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Quantum Cryptography. 1e29 = 29996224275833 x 3475385758524527 (LOCK) (MESSAGE/CODE) (KEY) Classical computer doing 20 GFLOPS needs 10 15 /20x10 9 ~ 0.5 day to solve this Largest prime has 22,338,618 digits 4000 years !.
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Quantum Cryptography 1e29 = 29996224275833 x 3475385758524527 (LOCK) (MESSAGE/CODE) (KEY) Classical computer doing 20 GFLOPS needs 1015/20x109 ~ 0.5 day to solve this Largest prime has 22,338,618 digits 4000 years!
Inverter/NOT gate in out out in NOT 0 1 1 0
2-input Logic Gates 0 0 0 0 0 0 0 0 1 1 1 0 AND OR 1 1 0 0 0 1 1 1 1 1 1 1
2-input Logic Gates 0 0 0 0 0 0 1 1 0 0 0 0 1 1 1 1 1 0 NAND NOR XOR 1 1 1 0 0 0 1 1 0 1 1 1 1 1 1 0 0 0 SUM
NAND is universal A A A B NAND NOT AND = NOT(NAND) OR made of NAND and NOT A + B = NOT[NOT(A+B)] = NOT[NOT(A) and NOT(B)]
NAND is universal A.B A A.B B NAND NOT A A A.B = A + B NOT B B NOT
2 Bit Half-Adder A 0 0 0 CARRY = AND B 0 1 0 SUM = XOR 1 0 0 A B CARRY SUM 1 1 1 0 0 1 1 1 1 0 2
Full-Adder A Sum S B Carry Cout Cin S = A + B + Cin Cout = A.B + B.Cin + Cin.A
Encoder: n inputs, 2n outputs (Identifies each A-B possibility) Enable D0 A D1 B D2 D0 = A NOR B = NOT(A) AND NOT(B) D1 = NOT(A) AND B D2 = A AND NOT(B) D3 = A AND B D3
Decoder: 2n inputs, n outputs Collapses possibilities Enable D0 A D1 B D2 A = D2 + D3 B = D1 + D3 D3
2x1 and 4x 1 Multiplexer (Selector) S1 S0 Control S A A B F F C B D S=0, F=A S=1, F=B F = S.A + S.B (S0,S1) = (0,0), F=A (0,1), F=B (1,0), F=C (1,1), F=D F = S0.S1.A + S0.S1.B + S0.S1.C + S0.S1.D
8 x 1 Multiplexer DEMUX S1 S0 S2 S1 S0 S2 D0 D1 D0 D2 D1 D3 D2 F D4 D3 F D5 D4 D6 D5 D7 D6 D7
STORAGE/MEMORY (FLIP-FLOPS) Need feedback S Q R Q
Build a “NOT” with CMOS inverters PMOS NMOS • voltage turns • PMOS on, shorts • to power supply voltage + voltage turns NMOS on, shorts to ground ON Gain OFF
Build a NAND with CMOS inverters 0 0 1 0 1 1 NAND 1 0 1 1 1 0
Build a NOR with CMOS inverters 0 0 1 0 1 0 NOR 1 0 0 1 1 0
Build memory with CMOS (DRAM, SRAM, Flash)
Dissipation in CMOS PMOS Pdiss = IOFFVDD + aCVDD2f NMOS • Leakage high with scaling IOFF • Reduce VDD Steep devices • Reduce C/ION ratio FinFETs • F2D = 0.5√L • F3D = 0.5√(L/Nlayers)
Dissipation in binary switching “0” Barrier large enough to prevent spontaneous backflip “1”
Dissipation when electron falls downhill eVAB A VAB B N els: (N+2)kBTln2 We’re already pretty efficient for on-chip dissipation! (Zhirnov) 3kBTln2 Perr = e-qDV/kT < ½ qDV > kBTln2 kBTln2
Reversible Switching (“Moonwalking potential”) When we know where electron sits SLIDE When we don’t know where electron sits OOPS !! KNOWLEDGE CONNECTED TO DISSIPATION !
