1/22 Quantum Algorithms Deutsch-Josza Algorithm: quantum computers are more powerful than classical...

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Quantum Algorithms• Deutsch-Josza Algorithm: quantum computers are more powerful than classical ones.f :b∈ 0,1( )→ b∈ 0,1( ) four possible functions

b f(b)

balanced functions

1 2 4 4 4 3 4 4 4 unbalanced functions1 2 4 4 4 3 4 4 4

• Is f(b) balanced or unbalanced?

classically: need to evaluate f(b) twice; quantum algorithm: only once.

•Inverse Database search Example: telephone book with N entries

AARDVAARK, A…5-2346AARONSON, SJ…3-2349….JAMES, DFV…7-5436JOHANSSON, Scarlett…(you’ll be lucky…)KWIAT, PG…3-9116….ZEILINGER, A…5-1201ZYXT, X…4-8719

who has the phone number 7-5436 ?

classically: ~ N/2 queriesquantum algorithm: ~

Grover’s Algorithm

H2L 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

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 = Me mod n

encrypted message, E

• calculate: Ed 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

• Example: n = 77; c = 8;

0 10 20 30 40a

fn,c(x)

• The function fn,c(x) = cx mod n , is

periodic, (period r). • Either or is a factor of n.

GCD cr / 2 −1,n( )

GCD cr / 2 +1,n( )

• Chose a number, c, which is coprime with n i.e. GCD(c,n) =1

GCD 85 +1,77( )=11

GCD 85 −1,77( ) =7

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).

Large number of evaluations are replaced

by one

Quantum Factoring

then you can do this : ˆ U f x 0x∑ ⎛

⎝ ⎜

⎠ ⎟= x fN ,C x( )

• Classical factoring: evaluate fn,c(x) for

a large number of values of x until you can find r.

if you can do this : ˆ U f x 0 = x fN ,C x( )

• Quantum parallelism:

i.e. the state of multiple qubits corresponding to x; e.g.if x=29, ⏐x⟩=⏐11101⟩

φ x( ) =1

2L / 2fN ,C y( ) exp i

⎝ ⎜

⎠ ⎟

2 L -1

Quantum Factoring (cont.)• Quantum Fourier Transform to argument register:

UQFT ⊗ I( ) x fN ,C x( )x∑ = x φ x( )

• If 2L/r=M, number of periods in the argument register:

φ x( ) = M ϕ s if x = sM s = 0,1,2....r −1( )

0 otherwise

⎧ ⎨ ⎩

0 T 2T

Periodic Function

... 0 1/T 2/T 3/T...

Fourier Transform

Quantum Factoring (cont., again)

x φ x( )x∑ = s2L /r ϕ s

r-1∑

• Thus the state after the QFT is:

• Discard the function register: the argument register is in a mixed state:

ρ final = s2L r s2L rs =0

r-1∑

• 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.

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).

modular exponentiation Fourier trans.initiation

Circuit Diagram

argument

function

x 0x=0

2 L-1∑

x fN ,C x( )x=0

2 L-1∑€

C2L−1

sN /r ϕ ss =0

r-1∑

readout

RSA cryptosystem:

•polynomial work to encrypt/decrypt

•exponential work to break = factoring

•BUT quantum factoring is only polynomial work

27997833911221327870829467638722601621070446786955428537560009929326128400107609345671052955360856061822351910951365788637105954482006576775098580557613579098734950144178863178946295187237869221823983

RSA 200:3532461934402770121272604978198464368671197400197625023649303468776121253679423200058547956528088349

7925869954478333033347085841480059687737975857364219960734330341455767872818152135381409304740185467

Vulnerability of RSA to Quantum Computers?

RSA200

0 200 400 600 800 10001

10001 10

101 10

111 10

121 10

131 10

141 10

151 10

161 10

171 10

181 10

191 10

201 10

211 10

221 10

231 10

Classical ~ exp{AL}

# of instructions

# of bits, L, factored

~ 1020

instructions: 16 months (2003-05)

Shors Algorithm~ L3

~ 1012 operations:Hours ?

