The document presents a probability loophole in the CHSH inequality. It shows that if measurement settings for Alice (A) and Bob (B) are restricted to intervals on the real number line, then the probability that a local hidden variable (LHV) model reproduces the quantum prediction of E(a,b) = -1 for the CHSH setting (a,b) = (1,1) is greater than zero. This is due to the existence of LHV models where the integrals evaluating E(x,y) for different settings are equal to each other, violating the CHSH inequality.

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. 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. 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. { }
{ }
{ }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. 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. 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. 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
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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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
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. 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. 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. 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. 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
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
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
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. 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. 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