Information entropy DF = -TDS • S = kBlnW • = <kBln(1/p) > • = -SikB(pilnpi)
Dissipation in CMOS DF = -TDS S = -SikB(pilnpi) When we know where electron sits (read and copy) Si = 0 Sf = 0 When we don’t know where electron sits Sf = 0 Si = -kBln1/2 = kBln2
Dissipation in AND AND 0 0 0 DF = -TDS 0 1 0 S = -SikB(pilnpi) 1 0 0 1 1 1 Si = -4kB(1/4ln1/4) Sf = -kB(3/4ln3/4 + 1/4ln1/4) Squeezing of phase space (2 inputs, 1 output)
No Dissipation (CNOT) 0 0 0 A A A B Aout Bout B Flip B iff A = 1 0 1 0 0 1 0 1 1 Preserve phase space Bout is XOR ! 1 1 1 1 0 SUM
CNOT is reversible ! Aout A A Bout B B
CSWAP (Fredkin gate) A A B Flip B and C if A = 1 C A B Aout Bout C Cout 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 1 0 0 1 1 0 1 1 1 0 0 1 0 0 0 1 0 1 0 1 1 0 1 1 1 0 0 1 1 1 1 1
CCNOT (Toffoli gate) A Flip A iff B = 1 and C = 1 B B A B Aout Bout C Cout C C 0 0 0 0 0 0 0 0 1 0 0 1 0 1 0 Set C = 1 Get A XOR B ! (Same as SUM bit) 0 1 0 1 1 1 0 1 1 1 0 0 1 0 0 1 0 1 1 0 1 1 1 0 1 1 0 0 1 1 1 1 1
PRIME FACTORIZATION Use Superposition + Entanglement + reversible gates for one-shot computation • Shor’s Algorithm • Grover’s Algorithm • Kane’s Nuclear spin QC
What is Entanglement ?
EPR Paradox: Pion decay two entangled states in a singlet (e and e+ with opposite spins) |00> + |11> Move far far away Measure bit 1 0 Bit 2 must collapse to 0 instantly !! (Nonlocality since light needs finite time)
Is entanglement real? Could uncertainty accommodate a hidden variable? Is uncertainty just in knowledge or fundamental ??
Bell’s Inequality |00> + |11> Separate particles Run each through a rotation Uq1 x Uq2 Uq1= Uq2 = U+(p/2-q2) cosq1 -sinq1 sinq1 cosq1 1 0 |0> = |1> = 0 1
Result -sin(q1-q2) [|00> + |11>] Coincidence + cos(q1-q2) [ |10> - |01>] Non-coinc P(AC = 1) = sin2(q1-q2) P(AC = -1) = cos2(q1-q2) P(AC = -1) – P(AC = 1) = P(01) + P(10) – P(00) – P(11) = correlation = cos[2(q1-q2)]
Proof: Bell’s inequality Own vs Rent: O and O Car vs No Car: C and C TV vs No TV: T and T OCT + OCT + OCT + OCT + OCT + OCT + OCT + OCT = 1 If only two questions on a survey – coalesce info thus O O T T C C C C O O T T ? ? ? ? ? ? ? ? ? ? ? ? Jay Sulzberger
Proof: Bell’s inequality Test for accuracy: OCT + OCT + OCT + OCT + OCT + OCT + OCT + OCT = 1 TO + TO + OC + OC + CT + CT ≤ 2 Proof: O O T T C C C C O O T T ? ? ? ? ? ? TOC + TOC + TOC + TOC + OCT + OCT + OCT + OCT + CTO + CTO + CTO + CTO = 2 [ 1 – OCT – OCT] ≤ 2 LHS: ? ? ? ? ? ?
P ( A & [not B] ) + P ( B & [not C] ) ≥ P ( A & [not C] ) TO + TO + OC + OC + CT + CT ≤ 2 http://arxiv.org/pdf/0704.2529v2.pdf
Entanglement is real Electrons are in superposition A classical measurement decouples them by entangling with an apparatus. Decoherence kills phase information
Entanglement is real How can we use this for QC ?
Useful for Quantum Cryptography 1e29 = 29996224275833 x 3475385758524527 (LOCK) (MESSAGE/CODE) (KEY) Classical computer doing 20 GFLOPS needs 1015/20x109 ~ 0.5 day to solve this Largest prime has 22,338,618 digits 4000 years! Create superposition of all states related to key as input Entangle with gateable output Measure output input collapses (don’t peek yet !) Use gates to do Fourier Transform on input input further collapses Now Read input From input + output reads, we can get factor !!!