Summary Introduction: the EPR argument and the J. Bell test EPR - PowerPoint PPT Presentation

Measurement of EPR- type flavour entanglement in _ ϒ (4S)  B 0 B 0 decays A. Bay Ecole Polytechnique Fédérale de Lausanne quant-ph/0702267 accepted for publication in PRL 1 TaM August 2007

Summary Introduction: the EPR argument and the J. Bell test EPR correlations with neutral B mesons (and Kaons) - production of entangled B 0 B 0 states - is it possible to Bell test ? EPR correlation studies by the BELLE experiment - tests of two specific "local realistic" models - "New Physics" searches: decoherence of entangled states 2 TaM August 2007

The EPR argument (1935) •A "complete theory" contain an element for each element of reality • EPR consider an entangled two particle system and the measurement Ψ of two non-commuting observables (position and momentum) • Entanglement to transport the information from one sub-system to the other • EPR identify a contradiction with the QM rule for non-commuting observables. 3 TaM August 2007

Bohm (1951), entangled states Bohm analysis of the EPR : two spin 1/2 particles from singlet state y | Ψ 1,2 〉 = (1/ √ 2) ( | ↑〉 1 | ↓〉 2 − | ↓〉 1 | ↑〉 2 ) spin I II z analyser Particle 1 Particle 2 Source x * The two spin are entangled: a measurement S x = +1/2 of the spin projection //x for particle 1 implies that we can predict the outcome of a measurement for 2 : S x = − 1/2. * This will happen even if the decision to orient the polarizer for particle 1 is done at the very last moment => no causal connection. How to explain this ? - with the introduction of a new instantaneous communication channel between the two sub-systems ... - or with the introduction of some new hidden information for particle 2 , so that the particle knows how to behave. => QM is incomplete. 4 TaM August 2007

J. S. Bell theorem (1964) This problem was revitalized in 1964, when Bell suggested a way to distinguish QM from local models featuring hidden variables (J. S. Bell, Physics 1 , 195 (1964)). The Bell theorem : Local hidden variable theories cannot reproduce all possible (statistical) results of QM. Extended by J. Clauser, M. Horne, A. Shimony, and R. Holt, Phys. Lett. 23 , 880 (1969). Several experiments have been done with photon pairs, atoms,... 5 TaM August 2007

Bell-CHSH J. Clauser, M. Horne, A. Shimony, and R. Holt, Phys. Lett. 23 , 880 (1969) a, b (and b',c) : orientations of the two analyzers λ : additional information, i. e. hidden variables A( a , λ ), B( b , λ ) : results of the measurements (+1 or − 1) Correlation function : ∫ E(a,b) ≡ A(a, λ )B(b, λ ) ρ ( λ )d λ Γ ρ ( λ ) is the normalized probability distribution . . . with products like B( b , λ )B( c , λ ) . . . can be violated by QM gives E(a,b) − E(a,c) + E(b',b) + E(b',c) ≤ 2 If E(a,b) depends only on α = a − b, with β = c − b , γ = b − b' E( α ) − E( α + β ) + E( γ ) + E( β + γ ) ≤ 2 6 TaM August 2007

Bell-CHSH experiment by Aspect et al. Experiment by A. Aspect, P. Grangier, G. Roger, PRL 4 9 , 91 (1981): measurement of the correlations of the linear polarization of two photons from a Ca40 source: r r r r r E r r r r r ( ) = P ( ) + P ( ) − P ( ) − P ( ) a , b a , b a , b a , b a , b ++ −− + − − + is the correlation coefficient of the measurements on the two photons with r r the directions of the two analyzers a , b r r ( ) the probabilities of obtaining a ±1 result along direction a (particle 1) P a , b ±± and b (particle 2). r r QM predicts E( δ ) = cos(2 δ ), with δ the angle between the two directions a , b Assuming local realism we have − 2 ≤ S ≤ 2 r r r r S = E ( r ) − E ( r ') + E ( r ) + E ( r a , b a , b a ', b a ', b ') where 7 TaM August 2007

Aspect et al. apparatus Two polarimeters with orientations a and b perform dichotomic measurements of linear polarization of the 2 photons ( ν 1 , ν 2 ) from a Ca40 source. The apparatus registers single rates and coincidence rates. 8 TaM August 2007

Bell-CHSH test by Aspect et al. r r a b r r They choose the following optimal configuration a ' b ' r r r r ( r ) = ( r ) = ( r ') ≡ φ ( r a , b a ', b a ', b a , b ') = 3 φ 22.5 0 S giving S( φ ) = 3E( φ ) − E(3 φ ) 2 2 QM LR limit The QM maximal value is S(22.5 o ) = 2 √ 2=2.83 The LR limit is |S| ≤ 2 φ [ deg ] Needs to account for detection efficiency : Aspect et al. estimate the correlations for a given angle ( δ = φ or δ =3 φ ): E ( δ ) = ( R ++ + R −− ) − ( R + − + R − + ) ( R ++ + R −− ) + ( R + − + R − + ) ( δ ) The experiment gives: S( φ =22.5 o )=3E (22.5 ο ) −Ε (67.5 ο ) = 2.697±0.015 > 2 9 TaM August 2007

