The genesis Coenhagen School (Bohr, eisenberg, ) Introduction to Quantum Comuting Frédéric Magniez The state of a quantum articule is only fied after a measurement Bennett, Brassard 84: erfectly secure quantum encrytion that can be used in ractice! Paradoe of Einstein, Podolsky, Rosen 35 INF554 Lectures 8 & 9 Quantum boes 3 Very distant articules remain linked!? Asect, Grangier, Roger, Dalibard 8: yes! Quantum encrytion of Ekert 9 can be certifiable Quantum key distribution 4 Classical information is encoded using bit (0/) Preskill The measure describes the state of the system A random bit is a hidden bit Quantum information is encoded using quantumbit Several ossible measures Outcome is determined during the measurement Problem Setting No rior shared secret information between Alice and Bob Authenticated classical channel Goal: Get a rivate key between Alice and Bob Alication: Onetime ad (Miller 88Shanon 945) Classical results Imossible: all the information is in the canal Possible (using randomized techniques): Amlify the rivacy of an imerfect rivate key WashingtonMoscow hotline (963)
The rotocol BB84 [BennettBrassard 84] 5 Qubit 6 Protocol: quantum art Key: Encoding: State dimensional unit vector i = cos +sin i i Decoding: Key: Protocol: classical art Reconciliation: Alice and Bob ublicly announce their coding choices A&B only kee key bits with same choices Security: Interceting and oening a bo errors A&B check few key bits at random ositions Privacy amlification: Perfect key using with few other more key bits Conclusion Secrete key generation using an authenticated classical channel Small initial rivate key large rivate key, with no authenticated channel general case (comle amlitudes): i = = + i, + = Randomized orthogonal rojection Evolution + i Unitary transformation G U() ( reversible) Definition: G C s.t. G G = Id i i G 0 i = G i 0 i = G i G i Eamles of transformations 7 Polarization of hotons 8 Reversible classical transformation Identity Negation adamard transformation Definition: halfwave blade at,5 Id Proerties: quantum coin fliing bi ( + i) = ( + ( ) b i) i does not commute! State Polarization: dimensional vector i = cos!i + sin "i Calcite crystal searates horizontal and vertical olarizations A measure modifies the system Transformation Well known transformation: halfwave blade orthogonal symmetry around its ais Any rotations (ossibly with comle angles) cos sin
Eercice : Quantum key distribution Imlementation Elain how to realize the boes of slide 3 Imlement the rotocol of slide 4 using random bits, adamard transformations, and measurements Analysis of a secific attack Assume a third arty Eves intercets a hoton with robability /0, observes it, and forwards the rojected hoton to Bob Assume furthermore that Alice & Bob check each bit of their key with robability /0 Comute The robability Eve learns a bit of the secret key The robability Eve is detected 9 Entanglement Princile: distant boes which remain entangled Outcomes are random but correlated if boes are oened similarly and uncorrelated otherwise Bell 64 inequality Cooerative random game Classical 75% of victory Quantum > 85% of victoiry Eerimental verification at Orsay in 98 Alication: quantum certification 0 BellCS inequality as a classical game Eercice : CS inequality Game Alice and Bob share random bits but cannot communicate Alice receives a random bit, Bob y Alice returns a bit a, Bob b a Goal: maimize 0 0 0 0 0 shared random bits = Pr(a b = ^ y),y CS inequality [969] The best robabilistic strategy achieves =3/4 0 0 0 0 y b Deterministic strategy Provide a deterministic strategy achieving =3/4 Show that no deterministic strategy can achieve = Conclude that 3/4 for every deterministic strategies Randomized strategy We assume that both layers have access to a shared source of randomness, called λ Note: Physicists call λ a hidden variable Justify why this is the most owerful model of random ressource Let λ be the winning robability when λ is fied Show that there must be some λ such that λ Conclude that the best robabilistic strategy achieves =3/4
