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Quantum Computing MAS 725 Hartmut Klauck NTU 5.3.2012.

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Presentation on theme: "Quantum Computing MAS 725 Hartmut Klauck NTU 5.3.2012."— Presentation transcript:

1 Quantum Computing MAS 725 Hartmut Klauck NTU

2 More about measurements(I) Some linear algebra: Vector space V (dimension d) Subspaces: U µ V and U is also a vector space (dim e < d) There is an orthonormal basis: v 1,...,v e,...,v d, span(v 1,...,v e )=U Projection onto U: P U =  i=1...e | v i ih v i | Example: Projector P=|v 1 ih v 1 | P |v 1 i = |v 1 ih v 1 | v 1 i = |v 1 i P (  |v 1 i +  |v 2 i )=  |v 1 i +  |v 1 ih v 1 | v 2 i =  |v 1 i

3 Measurements (II) A ? B iff v ? w for all v 2 A, w 2 B A © B=V iff A ? B and for all v 2 V: v =  u +  w, u 2 A, w 2 B, k u k, k v k = 1, |  | 2 + |  | 2 = k v k 2

4 Measurements (III) Hilbert spaces V, dim k Observable: System l · k of subspaces S 0,...,S l-1, pairwise orthogonal S 0 ©  © S l-1 = V Probability of measuring i is k Proj(S i ) |  i k 2 The state |  i collapses to Proj(S i ) |  i / k Proj(S i ) |  ik (renormalized) Observables correspond to a measurement device We only consider projections measurements

5 Measurements (IV) A matrix A is Hermitian, if A=A y Hermitian matrices have only real eigenvalues and are diagonalizable: U y A U is diagonal for some unitary U The eigenspaces of A decompose C n C n = © V i Let ¸ (i) denote an eigenvalue of A, and P i the projection onto its eigenspace Then A=  i ¸ (i) P i

6 Measurements (V) A Hermitian matrix is a concise representation of an observable Eigenvalues correspond to measurement outcomes Eigenspaces decompose C n

7 Measurements: example Two qubits, living in C 4 Observable: S 0 =span(|00 i,|01 i ) S 1 =span(|10 i,|11 i ) S 0 ? S 1 This observable corresponds to measuring the first qubit: S 0 indicates that it is 0, S 1 indicates 1

8 Measuring an EPR-Pair The state is 1/2 1/2 ¢ (|00 i +|11 i ) We measure the first qubit Result: If we measure 0, then the state collapses to |00 i If we measure 1 we get |11 i Each happens with probability ½ Qubit 2 collapses right after measuring qubit 1 The qubits act like a shared public coin toss. This is even true if the qubits are spatially separated

9 Summary Hilbert space: register holding a quantum state Vectors: states Unitary transformation: evolution of states (computation) Observables: for measuring the computation‘s output The probability distribution on results: output of the computation Projected and normalized vector: the remaining quantum state Measurement is the only way to extract information from a quantum state

10 More about qubits No-Cloning Theorem Bell states Quantum Teleportation

11 No Cloning Suppose we are given a quantum state |  i Can we make a copy? Copying classical information is trivial (ask the music industry about that…) I.e., there is a unitary transformation U n : U n |x i |0 i = |x i |x i for all x 2 {0,1} n But then by linearity U 1 1/2.5 (|0 i +|1 i ) |0 i = 1/2 1/2 (|00 i +|11 i )  1/2 1/2 (|0 i +|1 i ) times 1/2.5 (|0 i +|1 i )

12 No Cloning Theorem Theorem: There is no unitary U, such that for all quantum states |  i on n qubits: U |  i ­ |0 m i = |  i­ |  i­ |  (  ) i for some m und |  (  ) i (which is garbage) I.e. there is no universal way to copy unknown quantum states without error! [Dieks, Wootters/Zurek 82]

13 Proof: No Cloning Let a linear U be given (this also fixes m). Then U |0 n i |0 m i =|0 n i ­ |0 n i ­ |  0 i U |1 n i |0 m i =|1 n i ­ |1 n i ­ |  1 i Then also (due to linearity) U 1/2 1/2 ( |0 n i +|1 n i ) |0 m i = 1/2 1/2 ( U |0 n i |0 m i + U |1 n i |0 m i ) = 1/2 1/2 ( |0 2n i ­ |  0 i + |1 2n i ­ |  1 i ) But we wanted (for some |  2 i ) 1/2 (|0 n i +|1 n i ) ­ (|0 n i +|1 n i ) ­ |  2 i

