Docstoc

by entangled

Document Sample
by entangled Powered By Docstoc
					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 1013 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(mt )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 imT e   2
                                           K L K L  re imT 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
                                     103  e6  0.4                so     R  0.4

 J /        Charm factory     K  0.94             T  0.15 S
                             S
                                                                             R  103
                                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

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:5
posted:10/21/2012
language:English
pages:30