nqubit 3 Mathematical background: Tensor roduct 4 State ic {0,}n such that k ik = i = {0,} n Eamles Searated qubit: 00i + = ( + i) Entangled qubit: 00i + i 6= i i EPR state Randomized orthogonal rojection {0,} n Evolution Unitary transformation G U( n ) with i G 0 i = G i {0,} n = ( G C n n s.t. G G = Id) Vector saces V, W: vector saces V W is the free vector sace San ( v w : v V, w W ) with equivalence relations (v+v) w = v w + v w v (w+w) = v w + v w (c v) w = v (c w) = c (v w) Linear mas S: V, T: W Y : linear mas S T : V W Y is the linear ma satisfying S T (v w) = S(v) T(w) (and etended by linearity) Alications Joint robability distributions on saces V, W D( VW) = D(V) D(W) D(V)D(W) (: roduct distributions) Transformation c 5 Partial measure: qubit case 6 Definition c 0bi = 0bi c = i ( c abi = ai a Reresentation control bit target bit Bell basis change yi b)i bi yi 0 000 c = B000 C @ 000A 000 00i = ( 00i + i) 0i = ( + ) 0i = ( 00i i = ( i) ) of first qubit Projectors of first qubit = a 00 + b 0 + c 0 + d P 0 = 00 00 + 0 0 = 0 0 I P = 0 0 + = I P 0? P = Id P 0 i = a + b P0 i P 0 i = 0 P i P i P i = = c + d Generalization Partial measure roject to a subsace comatible with the observation Probability = square norm of the rojection Outcome = renormalization of the rojection a 0 + b a + b c 0 + d c + d
Eercice 3 7 BellCS inequality as a quantum game 8 Partial vs comlete measurement Consider any twoqubit state, and measure its first qubit and then its second qubit Comute the robability distribution of the outcome Conclude that observing the two qubits is equivalent to measuring each qubit individually in any order Note: This can be generalized to any number of qubits Noncloning Assume there is a unitary ma U such that, for every qubit i: U( i) = i i Comute U( i) for i = ( + i) using the definition of U using the linearity of U and then again the definition of U Get a contradiction and conclude Reminder Goal: maimize Quantumly Alain and Bob share an EPR state a = Pr(a b = ^ y),y 0 0 + Bob erforms a rotation of angle 8 If =, Alice erforms a rotation of angle 4 If y =, Bob erforms a rotation of angle 4 Alice et Bob observe their qubit and send their resective outcomes Theorem: = cos ( 8 ) 0.85 / / y b Realization: [AsectGrangierRogerDalibard: Orsay 8] Eercice 4: EPR state 9 Suerdense coding [99] 0 Entangles boes Imlement the entangled boes of slide 0 using EPR states Proerties Show that alying a unitary U on the first qubit of an EPR state is equivalent to alying the transosed matri of U on its second qubit Quantum game Prove the theorem of revious slide Problem Alice & Bob share an EPR state: Alice wants to send two bits y to Bob But Alice can only send one qubit to Bob Bell basis change yi 00i = ( 00i + i) 0 0 / + qubit / y y? yi 00i = ( 00i + i) 0i = ( + ) 0i = ( 00i i = ( i) ) Protocol Alice alies to its qubit, if y=; and FLIP, if = Alice sends its qubit to Bob Bob erforms the inverse of the Bell basis change, and observes y FLIP = 0 0
Quantum teleortation Realization of teleortation Problem Alice wants to transmit a qubit i to Bob Bob: far and unknown osition to Alice Circuit i = + i yi Realization Alice 0 0 Bob ψ Interaction quantique i i Interaction interne Interaction classique The quantum communication does not reveal anything on i! Alice ψ Bob Analysis P Final state with yi = () y (FLIP),y yi yi i By measuring,y, third qubit is rojected to yi After learning,y, Bob can correct yi to i Realizations hoton [Zeilinger et al : Innsbruck 97] hoton, 6 km [Gisin et al : Genève 0] atom [Blatt et al : Innsbruck 04] Today: over 00km yi Coin fliing 3 EPR based coin fliing 4 Problem Alice and Bob are fare away They want to fli a coin in a fair way but they don t trust each other Classically Solutions based on harness assumtions of combinatorial roblems No unconditionally secure solution Quantumly There eists a rotocol with maimal bias 0,5 [00] There is no rotocol with bias better than 0,07 [00] There eists a rotocol with maimal bias 0,07 [009] Weak version: election Alice wants head Bob wants tail There eists a rotocol with arbitrarily small bias [007] Main idea Assume Alice & Bob share an EPR state 0 0 / + / Alice & Bob observe their qubit and get bit a,b Fact a=b with robability a (res. b) is a uniform random bit Problems Who create the EPR state? If Alice does, Bob needs to check that is an EPR state: And for instance not 00i a=b=0 with robability In ordert o check the EPR state, Bob needs the qubits Then Alice needs to check that Bob gives back the correct qubit