2. Simplifications for a Few QubitsN=15, C=4

x fN,C(x)

i.e. period 2

N=15, C=2

x fN,C(x)

i.e. period 4

Simplifications for a Few QubitsN=15, C=4

x fN,C(x)

00 001

01 100

10 001

11 100

i.e. period 2

N=15, C=2

x fN,C(x)

000 0001

001 0010

010 0100

011 1000

100 0001

i.e. period 4

this is too profligate with qubits....

Simplifications for a Few QubitsN=15, C=4

xLogC[fN,C(x)]

11 01i.e. period 2

N=15, C=2

xLogC[fN,C(x)]

000 00

001 01

010 10

011 11

100 00i.e. period 4

cancelMinimalist Period 2 Circuit

• Top rail cancellation occurs for all r =2n.• Two qubits, one quantum gate

xLogC[fN,C(x)]

Slightly Less Minimalist Period 2 Circuitwithout Logarithm of fN,C(x)

• Three qubits, two gates.

What about Period 4?

T =exp iπ /8[ ] 0

0 exp −iπ /8[ ]

⎝ ⎜

⎠ ⎟

xLogC[fN,C(x)]

000 00

001 01

010 10

011 11

100 00

What about Period 4?

• 4 qubits, 2 quantum gates

xLogC[fN,C(x)]

000 00

001 01

010 10

011 11

100 00

*Lanyon, Weinhold, Langford, Barbieri, James, Gilchrist, and White, PRL 99, 250505 (2007)

4-photonsource

3. Factoring Experiment*

GCD of Cr/2±1 and NGCD of 41±1 and 15 = 3,5

P110 = 27 ± 2%

order-2 order-4

F = 99.9 ± 0.3%SL= 99.9 ± 0.6% F = 98.5 ± 0.6%

SL= 98.1 ± 0.8%

P000 = 27 ± 2%P100 = 24 ± 2%

GCD of Cr/2±1 and NGCD of 41±1 and 15 = 3,5

add redundantbit then reverseargument bits

P10 = 48 ± 3%

P00 = 52 ± 3% failure

GCD of 43±1 and 15 = 3,5

P010 = 23 ± 2%

algorithm works near perfectly …?

Order-finding algortihm uses mixed output state: non-deterministic

Measuring the output

order-2

order-4

FGHZ = 59 ± 4%WGHZ = 9 ± 4%

SL= 62 ± 4%

F2Bell = 98.5 ± 0.6%=

Tbd = 41 ± 5%Tce = 33 ± 5%

SL= 98.1 ± 0.8%

joint state of argumentand function registersis entangled & mixed

joint state of argumentand function registersis highly entangled

independent photons

dependent photons

Measuring the output

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.

xLogC[fN,C(x)]

5 2i.e. period 3

4. Where next: Period 3?N=21, C=4

xLogC[fN,C(x)]

000 00

001 01

010 10

011 00

100 01

101 10i.e. period 3

After modular exponentiation, a three qubit argument register plus two qubit function register will be in the state:⏐⟩= ⏐⟩⏐⟩⏐⟩⊗⏐⟩⏐⏐⟩⏐⟩⏐⟩⊗⏐⟩⏐⏐⟩⏐⟩⊗⏐⟩

A bit less scary...

• Can we make this state using established techniques for making W states and GHZ states?

where: ⏐⟩⏐⟩⏐⟩√

⏐⟩⏐⟩⏐⟩⏐⟩√

flip qubit #2:⏐⟩= ( ⏐⟩⊗⏐⟩⏐⟩⊗⏐⟩⏐⟩⊗⏐⟩

3 00 + 3 01 + 2 10 GHZ W W

Period 3 Circuit

• 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.

1/22 Quantum Algorithms Deutsch-Josza Algorithm: quantum computers are more powerful than classical...

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