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Explore the basics of quantum information processing, including quantum algorithms, quantum error-correcting codes, and quantum cryptography. Learn about the potential of quantum systems and how they can affect various fields.
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Introduction to Quantum Information ProcessingCS 467 / CS 667Phys 667 / Phys 767C&O 481 / C&O 681 Richard Cleve DC 2117 cleve@cs.uwaterloo.ca Lecture 1 (2008)
number of transistors 109 108 107 106 105 year 104 1975 1980 1985 1990 1995 2000 2005 Moore’s Law • Measuring a state (e.g. position) disturbs it • Quantum systems sometimes seem to behave as if they are in several states at once • Different evolutions can interfere with each other Following trend … atomic scale in 15-20 years Quantum mechanical effects occur at this scale:
Quantum mechanical effects Additional nuisances to overcome? or New types of behavior to make use of? [Shor, 1994]: polynomial-time algorithm for factoring integers on a quantum computer This could be used to break most of the existing public-key cryptosystems on the internet, such as RSA
START FINISH p1 p2 START p3 FINISH α1 α2 α3 START FINISH Quantum algorithms Classical deterministic: Classical probabilistic: Quantum: POSITIVE NEGATIVE
quantum information theory classical information theory Also with quantum information: • Faster algorithms for combinatorial search [Grover ’96] • Unbreakable codes with short keys [Bennett, Brassard ’84] • Communication savings in distributed systems [C, Buhrman ’97] • More efficient “proof systems” [Watrous ’99] … and an extensive quantum information theory arises, which generalizes classical information theory For example: a theory of quantum error-correcting codes
This course covers the basics of quantum information processing Topics include: • Quantum algorithms and complexity theory • Quantum information theory • Quantum error-correcting codes • Physical implementations* • Quantum cryptography • Quantum nonlocality and communication complexity
General course information • Background: • classical algorithms and complexity • linear algebra • probability theory • Evaluation: • 5 assignments (12% each) • project presentation (40%) Recommended texts: An Introduction to Quantum Computation, P. Kaye, R. Laflamme, M. Mosca (Oxford University Press, 2007). Primary reference. Quantum Computation and Quantum Information, Michael A. Nielsen and Isaac L. Chuang (Cambridge University Press, 2000). Secondary reference.
digital: analog: 1 1 1 r 0 0 0 0 1 r [0,1] Types of informationis quantum information digital or analog? probabilistic p q • Probabilities p, q0,p + q= 1 • Cannot explicitly extract p and q • (only statistical inference) • In any concrete setting, explicit • state is 0 or 1 • Issue of precision (imperfect ok) • Can explicitly extract r • Issue of precision for • setting & reading state • Precision need not be • perfect to be useful
1 1 0 0 0 1 Quantum (digital) information • Amplitudes , C, ||2+ ||2= 1 • Explicit state is • Cannot explicitly extract and • (only statistical inference) • Issue of precision (imperfect ok)
Ket:ψalways denotes a column vector, e.g. Dirac bra/ket notation Convention: Bra:ψalways denotes a row vector that is the conjugate transpose of ψ, e.g. [*1 *2 *d] Bracket:φψ denotes φψ, the inner product of φandψ
Rotation: NOT (bit flip): Hadamard: Phase flip: Basic operations on qubits (I) (0) Initialize qubit to |0 or to |1 (1) Apply a unitary operation U(U†U=I) Examples:
0+1 Basic operations on qubits (II) 1 ψ′ (3) Apply a “standard” measurement: ψ ||2 0 ||2 … and the quantum state collapses () There exist other quantum operations, but they can all be “simulated” by the aforementioned types Example: measurement with respect to a different orthonormal basis {ψ,ψ′}
Distinguishing between two states Let be in state or Question 1: can we distinguish between the two cases? • Distinguishing procedure: • apply H • measure This works because H+=0 andH−=1 Question 2: can we distinguish between 0 and +? Since they’re not orthogonal, they cannot be perfectly distinguished …
n-qubit systems Probabilistic states: Quantum states: Diracnotation: |000, |001, |010, …, |111are basis vectors, so
Unitary operations: ? Measurements: Operations on n-qubit states (U†U = I) … and the quantum state collapses
Product state (tensor/Kronecker product): Example of an entangled state: … can exhibit interesting “nonlocal” correlations: Entanglement
qubits: time U W #1 V #2 #3 #4 unitary operations measurements Structure among subsystems
0 1 1 0 1 1 0 0 1 1 0 1 Quantum computations Quantum circuits: “Feasible” if circuit-size scales polynomially
(do nothing) U The resulting 4x4 matrix is Maps basis states as: 00 0U0 01 0U1 10 1U0 11 1U1 Example of a one-qubit gate applied to a two-qubit system
Controlled-U gates U Resulting 4x4 matrix is controlled-U = Maps basis states as: 00 00 01 01 10 1U0 11 1U1
a a X b ab 0 +1 0−1 0−1 0−1 Controlled-NOT (CNOT) ≡ Note: “control” qubit may change on some input states
Introduction to Quantum Information ProcessingCS 467 / CS 667Phys 667 / Phys 767C&O 481 / C&O 681 Richard Cleve DC 2117 cleve@cs.uwaterloo.ca Lecture 2 (2008)
How much classical information in n qubits? • 2n1 complex numbers apparently needed to describe an arbitrary n-qubit pure quantum state: • 000000 + 001001 + 010010 + + 111111 • Does this mean that an exponential amount of classical information is somehow stored in n qubits? • Not in an operational sense ... • For example, Holevo’s Theorem (from 1973) implies: one cannot convey more than n classical bits of information in n qubits
b1 b1 U U ψ b2 ψ b2 b3 b3 n qubits n qubits bn bn 0 bn+1 0 bn+2 m qubits 0 bn+3 0 bn+4 0 bn+m Holevo’s Theorem Easy case: Hard case (the general case): b1b2 ... bncertainly cannot convey more than n bits! The difficult proof is beyond the scope of this course
Superdense coding (prelude) Suppose that Alice wants to convey two classical bits to Bob sending just one qubit ab Alice Bob ab By Holevo’s Theorem, this is impossible
Superdense coding In superdense coding, Bob is allowed to send a qubit to Alice first ab Alice Bob ab How can this help?
