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Vorlesung Quantum Computing SS 08 1 quantum parallelism a 1 F |00> + a 2 F |01> + a 3 F |10> + a 4 F |11> a 1 |00> + a 2 |01> + a 3 |10> + a 4 |11> input.

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Presentation on theme: "Vorlesung Quantum Computing SS 08 1 quantum parallelism a 1 F |00> + a 2 F |01> + a 3 F |10> + a 4 F |11> a 1 |00> + a 2 |01> + a 3 |10> + a 4 |11> input."— Presentation transcript:

1 Vorlesung Quantum Computing SS 08 1 quantum parallelism a 1 F |00> + a 2 F |01> + a 3 F |10> + a 4 F |11> a 1 |00> + a 2 |01> + a 3 |10> + a 4 |11> input b 1 |00> + b 2 |01> + b 3 |10> + b 4 |11> = output F Superposition for input created by Hadamard gates Functions are represented by unitary operators Quantum state tomography – but how to get a real result?

2 Vorlesung Quantum Computing SS 08 2 Deutsch algorithm: table of truth D. Deutsch, Proceedings of the Royal Society of London A 400 (1985), 97 simplest example: 1-bit-to-1-bit function f : x {0,1} case case case case constantbalanced f (1) f (0) real or false? falsetrue f(0) f(1)

3 Vorlesung Quantum Computing SS 08 3 strength of quantum parallelism calculate a function f(x) with x {0,…,K} for all x at once classicalquantum function balanced(f:function) a=f(0) b=f(1) return a b function balanced(U f :function) 1 = H 0 = H | 2 = U f 1 3 = H x 2 = f( f( |R return f( f( use quantum computing only if f(x) is expensive to calculate (time, memory), e. g., spin dynamics for huge structures problem: superposition cannot be read-out: it will always collapse to an eigenstate

4 Vorlesung Quantum Computing SS 08 4 Deutsch algorithm: principle input eigen- state read- out eigen- state super- position transfer to eigen- state or test- function ? ? one run only! 0123

5 Vorlesung Quantum Computing SS 08 5 quantum circuit UfUf x x y y f(x) data register target register addition modulo 2 = f(x) for y=0 0 2 f(0) f(1) 2 H 2 = 2 = n 1 x f (x) x evaluation of function f (x) with n data qubits x and 1 target qubit BUT: How to extract information? 2

6 Vorlesung Quantum Computing SS 08 6 Deutsch algorithm: quantum circuit 1 0 UfUf x x y y f(x) H H H H 2 = 2 = U f 1

7 Vorlesung Quantum Computing SS 08 7 true 2 U f 1 (|0 |1 ) (| f (0) - |1 f (0) ) 1 2 Deutsch algorithm: find U f case case case case f (1) f (0) UfUf U f x, y x, y f(x) U f : |0,0 |0,0, |0,1 |0,1, |1,0 |1,1, |1,1 |1,0 U f = CNOT U f : |0,0 |0,1, |0,1 |0,0, |1,0 |1,0, |1,1 |1,1 U f = Z-CNOT U f : |0,0 |0,1, |0,1 |0,0, |1,0 |1,1, |1,1 |1,0 U f = NOT U f : |0,0 |0,0, |0,1 |0,1, |1,0 |1,0, |1,1 |1,1 U f = ID 2 U f 1 (|0, f (0) - |0,1 f (0) + |1, f (1) - |1,1 f (1) ) U f 1 (|0 + |1 ) (| f (0) - |1 f (0) ) 1 2 false information encoded in phase of x-qubit

8 Vorlesung Quantum Computing SS 08 8 Deutsch algorithm: get the answer 3 H x false : H 2 x =H = = f (0) f (1) true : H 2 x = H = = f (0) f (1) | f (0) - |1 f (0) 2 read-out

9 Vorlesung Quantum Computing SS 08 9 Deutsch algorithm: summary evaluates a global property of a function f(x) with a single run Example: f(x) f(0) = 1, f(1) = UfUf x x y y f(x) H H H H 2 = 2 = H 2 x =H = = U f

10 Vorlesung Quantum Computing SS Deutsch algorithm: implementation 40 Ca S 1/2 4 2 P 1/2 4 2 P 3/2 3 2 D 3/2 3 2 D 5/2 397 nm 729 nm 854 nm 866 nm |0 |1 quadrupole transition used for Laser cooling detection D 5/2 occupation P D red sideband blue sideband Roos et al: Phys. Rev. Lett. 83, 4713 (1999)

11 Vorlesung Quantum Computing SS Deutsch algorithm: implementation Gulde et al: Nature 421, 48 (2003) case 1 : U f = ID case 2 : U f = NOT case 3 : U f = CNOT case 4 : U f = Z-CNOT

12 Vorlesung Quantum Computing SS composite pulses example: imprecise B 1 -field rotate from z to -z: 1) – pulse at y (180 y ) 2) three pulse sequence: 90 y 180 x 90 y M. H. Levitt: Prog. Nucl. Magn. Reson. Spectrosc. 18 (1986) for small deviations (180-2 ) y for small deviations (90- ) y (180-2 ) x (90- ) y