Google and NASA sna u quantum comuter : Nature News & Comment 0//03 :09 NATURE NEWS EPR based coin fliing 5 0 0 + Protocol 0 0 / Google and NASA sna u quantum comuter / DWave machine to work on artificialintelligence roblems. / / Initialization + Alice reares EPR states Alice send the corresonding first qubits to Bob Selection Bob select the EPR state that will be use for fliing The other EPR state will be use for checking the honesty of Alice Alice and Bob observe their resective qubit of the fliing EPR state Checking Alice sends to Bob her qubit of the checking EPR state Bob measures the checking EPR state Nicola Jones 6 May 03 DWave, the small comany that sells the world s only commercial quantum comuter, has just bagged an imressive new customer: a collaboration between Google, NASA and the nonrofit Universities Sace Research Association. The three organizations have joined forces to install a DWave Two, collaboration the Quantum Artificial Intelligence Lab at NASA's Theorem Sign In My Account Ames Research Center in Moffett Field, California. The lab will elore areas such as machine learning making comuters sort If both articiant are honest, the outcome is a erfect random bit If one of the articiants is dishonest, the maimal bias is /4 Attacks Print the comuter comany's latest model, in a facility launched by the If the measure outcomes is correct, Bob accets coin Otherwise, Bob declares that Alice has cheated 6 and analyse data on the basis of revious eerience. This is useful for functions such as language translation, image searches and Goal: increase the robability to get 0 voicecommand recognition. We actually think quantum machine may rovide the most creative roblemsolving rocess SUBSCRIBE: omemeasure Delivery Digital Subscritions Estate Rentals Service Alley Bob s attack: its Gift qubits, and select the EPRRealair giving 0 (ifcars any)today's Paer Going Out Guide Find&Save learning under the known laws of hysics, says a blog ost from Google 00 EPR state + EPR state 00 PostTV Politics Oinions Local Sorts National World Business Tech Lifestyle Entertainment Jobs More Alice s attack: 3 3 National Security The DWave Two quantum comuter has a 5qubit rocessor (ictured) that can do some calculations thousands of times faster than conventional comuters. DWAVE describing the deal. The Googleled collaboration is only the second customer to buy a comuter from DWave, which is based in Burnaby, Canada. Aerosace giant Lockheed Martin, headquartered in Bethesda, Maryland, was the first. Lockheed urchased a In the News Barack Obama Drones American Airlines Benghazi American Idol DWave quantum comuter in 0 and installed it in a new Quantum Comutation Center at the University of Southern 7 Safari Power Saver Click to Start Flash Plugin Suercomuter 8 California (USC) in Los Angeles. DWave declines to disclose the rice of their comuters. Both quantumcomuting centres the one at USC and the one at Ames have reserved 0% of their comuter time Feynman 8 for access by outside researchers. Judging by the thirdarty requests we've had, I'd say there should be lenty of Can systems besays robabilistically simulated demand robably more thanquantum can be accommodated, Daniel Lidar, director of the USCby centre. So far, eole Senate reort: Benghazi attack was reventable VIDEO To Sringsteen olitical moments NSA seeks to build quantum comuter that could crack most tyes of encrytion MAP The United States (of Pizza) a classical comuter? [...] the answer is certainly, No! Ferris wheel to join have