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Geurdes Monte Växjö

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Han Geurdes' slides of his talk at the Växjö conference “Quantum Theory: Advances and Problems”, Monday 10 June, 2013.

Reference: doi:10.1016/j.rinp.2014.06.002
Results in Physics Volume 4, 2014, pages 81–82
“A probability loophole in the CHSH” by Han Geurdes.

I analyse this alleged disproof of Bell's theorem in my own paper http://arxiv.org/abs/1506.00223

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Geurdes Monte Växjö

  1. 1. Han Geurdes 1 Talk at Växjö conference “Quantum Theory: Advances and Problems” Monday 10 June, 2013 Reference: doi:10.1016/j.rinp.2014.06.002 Results in Physics Volume 4, 2014, pages 81–82 “A probability loophole in the CHSH”
  2. 2. Bell Correlation and experiment ( ), ( ) { 1,1} ' ( , ) ( : ) ( ) A a B b LHV E a b A a s B b d λ λ λ λ λ λ ρ λ λ ∈Λ − Λ = ∈ ∈ ∫ {1, 2} Re ( ) { 1,1} Re ( ) { 1,1} {1, 2} A X Y Ba S S bAlice A S B Bob r r a s A s B b → ← → ← ↑ ↓ ↓ ↑ ∈ ∈ − ∈ − ∈ 2
  3. 3. CHSH contrast (1 ,1 ) (1 ,2 ) (2 ,1 ) (2 ,2 ) 2A B A B A B A BS E E E E= − − − ≤ { }Pr | | 2| 0S LHVs> = ⇔ ∴ { }Pr | | 2| 1.S LHVs≤ =
  4. 4. { } { } { }0 0 ( , , , ) | ( ) ( ) ( ) ( ) 1 ( , , , ) | ( ) ( ) ( ) ( ) 1 ( , , , ) | ( ) ( ) ( ) ( ) 1 ( , , , ) ( , , , ) ( , , , ) a b x y A a B b A x B y a b x y A a B b A x B y a b x y A a B b A x B y a b x y a b x y a b x y λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ + − + − Ω = ∈Λ = = + Ω = ∈Λ = = − Ω = ∈Λ = − = ± Λ = Ω Ω ΩU U { }( , ) ( , ) ( ) ( ) ( ) ( ) (1)E a b E x y A a B b A x B y dλ λ λ λ λ λ ρ λ ∈Λ − = −∫ ( , )&( , )a b x y 4
  5. 5. Integral forms { } 0 ( , , , ) ( , ) ( , ) ( ) ( ) ( ) ( ) a b x y E a b E x y A a B b A x B y dλ λ λ λ λ λ ρ λ ∈Ω − = −∫ 0 0( , ) 0E a b = 0 0( , ) ( , )a b a b= 0 0 0 1 2 ( , , , ) ( , ) ( ) ( ) (2) a b x y E x y A x B y dλ λ λ λ ρ λ ∈Ω = ∫ 5
  6. 6. Integral forms: consistency condition Hence, 0 0( , ) 0E a b = 6 0 0 0 0 1 2 ( , , , ) ( , , , ) ( , ) (3) a b x y a b x y E x y d dλ λ λ λ ρ λ ρ λ + −∈Ω ∈Ω = −∫ ∫ 0 0 0 0 0 0 0 0 0 0 0 ( , , , ) ( , , , ) ( , , , ) ( , ) ( ) ( ) 0 a b x y a b x y a b x y E a b A a B b d d d λ λ λ λ λ λ λ λ ρ λ ρ λ ρ λ + − ∈Ω ∈Ω ∈Ω = + + − = ∫ ∫ ∫