if a=1then apply X to qubit if b=1then apply Z to qubit send the qubit back to Bob • Alice: How superdense coding works • Bob creates the state 00+11 and sends the first qubit to Alice Bell basis • Bob measures the two qubits in the Bell basis
H Measurement in the Bell basis Specifically, Bob applies to his two qubits ... and then measures them, yielding ab This concludes superdense coding
Introduction to Quantum Information ProcessingCS 467 / CS 667Phys 667 / Phys 767C&O 481 / C&O 681 Richard Cleve DC 2117 cleve@cs.uwaterloo.ca Lecture 3 (2008)
Recap • n-qubit quantum state:2n-dimensional unit vector • Unitary op: 2n2n linear operation U such that U†U=I(where U† denotes the conjugate transpose of U) U0000 =the 1st column of U U0001 =the 2nd column of U the columns of U : : : : : :are orthonormal U1111 =the (2n)th column of U
2 2 span of 0 and 1 0 1 Incomplete measurements (I) Measurements up until now are with respect to orthogonal one-dimensional subspaces: The orthogonal subspaces can have other dimensions: (qutrit)
Incomplete measurements (II) Such a measurement on 0 0 +1 1+2 2 (renormalized) results in 00+11 with prob 02+12 2with prob 22
Result is the mixture 0000+0101 with prob 002+012 1010+1111 with prob 102+112 Measuring the first qubit of a two-qubit system 0000+0101 +1010 +1111 • Defined as the incomplete measurement with respect to the two dimensional subspaces: • span of 00 & 01 (all states with first qubit 0), and • span of 10 & 11 (all states with first qubit 1)
Easy exercise: show that measuring the first qubit and then measuring the second qubit gives the same result as measuring both qubits at once
0+1 0+1 Teleportation (prelude) Suppose Alice wishes to convey a qubit to Bob by sending just classical bits If Alice knows and , she can send approximations of them ―but this still requires infinitely many bits for perfect precision Moreover, if Alice does not know or , she can at best acquire one bit about them by a measurement
Teleportation scenario In teleportation, Alice and Bob also start with a Bell state (1/2)(00+11) 0+1 and Alice can send two classical bits to Bob Note that the initial state of the three qubit system is: (1/2)(0+1)(00+11) = (1/2)(000+ 011+ 100+ 111)
Initial state: (0+1)(00+11)(omitting the 1/2 factor) = 000+ 011+ 100+ 111 = ½(00+ 11)(0 +1) + ½(01+ 10)(1 +0) + ½(00−11)(0 −1) + ½(01−10)(1 −0) How teleportation works Protocol: Alice measures her two qubits in the Bell basis and sends the result to Bob (who then “corrects” his state)
½00(0 +1) + ½01(1 +0) +½10(0 − 1) + ½11(1 −0) (00, 0 +1) with prob. ¼ (01, 1 +0) with prob. ¼ (10, 0 −1) with prob. ¼ (11, 1 −0) with prob. ¼ H What Alice does specifically Alice applies to her two qubits, yielding: Then Alice sends her two classical bits to Bob, who then adjusts his qubit to be 0 +1 whatever case occurs
00, 0 +1 01, X(1 +0)=0 +1 10, Z(0 −1) =0 +1 11, ZX(1 −0)= 0 +1 Bob’s adjustment procedure Bob receives two classical bits a, b from Alice, and: if b=1he applies X to qubit if a=1he applies Z to qubit yielding: Note that Bob acquires the correct state in each case
Summary of teleportation 0+1 H Alice a 00+11 b Bob 0+1 X Z Quantum circuit exercise: try to work through the details of the analysis of this teleportation protocol
a a a 0 ψ ψ ψ 0 Classical information can be copied a a a 0 What about quantum information? ?
Candidate: works fine for ψ =0and ψ =1 ... but it fails for ψ =(1/2)(0+1) ... ... where it yields output (1/2)(00+11) instead of ψψ =(1/4)(00+01+10+11)
ψ ψ U 0 ψ 0 g No-cloning theorem Theorem: there is no valid quantum operation that maps an arbitrary state ψ to ψψ Proof: Let ψ and ψ′ be two input states, yielding outputs ψψg and ψ′ψ′g′ respectively Since U preserves inner products: ψψ′ = ψψ′ψψ′gg′ so ψψ′(1− ψψ′gg′) = 0 so ψψ′ =0or1