EPR correlations in high energy experiments In the '60 Lee and Yang recognized the EPR behaviour of the neutral Kaon doublet in a J PC = − 1 state: here the "strangeness" S= + 1 or S= − 1 of the two Kaons plays the role of spin up or down. K 0 K 0 * Tests have been carried out on correlated Apostolakis et al. , CPLEAR collab., Phys. Lett. B 422 , 339 (1998) Ambrosino et al. , KLOE collab., Phys. Lett. B 642 , 315 (2006). In B factories we have the opportunity to study the flavour entanglement in B 0 pairs from ϒ (4S) → B 0 B 0 10 TaM August 2007

K 0 and B 0 K 0 = ds B 0 = db m~0.5 GeV/c 2 m~5 GeV/c 2 0 = d 0 = d K s B b 1 0 = Normalization p K 0 − q K [ ] 0 K S ~ 0 = ... B H ~ 1 0 = Normalization p K 0 + q K [ ] ~ 0 0 = ... K L B L ~ p = 1 + ε q = 1 − ε ε is the CP violating parameter ~10 − 3 lifetime K 0 Short τ = 0.9 10 − 10 s ~identical lifetimes τ = 1.5 10 − 12 s K 0 Long τ = 5.2 10 − 8 s c τ ~ 500 µ m flavour oscillation parameter: quite fast oscillation: 0 ≈ 3 × 10 − 10 MeV 0 ≈ 3 × 10 − 12 MeV Δ m K = m K L 0 − m K S Δ m d = m B H 0 − m B L ~ 0.005 1/ps ~ 0.5 1/ps 11 TaM August 2007

B 0 B 0 oscillation d b t 0 W + B W + B 0 d t b Suppose to produce a pure beam of B 0 at t=0, then = 1 } ≈ 1 2 B 0 B 0 (t) 4 e −Γ H t + e −Γ L t + 2e −Γ t cos(t Δ m d ) 2 e −Γ t 1 + cos(t Δ m d ) { { } = 1 } ≈ 1 2 0 B 0 (t) 4 e −Γ H t + e −Γ L t − 2e −Γ t cos(t Δ m d ) 2 e −Γ t 1 − cos(t Δ m d ) { { } B 0 ≈ 3 × 10 − 10 MeV Δ m d = m B H 0 − m B L Γ L ≈ Γ H Γ = 1 ( ) 2 Γ L + Γ H 12 TaM August 2007

Analogy with spin measurement The analogy proposed (by many authors) is Meson spin1/2 photon K or B B 0 V ⇑ z 0 B H ⇓ z 1 { } B H L = V − i H ⇒ y 2 1 { } B L R = V + i H ⇐ y 2 Differences with photons: K and B are unstable, and we cannot test for an arbitrary superposition state a B 0 + b B 0 Kaon case will be presented by Di Domenico, and Sozzi, ... 13 TaM August 2007

Production of flavour entangled states ( 1/2) e + e − → Υ (4 S ) → B 0 B 0 24% Y(4S) 76% continuum Interaction Point extension: Interaction Point extension: 77 µm m σ 2 µm m σ 4 mm σ x ≈ 77 µ σ y ≈ 2 µ σ z ≈ 4 mm σ x ≈ y ≈ z ≈ 8 GeV electrons 3.5GeV positrons KEK-B ⇒ production of Υ (4S) (10.58GeV/c 2 ) βγ = 0.425 Υ (4S) → B 0 B 0 → B + B − 14 TaM August 2007

Production of flavour entangled states ( 2/2) e + e − → Υ (4 S ) → B 0 B 0 From Y(4S) is a resonance J CP = 1 − − . The strong decay conserves C which is transferred to the final state: | Ψ〉 = (1/ √ 2) ( |B 0 〉 1 |B 0 〉 2 − |B 0 〉 1 |B 0 〉 2 ) Because of this a measurement of the flavour (B 0 or B 0 ) of particle 1, tells with 100% probability that, at the same proper time , particle 2 is of the opposite flavour (B 0 or B 0 ). Experimentally, the flavour of a B can only be determined when it decays, B 0 → D * − l + ν and only if it decays into a “flavour-specific” mode, like l + W + the sign of the lepton ν b tells us the neutral B c B 0 flavour d D * − 15 TaM August 2007

Experiment with of correlated B 0 B 0 ϒ (4S) produced with βγ = 0.425 l + → B 0  by the asymmetric collider e, µ  l − → B 0  ϒ (4S) QM: region of B 0 & B 0 Δ z D coherent evolution z 1 z 0 z 2 z B 0 and B 0 oscillate coherently. Than the other B 0 oscillates When the first decays, the other is freely before decaying known to be of the opposite after a time given by flavour, at the same proper time Δ t ≈ Δ z /c βγ N.B. : production vertex position z 0 not very well known : only Δ z is available ! 16 TaM August 2007

QM predictions for entangled pairs Time ( Δ t)-dependent decay rate into two O pposite F lavour (OF) states R OF ∝ e −Γ (t 1 + t 2 ) (1 + cos( Δ m d Δ t)) Δ m d is the mass difference idem, into two S ame F lavour (SF) states of the two mass eigenstates R SF ∝ e −Γ (t 1 + t 2 ) (1 − cos( Δ m d Δ t)) => we obtain the time-dependent asymmetry A QM ( Δ t) = R OF − R SF ( Δ t) = R OF + R SF = cos( Δ m d Δ t) Δ m d = 0.507ps − 1 ( ignoring CP violation effects O(10 -4 ), and taking Γ H = Γ L ) 17 TaM August 2007