  7. 7. Local HVs in 1 2λ λ λρ ρ ρ= 1 2( , )λ λ λ= 1λ 2λ 0 0 0 1 2 ( , , , ) ( , ) ( ) ( ) a b x y E x y A x B y dλ λ λ λ ρ λ ∈Ω = ∫ 1 1 1 2 2 2 1 1 2 2 , [ , ] (4) 0, [ , ]j j j λ λ ρ λ ∈ −⎧⎪ = ⎨ ∉ −⎪⎩ 7
  8. 8. 8 1 2 1 2 0 0 0 1 2 ( , ) ( , , , ) ( , ) ( ) ( ) a b x y E x y A x B y d dλ λ λ λ λ λ ∈Ω = ∫∫ 1 1 1 22 2 [ , ],j − Λ = Λ = Λ ×Λ 1 2 0 0 1 2 0 0 1 2 1 2 ( , ) ( , , , ) ( , ) ( , , , ) ( , ) a b x y a b x y E x y d d d d λ λ λ λ λ λ λ λ + −∈Ω ∈Ω = −∫∫ ∫∫
  9. 9. Settings for 1 2 1 2 0 0 0 1 2 ( , ) ( , , , ) ( , ) ( ) ( ) a b x y E x y A x B y d dλ λ λ λ λ λ ∈Ω = ∫∫ 1 (1,0,0)A = 2 (0,1,0)A = ( )1 1 2 2 1 , ,0B − = ( )1 1 2 2 2 , ,0B − − = { }(1 ,1 ),(1 ,2 ),(2 ,1 ),(2 ,2 )A B A B A B A B = ϒ { } { }0 01 ,2 ,1 ,2 , 1 ,2 ,1 ,2A A B B A A B Ba b∉ ∉ 9
  10. 10. Locality 1 2 1 2 0 0 0 1 2 ( , ) ( , , , ) ( , ) ( ) ( ) a b x y E x y A x B y d dλ λ λ λ λ λ ∈Ω = ∫∫ Se#ng  for  A   Interval  for  the  hidden  variable   1A 1 1 1 ,1 2 2 I −⎡ ⎤ = −⎢ ⎥ ⎣ ⎦ 2A 2 1 1 1 , 2 2 I ⎡ ⎤ = − +⎢ ⎥ ⎣ ⎦ Se#ng  for  B   Interval  for  the  hidden  variable   1B 1 1 ,0 2 J −⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ 2B 2 1 0, 2 J ⎡ ⎤ = ⎢ ⎥ ⎣ ⎦ 1λ 2λ 10
  11. 11. Measurement functions for [ ] 1 1 1 1 11 ( ) { 1,1}, ( ) ( ) , I I x A x sign x λ λ λ λ α ζ λ ∈ ∈Λ ∈ − ∀⎧⎪ = ⎨ − ∀⎪⎩ g g [ ] 2 2 2 2 22 ( ) { 1,1}, ( ) ( ) , J J y B y sign y λ λ λ λ β η λ ∈ ∈Λ ∈ − ∀⎧⎪ = ⎨ − ∀⎪⎩ g g 1 ( )xλα 2 ( )yλβ 11 ( , )x y ∈ϒ ( )xζ ( )yη
  12. 12. Probability of LHV E for ( )1 2 Pr (1 ) (1 ) 1 0A Bλ λα β = − > ( )0 0 0 0 1 1Pr ( , , , ) & ( , , , ) 0a b x y a b x y I J+ −Ω = ∅ Ω = × > ( )0 0 0 1 1 1 1 1 2 1 2Pr ( , , , ) (( ) ) (( ) ) ( ) 0a b x y I J I J I JΩ = Λ × Λ × × >U U ( , ) (1 ,1 )A Bx y = 12
  13. 13. The integral 1 2 1 2 0 0 0 1 2 ( , ) ( , , , ) ( , ) ( ) ( ) a b x y E x y A x B y d dλ λ λ λ λ λ ∈Ω = ∫∫ ( , ) (1 ,1 )A BE x y E= 1 2 1 2 1 2 1 2 1 2 1 1 1 1 2 1 2 1 1 2 1 2 ( , ) 1 2 ( , ) ( ) 1 2 ( , ) ( ) ( , ) ( ) ( ) ( ) ( ) ( ) ( ) I J I J I J E x y A x B y d d A x B y d d A x B y d d λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ λ ∈ × ∈ Λ × ∈ Λ × = + + ∫∫ ∫∫ ∫∫ 13
  14. 14. A picture ( )1 1 2 2 ,2λ α ( )1 1 2 2 , − ( )1 1 2 2 ,− ( )1 1 2 2 ,− − 0 0 1 1( , , , )a b x y I J−Ω = × 0 0 0 1 1( , , , ) ( )a b x y I JΩ = Λ × ( ), (1 ,1 )A Bx y = 0 0 0 1 2( , , , )a b x y I JΩ ⊃ × 1 1 2( )I JΛ × 1 1 1( )I JΛ × 14 1λ ( )1( )sign xζ λ− β ( )2( )sign yη λ−
