Quantum Optics II Cozumel, Mexico, December 5-8, 2004 ”Entanglement in Time and Space” J.H. Eberly, Ting Yu, K.W. Chan, and M.V. Fedorov University of Rochester / Prokhorov Institute We consider entanglement as a dynamic property of quantum states and examine its behavior in time and space. Some interesting findings: (1) adding more noise helps fight phase-noise disentanglement, and (2) high entanglement induces spatial localization, equivalent to a quantum memory force. • Ting Yu & JHE, Phys. Rev. Lett. 93, 140404 (2004). • K.W. Chan, C.K. Law and JHE, Phys. Rev. Lett. 88, 100402 (2002) • JHE, K.W. Chan and C.K. Law, Phil. Trans. Roy. Soc. London A 361, 1519 (2003). • M.V. Fedorov, et al., Phys. Rev. A 69, 052117 (2004). 1 Entanglement means a superposition of conflicting information about two objects. Superposition of conflicting information, but only one object. Can you handle the conflicting information here? Which face is in the back? 2 A pair of conflicts can be “entangled” Try to see both at the same time. Do they “flip” together? 3 Measurement cancels contradiction A pair of boxes, but only one view of them 4 Bell States provide a simple example Schrödinger-cat “Bell State”: | - > =|C*>|N*> + |C>|N> excited cat = C*, dead cat = C, excited nucleus = N*, ground state = N 5 Bell States provide a simple example Schrödinger-cat “Bell State”: | - > =|C*>|N*> + |C>|N> excited cat = C*, dead cat = C, excited nucleus = N*, ground state = N <N| - > = |C> (sorry, Cat) 6 Overview Issue -- entanglement in time and space. Illustration #1 -- two atoms are excited and entangled but not communicating with each other. Result #1 -- both atoms decay, diag e-2gt and offdiag e-gt , just as expected; but entanglement of the atoms behaves qualitatively differently. Illustration #2 -- two atoms fly apart in molecular dissociation. Result #2 -- entanglement means localization in space (a “quantum memory force”). For detailed treatments: Ting Yu & JHE, PRL 93 140404 (2004), and M.V. Fedorov, et al., PRA 69, 052117 (2004). 7 Illustration #1: A B HAT = (1/2)wAsZA + (1/2) wBsZB , HCAV = kwkak†ak + knkbk†bk HINT = k(gk*s-Aak† + gks+Aak) + k(fk*s-Bbk† + fks+Bbk) At t=0 the joint initial state rAB is entangled and mixed (not pure). The atoms only decay (no cavity feedback). After t=0, what happens to entanglement? 8 mixed initial states, C = 2/3 r = ++ +-+ -- ++ a 0 0 0 +- -+ 0 0 1 1 1 1 0 0 -0 0 0 d Initial state, entangled and mixed, where d = 1-a. C = concurrence EOF concurrence 1≥C≥0 9 Time Evolution Result: r = a 0 0 0 0 0 1 1 1 1 00 0 0 0 d a(t) 0 0 0 0 b(t) z(t) 0 0 z*(t) c(t) 0 0 0 0 d(t) Obvious point: the atoms go to their ground states, so d 3, and the other elements decay to zero. No surprise: the decay of r(t) is smooth, and exponential, measured by usual natural lifetime: s±A(t) = s±A(0) exp[-GAt/2 ± iwAt] Reminder: the atoms decay independently. 10 Sol’n. in Kraus representation: ˜ (t) = † r K (t) r(0) K (t) Kraus operators: g 0 g 0 A B K1 0 1 0 1 0 K 3 w A 0 g B 0 0 0 1 g A K 2 0 0 0 1 w B 0 0 0 K 4 w A 0 0 0 w B 0 0 gA = exp[-G t / 2 ] wA2 = 1- exp[-Gt ] , same for B. for details, see Ting Yu and JHE, PRB 68, 165322 (2003) 11 Kraus matrix evolution: r a 0 0 0 0 0 b z z* c 0 0 0 0 0 d Decays all depend on gA(t) = exp(-GAt/2) and gB(t) = exp(-GBt/2). a(t) = gAgB, b(t) = gB2 +gA2 wB2, c(t) = gA2 +gB2 wA2, d(t) = wA2 +wA2 +wA2wA2, z(t) = gAgB , where wA2 = 1 - gA2, etc. Usual Born-Markov solutions for the separate atoms: s±A(t) = s±A(0) exp[-GAt/2 ± iwAt] For detailed treatment: Ting Yu & JHE, PRL 93, 140404 (2004). 