# 1 ”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.

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1 ”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). Quantum Optics II Cozumel, Mexico, December 5-8, 2004

2 Entanglement means a superposition of conflicting information about two objects. Can you handle the conflicting information here? Which face is in the back? Superposition of conflicting information, but only one object.

3 Try to see both at the same time. Do they “flip” together? A pair of conflicts can be “entangled”

4 Measurement cancels contradiction A pair of boxes, but only one view of them

5 Schrödinger-cat “Bell State”: |   -  > =|C * >|N * >  |C>|N> excited cat = C*, dead cat = C, excited nucleus = N*, ground state = N Bell States provide a simple example

6 Schrödinger-cat “Bell State”: |   -  > =|C * >|N * >  |C>|N> excited cat = C*, dead cat = C, excited nucleus = N*, ground state = N = |C> (sorry, Cat) Bell States provide a simple example

7 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 -2  t and off- diag  e -  t, 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”). Overview For detailed treatments: Ting Yu & JHE, PRL 93 140404 (2004), and M.V. Fedorov, et al., PRA 69, 052117 (2004).

8 Illustration #1: H AT = (1/2)  A  Z A + (1/2)  B  Z B, H CAV =  k  k a k † a k +  k k b k † b k H INT =  k (g k *  - A a k † + g k  + A a k ) +  k (f k *  - B b k † + f k  + B b k ) At t=0 the joint initial state  AB is entangled and mixed (not pure). The atoms only decay (no cavity feedback). After t=0, what happens to entanglement? A B

9 mixed initial states, C = 2/3  = ++ +- -+ -- ++ a 0 0 0 +- 0 1 1 0 -+ 0 1 1 0 -- 0 0 0 d Initial state, entangled and mixed, where d = 1-a. C = concurrence EOF concurrence 1 ≥ C ≥ 0

10  = a 0 0 0 0 1 1 0 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  (t) is smooth, and exponential, measured by usual natural lifetime: Time Evolution Result:  ± A (t)  =  ± A (0)  exp[-  A t/2 ± i  A t] Reminder: the atoms decay independently.

11 Kraus operators:  A = exp[-  t / 2 ]  A 2 = 1- exp[-  t ], same for B. (t) =    K  (t)  (0) K  † (t) Sol’n. in Kraus representation: for details, see Ting Yu and JHE, PRB 68, 165322 (2003)

12 Kraus matrix evolution:  a 0 0 0 0 b z 0 0 z* c 0 0 0 0 d Decays all depend on  A (t) = exp(-  A t/2) and  B (t) = exp(-  B t/2). a(t) =  A  B, b(t) =  B 2 +  A 2  B 2, c(t) =  A 2 +  B 2  A 2, d(t) =  A 2 +  A 2 +  A 2  A 2, z(t) =  A  B, where  A 2 = 1 -  A 2, etc.  ± A (t)  =  ± A (0)  exp[-  A t/2 ± i  A t] Usual Born-Markov solutions for the separate atoms: For detailed treatment: Ting Yu & JHE, PRL 93, 140404 (2004).

13 art by Curtis Broadbent Entanglement has its own rules, and follows the atom decay law only exceptionally. Entanglement can be completely lost in a finite time! Entanglement evolution: Ting Yu & JHE, Phys. Rev. Lett. 93, 140404 (2004).

14 Noise + Entanglement 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. Werner state density matrix

15 Two Noises + Entanglement Add some diagonal relaxation, for example via vacuum fluctuations. Ting Yu & JHE (in preparation). Werner entanglement gets some protection from added noise! Pure off-diagonal relaxation of qubits Add diag. to off-diag. relaxation of qubits

16 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.

Physical examples of breakup Spontaneous emission (K ≈ 1) Chan, Law and Eberly, PRL 88, 100402 (2002) Fedorov, et al. (in preparation, 2004) Raman scattering (K > 100) Chan, Law, and Eberly, PRA 68, 022110 (2003) Chan, et al., JMO 51, 1779 (2004). 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) Ionization/Dissociation (K > 10) Fedorov, et al., PRA 69, 052117 (2004). Chan and Eberly, quant-ph 0404093 17

18 EPR system is created by break-up The variables entangled are positions ( x 1 and x 2 ), or momenta ( k 1 and k 2 ). Perfect correlation is implied in the EPR wavefunction: Original paper: Einstein, Podolsky and Rosen, Phys. Rev. 47, 777 (1935). How much correlation is realistic? How to measure it?

19 Localization - Entanglement All information is in . We can plot the two-particle density |  | 2 vs. x 1 and x 2. Knowledge of one particle gives information about the other particle. Joint localization information is packet entanglement. ||2||2 Fedorov ratios for particle localization: We can calculate these for a simple dissociation model.

20 relative partCM part Time-dependent EPR example Given a dissociation rate  d, a post-breakup diatomic  is: M.V. Fedorov, et al., PRA 69, 052117 (2004) / quant-ph/0312119.

21 Joint Probability Density |  total | 2

22 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: Dynamics of localization What do we know and when do we know it?

23 Dynamics of localization - F ratios Experiments track localization via packet spreading (i.e.,spatial variances). The two-particle ratio  (t) = ∆x/2∆X is a convenient parameter [*] connected with dynamical evolution. Plots of |  (t)| 2 vs. x 1 and x 2 : CalculationInferred dependence * Chan, Law and Eberly, PRL 88, 100402 (2002).

24 Universal man-in-street theory This makes it easy to calculate the Fedorov ratios (F 1 ~ F 2 ) at t=0 and for later times. Question: can we guess what happens to localization? * K.W. Chan and JHE, quant-ph 0404093 Model the breakup state as double-Gaussian [*].

25 Therefore P(x, X; t) =  (x, X; t)  2 has two similar real exponents. Entanglement migration to phase When these exponents are added, the nonseparable x 1 x 2 term  0 at a specific time t 0 :

26 Quantum memory force (QMF) 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.Fe dorov, et al. (in preparation). Chan-Law-Eberly, PRL 88, 100402 (2002) Atom Photon

27 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  Man-in-street theory and phase entanglement Memory effects enforce spatial configurations (QMF) Summary / dynamics of entanglement

28 Acknowledgement Research supported by NSF grant PHY-00-72359, MURI Grant DAAD19-99-1- 0215, 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).

29 Any bipartite pure state can be written as a single discrete sum: Continuous basis  discrete basis Unique association of system 1 to system 2 Size of ent. e.g., More sophisticated Schmidt analysis

30 Discretization of continuum information, the Schmidt advantage Continuous-mode basisSchmidt-mode basis Pure-state non-entropic measure of entanglement: Schmidt number counts experimental modes, provides practical metric Unique mode pairs

31 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. Interpreting K, the Schmidt number

32 Comparing the photodissociation process with the double- Gaussian model, we identify  x 0 = v ⁄  d and  X 0 =  R 0. If we take  R 0 = 10 nm,, and define  d =  d -1, then with  d in sec, Estimation of K for photodissociation

33 Retreat of entanglement into phase The Fedorov ratios for double-Gaussian : where, so. Note non-equivalence of k-space and x-space for these experimentally measurable quantities. Position: Momentum:

34

35 The Schmidt modes are the number states and one finds: Double-Gaussian Schmidt analysis while from the actual wave function we had inferred For the man-in-street double-Gaussian model (with m 1 = m 2 ) with  (t) = ∆x(t)/2∆X(t). These are the same, except for spreading!

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