  15. 15. The covariance integral [ ] [ ] [ ] [ ] 1 2 1 2 1 2 1 1 1 1 2 1 1 2 2 1 2 ( , ) 1 1 2 ( , ) ( ) 1 2 1 2 ( , ) ( ) (1 ,1 ) (1 ) (1 ) (1 ) (1 ) A B B I J A I J A B I J E sign d d sign d d sign sign d d λ λ λ λ λ λ α η λ λ λ ζ λ β λ λ ζ λ η λ λ λ ∈ × ∈ Λ × ∈ Λ × = − + − + − − ∫∫ ∫∫ ∫∫ 1 2 ( , ) ( ) ( ) ( ) ( ) (5)E x y U y V x U y V xα β= + + 15
  16. 16. V integral [ ] 1 2 1 1 1 2 2 1(2 ) 1 1 1 1 (2 ) (2 ) (2 ) 2 (2 ) 1. A A A A A I V sign d d d ζ λ ζ ζ λ λ λ λ ζ − + ∈Λ − = − = − = +∫ ∫ ∫ 2 (1 [1 2, 2 1] ( 0.414214 , 0.414 ) 1 [1 2, 2 1] 2 (2 ) 1 [1 2, 2 1] 214) A A V ζ ζ ∈ − − ∈ − − + ∈ − − − − ≈ 16 [ ] 1 2 1 1 1 1 2 (1 ) 1 1 1 1 1 (1 ) (1 ) (1 ) 2 (1 ) 1 A A A A A I V sign d d d ζ λ ζ ζ λ λ λ λ ζ ∈Λ − = − = − = −∫ ∫ ∫ 1 1 1 1 1 1 1 ,1 1 , 2 2 2 2 I I −⎡ ⎤ ⎛ ⎤ = − ⇒ Λ = −⎜⎢ ⎥ ⎥ ⎣ ⎦ ⎝ ⎦ 2 1 2 1 1 1 1 1 , , 1 2 2 2 2 I I ⎡ ⎤ ⎡ ⎞ = − + ⇒ Λ = − − + ⎟⎢ ⎥ ⎢ ⎣ ⎦ ⎣ ⎠ { }1 ,2 ( )A Ax xζ∈ ∧
  17. 17. U integral 17 [ ] 1 2 2 2 (1 ) 1 2 2 2 2 2 0 (1 ) (1 ) (1 ) 2 (1 ) . B B B B B J U sign d d d η λ η η λ λ λ λ η ∈ = − = − = −∫ ∫ ∫ [ ] 1 2 1 2 (2 ) 0 1 2 2 2 2 2 (2 ) (2 ) (2 ) 2 (2 ) . B B B B B J U sign d d d η λ η η λ λ λ λ η ∈ − = − = − = +∫ ∫ ∫ 1 1 2 2 [ , ] ( 0.70711 , 0.70711)U − ∈ ≈ −{ }1 ,2 ( )B By yη∈ ∧
  18. 18. Numerical analysis 1 2 Pr (1 ,1 ) 0A BE −⎡ ⎤= >⎣ ⎦ 1 2 2 (1 ) (1 ) (1 ) (1 )B A B AU V U V α α− + = − 1α = U   V   Step  size  h   -­‐0.60711   0.075786   0.01   0.0004   -­‐0.45511   0.215786   0.001   9.1x10-­‐6   -­‐0.45371   0.218186   0.0001   9.9x10-­‐7   ( )1 2 2 ( , )U V U V UV α δ α − = − + − ( , )U Vδ (1 ,1 )A B 18
  19. 19. Numerical analysis 1 2 Pr (1 ,1 ) 0A BE −⎡ ⎤= >⎣ ⎦ 1α = − U=U’   V=V’   Step  size  h   0.31000   -­‐0.40421   0.01   2.1x10-­‐5   0.32300   -­‐0.37421   0.001   4.7x10-­‐6   0.32760   -­‐0.36691   0.0001   8.0x10-­‐7   ( )1 2 2 ( , )U V U V UV α δ α − = − + − ( , )U Vδ (1 ,1 )A B 19 1 2 2 '(1 ) '(1 ) '(1 ) '(1 )B A B AU V U V α α− + = −
  20. 20. 20 Hence: 1 02 Pr (1 ,1 ) | 0A BE LHV−⎡ ⎤≈ Ω >⎣ ⎦ U V -­‐0.453710      0.218186   U’ V’    0.32760    -­‐0.366910   1α = 1α = −
  21. 21. Can we have: 1 2 Pr (1 ,2 ) 0A BE⎡ ⎤= >⎣ ⎦ ( )1 2 Pr ( ) ( ) 1 0x yλ λα β = > ( )0 0 1 2 0 0Pr ( , , , ) & ( , , , ) 0a b x y I J a b x y+ −Ω = × Ω = ∅ > ( )0 0 0 1 1 1 1 1 2 1 1Pr ( , , , ) (( ) ) (( ) ) ( ) 0a b x y I J I J I JΩ = Λ × Λ × × >U U ( , ) (1 ,2 )A Bx y = 21
