1/22 Quantum Algorithms • Deutsch-Josza Algorithm: quantum computers are more powerful than classical ones. f : b ∈ 0,1 ( )→ b ∈ 0,1 ( ) four possible fun b f(b) 0 0 1 0 b f(b) 0 1 1 1 b f(b) 0 0 1 1 b f(b) 0 1 1 0 balanced functions 1 2 4 4 4 3 4 4 4 unbalanced functions 1 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.
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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.
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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
periodic, (period r). • Either or is a factor of n.
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GCD cr / 2 −1,n( )
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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).
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Large number of evaluations are replaced
by one
Quantum Factoring
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then you can do this : ˆ U f x 0x∑ ⎛
⎝ ⎜
⎞
⎠ ⎟= x fN ,C x( )
x∑
• Classical factoring: evaluate fn,c(x) for
a large number of values of x until you can find r.
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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⟩
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φ x( ) =1
2L / 2fN ,C y( ) exp i
2π
2Lx.y
⎛
⎝ ⎜
⎞
⎠ ⎟
y =0
2 L -1
∑
Quantum Factoring (cont.)• Quantum Fourier Transform to argument register:
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UQFT ⊗ I( ) x fN ,C x( )x∑ = x φ x( )
x∑
• If 2L/r=M, number of periods in the argument register:
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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
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Quantum Factoring (cont., again)
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x φ x( )x∑ = s2L /r ϕ s
s =0
r-1∑
• Thus the state after the QFT is:
• Discard the function register: the argument register is in a mixed state:
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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.
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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).
joint state of argumentand function registersis entangled & mixed
joint state of argumentand function registersis highly entangled
independent photons
dependent photons
Measuring the output
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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.
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xLogC[fN,C(x)]
0 0
1 1
2 2
3 0
4 1
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:⏐⟩= ⏐⟩⏐⟩⏐⟩⊗⏐⟩⏐⏐⟩⏐⟩⏐⟩⊗⏐⟩⏐⏐⟩⏐⟩⊗⏐⟩
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A bit less scary...
• Can we make this state using established techniques for making W states and GHZ states?
where: ⏐⟩⏐⟩⏐⟩√
⏐⟩⏐⟩⏐⟩⏐⟩√
⏐⟩⏐⟩⏐⟩⏐⟩√
flip qubit #2:⏐⟩= ( ⏐⟩⊗⏐⟩⏐⟩⊗⏐⟩⏐⟩⊗⏐⟩
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3
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3
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2
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3 00 + 3 01 + 2 10 GHZ W W
Period 3 Circuit
H
X
X
X
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.