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”

Bell Correlation and experiment

( ), ( ) { 1,1} '

( , ) (

:

) ( )

A a B b LHV

E a b A a

s

B b d

λ

λ λ λλ

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 ra s A s B b

→ ← → ←

↑ ↓ ↓ ↑

∈ ∈ − ∈ − ∈

2

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≤ =

{ }{ }{ }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 ya 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 λλ

ρ λ∈Λ

− = −∫

( , )&( , )a b x y

4

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 λλ

ρ λ∈Ω

− = −∫

0 0( , ) 0E a b = 0 0( , ) ( , )a b a b=

0 0 0

12

( , , , )

( , ) ( ) ( ) (2)a b x y

E x y A x B y dλ λ λλ

ρ λ∈Ω

= ∫

5

Integral forms: consistency condition

Hence,

0 0( , ) 0E a b =

6

0 0 0 0

12

( , , , ) ( , , , )

( , ) (3)a b x y a b x y

E x y d dλ λλ 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 λλ

λ λλ λ

ρ λ

ρ λ ρ λ+ −

∈Ω

∈Ω ∈Ω

= +

+ − =

∫

∫ ∫

Local HVs in

1 2λ λ λρ ρ ρ= 1 2( , )λ λ λ=

1λ 2λ

0 0 0

12

( , , , )

( , ) ( ) ( )a b x y

E x y A x B y dλ λ λλ

ρ λ∈Ω

= ∫

1 1 12 2 2

1 12 2

, [ , ](4)

0, [ , ]j

j

jλ

λρ

λ

∈ −⎧⎪= ⎨

∉ −⎪⎩

7

8

1 2

1 2 0 0 0

1 2( , ) ( , , , )

( , ) ( ) ( )a b x y

E x y A x B y d dλ λλ λ

λ λ∈Ω

= ∫∫

1 11 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 dλ λ λ λ

λ λ λ λ+ −∈Ω ∈Ω

= −∫∫ ∫∫

Settings for

1 2

1 2 0 0 0

1 2( , ) ( , , , )

( , ) ( ) ( )a b x y

E x y A x B y d dλ λλ λ

λ λ∈Ω

= ∫∫

1 (1,0,0)A = 2 (0,1,0)A = ( )1 12 2

1 , ,0B−= ( )1 1

2 22 , ,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

Locality

1 2

1 2 0 0 0

1 2( , ) ( , , , )

( , ) ( ) ( )a b x y

E x y A x B y d dλ λλ λ

λ λ∈Ω

= ∫∫

Se#ng  for  A   Interval  for  the  hidden  variable  

1A 11 1,12 2

I −⎡ ⎤= −⎢ ⎥⎣ ⎦

2A 21 11 ,2 2

I ⎡ ⎤= − +⎢ ⎥⎣ ⎦

Se#ng  for  B   Interval  for  the  hidden  variable  

1B 11 ,02

J −⎡ ⎤= ⎢ ⎥⎣ ⎦

2B 210,2

J ⎡ ⎤= ⎢ ⎥⎣ ⎦

1λ 2λ

10

Measurement functions for

[ ]1 1

1

1 11 \

( ) { 1,1},( )

( ) ,I

I

xA x

sign xλ λ

λλ

α

ζ λ∈

∈Λ

∈ − ∀⎧⎪= ⎨

− ∀⎪⎩

g

g

[ ]2 2

2

2 22 \

( ) { 1,1},( )

( ) ,J

J

yB y

sign yλ λ

λλ

β

η λ∈

∈Λ

∈ − ∀⎧⎪= ⎨

− ∀⎪⎩

g

g

1( )xλα

2( )yλβ

11

( , )x y ∈ϒ

( )xζ ( )yη

Probability of LHV E for

( )1 2Pr (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

The integral 1 2

1 2 0 0 0

1 2( , ) ( , , , )

( , ) ( ) ( )a b x y

E x y A x B y d dλ λλ 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 λλ λ

λ λλ λ

λ λλ λ

λ λ

λ λ

λ λ

∈ ×

∈ Λ ×

∈ Λ ×

= +

+

∫∫

∫∫

∫∫

13

A picture

( )1 12 2,2λ

α ( )1 12 2, −

( )1 12 2,−

( )1 12 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−

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 BI J

AI J

A BI 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 ×

= − +

− +

− −

∫∫

∫∫

∫∫

12

( , ) ( ) ( ) ( ) ( ) (5)E x y U y V x U y V xα β= + +

15

V integral

[ ]12

11 1 2 2

1(2 )

1 1 1 1\ (2 )

(2 ) (2 ) 2 (2 ) 1.A

A

A A AI

V sign d d dζ

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

[ ]12

11 1 1 2

(1 )

1 1 1 1\ 1 (1 )

