1. 1/22 Quantum Algorithms Deutsch-Josza Algorithm: quantum computers are more powerful than classical ones. b f(b) 0 1 0 b f(b) 0 1 1 b f(b) 0 1 b f(b) 0 1 1 0 Is f(b) balanced or unbalanced? classically: need to evaluate f(b) twice; quantum algorithm: only once.

2. 2/22 Inverse Database search Example: telephone book with N entries AARDVAARK, A…5-2346 AARONSON, SJ…3-2349 …. JAMES, DFV…7-5436 JOHANSSON, Scarlett…(you’ll be lucky…) KWIAT, PG…3-9116 …. ZEILINGER, A…5-1201 ZYXT, X…4-8719 who has the phone number 7-5436 ? classically: ~ N/2 queries quantum algorithm: ~ Grover’s Algorithm

3. 3/22 H 2 L Algorithm* *Harrow, Hassidim and Lloyd, 2009 Given a matrix A and a vector b, find a vector x such that A x =b A is sparse (lots of zeros on off diagonal) not interested in x, rather in x †.M.x for some M. exponential speed-up demonstrated over classical algorithms

4. 4/22 Alice Bob RSA* Encryption and Factoring *Rivest, Shamir & Adelman, 1978; (also Clifford Cocks, 1973). select two prime numbers: p,q calculate: n = p.q;  = (p-1).(q-1) select e, with GCD(e,  ) = 1 calculate d, with e.d = 1 mod  public key: n,e Message: M calculate: E = M e mod n encrypted message, E calculate: E d mod n = M mod n Easy to find the message if you know p and q Security relies on difficulty of factoring n Message M

5. 5/22 Example: n = 77; c = 8; f n,c (x) x The function f n,c (x) = c x mod n, is periodic, (period r). Either or is a factor of n. Chose a number, c, which is coprime with n i.e. GCD(c,n) =1 From data, r = 10; Period Finding  Factoring Factoring Numbers* *P. Shor, Proc. 35th Ann. Symp. Found. Comp. Sci. 124-134 (1994); also: Preskill et al., Phys Rev A 54, 1034 (1996).

6. 6/22 Large number of evaluations are replaced by one Quantum Factoring Classical factoring: evaluate f n,c (x) for a large number of values of x until you can find r. Quantum parallelism: i.e. the state of multiple qubits corresponding to x ; e.g.if x =29,  x 〉 =  11101 〉

7. 7/22 Quantum Factoring (cont.) Quantum Fourier Transform to argument register: If 2 L /r=M, number of periods in the argument register: 0 T2T Periodic Function... 01/T 2/T 3/T... Fourier Transform 

8. 8/22 Quantum Factoring (cont., again) Thus the state after the QFT is: Discard the function register: the argument register is in a mixed state: Measurement of the function register yields, with high probability a number which is a multiple of N/r ; extracting r, you can find the factors.

9. 9/22 BUT.... Unless you get real lucky, N is not a multiple of r How actually do you implement the unitary operations for modular exponentiation and quantum Fourier transform? - you can fix this using bigger registers, so the ‘periodic’ signal swamps the rest. -both can be done efficiently (i.e. in a polynomial number of operations) -break down complicated operations into simpler operations (e.g. multiplexed adders and repeated squaring), which can be performed by CNOTs and related multi-qubit quantum gates. -QFT can be simplified by dropping some operations, and by doing it ‘semi-classically’ by measurement and feed-forward* *R. B.Griffiths and C.-S. Niu Phys. Rev. Lett. 76 3228 (1996).

10. 10/22 modular exponentiation Fourier trans. initiation Circuit Diagram argument function .... QFT readout

11. 11/22 RSA cryptosystem: polynomial work to encrypt/decrypt exponential work to break = factoring BUT quantum factoring is only polynomial work 27997833911221327870829467 63872260162107044678695542 85375600099293261284001076 09345671052955360856061822 35191095136578863710595448 20065767750985805576135790 98734950144178863178946295 187237869221823983 RSA 200: 35324619344027 70121272604978 19846436867119 74001976250236 49303468776121 25367942320005 85479565280883 49 79258699544783 33033347085841 48005968773797 58573642199607 34330341455767 87281815213538 14093047401854 67 x = Vulnerability of RSA to Quantum Computers? RSA200 Classical ~ exp{AL} # of instructions # of bits, L, factored ~ 10 20 instructions: 16 months (2003-05) Shors Algorithm~ L 3 ~ 10 12 operations: Hours ?

