Presentation on theme: "1 The Feasibility of Testing LHVTs in High Energy Physics 李军利 中国科学院 研究生院 桂林 2006.10.27-11.01 In corporation with 乔从丰 教授 Phys.Rev. D74,076003, (2006)"— Presentation transcript:
1 The Feasibility of Testing LHVTs in High Energy Physics 李军利 中国科学院 研究生院 桂林 2006.10.27-11.01 In corporation with 乔从丰 教授 Phys.Rev. D74,076003, (2006)
2 Content EPR-B paradox. Bell inequality. Bell Inequality in Particle physics. The Feasibility of Testing LHVTs in Charm factory.
3 1.EPR-B paradox In a complete theory there is an element corresponding to each element of reality. Physical reality: possibility of predicting it with certainty, without disturbing the system. Non-commuting operators are mutually incompatible. I. The quantities correspond to non-commuting operators can not have simultaneously reality. or II. QM is incomplete. Einstein, Podolsky, Rosen. 1935
4 Two different measurements may performe upon the first particle. Due to angular momentum conservation and Einstein’s argument of reality and locality, the quantities of Non-commuting operators of the second particle can be simultaneously reality. EPR: ( Bohm’s version) So QM is incomplete !
5 Bohr ’ s reply Bohr contest not the EPR demonstration but the premises. An element of reality is associated with a concretely performed act of measurement. This makes the reality depend upon the process of measurement carried out on the first system. That it is the theory which decides what is observable, not the other way around. ---- Einstein
6 2.Bell Inequality (BI) EPR-(B) Hidden variable theory Von Neuman (1932) : the hidden variable is unlikely to be true. Gleason(1957), Jauch(1963), Kochen-Specher(1967) D>=3 paradox. Bell D=2. Bell inequality(1964). D>=3. contextual dependent hidden variable theorem would survive.
7 Hidden variable and Bell inequality a b c d QM: Bell, physics I,195-200, 1964 CHSH, PRL23,880(1969) LHVT:
8 Optical experiment and result Aspect 1982 two channel polarizer. PRL49,91(1982) Experiment with pairs of photons produced with PDC PRL81,3563(1998) All these experiments conform the QM! W. Tittel, et al
9 3.Bell Inequality in Particle physics Test BI with fermions or massive particles. Test BI with interactions other than electromagnetic interactions. Strong or Weak actions. Energy scale of photon case is eV range. Nonlocal effects may well become apparent at length scale about cm. S.A. Abel et al. PLB 280,304 (1992)
10 Bell Inequality in Particle physics In spin system: the measurement of spin correlation in low-energy proton proton scattering. [M.Lamehi-Rachti,W.Mitting,PRD,14,2543,1976]. Spin singlet state particle decay to two spin one half particles. [N.A. Tornqvist. Found.Phys.,11,171,1981]. With meson system: Quasi spin system.
11 Mass eigenstatesCP eigenstates S eigenstates Like the photon case they don’t commutate Are regard as the quasi-spin states. Berltamann, Quant-ph/0410028 Note that the H is not a observable [not hermitian] Fix the quasi-spin and free in time. 1 2
12 Experiment of system A.Go. J.Mod.Opt. 51,991. They use to test the BI: as the flavor tag. However, debates on whether it is a genuine test of LHVTs or not is still ongoing. PLA332,355,(2004) R.A. Bertlmann
13 Other form of nonlocality Nonlocality without using inequalities. GHZ states: three spin half particles. (1990) Kochen-Specher: two spin one particles. (80) L. Hardy: two spin one half particles. Dimension-6 Hardy’s proof relies on a certain lack of symmetry of the entangled state. PRL71, 1665 (1993)
14 GHZ states: three spin half particles. “No reasonable definition of reality could be expected to permit this.” (1) (2) (3) (4) which contradict (4).
15 Kochen-Specher: two spin one particles. (1). Any orthogonal frame (x, y, z), 0 happens exactly once. (2). Any orthogonal pair (d, d’), 0 happens at most once. If h(a0)=h(a7)=0, then h(a1)=h(a2)=h(a3)=h(a4)=1. So that h(a5)=h(a6)=0, by (1), which contradict (2). A set of eight directions represented in following graph: Consider a pair of spin 1 particle in singlet state. So can determine the value for Si without disturbing that system, leading the non-contextuality. R.A. Bertlmann & A. Zeilinger quantum [un]speakables from Bell to quantum information
16 L. Hardy: two spin one half particles. But because of 2 & 3. If D=1 then G=1. If E=1 then F=1. So if the probability that D=E=1 is not 0, then the probability for F=G=1 won’t be 0. Jordan proved that for the state like: PRA50, 62 (1994) Jordan Ifthere exist four projection operators satisfy:
18 Hardy type experiment with entangled Kaon pairs Generate a asymmetric state. Eberhard’s inequality (EI). PRL88, 040403 (2002), PRL89, 160401 (2002). Quant-ph/05011069. A. Bramon, and G. Garbarino: A. Bramon, R. Escribano and G. Garbarino:
19 To generate the asymmetric state, fix a thin regenerator on the right beam close to decay points. Then the initial state: Becomes: Let this state propagate to a proper time T:
20 Normalize it to the surviving pairs leads to: where component has been enhanced. has been further suppressed.
21 4.The Feasibility of Testing LHVTs in Charm factory Easy to get space-like separation. Can test the phenomena: less entangled state leads to larger violation of inequality. In the charm factory the entanglement state formed as: where
22 The four joint measurement of the transition probability needed in the EI predicted by QM take the following form:
23 Take into EI: See Figure 1 (for See figure 3) where is the violation degree of the inequality. From QM we have: First assume
25 Actually has non-zero magnitude : The shaded region is the requirement of the real and imagine part of R when violation between QM and LHVTs can be seen from inequality.
26 The advantage of over factory Charm factory Space-like separation required: In To make sure the misidentification of is of order per thousand Properly choose PRL88, 040403 (2002) so has a wider region of R in discriminating QM from LHVT. There is phenomena can be test due to this advantage.
27 Quantify the entanglement PRL80, 2245 (1998)W. Wootters Where: This mean the state become less entangled during time evolution ! Historically the amount of the violation was seen as extent of entanglement. This may not be the case in EI. As indicated in Figure 1 & 2. To see this we must quantify the degree of entanglement Take concurrence as a measure of this quantity. C changes between 0 to 1 for no entanglement and full entanglement.
28 Express the violation in degree of entanglement PLA154,201(1991) Abouraddy et al. N.Gisin PRA64,050101,(2001) 1.The usual CHSH inequality: 2. The Hardy state using Eberhard’s inequality :See the figure next page Note we make a trick in the figure that substitute C with
29 The Entanglement and Bell inequality violation Magnitudes below zero of VD is the range of violation