mostly used these machines to elore ossible alications of quantum comuting and to investigate how the D.C. area skyline comuter behaves, rather than to solve reviously unanswered roblems. Deutsch 85 Alternative model The DWave comuterquantum is unusual because uses quantum bits (qubits) bits that Turingit Machine can eist in two states, on and off, simultaneously to seed u calculations, and Eistence of a universal Turing Machine By Steven Rich and Barton Gellman, Published: January Email the writers because it does not oerate on the normal 'gate' model of comuting, whereby logic In roomsize metal boes secure against electromagnetic leaks, the National Security Agency is racing to build a comuter that could break nearly every kind of encrytion used to rotect banking, medical, business and government records around the world. gates are used to maniulate those bits. Instead, it is an 'adiabatic' comuter, which reads out the ground state of its qubits to find a solution. The academic community Simon, Shor 94 Quantum algorithms with eonential seedu Quantum attack of ublickey crytosystems htt://www.nature.com/news/googleandnasasnauquantumcomuter.999 According to documents rovided by former NSA contractor Edward Snowden, the effort to build a crytologically useful quantum comuter a machine eonentially faster than classical comuters is art of a $79.7 million research rogram titled Penetrating ard Targets. Much of the work is hosted under classified contracts at a laboratory in College Park, Md. Vous Vendez Un Bien? 7 www.artenaireeuroeen.fr Vendez 00% Entre Particuliers Vendez Sans Commission à la Vente Related stories Quantum comuter asses seed test Further roof for Page of 4
Quantum arallelism 9 Logical comuting 30 nqubit Suerosition of all ossible values n ossible values Parallel comutation In one ste, n comutations But only one outcome can be (randomly) observed! Strategy Nb of articules in the Universe 300 Combine cleverly those values before measuring them 4 bits can take 4 =6 values 0000 000 000 00 000 00 00 0 000 00 00 0 00 0 0 Gates A gate C is a function on at most 3 qubits Eamle: AND, OR,,... Circuit A circuit is a sequence of gates The size of C is its number L of gates C comutes a function f if for all inut : C = C L...C C C(, 0 k )=(f(),z) { f() OR } 0 0 AND Theorem Any function can be comuted by a circuit using only, OR, AND gates Quantum gates and circuits Gates U U( k ), k =,, 3 A quantum gate is a unitary ma that acts uon at most 3 qubits Tensor roduct of gates Circuit A quantum circuit is a sequence of gates (etended by Id) G G i i (G G ) i i =(G i)(g i) G R 4 Theorem Any unitary can be realized eactly by a circuit and aroimated using only gates c and 3 Reversible comuting Reversible circuit A logical circuit is reversible if each gate is reversible A reversible circuit is also a quantum circuit (since it ermutes logical states) Embedding f where: 0 = 0 = 0 0 = = 0 u v=(u v,u v,...) Theorem If a function f can be comuted by a logical circuit of size L, then f can also be comuted by a reversible circuit of size O(L) Universality f : {0, } n! {0, } m : {0, } n+m! {0, } n+m f (, y) =(, y f()) The Toffoli gate (cc) is universal for reversible comutating T (a, b, c) =(a, b, c (a ^ b)) 3
Quantum imlementation of classical functions 33 A first quantum algorithm [99] 34 Normal form Function: f : {0, } n! {0, } m Circuit: U f : 7! f()i yi 7! y f()i Alternative form Sf Boolean function: Circuit: i = ( Conclusion: 8 < : i) f : {0, } n! {0, } Uf U f ( i) =S f () i 8 < : ( f()i f()i) = ( )f() ( i) DeutschJozsa roblem Oracle inut: f : {0, } n {0, } a blackbo function such that f is either constant or balanced Outut: 0 iff f is constant Query comleity Deterministic: n + Quantum: f(3) =? f(3) = Secial case n= No restriction on f Deterministic vs quantum: queries vs query Quantum solution ( n= ) 35 Analysis ( n= ) 36 7! f() can be nonreversible! Reversible imlementation of f Initialization: S f f constant? f balanced i S f + i ( ( ) ) f(b) f(0) + ( ) f() i Parallelization: ( + i) adamard gate: halfwave blade at,5 Quantum circuit ( + ( ) b i) S f? Query to f: (( ) f(0) + ( ) f() i) Interferences: ( )f(0) ( + i) +( ) f() ( i) Final state: (( )f(0) + ( ) f() ) +(( ) f(0) ( ) f() ) i