12. 12/22 2. Simplifications for a Few Qubits N=15, C=4 xf N,C (x) 01 14 21 34 i.e. period 2 N=15, C=2 xf N,C (x) 01 12 24 38 41 i.e. period 4

13. 13/22 Simplifications for a Few Qubits N=15, C=4 xf N,C (x) 00001 01100 10001 11100 i.e. period 2 N=15, C=2 xf N,C (x) 0000001 0010010 0100100 0111000 1000001 i.e. period 4 this is too profligate with qubits....

14. 14/22 Simplifications for a Few Qubits N=15, C=4 xLog C [f N,C (x)]0001 1000 1101 i.e. period 2 N=15, C=2 xLog C [f N,C (x)] 00000 00101 01010 01111 10000 i.e. period 4

15. 15/22 cancel Minimalist Period 2 Circuit   X QFT  Z Top rail cancellation occurs for all r =2 n. Two qubits, one quantum gate xLog C [f N,C (x)]0001 1000 1101

16. 16/22  X X X Slightly Less Minimalist Period 2 Circuit without Logarithm of f N,C (x) Three qubits, two gates.

17. 17/22   X X What about Period 4?  QFT   ZZ T xLog C [f N,C (x)] 00000 00101 01010 01111 10000

18. 18/22   XX What about Period 4?  Z 4 qubits, 2 quantum gates xLog C [f N,C (x)] 00000 00101 01010 01111 10000  Z

19. 19/22 *Lanyon, Weinhold, Langford, Barbieri, James, Gilchrist, and White, PRL 99, 250505 (2007) 4-photon source 3. Factoring Experiment*

20. 20/22 GCD of C r/2 ±1 and N GCD of 4 1 ±1 and 15 = 3,5 P 110 = 27 ± 2% order-2order-4 F = 99.9 ± 0.3% S L = 99.9 ± 0.6% F = 98.5 ± 0.6% S L = 98.1 ± 0.8% P 000 = 27 ± 2% P 100 = 24 ± 2% GCD of C r/2 ±1 and N GCD of 4 1 ±1 and 15 = 3,5 r=2 add redundant bit then reverse argument bits P 10 = 48 ± 3% P 00 = 52 ± 3% failure r=6 GCD of 4 3 ±1 and 15 = 3,5 r=2 P 010 = 23 ± 2% algorithm works near perfectly …? Order-finding algortihm uses mixed output state: non-deterministic Measuring the output

21. 21/22 order-2 order-4 F GHZ = 59 ± 4% W GHZ = 9 ± 4% S L = 62 ± 4% F 2Bell = 98.5 ± 0.6%= T bd = 41 ± 5% T ce = 33 ± 5% S L = 98.1 ± 0.8% joint state of argument and function registers is entangled & mixed joint state of argument and function registers is highly entangled independent photons dependent photons Measuring the output

22. 22/22 Conclusions Simplified versions of Shor’s Algorithm are accessible with today’s quantum technology technology. Improving these results, step-by-step, is as good a route to practical quantum computers. Complexity of quantum circuit depends on period r, rather than size of number.

23. 23/22 xLog C [f N,C (x)]012 30 41 52 i.e. period 3 4. Where next: Period 3? N=21, C=4 xLog C [f N,C (x)] 00000 00101 01010 01100 10001 10110 i.e. period 3 After modular exponentiation, a three qubit argument register plus two qubit function register will be in the state:  〉 =  ⊗   ⊗   ⊗ 

24. 24/22 A bit less scary... Can we make this state using established techniques for making W states and GHZ states? where:     √√     √√     √√ flip qubit #2:  〉 = (    ⊗    ⊗    ⊗  

25. 25/22 GHZWW Period 3 Circuit H X X X Ry()Ry() 5 qubits, 8 gates + QFT (still pretty scary) Period is not a power of 2; full QFT needed. Size of the argument register will not be a factor of the period. QFT X X X X X H R( 