12 Entanglement evolution: Entanglement has its own rules, and follows the atom decay law only exceptionally. Entanglement can be completely lost in a finite time! art by Curtis Broadbent Ting Yu & JHE, Phys. Rev. Lett. 93, 140404 (2004). 13 Noise + Entanglement 1 1 r (1 F) I + (4F 1) | ( ) ( ) | Werner state density matrix 3 3 W Werner qubits undergo only off-diagonal relaxation under the influence of phase noise. All entanglement of a Werner state is destroyed in a finite time by pure phase noise. 14 Two Noises + Entanglement Pure off-diagonal relaxation of qubits Add some diagonal relaxation, for example via vacuum fluctuations. Add diag. to off-diag. relaxation of qubits Werner entanglement gets some protection from added noise! Ting Yu & JHE (in preparation). 15 Spatial localization and entanglement With just two objects, high entanglement can be reached by allowing each object a wide variety of different states. The idealized two-particle wave function used by Einstein, Podolsky and Rosen in their famous 1935 EPR paper used continuous variables (infinite number of states) to get the maximum degree of entanglement. Experiments following the original EPR “breakup” scenario have not been done yet. 16 Physical examples of breakup Down conversion (K ≈ 4.5 — 1000’s) Huang and Eberly, JMO 40, 915 (1993) Law, Walmsley and Eberly, PRL 84, 5304 (2000) Law and Eberly, PRL 92, 127903 (2004) Raman scattering (K > 100) Chan, Law, and Eberly, PRA 68, 022110 (2003) Chan, et al., JMO 51, 1779 (2004). Spontaneous emission (K ≈ 1) Chan, Law and Eberly, PRL 88, 100402 (2002) Fedorov, et al. (in preparation, 2004) Ionization/Dissociation (K > 10) Fedorov, et al., PRA 69, 052117 (2004). Chan and Eberly, quant-ph 0404093 17 EPR system is created by break-up The variables entangled are positions (x1 and x2), or momenta (k1 and k2). Perfect correlation is implied in the EPR wavefunction: (x , x ) (x x ) 0 1 2 1 2 How much correlation is realistic? How to measure it? Original paper: Einstein, Podolsky and Rosen, Phys. Rev. 47, 777 (1935). 18 Localization - Entanglement • All information is in . We can plot the two-particle density ||2 vs. x1 and x2. • Knowledge of one particle gives information about the other particle. • Joint localization information is packet entanglement. Fedorov ratios for particle localization: x1 x2 F1 F2 cond x1 x2cond ||2 We can calculate these for a simple dissociation model. 19 Time-dependent EPR example Given a dissociation rate gd, a post-breakup diatomic is: R 2 r r (r1,r2 ;t) ~ 2 (vt r)expg d t exp 2 r v 4R 0 relative part rel CM part 2 (r vt)g d v CM 2 R R0 M.V. Fedorov, et al., PRA 69, 052117 (2004) / quant-ph/0312119. 20 Joint Probability Density |total|2 rel 2 CM 2 21 Dynamics of localization What do we know and when do we know it? • Massive particles spreading wavepackets: x x(t) and X X(t) • EPR pairs: [x, P] = 0 and [X, p] = 0 nonlocality • Spreading is governed by the free-particle Hamiltonian. • Time evolution is merely via phase in the momentum picture: k12 ˜ ( k1;0) exp 2 4k 1 k12 ht 2 ˜ ( k1;t) exp i k1 2 4m1 4k 1 22 Dynamics of localization - F ratios Plots of |(t)|2 vs. x1 and x2 : Experiments track localization via packet spreading (i.e.,spatial variances). The two-particle ratio h(t) = ∆x/2∆X is a convenient parameter [*] connected with dynamical evolution. Calculation x1 F 1 x1cond Inferred dependence B F(t) ~ A h(t) + h(t) * Chan, Law and Eberly, PRL 88, 100402 (2002). 23 Universal man-in-street theory Model the breakup state as double-Gaussian [*]. 