13 Vorlesung Quantum Computing SS Deutsch algorithm: Fidelity Gulde et al: Nature 421, 48 (2003)

14 Vorlesung Quantum Computing SS more than 1 ion Nägerl et al: Phys. Rev. A 60, 145(1999) switching between ions within 14 s

15 Vorlesung Quantum Computing SS CNOT with 2 ions Schmidt-Kaler et al: Nature 422, 408(2003) 3 m n=0n=1 n=0n=1 ControlTarget 0 0 R + (,0) 11 CNOT: spin state changed for n=0 R + (, ) e e

16 Vorlesung Quantum Computing SS CNOT with 2 ions Schmidt-Kaler et al: Nature 422, 408(2003) 3 m n=0n=1 n=0n=1 ControlTarget 0 0 R + (,0) 2 1 CNOT: spin state changed for n=0 R + (, ) e e R(,0) 1

17 Vorlesung Quantum Computing SS error sources Schmitd-Kaler et al: Nature 422, 408 (2003) CNOT with 2 40 Ca + ions as qubits and vibration mode as coupling source contribution laser frequency noise ~100 Hz (phase coherence) ~10% laser intensity fluctuations ~1% laser detuning error ~2% residual thermal excitation ~4% addressing error ~3% off-resonant excitations (for t gate = 600 s ) ~4%

18 Vorlesung Quantum Computing SS Needle in a haystack Grover algorithm: search in an unsorted database Quantum mechanics helps in searching for a needle in a haystack Phys. Rev. Lett. 79, 325 (1997) Lov Grover, Bell labs

19 Vorlesung Quantum Computing SS Search in a database Example: search for a specific number in a phonebook N = 2 n entries with index x = 0…N-1. A detector function f(x): f(x) = 0 - Entry x is no solution: f(x) = 1 - Entry x is a solution: We need:

20 Vorlesung Quantum Computing SS The oracle Grovers algorithm minimizes calls to oracle Classical: on average N/2 calls to oracle. Quantum: number of calls N |x : adresses of data register f(x)=1 if entrance is solution

21 Vorlesung Quantum Computing SS H n H H n 01 1 Quantum circuit Target register: oracle qubit |q prepared in 1 = |1 Data register: superposition of n=2 qubits |x 1 = |0 123 H 2 = 2 = f(x) = 1 R iterations

22 Vorlesung Quantum Computing SS Oracle operator U O x q |x |q f (x) |q is flipped, if |x points to register with solution |q 0 2 H = 2 U O |x |q 0 = (-1) f(x) |x |q 0 2 |x 2 |x 2 |x 2 = |x UOUO f(x) = 0: f(x) = 1:

23 Vorlesung Quantum Computing SS Grovers algorithm H 2 = 2 = Oracle qubit does not change: Look at data register only. 2 = 3.1 = U O 2 = (|00 - |01 + |10 + |11 ) = H = = (|00 + |01 - |10 + |11 ) 3.3 = C 3.2 = (-1) x-0) =(|00 - |01 + |10 - |11 ) = H =|01

24 Vorlesung Quantum Computing SS Grovers algorithm H H H H U O |x |q 0 = (-1) f(x) |x |q H = C C

25 Vorlesung Quantum Computing SS Geometrical analysis C = |0 0| with 2 = H n and 2 = H n G= H n C H n U O = H n (2|0 0|-1)H n U O = (2| 2 2 |-1) U O | = N-M 1 (1-f(x)) x x M 1 f(x) x x | = Superpostion of solutions Superpostion of no-solutions | 2 = | + | N N-M N M = cos | + sin | 2 2 | | | 2 2

26 Vorlesung Quantum Computing SS Geometrical analysis 2 | | | 2 U O 2 = cos | - sin | 2 2 U O | 2 | 2 U O | 2 (2| 2 2 |-1)=| 2 2 | - (1- | 2 2 |) = P 2 – P 2 :relexion at | 2 G| 2 G 2 = cos | + sin | G k 2 = cos | + sin | k G = cos - sin sin

27 Vorlesung Quantum Computing SS Iterations Necessary number R of iterations is CI to 2 2 = = 2 R CI = 2 2 arcsin N M 4 M N N M << 1 with error probability p sin 2 = 2 N M For more iterations than R, error increases => One needs to know number M of solutions.

28 Vorlesung Quantum Computing SS ion trap quantum computing: a summary qubit representation –hyperfine states ( 9 Be +, 43 Ca + ) –electronic states ( 40 Ca + ) –vibrational modes qubit manipulation: laser irradiation initial state preparation: –Doppler and sideband cooling read-out: fluoresence

29 Vorlesung Quantum Computing SS relaxation and operation electronic states: –energy relaxation time T 1 ~ 1s –phase relaxation time T 2 ~ 10 ms –gate operation time T gate ~ 200 s –T 2 /T gate ~ 50 hyperfine states: –phase relaxation time T 2 ~ 10s –gate operation time T gate ~ 10 s –T 2 /T gate ~ 10 6 source: Homepage group A. Steane

30 Vorlesung Quantum Computing SS scaling to more than 10 ions Kielpinski, Monroe, Wineland: Nature 417, 709 (2002) C. Monroe group, Michigan 06


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