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1 Introduction to Quantum Information Processing CS 467 / CS 667 Phys 667 / Phys 767 C&O 481 / C&O 681 Richard Cleve DC 2117 cleve@cs.uwaterloo.ca Lecture 1 (2008)
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2 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: 1975198019851990199520002005 10 4 10 5 10 6 10 7 10 8 10 9 number of transistors year
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3 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
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4 FINISH FINISH Quantum algorithms Classical deterministic: Classical probabilistic: STARTFINISH Quantum: POSITIVENEGATIVE α1α1α1α1 α3α3α3α3 α2α2α2α2 p1p1p1p1 p3p3p3p3 p2p2p2p2 START START
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5 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 classical information theory quantum information theory For example: a theory of quantum error-correcting codes
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6 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
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7 General course information Background: classical algorithms and complexity classical algorithms and complexity linear algebra linear algebra probability theory probability theory Evaluation: 5 assignments (12% each) 5 assignments (12% each) project presentation (40%) 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.
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8 Basic framework of quantum information
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9 qp Types of information is quantum information digital or analog? digital: 0 1 0 0 1 1 analog: 0 1 r [0,1] r Can explicitly extract r Issue of precision for setting & reading state Precision need not be perfect to be useful Probabilities p, q 0, 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) probabilistic
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10 Quantum (digital) information 0 1 0 0 1 1 Amplitudes , C, | | 2 + | | 2 = 1 Explicit state is Cannot explicitly extract and (only statistical inference) Issue of precision (imperfect ok)
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11 Dirac bra/ket notation Ket: ψ always denotes a column vector, e.g. Bracket: φ ψ denotes φ ψ , the inner product of φ and ψ Bra: ψ always denotes a row vector that is the conjugate transpose of ψ , e.g. [ * 1 * 2 * d ] Convention:
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12 Basic operations on qubits (I) Rotation: Hadamard:Phase flip: NOT (bit flip): (0) Initialize qubit to |0 or to |1 (1) Apply a unitary operation U ( U † U = I ) Examples:
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13 Basic operations on qubits (II) (3) Apply a “standard” measurement: 0 + 1 ( ) 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 ′ } ||2||2 ||2||2 00 11 ψψ ψ′ψ′ … and the quantum state collapses
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14 Distinguishing between two states Question 1: can we distinguish between the two cases? Let be in state or Distinguishing procedure: 1.apply H 2.measure This works because H + = 0 and H − = 1 Question 2: can we distinguish between 0 and + ? Since they’re not orthogonal, they cannot be perfectly distinguished …
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15 n -qubit systems Probabilistic states:Quantum states: Dirac notation: |000 , |001 , |010 , …, |111 are basis vectors, so
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16 Operations on n -qubit states Unitary operations: … and the quantum state collapses Measurements: ( U † U = I ) ?
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17 Entanglement Product state (tensor/Kronecker product): Example of an entangled state: … can exhibit interesting “nonlocal” correlations:
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18 Structure among subsystems V U W qubits: #2 #1 #4 #3 time unitary operationsmeasurements
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19 Quantum computations 00 11 11 00 11 00 1 0 1 0 1 1 Quantum circuits: “Feasible” if circuit-size scales polynomially
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20 Example of a one-qubit gate applied to a two-qubit system U (do nothing) The resulting 4x4 matrix is 00 0U001 0U110 1U011 1U100 0U001 0U110 1U011 1U1 Maps basis states as:
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21 Controlled- U gates U 00 0001 0110 1U011 1U100 0001 0110 1U011 1U1 Maps basis states as: Resulting 4x4 matrix is controlled- U =
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22 Controlled-NOT (CNOT) X aa bb abab aa ≡ Note: “control” qubit may change on some input states 0 + 10 + 1 0 − 10 − 10 − 10 − 1 0 − 10 − 1
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23 Introduction to Quantum Information Processing CS 467 / CS 667 Phys 667 / Phys 767 C&O 481 / C&O 681 Richard Cleve DC 2117 cleve@cs.uwaterloo.ca Lecture 2 (2008)
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24 Superdense coding
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25 How much classical information in n qubits? 2 n 1 complex numbers apparently needed to describe an arbitrary n -qubit pure quantum state: 000 000 + 001 001 + 010 010 + + 111 111 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