General solution for DeutshJozsa 37 Analysis 38 Reversible imlementation of f S f ( ) f() ( ) f() {0,} n {0,} n Quantum Fourier transform QFT n QFT n = Quantum circuit n/ ( ) y yi y S f ( + ( y = where i y i mod i QFT QFT? ) b i) QFT S f QFT Initialization: Query to f: 00...0i Parallelization: n/ {0,} n Interferences: Final state: n/ ( ) f() {0,} n n ( ) f()+ y yi,y{0,} n n {0,} n ( ) f constant f() 00...0i +? f balanced y6=00...0 00...0i yi, y6=00...0 y yi BernsteinVazirani 39 Eercice: Analysis 40 Problem Oracle inut: f : {0, } n {0, } a blackbo function such that f() =a for some fied a {0, } n Outut: a Query comleity Randomized: n Query f(0 i 0 ni )=ai, for i=,,...,n Quantum: QFT S f QFT Initialization: Parallelization: Query to f: Interferences: ai Quantum circuit S f QFT QFT ai Final state:
On the difficulty of factorization 4 Asymmetric encrytion 4 RSA Challenges htt://www.rsasecurity.com/rsalabs Oneway functions Eamle: multilication / factorization Bases of modern encrytion (Rivest, Shamir, Adleman 77) RSA640 (93 digits) : 3074840490043735075003588856793003734608477545706948830644058085045563468967738678437967838 033454707308509954859007337748783557438645404697366047765346609 = 6347336458095384844333883865090859847836700330938085389333000450858675579 90087866483685573935439754789678996855493666638539088070380044989579646557 RSA Algorithm (allows rivate communication) security based on the difficulty of factorizing RSA challenges (99007) RSA00, $,000, 99 RSA640, $0,000, 005 7 9 =? 667 =?? 3074840490043735075003588856 7930037346084775457069488 30644058085045563468967738 67843796783803345470730850 9954859007337748783557438 645404697366047765346609 =?? Quantum algorithm for factorization 43 From eriod finding to factorization 44 Classical reduction Factorization can be reduced to eriod finding (of some arithmetic function) Quantum tool: Fourier Transform FT reveals the eriod of a signal FT is (very) fast on a quantum suerosition Shor 94 307484049004373507500358885679300373460847754570694883064405808504556346896773 8678437967838033454707308509954859007337748783557438645404697366047765346609 = 6347336458095384844333883865090859847836700330938085389333000450858675579 90087866483685573935439754789678996855493666638539088070380044989579646557 Theorem [SimonShor 94] Finding the eriod of any function on an abelian grou can be done in quantum time oly (log G ) Order finding Inut: integers n and a such that gcd(a,n)= Outut: the smallest integer q 0 such that a q = mod n Reduction to eriod finding: the eriod of a mod n is q Factorization Inut: integer n Outut: a nontrivial divisor of n Reduction: Factorization R Order finding Check that gcd(a,n)= Comute the order q of a mod n Restart if q is odd or a q/ mod n Otherwise (a q/ ) (a q/ + ) = 0 mod n Return gcd(a q/ ±, n)
Simon s roblem 45 Quantum solution 46 Problem Oracle inut: f : {0, } n!{0, } n a blackbo function such that 9s 6= 0 n : 8 6= y, f() =f(y) () y = Outut: the eriod s Comleity Randomly: (n) queries Quantumly: O(n) queries and time O(n 3 ) Idea Use a Fourier transformation: QFT n = y = where i y i mod i Realization of QFTn using adamard gates: 0 f() U f ( + ( ) b i) n/ ( ) y yi y QFT n U f wi w f()i s 0 n i QFT n U f 0 n i Initialization: Parallelization: Query to f: Filter: 0 n i 0 n i n/ n/ 0 n i QFT n Partial measure: roject to a subsace comatible (( ) y with + ( ) ( the s) y observation ) yi f()i Probability = square norm of the rojection Outcome = renormalization of ( ) y the rojection ( + ( ) s y ) yi f()i Interferences: f()i ( + (n+)/ y (n+)/ y (n )/ y:s y=0 f()i si) f()i yi f()i yi : y s? Finding the eriod 47 More difficult... 48 Construction of a linear system After n + k iterations: y,y,...,y n+k s s 0 n is solution of the linear system in t: y t = 0 y t = 0. y n+k t = 0 The y i are of rank n with roba / k+ System solutions: 0 n and s Comleity Constructing the system: O(n) queries, time O(n ) Solving the system: no query, time O(n 3 ) y t + y t +...