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meow!. Near-deterministic teleportation using linear optics. M. S. Tame. KIAS Journal Club. Topics today. Brief review from last time Mode Teleportation Loss detection Near-deterministic teleportation Fast Fourier Transform CNOT with success n 2 /(n+1) 2.
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meow! Near-deterministic teleportation using linear optics M. S. Tame KIAS Journal Club
Topics today • Brief review from last time • Mode Teleportation • Loss detection • Near-deterministic teleportation • Fast Fourier Transform • CNOT with success n2/(n+1)2
Review from last time… b’ a’ b a Qubit: As opposed to:
Review from last time… b | 1 > | 0 > a b | 0 > a | 1 > Qubit:
B Bt b b b’ b’ a a a’ a’ Review from last time…
Review from last time… | yc > = c2 | 2 > c1 | 1 > c0 | 0 > + + Btcb c’ c Bba Bba b’ b | 1 > q2 a’ a | 0 > q1 q1 | y 1 | y 2 | y 3 | y 4 > > > > c c’ NS b b’ | 1 > | 1 > Detection Condition (D.C) a a’ | 0 > | 0 >
Review from last time… c c’ NS NS = b b’ | 1 > P = 1/4 | 1 > Detection Condition (D.C) a a’ | 0 > | 0 >
Review from last time… a’ a C-Z1/16 b b’ P = 1/4 Btac Bac NS NS c’ c P = 1/4 d d’
Review from last time… b b’ P = 1/4 Btac Bac NS a’ a NS c’ c P = 1/4 d d’ C-Z1/16
Review from last time… C-Z1/16 b b’ P = 1/4 Btac Bac NS a’ a NS c’ c p p P = 1/4 d d’ Btcd Btcd = C-NOT1/16
Mode Teleportation Teleportation to double success probability !
Mode Teleportation 1 2 3 B12 1 ( R1 ) n 2 n ( R2 ) 3
Mode Teleportation B12 1 ( R1 ) n 2 n ( R2 ) 3
Mode Teleportation B12 1 ( R1 ) n 2 n ( R2 ) 3
Mode Teleportation B12 1 ( R1 ) n 2 n ( R2 ) 3
Mode Teleportation teleport B12 1 ( R1 ) n B23 2 ( R2 ) n C 3
2 qubit teleportation Mode Teleportation teleport teleport
CNOT CS1/16 C-Z1/16 b b’ P = 1/4 Btac Bac NS a’ a NS c’ c p p P = 1/4 d d’ C-NOT1/16
2 qubit teleportation Mode Teleportation teleport CS1/16 teleport
C-Z using mode teleportation Mode Teleportation Partial teleport (needs C1) Partial teleport (needs C1) CS1/16
C-Z reduced to a preparation problem Mode Teleportation Problem: How do we slow down photons for Qubits 1 and 2 until state is prepared? wow! C-Z1/4 (16 state preparation attempts needed, which corresponds to 32 attempted NS operations. But these can be achieved in negligable time dependent on detectors)
C-NOT1/16 Mode Teleportation C-Z1/16 b b’ P = 1/4 Btad Bad NS a’ a NS d’ d p p P = 1/4 c c’ Btdc Btdc C-NOT1/16
C-NOT1/4 Mode Teleportation C-Z1/4 2 2’ P = 1/4 Btba Bba NS 1’ 1 NS 3’ 3 p p P = 1/4 4 4’ Bt34 Bt34 C-NOT1/4
Mode Teleportation Loss detection needed for scalability !
Qubit teleportation with loss detection Mode Teleportation State preparation takes 16 attempts CS1/16 needs to be introduced...
Teleportation of 2 qubits with loss detection Mode Teleportation
C-Z1/4 using teleportation of 2 qubits with loss detection Mode Teleportation State preparation takes 4096 attempts !
Mode Teleportation What about this very bad ¼ probability !
Near-deterministic teleportation ‘CLAIM’ teleport ( R ) n 0 ( R ) n 1 P2pk/(n+1) Fn+1 (n+1) modes : 0 to n ( R ) n n n+1 | 0 > (2n-L) modes : n+1 to n+k-1 | 0 > n+k-1 n+k P (1) mode : n+k n+k+1 | 1 > (2n-L) modes : n+k+1 to 2n | 1 > 2n Probability n/(n+1)
Near-deterministic teleportation ‘PROOF’ Far too long for this presentation !
Near-deterministic teleportation ‘SHORT PROOF’ Easier to look at failure probability ( R ) n 0 | 0 > ( R ) n 1 | 0 > P2pk/(n+1) Fn+1 ( R ) n n | 0 > ( R ) n 0 | 1 > ( R ) n 1 | 1 > P2pk/(n+1) Fn+1 ( R ) n n | 1 > But how does this correspond to teleportation probability ?
Near-deterministic teleportation ‘SHORT PROOF’ ( R ) n 0 ( R ) n 1 Fn+1 ( R ) n n n+1 n+k-1 n+k n+k+1 2n
You asked for it ? OK then… help! Therefore
You asked for it ? OK then… Therefore for mode n+k we have
Near-deterministic teleportation ‘SHORT PROOF’ q1 q2 q3 qn ( R ) n 0 ( R ) n 1 P2pk/(n+1) Fn+1 ( R ) n n n+1 n+k-1 n+k n+k+1 2n
Mode Teleportation What’s in that ‘Magic Box’ Mark ? Fn+1 T.H. Cormen, C.E. Leiserson, and R.L. Rivest, “Introduction to Algorithms”, p795 (MIT Press, Cambridge, MA, 1990).
Fast Fourier Transform using linear optics Butterfly operation How is this done with linear optics ?
> > > > | Q1 | Q2 | Q1 | Q2 Teleporting 2 qubits with probability n2/(n+1)2
> > | Q1 | Q2 C-Z with probability n2/(n+1)2 > | Q’1 State preparation = cs1/16 > | Q’2
Mode Teleportation All that trouble for a C-NOT gate !
Efficiency using available components Not including the case where the other cs operations can be aborted if the previous one fails Ppreparation = (1/16)n For Psuccess = n2/(n+1)2 = 0.924 n = 25 Ppreparation = 7.8 X 10-31 But does this effect the preparation time ? tprep
References meow! References: E. Knill, R. Laflamme and G. J. Milburn, "A scheme forefficientquantum computation with linear optics",Nature 409, 46 (2001). J. L. O’Brien, G. J. Pryde, A. G. White, T. C. Ralph and D. Branning, “Demonstration of an all-optical quantum computer controlled-NOT gate",Nature 426, 264 (2003). M. A. Nielsen and I. L. Chuang, “Quantum Computation and Quantum Information”, Cambridge University Press, Cambridge (2000). T.H. Cormen, C.E. Leiserson, and R.L. Rivest, "Introduction to Algorithms", p795 (MIT Press, Cambridge, MA, 1990). And of course C. H. Bennett, G. Brassard, C. Crepeau, R. Jozsa, A.Peres, and W. K. Wootters, "Teleporting an Unknown Quantum State via Dual Classical and Einstein-Podolsky-Rosen Channels“, Phys. Rev. Lett., vol 70: 1895-1899 1993. Next time: “Boosting success with Quantum Codes” (after I’ve read Preskill’s notes….)