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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)

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Content

EPR-B paradox. Bell inequality. Bell Inequality in Particle physics. The Feasibility of Testing LHVTs in Charm facto

ry.

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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

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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 !

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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

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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.

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Hidden variable and Bell inequality

( , ) ( ) ( , ) ( , )E a b d A a B b

2 ( , ) - ( , ) ( , ) ( , ) 2S E a d E a b E c b E d c

ab

c

d

(a,b)= a bE a b QM:

Bell, physics I,195-200, 1964

CHSH, PRL23,880(1969)

LHVT:

1),(|),(),(|1 cbEbaEcaES

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Optical experiment and result

Aspect 1982 two channel polarizer.

PRL49,91(1982)2.697 0.015S

Experiment with pairs of photons produced with PDC

PRL81,3563(1998)

All these experiments conform the QM!

22)009.085.0( S W. Tittel, et al

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3.Bell Inequality in Particle physics Test BI with fermions or massive particles. Test BI with interactions other than electromagnet

ic interactions. Strong or Weak actions. Energy scale of photon case is eV range. Nonlocal

effects may well become apparent at length scale about cm.

1310 S.A. Abel et al. PLB 280,304 (1992)

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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.

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}{1 00 KqKpN

K s

}{1 00 KqKpN

KL

}{2

1 001 KKK

}{2

1 002 KKK

Mass eigenstates CP eigenstates S eigenstates

00 KKS

00 KKS

Like the photon case they don’t commutate

}{2

11 0000, KKKK 0K 0K Are regard as the quasi-spin states.

)()cos(),( rl ttlr etmttE

2

iMH LSLSLS KKH SLSLLS

im

2

Berltamann, Quant-ph/0410028

Note that the H is not a observable [not hermitian]

Fix the quasi-spin and free in time.

1

2

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Experiment of 00BB system

A.Go. J.Mod.Opt.51,991.

lDBlDB *0*0 ,

They use 2S 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

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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)

.,

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GHZ states: three spin half particles.

)111000(2

1CBACBAABC

ABCABC

Cx

Bx

Ax

ABCABC

Cx

By

Ay

ABCABC

Cy

Bx

Ay

ABCABC

Cy

By

Ax

1

1

1

1

Cx

Bx

Ax

Cx

By

Ay

Cy

Bx

Ay

Cy

By

Ax

mmm

mmm

mmm

mmm

“No reasonable definition of reality could be expected to permit this.”

(1)

(2)

(3)

(4)

)3()2()1( 1)()()( 222 Cy

By

Ay

Cx

Bx

Ax mmmmmm

which contradict (4).1 Cx

Bx

Ax mmm

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Kochen-Specher: two spin one particles.

2222 zyx SSS(1). Any orthogonal frame (x, y, z), 0 happens exactly once.

(2). Any orthogonal pair (d, d’), 0 happens at most once.

)1,1,1()1,0,0(

)0,0,1()0,1,1()1,1,0(

)0,1,1()1,1,0()1,1,1(

76

543

210

aa

aaa

aaa

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

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L. Hardy: two spin one half particles.

0.1 FG 0)1(.2 GD

0)1(.3 EF 0.4 DE

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

mep i sincos

1tan If there exist four projection operators satisfy:

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)]'()['()'()()()]([)()'( bBIaAbBaAbBaAIbBaA

1 234

0.1 FG 0)1(.2 GD

0)1(.3 EF 0.4 DE

Compare to previous page:

0)(*,,*)'()','(),'()',(),(1 bPaPbaPbaPbaPbaP

0)()'()'()'(

)()'()'()()()(1

bIBIaAbBaA

bBaAbBaAbBaA

Consider the CH inequality: PRD10, 526 (1974)

PRA52, 2535 (1995) Garuccio

Eberhard Inequality

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Hardy type experiment with entangled Kaon pairs Generate a asymmetric state. Eberhard’s inequality (EI).

][2

1)(

2 LLSLLS KRKKKKKR

T

)()()()( 0000 KKPKKPKKPKKP LLRSSLRLLRLR

PRL88, 040403 (2002),

PRL89, 160401 (2002).

Quant-ph/05011069.

A. Bramon, and G. Garbarino:

A. Bramon, R. Escribano and G. Garbarino:

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][2

1)( LLSSSLLS KrKKrKKKKKt

To generate the asymmetric state, fix a thin regenerator on the right beam close to decay points. Then the initial state:

][2

1SLLS KKKK

Becomes:

Let this state propagate to a proper time T:

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Normalize it to the surviving pairs leads to:

][2

12 LLSLLS KRKKKKK

R

.Re 2)( TTmmii

LSSL

reR

where

]

[2

)()(

)(2

1)(

2

1

SS

TmTiLL

TmTi

SLLS

KKereKKere

KKKKTN

T

SLLS

component has been enhanced.

has been further suppressed.

LLKK

SS KK

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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.

][2

1)(/ 2

)(

2 LL

TTmmi

SLLS KKreKKKKR

TJLS

SL

][2

1)0(/ 0000 KKKKJ

In the charm factory the entanglement state formed as:

.Re 2)( TTmmii

LSSL

re

where

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)2(4

Re2),( 2

2

00

RKKP

i

QM

)2(2

Re1),( 2

2

0

RKKP

i

LQM

)2(2

Re1),( 2

2

0

RKKP

i

LQM

0),( SSQM KKP

The four joint measurement of the transition probability needed in the EI predicted by QM take the following form:

20000 )(/),( TJKKKKPQM

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.)2(2

)1(0

)2(2

)1(

)2(4

)2(2

2

2

2

2

2

R

R

R

R

R

R

Take into EI:

)]()()([)( 0000 KKPKKPKKPKKPVD LLRSSLRLLRLREI

See Figure 1

0 0(for See figure 3)

where VDis the violation degree of the inequality.

From QM we have: First assume

.)2(4

432

2

R

RRVDEI

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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.

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The advantage of over

22.0K

94.0K

factory

/J Charm factory

Space-like separation required:

/J

4.010 632

eerRTS

31.032 1010

eerRTS

11 ST In

To make sure the misidentification of is of order per thousandS LK with K

ST 5Properly choose PRL88, 040403 (2002)

0.15 ST

4.0R

310R

so

so

has a wider region of R in discriminating QM from LHVT./J

There is phenomena can be test due to this advantage.

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Quantify the entanglement

TLSerRJJJC

)(222

2

2

2~/

~/)/(

PRL80, 2245 (1998)W. Wootters*21 )/(

~/

~ JJ yyWhere:

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.

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2 ( , ) - ( , ) ( , ) ( , ) 2S E a d E a b E c b E d c

)1(22 CS

Express the violation in degree of entanglement

PLA154,201(1991)

Abouraddy et al.

N.Gisin

PRA64,050101,(2001)

)2(4

432

2

R

RRVDEI

2

2

2

RC

4

22)1(3 2CCCVDEI

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 .1 2x

2)1(2 CVDCHSH

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The Entanglement and Bell inequality violation

Magnitudes below zero of VD is the range of violation

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Thank you for your patience.