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26 Holevo’s Theorem U ψψ n qubits b1b1 b2b2 b3b3 bnbn Easy case: b 1 b 2... b n certainly cannot convey more than n bits! Hard case (the general case): ψψ n qubits b1b1 b2b2 b3b3 bnbn U 00 00 00 00 00 m qubits bn+1bn+1 bn+2bn+2 bn+3bn+3 bn+4bn+4 bn+mbn+m The difficult proof is beyond the scope of this course
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27 Superdense coding (prelude) By Holevo’s Theorem, this is impossible AliceBob ab Suppose that Alice wants to convey two classical bits to Bob sending just one qubit ab
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28 Superdense coding How can this help? AliceBob ab In superdense coding, Bob is allowed to send a qubit to Alice first ab
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29 How superdense coding works 1.Bob creates the state 00 + 11 and sends the first qubit to Alice 2.Alice: if a = 1 then apply X to qubit if b = 1 then apply Z to qubit send the qubit back to Bob ab state 00 00 + 11 01 00 − 11 10 01 + 10 11 01 − 10 3.Bob measures the two qubits in the Bell basis Bell basis
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30 Measurement in the Bell basis H Specifically, Bob applies to his two qubits... inputoutput 00 + 11 00 01 + 10 01 00 − 11 10 01 − 10 11 and then measures them, yielding ab This concludes superdense coding
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31 Introduction to Quantum Information Processing CS 467 / CS 667 Phys 667 / Phys 767 C&O 481 / C&O 681 Richard Cleve DC 2117 cleve@cs.uwaterloo.ca Lecture 3 (2008)
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32 Teleportation
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33 Recap n -qubit quantum state: 2 n -dimensional unit vector Unitary op: 2 n 2 n linear operation U such that U † U = I (where U † denotes the conjugate transpose of U ) U 0000 = the 1 st column of U U 0001 = the 2 nd column of U the columns of U : : : : : : are orthonormal U 1111 = the (2 n ) th column of U
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34 Incomplete measurements (I) Measurements up until now are with respect to orthogonal one-dimensional subspaces: 00 11 22 The orthogonal subspaces can have other dimensions: span of 0 and 1 22 (qutrit)
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35 Incomplete measurements (II) Such a measurement on 0 0 + 1 1 + 2 2 results in 0 0 + 1 1 with prob 0 2 + 1 2 2 with prob 2 2 (renormalized)
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36 Measuring the first qubit of a two-qubit system Result is the mixture 00 00 + 01 01 with prob 00 2 + 01 2 10 10 + 11 11 with prob 10 2 + 11 2 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) 00 00 + 01 01 + 10 10 + 11 11
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37 Easy exercise: show that measuring the first qubit and then measuring the second qubit gives the same result as measuring both qubits at once
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38 Teleportation (prelude) Suppose Alice wishes to convey a qubit to Bob by sending just classical bits 0 + 1 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
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39 Teleportation scenario 0 + 1 ( 1 / 2 )( 00 + 11 ) In teleportation, Alice and Bob also start with a Bell state 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 )
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40 How teleportation works ( 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 ) Initial state: Protocol: Alice measures her two qubits in the Bell basis and sends the result to Bob (who then “corrects” his state)
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41 What Alice does specifically Alice appliesto 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 ½ 00 ( 0 + 1 ) + ½ 01 ( 1 + 0 ) + ½ 10 ( 0 − 1 ) + ½ 11 ( 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
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42 Bob’s adjustment procedure if b = 1 he applies X to qubit if a = 1 he applies Z to qubit Bob receives two classical bits a, b from Alice, and: 00, 0 + 1 01, X ( 1 + 0 ) = 0 + 1 10, Z ( 0 − 1 ) = 0 + 1 11, ZX ( 1 − 0 ) = 0 + 1 yielding: Note that Bob acquires the correct state in each case
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43 Summary of teleportation H XZ 0 + 1 00 + 11 0 + 1 b a Alice Bob Quantum circuit exercise: try to work through the details of the analysis of this teleportation protocol
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44 No-cloning theorem
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45 Classical information can be copied 0 aa a 0 aa a What about quantum information? ψψ 00 ψψ ψψ ?
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46 works fine for ψ = 0 and ψ = 1 ... but it fails for ψ = ( 1 / 2 )( 0 + 1 )...... where it yields output ( 1 / 2 )( 00 + 11 ) instead of ψ ψ = ( 1 / 4 )( 00 + 01 + 10 + 11 ) Candidate:
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47 No-cloning theorem Theorem: there is no valid quantum operation that maps an arbitrary state ψ to ψ ψ Proof: Let ψ and ψ′ be two input states, yielding outputs ψ ψ g and ψ ′ ψ ′ g ′ respectively Since U preserves inner products: ψ ψ ′ = ψ ψ ′ ψ ψ ′ g g ′ so ψ ψ ′ ( 1 − ψ ψ ′ g g ′ ) = 0 so ψ ψ ′ = 0 or 1 ψψ 00 00 ψψ ψψ gg U
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