(1 ) (1 ) 2 (1 ) 1A

A

A A AI

V sign d d dζ

λ ζ

ζ λ λ λ λ ζ∈Λ −

= − = − = −∫ ∫ ∫

1 1 11 1 1 1,1 \ 1 ,2 2 2 2

I I−⎡ ⎤ ⎛ ⎤= − ⇒Λ = −⎜⎢ ⎥ ⎥⎣ ⎦ ⎝ ⎦

2 1 21 1 1 11 , \ , 12 2 2 2

I I⎡ ⎤ ⎡ ⎞= − + ⇒Λ = − − + ⎟⎢ ⎥ ⎢⎣ ⎦ ⎣ ⎠

{ }1 ,2 ( )A Ax xζ∈ ∧

U integral

17

[ ]12

2 2

(1 )1

2 2 2 2 20 (1 )

(1 ) (1 ) 2 (1 ) .B

B

B B BJ

U sign d d dη

λ η

η λ λ λ λ η∈

= − = − = −∫ ∫ ∫

[ ]1

2 1 2

(2 ) 01

2 2 2 2 2(2 )

(2 ) (2 ) 2 (2 ) .B

B

B B BJ

U sign d d dη

λ η

η λ λ λ λ η∈ −

= − = − = +∫ ∫ ∫

1 12 2

[ , ] ( 0.70711 , 0.70711)U −∈ ≈ − { }1 ,2 ( )B By yη∈ ∧

Numerical analysis 1

2Pr (1 ,1 ) 0A BE −⎡ ⎤= >⎣ ⎦

12 2

(1 ) (1 ) (1 ) (1 )B A B AU V U V αα− + = −

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  

( )12 2

( , )U V U V UV αδ α −= − + −

( , )U Vδ

(1 ,1 )A B

18

Numerical analysis 1

2Pr (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  

( )12 2

( , )U V U V UV αδ α −= − + −

( , )U Vδ

(1 ,1 )A B

19

12 2

'(1 ) '(1 ) '(1 ) '(1 )B A B AU V U V αα− + = −

20

Hence:

102

Pr (1 ,1 ) | 0A BE LHV−⎡ ⎤≈ Ω >⎣ ⎦

U V -­‐0.453710      0.218186  

U’ V’    0.32760    -­‐0.366910  

1α = 1α = −

Can we have: 12

Pr (1 ,2 ) 0A BE⎡ ⎤= >⎣ ⎦

( )1 2Pr ( ) ( ) 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

Picture on (1 ,2 )A B

( )1 12 2,2λ

1λ

( )1 12 2, −

( )1 12 2,−

( )1 12 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−

Numerical analysis 1

2Pr (1 ,2 ) 0A BE⎡ ⎤= >⎣ ⎦

12 2

''(2 ) ''(1 ) ''(2 ) ''(1 )B A B AU V U V αα+ + =

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  

( )12 2

( , )U V U V UV αδ α= + + −

( , )U Vδ

(1 ,2 )A B

23

Numerical analysis 1

2Pr (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  

( )12 2

( , )U V U V UV αδ α= + + −

( , )U Vδ

(1 ,2 )A B

24

12 2

'''(2 ) '''(1 ) '''(2 ) '''(1 )B A B AU V U V αα+ + =

25

Hence:

102

Pr (1 ,2 ) | 0A BE LHV⎡ ⎤≈ Ω >⎣ ⎦

U’’ V’’ 0.300100     0.404200  

U’’’ V’’’ -­‐0.677100   -­‐0.02160  

1α = 1α = −

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 ) 1A A

A A

VV

ζ

ζ

= −

= +

12

12

(1 ) 2 (1 )

(2 ) 2 (2 )B B

B B

U

U

η

η

= −

= +

V and U dice.

27

1α = −1α =

1A

1V V= 2V V ʹ′=

1α = −1α =

2 ( )1A A

3V V ʹ′ʹ′= 4V V ʹ′ʹ′ʹ′=

1β = −1β =

1B

2U U ʹ′= 1U U=

1β = −1β =

2 (1 )B B

3U U ʹ′ʹ′= 4U U ʹ′ʹ′ʹ′=

Coin-1

Coin-1

Coin-2

Coin-2

4–sided Dice

4–sided Dice

12 2

ˆ ˆ ˆ ˆPr ( ) ( ) ( ) ( ) |( , ) (1 ,1 ), 1 0A BU y V x U y V x x yαα α⎡ ⎤− + = − = = ± >⎣ ⎦12 2

ˆ ˆ ˆ ˆPr ( ) ( ) ( ) ( ) |( , ) \{(1 ,1 )}, 1 0A BU y V x U y V x x yαα α⎡ ⎤+ + = ∈ϒ = ± >⎣ ⎦

Operational Test

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′= = =

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