability 1/4:

1. both photons are reflected

2. both photons are transmitted

3. a is reflected, b is transmitted

4. b is reflected, a is transmitted


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The indistinguishability is necessary for the inteference of the two particles in the beamsplitter. Now it is not possible to decide which of the initial photons left the beam splitter inthe given output ports. The initial state is given by

|i =

|H1 + |V1|a1|H2 + |V2|b2 .

Considering coherent superpositions in the beam splitter this initial state is transformed asdescribed in (6)

|f12 =

|H1 + |V1 i

2|c + 1

2|d

|H2 + |V2 1

2|c + i

2|d

.

As a result of the indistinguishability the total two-photon state including the spatial and thespin part has to obey bosonic quantum statistics. Therefore the outgoing physical state mustbe symmetric under exchange of labels 1 and 2.

We symmetrize the final state

|f = 12

|f12 + |f21 (7)and receive

|f = 12

2

(+ )

|H1|H2 + |V1|V2 i|c1|c2 + |d1|d2+ ( )|H1|H2 |V1|V2 i|c1|c2 + |d1|d2+ (+ )|H1|V2 + |V1|H2 i|c1|c2 + |d1|d2+ ( )|H1|V2 |V1|H2 |d1|c2 |c1|d2 (8)

We are able to project this two-photon state into two of the four Bell states

|12 = 12

|H1|V2 |V1|H2 , |12 = 12

|H1|H2 |V1|V2 .The last line of equation (8) describes the state |12 and indicates that if and only if thetwo outgoing photons are in this state they emerge the beam splitter in different beams.

For a full analysis of the Bell states we need to find a way to distinghuish between theother three states |+12, |+12, and |12. |+12 is the only state where the outgoing

beams have different polarizations. The identification of the remaining states |12 is moredifficult, because the two photons in this state always share the same polarization. For now itis possible to identify three of the four Bell states but we will just focus on the state |12.

3.4 The experiment by A. Zeilinger for 45-polarized photons

In this experiment A. Zeilinger only analyzes |12 which leads to the possibility of provingthe teleportation just in a quarter of the cases.

Before we learn the way to verify the teleportation we need to have a look on the pre-cautions that have to be made to accomplish a successful Bell state measurement. The Bellstate analysis is an extremely sensitive point and collapses as soon as the two photons can bedistinghuised in any way. Generally there are two necessary arrangements to make:


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3. EXPERIMENT 49

Figure 6: Experimental setup for the teleportation of

45-polarized photons: An UV-pulse passes

through the BBO crystal and decays into the first entangled pair of photons 2 and 3. The transmittingphotons are reflected by a mirror and pass the crystal again which causes another decay leading to anentangled pair of photons 1 and 4. Photon 4 is only a trigger which indicates that photon 1 is on itsway. The initial photon 1 is polarized and then superposed with photon 2 at the beam splitter. Aliceowns the detectors f1 and f2 to measure a coincidence of photon 1 and 2. The measuring leads tothe change of the polarization of photon 3 which can be checked with an polarized beam splitter andtwo detectors d1 and d2 belonging to Bob.

Good spatial and temporal overlap of the two photons needs to be guaranteed; any kind of Welcher-Weg-Information must be erased.

There is one effective way to destroy the path information of photon 1 and photon 2. For thispurpose we need to remember the correlation between the coherence time t and the widthof the frequency :

t 1

. (9)

The coherence time is the time interval within the two photons are detected as coincident.Increasing the coherence time of the photons to become much longer than the time intervalwithin they were created (= pulse duration) would lead to the impossibility of distinguishingbetween the particles as photon 1 or photon 2. According to equation (9) this can easily bedone by sharpening the frequency. Therefore Zeilinger uses bandwidth filters ( = 4.6 nm)to increase the coherence time to

500 fs. Compared to the pulse duration of 200 fs this is

sufficiently longer and guarantees the indistinguishablility.As we discussed in the last part, it is easy to identify |12 via a two-fold coincidence

measurement, here realized with detectors f1 and f2 (see figure 6). We already learned that ameasuring on photon 1 and photon 2 instantaneously changes the quantum state of photon 3.As shown in figure 6 Alice owns the detectors f1 and f2 to measure a coincidence of photon 1and 2 and thus to identify |12. Iff1f2 (two-fold conincidence) is detected then photon 3should be polarized instantaneously to +45. The polarization of photon 3 is checked by apolarized beam splitter which reflects in 90 or transmits photons whether they are polarized45 or +45.

Therefore a teleportation is verified by recording a three-fold coincidence d2f1f2 (+45)together with the absence of a three-fold coincidence d1f1f2 (

45) [5]. Thus the classical


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Figure 7: Theoretical prediction of three-fold coincidence. The signature of teleportation of a photonpolarization state +45 is a dip to zero at zero delay in the three fold coincidence rate with the 45analyzing detector d1f1f2 (a) and a constant value for the +45 analyzing detector d2f1f2 (b). Theshaded area is the region of teleportation.

Figure 8: Experimental results. Measured three-fold coincidence d2f1f2 (+45) with the absence ofd1f1f2 shown in (a) and measured three-fold coincidence d1f1f2 (45) with the absence of d2f1f2(b). The coincidence rates are plotted versus the delay in m which is varied by translating thereflection mirror given in figure 6.

information that both detectors f1 and f2 have fired simoultaneously must be sent from Aliceto Bob. This confirms that the teleportation and utilization of instantaneous interaction isno contradiction to Einsteins theory of relativity which limits the speed of all matter andinformation to the speed of light.

