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The Feasibility of Testing LHVTs in High Energy Physics 李军利 Phys.Rev. D74,076003, (2006) 中国科学院 研究生院 In corporation with 乔从丰 教授 桂林 2006.10.27-11.01 1 Content EPR-B paradox. Bell inequality. Bell Inequality in Particle physics. The Feasibility of Testing LHVTs in Charm factory. 2 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 3 EPR: ( Bohm’s version) 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. So QM is incomplete ! 4 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 5 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. 6 a Hidden variable and Bell inequality b LHVT: E(a, b) d ( ) A(a, ) B(b, ) c d QM: E(a,b)= a b a b S1 | E (a, c) E (a, b) | E (b, c) 1 Bell, physics I,195-200, 1964 S2 E(a, d ) - E(a, b) E(c, b) E(d , c) 2 CHSH, PRL23,880(1969) 7 Optical experiment and result Aspect 1982 two channel polarizer. S 2.697 0.015 PRL49,91(1982) Experiment with pairs of photons produced with PDC S (0.85 0.009 )2 2 W. Tittel, et al PRL81,3563(1998) All these experiments conform the QM! 8 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 1013 cm. S.A. Abel et al. PLB 280,304 (1992) 9 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. 10 Mass eigenstates CP eigenstates S eigenstates 1 1 1 Ks {p K 0 q K 0 } K1 { K0 K 0 } N 2 S K0 K0 1 1 KL {p K 0 q K 0 } K2 { K0 K 0 } S K0 K0 N 2 Like the photon case they don’t commutate 1 K0 1 , { K0 K 0 K 0 K0 } K0 Are regard as the quasi-spin states. 2 i i 2 H M H K LS LS KLS LS mSL SL 2 2 Note that the H is not a observable [not hermitian] ( tl t r ) E (t r , tl ) cos(mt )e Berltamann, Quant-ph/0410028 Fix the quasi-spin and free in time. 11 Experiment of B 0 B 0 system A.Go. J.Mod.Opt.51,991. They use S 2 to test the BI: B 0 D*l , B 0 D* l as the flavor tag. However, debates on whether it is a genuine test of LHVTs or not is still ongoing. R.A. Bertlmann PLA332,355,(2004) 12 Other form of nonlocality Nonlocality without using inequalities. GHZ states: three spin half particles.(1990) Dimension-6 Kochen-Specher: two spin one particles.(80) L. Hardy: two spin one half particles. Hardy’s proof relies on a certain lack of symmetry of the entangled state. , . PRL71, 1665 (1993) 13 GHZ states: three spin half particles. 1 ABC (0 A 0 B 0 C 1 A 1 B 1 C) 2 xA y y B C mxA m y mC B y 1 (1) A B C A B C ABC ABC y x y ABC ABC m y m x m y 1 (2) A B C A B C y y x ABC ABC m y m y m x 1 (3) A B C A B C x x x 1 (4) ABC ABC m x m x m x (1) (2) (3) mx mx mx (my )2 (my )2 (mC )2 1 A B C A B y mxA mx mx 1 B C which contradict (4). “No reasonable definition of reality could be expected to permit this.” 14 Kochen-Specher: two spin one particles. S x S y S z2 2 2 2 (1). Any orthogonal frame (x, y, z), 0 happens exactly once. (2). Any orthogonal pair (d, d’), 0 happens at most once. A set of eight directions represented in following graph: a0 (1,1,1) a1 (0,1,1) a2 (1,1,0) a3 (0,1,1) a4 (1,1,0) a5 (1,0,0) a6 (0,0,1) a7 (1,1,1) 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). 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 15 L. Hardy: two spin one half particles. Jordan proved that for the state like: cos p ei sin m If tan 1 there exist four projection operators satisfy: 1. FG 0 2. D(1 G) 0 3. (1 F ) E 0 4. DE 0 PRA50, 62 (1994) Jordan 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. 