+ y n t n =0 y t + y t +...+ y n t n =0. y n+k t + y n+k t +...+ yn n+k t n =0 Period Finding(G) Oracle inut: function f on G such that f is strictly eriodic for some unknown G: f() =f(y) () y Outut: generator set for Eamles Simon Problem: Factorization : Discrete logarithm: Pell s equations: Grah Isomorhism: Quantum olynomial time algorithms (in log G ) Abelian grous G: QFTbased algorithm [995] Normal eriod grous : QFTbased algorithm [000] Solvable grous G of constant eonent and constant length [003]... G a a t G =(Z ) n,= {0,s} G = Z, = rz G = Z,= {(r, ) : Z} G = R G = S n f
ard instances 49 Grover search algorithm 50 Shift roblem Z N oz Dihedral grou : subeonential time O( log N) [003] f(, 0) 5 3 9 7 6 0 5 4 shift = 3 f(, ) 3 9 7 6 0 5 4 5 Grah Isomorhism A : B : Instance of Period Finding on the symmetric grou where we just know how to imlement QFT... [997] General case Polynomial number of queries to f, but eonential ostrocessing time [999] A B a b 6 c 8 d 3 e 5 f g 4 h 7 Grover roblem Oracle inut : f : {0, } n!{0, } such that Outut : 0 Constraint : f is a blackbo Query comleity Randomized: ( n ) Quantum: ( n ) n = =) query 9! 0 : f( 0 )= Preliminary remarks 5 Quantum solution ( n =) 5 Imlementation of S f f ( ) f() = 0 0 i S f S 0 0 i Double adamard gate i i = i with ( + ( ( + ( y ) i) ) i) ( ) y yi y = y + y mod Initialization: 00i Parallelization: ( 00i + + + i) Query to : Interferences: Final state: f Query to : _ 0 00i 00i 0 i y 0 i ( ) 0 y yi ( ) 0 y yi 00i = 0 i y
Geometrical analysis 53 Geometrical analysis, general case 54 Grover oerator Grover oerator G def = S f S 0 G def = S f S 0 Vect R ( 0 i, unifi) 0 i Vect R ( 0 i, unifi) 0 i S f = S = S? S f = S = S? S 0 = S 00i S 0 = S 00i S 00i = S unifi G = S unifi S? = R unifi S 00i = S unifi G = S unifi S? = R unifi 0i? with sin = hunif 0 i = 0i? with sin = hunif 0 i = n After iteration unifi 7! G unifi = 0 i After T = /π ( n ) iterations unifi 7! G T unifi 0 i ow many quantum algorithms eist? Unstructured roblems Grover algorithm [996] Algebraic roblems SimonShor algorithm [994] Well structured roblems Classical algorithms are otimal! Problems with few structures Quantum walk based algorithms [003] quantum analogy of random walks Eamles Element Distinctness, Commutativity: N /3 [004] Triangle Finding: N 9/7 (lower bound N) [03] Square Finding: N.5 (lower bound N) [00] Matri Multilication: N 5/3 (lower bound N 3/ ) [006] ANDOR Tree evaluation: N [007] 55 To continue... 56 An Introduction to Quantum Comuting Authors: Philli Kaye, Raymond Laflamme, Michele Mosca Editor: Oford University Press Quantum Comutation and Quantum Information Authors: Michael A. Nielsen, Isaac L. Chuang Editor: Cambridge University Press Classical and Quantum Comutation Authors: A. Yu. Kitaev, A.. Shen, M. N. Vyalyi Editor: American Mathematical Society Collection: Graduate Studies in Mathematics Lecture Notes for Quantum Comutation Author: John Preskill Website: htt://www.theory.caltech.edu/~reskill/h9/ Quantum roofs for classical theorems Author: Andrew Drucker, Ronald de Wolf Website: htt://ariv.org/abs/090.3376
Where does the quantum sueriority come from? 57 Entanglement? No: they can be simulated using only real amlitude Yes: they can induce destructive interferences 0 0 / + / Negative amlitudes? Unfalsifiable money, artificial intelligence, Quantum comuting For a better understanding of quantum henomenon New mathematical tool for roving results in classical comuting! Technology ardness of amlitudes? No: amlitudes must be easily comutable for being hysically realizable Some quantum centers in the world CQC IQIS 59 RQC IQC PCQC QIS IQI 58 Alications Classical entanglement eists: shared randomness But quantum entanglement is stronger BellCS inequality and alications Comle amlitudes? Future QISC CQI CQT CQCT Comuter, intermediate models: boson samling Certification : encrytion, random generator, comutation www.cqc.fr 60