The temporal overlap between the photons 1 and 2 is changed by translating the reflectionmirror behind the BBO crystal shown in figure 6. Outside the region of teleportation (notshaded area in figure 7) photon 1 and photon 2 each will go to f1 or f2 independent of oneanother. Thus the probability for f1f2 coincidence is 1/2 which is twice as high as inside theregion of teleportation. The probability of receiving photon 3 at d1 and d2 is in each case 1/2as a result of its not well defined polarization state. This yields to the probability of 1/2 forthe

45 analysis (d1f1f2 coincidence) and the same for the +45 analysis outside the region


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REFERENCES 51

of teleportation. This theoretical prediction is shown in figure 7. The experimental results ofA. Zeilinger are in very good agreement to the theoretical prediction shown in figure 8.

References

[1] C. H. Benett, G. Brassard, C. Crepeau, R. Josza, A. Peres, and W. Woot-ers: Teleporting an unknown quantum state via dual classical and Einstein-Podolsky-Rosen channels, Phys. Rev. Lett. 70(13):18951899, March 1993

[2] Bouwemeester, Pan, Mattle, Eibl, Weinfurter, and Zeilinger: Experimen-tal quantum teleportation, Nature, 390:575579, December 1997

[3] C. Grey and P. Knight: Introductory Quantum Optics, pages 214217

[4] M. S. J.-W. Pan: Quantum Teleportation and Multi-photon Entanglement, PhDthesis, University of Vienna, 1999

[5] S. Will: Quantum teleportation, seminar on quantum optics.


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Talk 5

Quantum Cryptographywith Single Photons

Jan Klemmer(May 29th, 2006)

Cryptography has become more important than it ever was regarding all kindsof electronic communication. The security of the cryptosystems used today ismostly based on the difficulty of factoring big numbers. But with the realizationof a quantum computer, the factoring could be done with very little effort. Thesecurity of quantum cryptography is based on the rules of quantum mechanics andtherefore would not be threatened by a quantum computer.

1 Classical cryptography

Cryptography is the art of rendering a message unintelligible to any unauthorized party. Thealgorithms used to achieve this goal are called cryptosystems. Modern methods always involvethe exchange of a random number or binary string called the key which is used to makethe message unintelligible to third parties. This procedure is called encryption, the reverseprocess is called decryption. Once a message is encrypted, it should be impossible to decrypt itwithout the key. In reality, this condition is often weakened so that the decryption is extremelydifficult. Using common asymmetrical cryptosystems, modern computers would need many

years to decrypt a message, so the information is said to be secure for this period of time.

1.1 Asymmetrical (public key) cryptosystems

Asymmetrical cryptosystems use two different keys for the encryption and the decryptionprocess. One of the most popular systems of this kind is the RSA system, named after itsinventors Ronald Rivest, Adi Shamir, and Leonard Adleman. If Alice wants to send a messageto Bob (this is how sender and receiver are typically named), Bob has to generate two differentkeys: the public key which is used by Alice to encrypt the message and the private key whichhe keeps secret and which is used to decrypt the message. The security of RSA now relies onthe difficulty of factoring large numbers. This means, that for the factorization of a largenumber N, the number of steps a computer has to perform becomes exponential with N. The

53


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54 TALK 5. QUANTUM CRYPTOGRAPHY WITH SINGLE PHOTONS

basic idea of RSA is as follows: To create his private key, Bob uses two large prime numbers pand q (of size greater than 101000). In his public key, he publishes the product N = p q. Foran eavesdropper it would be sufficient to know p and q to reconstruct the private key. Thus,

the encrypted information is secure as long as N cannot be factorized in an appropriate time.There are two main problems with RSA:

The difficulty of factorization has not been proven yet. Hence, it is possible thatsomeone finds a fast algorithm for factorization.

In 1994, Peter Shor discovered a polynomial algorithm allowing fast factorization with aquantum computer. Thus, the realization of a quantum computer would make the RSAsystem useless.

1.2 Symmetrical (secret key) cryptosystems

These systems use the same key for both encryption and decryption. The one time padbelongs to this category. Here, Alice and Bob have to share a secret key consisting of arandom sequence of the binary values 0 and 1. The key has to fulfill the following conditions:

It has to be at least as long as the message.

It only may be used once (consequently the name one time pad).

If Alice wants to send a message to Bob, she just has to calculate the bit by bit XOR valueof the message and the key to obtain the encrypted message. Having received the encrypted

message, Bob only has to calculate the bit by bit XOR value again to gain the originalinformation.

The big advantage of this cryptosystem is that it is the only provable secure cryptosystem

known today. Yet, the key distribution is the main point of attack. The only practical (classic)way to exchange the keys is to use a public channel which can be eavesdropped. This problemcan be solved by using quantum cryptography (or quantum key distribution) which offers asecure possibility for the key exchange process.

2 Quantum cryptography with single photons

The security of quantum cryptography is based on the laws of quantum mechanics. One ofthese laws says that it is impossible to perform a measurement on a system without perturbingit. If now quantum systems like single photons are used to exchange a key, eavesdropping canbe identified due to the changes it has caused.


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2. QUANTUM CRYPTOGRAPHY WITH SINGLE PHOTONS 55

2.1 Polarized photons

Consider a single qubit in the superposition state

| = c0|0 + c1|1 .

Using photons, one can choose e. g. horizontal and vertical polarization as basis states:

|0 = | , |1 = | .

If the polarization state of a photon is measured in this basis (the rectilinear basis), themeasurement can have two different outcomes, | or |, respectively. If the initial state was

| = c0| + c1| ,

the probabilities for these two outcomes are given by |c0|2 and |c1|2, respectively. Here, themeasurement changes the state of

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