14 Proof: No Cloning (II) We have 1/2 1/2 ( |0 2n i ­ |  0 i + |1 2n i ­ |  1 i ) We want 1/2 (|0 n i +|1 n i ) ­ (|0 n i +|1 n i ) ­ |  2 i Claim: This is not the same! Proof of the claim: we compute the inner product. Same state ) inner product = 1 h  ­  |  ­  i = h  |  i ¢ h  |  i h  +  |  i = h  |  i + h  |  i

15 Proof No Cloning (III) We get 1/2 1/2 ( |0 2n i ­ |  0 i + |1 2n i ­ |  1 i ) We expect 1/2 (|0 n i +|1 n i ) ­ (|0 n i +|1 n i ) ­ |  2 i Inner product: 1/2 3/2 ¢ ( ( h 0 n |0 n i + h 0 n |1 n i ) ¢ ( h 0 n |0 n i + h 0 n |1 n i ) ¢ h  0 |  2 i ) +( h 1 n |0 n i + h 1 n |1 n i ) ¢ ( h 1 n |0 n i + h 1 n |1 n i ) ¢ h  1 |  2 i ) ) = 1/2 3/2 ¢ (1 ¢ 1 ¢ a 0,2 + 1 ¢ 1 ¢ a 1,2 ). Then |1/2 3/2 ¢ ( a 0,2 +a 1,2 )| · 2/2 3/2 < 1

16 No Cloning No unitary can map both a state and another state that are not orthogonal in a way that wakes a copy It is possible to clone quantum states with small success probability

17 Bell States Consider the following basis (John Bell) This is orthonormal in C 4

18 Generating Bell states H x y |  x,y i CNOT Gate: CNOT |0, y i  =|0, y i ; CNOT |1,y i =|1, 1-y i |  0,0 i = 1/2 1/2 (|00 i  +|11 i )=|  + i

19 Quantum Teleportation We are given a quantum state |  i Can we reproduce that state in another location? We cannot just send them over…. And we cannot copy them Classical communication is possible This problem seems hard since a single qubit   |0 i +  1  |1 i might contain a lot of information due to possibly irrational amplitudes

20 Quantum Teleportation |i|i Classical Communication 1/2 1/2 (|00 i +|11 i ) Local operations

21 Quantum Teleportation [Bennett et al. 93] Alice and Bob share an EPR pair (Alice has the first qubit, Bob the other) Alice has |  i (1 qubit quantum state) Alice applies CNOT to her qubit q 0 with |  i (control) and her EPR qubit q 1 Alice applies H to q 0 and measures q 0 and q 1 Alice sends the result of the measurement (2 bits) Bob applies a unitary transformation depending on the message to his EPR qubit q 2

22 Quantum Teleportation q 1, q 2 : EPR Pair; q 0 : |  i ; q 0,q 1 with Alice, q 2 with Bob Measurement

23 Pauli Transformations X: (NOT, Bit Flip) Z: (Phase Flip) Y:

24 Quantum Teleportation Create EPR Pair

25 Quantum Teleportation Then: CNOT on q 0 and q 1  x=0,1  x |x i­  y=0,1 1/2 1/2 |y,y i CNOT:  x,y  x /2 1/2 |x, x © y, y i Measure q 1 : Result is a Prob. for 0/1 is 0.5 each: |  0 | 2 /2+|  1 |  /2=1/2 Remaining state:  y  y © a |y © a, y i on q 0,q 2 Send a to Bob a=0 ) Bob does nothing a=1 ) Bob applies X-Gate (Bit Flip) Result in both cases:  y  y © a |y © a, y © a i =  y  y |y,y i

26 Quantum Teleportation State  y  y |y,y i Problem: Bob’s qubit q 2 is still entangled with q 0  y  y |y,y i

27 Quantum Teleportation Alice applies H to q 0 Alice measures q 0, assume the result is b, she sends b to Bob Applying H:  y  y |y,y i  1/2 1/2  z,y  y (-1) z ¢ y |z,y i Measuring q 0 (Result is b; 0/1 with Prob. 1/2):  y  y (-1) b ¢ y |y i on q 2 Bob corrects by applying a Z-Gate, if necessary

28 Quantum Teleportation  y  y |y i

29 Remarks The EPR-pair is consumed The original state |  i is destroyed [no cloning!] States with many qubits can be teleported one by one (we need to make sure entanglement between the teleported qubits is preserved)


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