  22. 22. Picture on (1 ,2 )A B ( )1 1 2 2 ,2λ 1λ ( )1 1 2 2 , − ( )1 1 2 2 ,− ( )1 1 2 2 ,− − 0 0 1 2( , , , )a b x y I J+Ω = × 0 0 0 1 2( , , , ) ( )a b x y I JΩ = Λ × ( ), (1 ,2 )A Bx y = 1 1 2( )I JΛ × 1 1 1( )I JΛ ×0 0 0 1 1( , , , )a b x y I JΩ ⊃ × 22 α β ( )2( )sign yη λ− ( )1( )sign xζ λ−
  23. 23. Numerical analysis 1 2 Pr (1 ,2 ) 0A BE⎡ ⎤= >⎣ ⎦ 1 2 2 ''(2 ) ''(1 ) ''(2 ) ''(1 )B A B AU V U V α α+ + = 1α = U=U’’   V=V’’   Step  size  h   0.3700   0.3258   0.01   5.4x10-­‐4   0.3001   0.4042   0.0001   3.4x10-­‐5   ( )1 2 2 ( , )U V U V UV α δ α= + + − ( , )U Vδ (1 ,2 )A B 23
  24. 24. Numerical analysis 1 2 Pr (1 ,2 ) 0A BE⎡ ⎤= >⎣ ⎦ 1α = − U=U’’’   V=V’’’   Step  size  h   -­‐0.67711   -­‐0.0142   0.01   1.8x10-­‐4   -­‐0.67711   -­‐0.0217   0.0001    4.1x10-­‐5   -­‐0.67710   -­‐0.0216   0.00001   8.0x10-­‐7   ( )1 2 2 ( , )U V U V UV α δ α= + + − ( , )U Vδ (1 ,2 )A B 24 1 2 2 '''(2 ) '''(1 ) '''(2 ) '''(1 )B A B AU V U V α α+ + =
  25. 25. 25 Hence: 1 02 Pr (1 ,2 ) | 0A BE LHV⎡ ⎤≈ Ω >⎣ ⎦ U’’ V’’ 0.300100     0.404200   U’’’ V’’’ -­‐0.677100   -­‐0.02160   1α = 1α = −
  26. 26. 26 U -0.45371 0.13669 -0.58041 0.32760 0.51735 -0.18975 0.3001 0.50360 -0.20350 -0.67710 0.01500 -0.69210 (1 )Bη (2 )Bη V 0.218186 0.60914 -0.39086 -0.36691 0.31654 -0.68345 0.4042 0.7021 -0.2979 -0.00216 0.50108 -0.49892 (1 )Aζ (2 )Aζ (1 ) 2 (1 ) 1 (2 ) 2 (2 ) 1 A A A A V V ζ ζ = − = + 1 2 1 2 (1 ) 2 (1 ) (2 ) 2 (2 ) B B B B U U η η = − = + V and U dice.
  27. 27. 27 1α = −1α = 1A 1V V= 2V Vʹ′= 1α = −1α = 2 ( )1A A 3V Vʹ′ʹ′= 4V Vʹ′ʹ′ʹ′= 1β = −1β = 1B 2U Uʹ′= 1U U= 1β = −1β = 2 (1 )B B 3U Uʹ′ʹ′= 4U Uʹ′ʹ′ʹ′= Coin-1 Coin-1 Coin-2 Coin-2 4–sided Dice 4–sided Dice 1 2 2 ˆ ˆ ˆ ˆPr ( ) ( ) ( ) ( ) |( , ) (1 ,1 ), 1 0A BU y V x U y V x x yα α α⎡ ⎤− + = − = = ± >⎣ ⎦ 1 2 2 ˆ ˆ ˆ ˆPr ( ) ( ) ( ) ( ) |( , ) {(1 ,1 )}, 1 0A BU y V x U y V x x yα α α⎡ ⎤+ + = ∈ϒ = ± >⎣ ⎦ Operational Test
  28. 28. Conclusion. [ ]Pr | | 2| 0 (6)S LHVs> > This confirms the two coin conclusion from the consistency condition. We may use The probability has got nothing to do with measurement error. 28 0 0 0 0( , ) ( , ) ... 0.E a b E a bʹ′ ʹ′= = =
  29. 29. Appendix. { } { } ( ) , (1 ) (2 ), (2 ) (1 ), (2 ) (2 ) ( ) , (1 ) (1 ), , (1 ) (1 ) A B A B A B A B A B dice I J I J I J dice I J I J + − Ω = ∅ × × × Ω = ∅ × ∅ × 29

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