More measurement & computing (Clusters, N00N states, Zeno gates,...) 13 Mar 2012
The cost of postselection Of course, if each gate only succeeds some fraction p of the time... the odds of an N-gate computer working scale as p N. Exponential cost cancels exponential gain in quantum computing. But, clever observation: gates commute with teleportation. Gottesmann & Chuang, Nature 402, 390 (1999) Perform the gates first, on blank registers (photons from entangled pairs, which in some sense could be in any state at all), and save up the gates that worked [linear cost!]. Only now teleport the input qubits into the already-successful gates! Alternate picture: the gates generated some interesting entangled states as a resource, and joint measurements with those states enable quantum computation this is more explicitly the idea of clusterstate ( one-way ) quantum computation.
The cost of postselection Of course, if each gate only succeeds some fraction p of the time... the odds of an N-gate computer working scale as p N. Exponential cost cancels exponential gain in quantum computing. But, clever observation: gates commute with teleportation. Gottesmann & Chuang, Nature 402, 390 (1999) Perform the gates first, on blank registers (photons from entangled pairs, which in some sense could be in any state at all), and save up the gates that worked [linear cost!]. Only now teleport the input qubits into the already-successful gates! Alternate picture: the gates generated some interesting entangled states as a resource, and joint measurements with those states enable quantum computation this is more explicitly the idea of clusterstate ( one-way ) quantum computation.
1 One-way quantum computation (cluster or graph states) (another type of measurement-induced computation)
1 2 3 H H H Cluster state notation means: (1) prepare all qubits in 0+1 (2) implement CPHASE (CQ) on any pairs connected by edges. 000 +001 +010-011 +100 +101-110 +111
Line clusters as wires 1 2 3 Suppose we measure X 2 = 1. Then Z 1 Z 3 =1, i.e., Z 1 = Z 3. Z 1 X 2 Z 3 = 1 Otherwise, we find X 2 = 1. Then Z 1 Z 3 = 1, i.e., Z 3 = Z 1, which we can fix with a spin flip. We have teleported bit 1 to bit 3.
Computing with clusters c t G 4 X G Z t Z c Z 4 = 1 If I measure X G =1, then Z c Z t Z 4 = 1, so Z 4 =Z c Z t ; this is basically a CNOT. Embedding into circuits : c(in) c(out) t(in) t(out)
Computing with clusters, big picture
Cluster-state QC is a parallel approach utilizing the power of measurement (as is the QND approach of Munro, Nemoto, et al.) Optical circuit for generation of a 4-photon cluster state. Rausssendorf &Briegel, PRL 86, 5188 (01) Walther et al., Nature 434, 169 (05) Amplifying weak(-ish) nonlinearities by using them to make measurements, which in turn lead to gates: Nemoto & Munro, PRL 93, 250502 (04)
2 Another interesting use of entangled resources
Jon Dowling s Slide of Magic BS Oscillates N times as fast
Highly number-entangled states ( 3003 experiment). Important factorisation: M.W. Mitchell et al., Nature 429, 161 (2004) States such as N,0> + 0,N> ( N00N states) could revolutionize metrology (from atomic clocks to optical-interferometric sensing), and have been proposed for lithography as well. But how to make them? + = A "noon" state A really odd beast: one 0 o photon, one 120 o photon, and one 240 o photon... but of course, you can't tell them apart, let alone combine them into one mode! Theory: H. Lee et al., Phys. Rev. A 65, 030101 (2002); J. Fiurásek, Phys. Rev. A 65, 053818 (2002)
How does this get entangled? H 2H H Non-unitary
How does this get entangled? V H & V H Perfectly unitary
How does this get entangled? 60 H & 60 H Non-unitary (Initial states orthogonal due to spatial mode; final states non-orthogonal.)
Trick #1 How to combine three non-orthogonal photons into one spatial mode? "mode-mashing" Yes, it's that easy! If you see three photons out one port, then they all went out that port. Post-selective nonlinearity
Trick #2 Okay, we don't even have single-photon sources *. But we can produce pairs of photons in down-conversion, and very weak coherent states from a laser, such that if we detect three photons, we can be pretty sure we got only one from the laser and only two from the down-conversion... SPDC laser 0> + ε 2> + O(ε 2 ) 0> + α 1> + O(α 2 ) εα 3> + O(α 3 ) + O(ε 2 ) + terms with <3 photons * But we re working on it (collab. with Rich Mirin s quantum-dot group at NIST)
Trick #3 But how do you get the two down-converted photons to be at 120 o to each other? More post-selected (non-unitary) operations: if a 45 o photon gets through a polarizer, it's no longer at 45 o. If it gets through a partial polarizer, it could be anywhere... (or nothing) (or nothing) (or <2 photons)
But do you really need non-unitarity? V H PBS Has this unitary, linear-optics, operation entangled the photons? Is HV> = + +> - > an entangled state of two photons at all, or merely an entangled state of two field modes? Can the two indistinguishable photons be considered individual systems? To the extent that they can, does bosonic symmetrization mean that they were always entangled to begin with? Is there any qualitative difference in the case of N>2 photons?
Where does the weirdness come from? HHH+VVV (not entangled by some definition) C A If a photon winds up in each of modes {A,B,C}, then the three resulting photons are in a GHZ state 3 clearly entangled subsystems. You may claim that no entanglement was created by the BS s and postselection which created the 3003 state... but then must admit that some is created by the BS s & postselection which split it apart. B
Making triphoton states... E.g., HV(H+V) R 3 + R 2 L + RL 2 + L 3 In HV basis, H 2 V + HV 2 looks number-squeezed ; in RL basis, phase-squeezed.
It works! Singles: Coincidences: Triple coincidences: Triples (bg subtracted): M.W. Mitchell, J.S. Lundeen, and A.M. Steinberg, Nature 429, 161 (2004)
Another entangling gate: root-swap In 00 01 10 11 Out 00 01 + i 10 10 + i 01 11
Double wells as entanglers
A two-particle double-well 2? What s 2?! But cf. the fascinating exchange gates of Anderlini et al., Nature 448, 452 (2007)!!
This is just like the HOM! A swap is just a mirror, or a Mach-Zehnder interferometer; a root(swap) is a beam splitter... or half a M-Z interferometer? in swapped entangled? if you post-select coincidences, yes. a valid logical state of 2 qubits? Not if both photons end up in the same arm... How to avoid illegal states, and postselection?
Some more references (incomplete!) Clusters Nielsen, "Universal quantum computation using only...", quant-ph/0108020 Raussendorf & Briegel, "A one-way quantum computer", PRL 86, 5188 (2001) Raussendorf & Briegel, PRA 68, 022312 (2003) Aliferis & Leung, "Computation by measurements: a unifying picture", quant-ph/0404082 Nielsen, "Cluster-state Quantum Computation", quant-ph/0504097 Walther et al, Nature 434, 169 (2005) Briegel et al, Nature Physics 19 (2009) {review of msmt-based computation} Anderlini et al, Nature 448, 452 (2007) {the exchange-interaction gate} Nemoto & Munro, PRL 93, 250502 (04) {weak nonlinearities / QND gate} N00N states & generation: H. Lee et al., Phys. Rev. A 65, 030101 (2002) J. Fiurásek, Phys. Rev. A 65, 053818 (2002) M.W. Mitchell et al., Nature 429, 161 (2004) Zeno gate: Franson, Jacobs, and Pittman, PRA 70, 062302 (2004) {Zeno gate} U.S. Patent 6995404 (www.patentstorm.us/patents/6995404/description.html)