1 x 2 X 2 * K.W. Chan and JHE, 0 (x1 , x 2 ) exp exp 2 2 quant-ph 0404093 ab 4x 4X 0 0 This makes it easy to calculate the Fedorov ratios (F1 ~ F2 ) at t=0 and for later times. Question: can we guess what happens to localization? x2 X2 (x, X;t) ~ exp exp 2 2 4(x 0 + i t /2) 4(X 0 + i t /2M) 24 Entanglement migration to phase x2 X2 (x, X;t) ~ exp exp 2 2 4(x 0 + i t /2) 4(X 0 + i t /2M) Therefore P(x, X; t) = (x, X; t)2 has two similar real exponents. x x 2 1 1 2 2 2 x 0 + ( t /2 ) 2 x 02 When these exponents are added, the nonseparable x1x2 term 0 at a specific time t0: t 2 M x X 0 0 0 m2 2 m1 X + X 1 M 1 M 2 2 X 02 + ( t /2M) 2 X 02 25 Quantum memory force (QMF) Atom Photon The dissociation example has a close analog in spontaneous emission. These atom-photon space functions show a “force” arising from shared quantum information, a “quantum memory force” (QMF). The first four bound states are shown for Schmidt number K = 3.5, which is slightly “beyond-Bell.,” i.e., K > 2. M.V.Fedorov, et al. (in preparation). Chan-Law-Eberly, PRL 88, 100402 (2002) 26 Summary / dynamics of entanglement • Entanglement dynamics are largely unknown (time or space) • Noisy environment kills entanglement but not intuitively • Individual atom decay is not a guide for entanglement • Diag. + off-diag. noise has a cancelling effect • EPR-type breakup is ubiquitous / creates two-party correlation • Conditional localization vs. entanglement ? • Packet dynamics, Fedorov ratio and control parameter h • Man-in-street theory and phase entanglement • Memory effects enforce spatial configurations (QMF) 27 Acknowledgement Research supported by NSF grant PHY-00-72359, MURI Grant DAAD19-99-10215, NEC Res. Inst. grant, and a Messersmith Fellowship to K.W. Chan. References Ting Yu & J.H. Eberly, PRL 93, 140404 (2004) and in preparation. M.V. Fedorov, et al., PRA 69, 052117 (2004). C.K. Law and J.H. Eberly, PRL 92, 127903 (2004). M.V. Fedorov, et al., PRA (in preparation). K.W. Chan, C.K. Law and J.H. Eberly, PRL 88, 100402 (2002). K.W. Chan and J.H. Eberly, quant-ph/0404093. A. Einstein, B. Podolsky and N. Rosen, Phys. Rev. 47, 777 (1935). 28 More sophisticated Schmidt analysis Any bipartite pure state can be written as a single discrete sum: • • 0 (x1, x 2 ) n n (x1 ) n (x 2 ) n 0 Continuous basis discrete basis Unique association of system 1 to system 2 e.g., (x1 x 2 ) n (x1 ) n (x 2 ) n Size of ent. K n 2 n 1 29 Discretization of continuum information, the Schmidt advantage Unique mode pairs Continuous-mode basis Pure-state non-entropic measure of entanglement: Schmidt-mode basis Schmidt number counts experimental modes, provides practical metric 30 Interpreting K, the Schmidt number K = 1, no entanglement. K = 2, perfect Bell states. K = 5, beyond Bell, more information. K = 10, still more info. Quantum info is always discrete and countable. 31 Estimation of K for photodissociation Comparing the photodissociation process with the doubleGaussian model, we identify x0 = v ⁄ gd and X0 = R0. If we take R0 = 10 nm, v 2E ~ c 1 eV ~ 10 3 m s , 10 GeV and define d = gd-1, then with d in sec, x 0 v h ~ ~ 1011 d 2X 0 g d R0 1 1 K h + 100 2 h K d (s) 32 Retreat of entanglement into phase The Fedorov ratios for double-Gaussian 0(x1,x2 ) : Position: Momentum: where x1 F1 co cosh r K C(t) x1 k1 G1 co cosh r K k1 e 2 r e 2r + (t /t 0 ) 2 , so C(t) 1 . 2r 2 1+ e (t /t 0 ) t0 2mx 0 X 0 Note non-equivalence of k-space and x-space for these experimentally measurable quantities. 33 34 Double-Gaussian Schmidt analysis For the man-in-street double-Gaussian model (with m1 = m2) 0 (x1, x 2 ) n 1n (x1 ) n2 (x 2 ) n 0 i The Schmidt modes are the number states n (xi ) xi n 1 x 0 2X 0 K n n + 2 2X0 x0 while from the actual wave function we had inferred B cond F1 x1 / x1 Ah(t) + h(t) with h(t) = ∆x(t)/2∆X(t). These are the same, except for spreading! and one finds: 2 1 35