16 Consider the CH inequality: PRD10, 526 (1974) 1 P(a, b) P(a, b' ) P(a' , b) P(a' , b' ) P(a' ,*) P(*,b) 0 1 A(a) B(b) A(a) B(b' ) A(a' ) B(b) A(a' ) B(b' ) A(a' ) I IB(b) 0 4 3 1 2 A(a' )B(b) [I A(a)]B(b) A(a)B(b' ) A(a' )[I B(b' )] Eberhard Inequality Compare to previous page: 1. FG 0 2. D(1 G) 0 PRA52, 2535 (1995) Garuccio 3. (1 F ) E 0 4. DE 0 17 Hardy type experiment with entangled Kaon pairs Generate a asymmetric state. Eberhard’s inequality (EI). 1 (T ) [ K S K L K L K S RKL K L ] 2 R 2 PLR ( K 0 K 0 ) PLR ( K 0 K L ) PLR ( K S K S ) PLR ( K L K 0 ) A. Bramon, and G. Garbarino: PRL88, 040403 (2002), PRL89, 160401 (2002). A. Bramon, R. Escribano and G. Garbarino: Quant-ph/05011069. 18 To generate the asymmetric state, fix a thin regenerator on the right beam close to decay points. Then the initial state: 1 [K S K L K L K S ] 2 Becomes: 1 (t ) [ K S K L K L K S rK S K S rK L K L ] 2 Let this state propagate to a proper time T: 19 N (T ) (T ) [KS K L K L KS 2 1 1 ( S L )T ( L S )T re imT e 2 K L K L re imT e 2 KS KS ] K L K L component has been enhanced. K S K S has been further suppressed. Normalize it to the surviving pairs leads to: 1 [ K S K L K L K S RKL K L ] 2 R 2 S L i ( m L mS ) T T where R Re i re 2 . 20 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: 1 J / (0) [K 0 K 0 K 0 K 0 ] 2 S L 1 i ( mL mS )T T J / (T ) [ K S K L K L K S re 2 KL KL ] 2 R 2 S L i ( m L mS ) T T where Re i re 2 . 21 The four joint measurement of the transition probability needed in the EI predicted by QM take the following form: 2 PQM ( K , K ) K K 0 0 0 0 J / (T ) i 2 2 Re i 2 1 Re PQM ( K , K ) 0 0 PQM ( K , K L ) 0 2(2 R ) 2 4(2 R ) 2 i 2 1 Re PQM ( K L , K ) 0 PQM ( K S , K S ) 0 2(2 R ) 2 22 Take into EI: VDEI PLR ( K 0 K 0 ) [ PLR ( K 0 K L ) PLR ( K S K S ) PLR ( K L K 0 )] where VD is the violation degree of the inequality. From QM we have: First assume 0 (for 0 See figure 3) (2 R) 2 (1 R ) 2 (1 R ) 2 0 . See Figure 1 4( 2 R ) 2( 2 R ) 2 2 2( 2 R )2 3R 2 4 R VDEI . 4( 2 R ) 2 23 24 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. 25 The advantage of J / over To make sure the misidentification of K S with K L is of order per thousand Properly choose T 5 S PRL88, 040403 (2002) Space-like separation required: In factory K 0.22 T 11 S S T R re 2 103 e6 0.4 so R 0.4 J / Charm factory K 0.94 T 0.15 S S R 103 T 3 3 R re 2 10 e 0.1 10 so J / has a wider region of R in discriminating QM from LHVT. There is phenomena can be test due to this advantage. 26 Quantify the entanglement 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. ~ ~ 2 2 C ( J / ) J / J / 2 R 2 r e( S L )T 2 2 ~ ~ Where: J / 1 y ( J / )* y 2 W. Wootters PRL80, 2245 (1998) C changes between 0 to 1 for no entanglement and full entanglement. This mean the state become less entangled during time evolution! 27 Express the violation in degree of entanglement 1.The usual CHSH inequality: S2 E(a, d ) - E(a, b) E(c, b) E(d , c) 2 S 2 2 (1 C ) N.Gisin PLA154,201(1991) VDCHSH 2 (1 C) 2 Abouraddy et al. PRA64,050101,(2001) 2. The Hardy state using Eberhard’s inequality : See the figure next page 3R 2 4 R C 2 VDEI 2 R 2 Note we make a trick in 4( 2 R 2 ) the figure that substitute C with 1 x 2 . 3(1 C ) 2 2 C C 2 VDEI 4 28 The Entanglement and Bell inequality violation Magnitudes below zero of VD is the range of violation 29 Thank you for your patience. 30

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