Today’s Outline - April 23, 2015segre/phys406/15S/lecture_23.pdfToday’s Outline - April 23, 2015 The Born Series The EPR paradox Problem 12.1 Bell’s inequality The EPR experiment

Today’s Outline - April 23, 2015

• The Born Series

• The EPR paradox

• Problem 12.1

• Bell’s inequality

• The EPR experiment

Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015

Final Exam: Monday, May 4, 201514:00-16:00, Herman Hall Ballroom

C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 1 / 22


• The Born Series

• The EPR paradox

• Problem 12.1







• The Born Series

• The EPR paradox

• Problem 12.1







• The Born Series

• The EPR paradox

• Problem 12.1







• The Born Series

• The EPR paradox

• Problem 12.1







• The Born Series

• The EPR paradox

• Problem 12.1







• The Born Series

• The EPR paradox

• Problem 12.1







• The Born Series

• The EPR paradox

• Problem 12.1






The impulse approximation

θF

b

In classical physics, the impulse approximation assumes that the particlekeeps going in a straight line and a transverse impulse is computed to givethe final scattering angle

I =

∫F⊥ dt −→ θ ∼= tan−1

(I

p

)this first-order correction is analgous to the first Born approximation wherethe incident particle is unaffected

following this analogy, we can generate a series of scattering correctionscalled the Born series



θF

b


I =

∫F⊥ dt −→ θ ∼= tan−1

(I

p





θF

b


I =

∫F⊥ dt

−→ θ ∼= tan−1(I

p





θF

b


I =

∫F⊥ dt −→ θ ∼= tan−1

(I

p

)

this first-order correction is analgous to the first Born approximation wherethe incident particle is unaffected




θF

b


I =

∫F⊥ dt −→ θ ∼= tan−1

(I

p





θF

b


I =

∫F⊥ dt −→ θ ∼= tan−1

(I

p




The Born series

Starting with the integral Schrodinger equation

ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr

rthis can be written as

now replace ψ on the rightside by it’s integral represen-tation

ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ

repeating the process, we have a recursive integral equation

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


The Born series


ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0,

g(r) = − m

2π~2e ikr



ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ


ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


The Born series


ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr

r

this can be written as


ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ


ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


The Born series


ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr



ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ


ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


The Born series


ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr



ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ


ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


The Born series


ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr



ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ


ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


The Born series


ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr



ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ


ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


The Born series


ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr



ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ


ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


The Born series


ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr



ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ


ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


The Born series


ψ(r) = ψ0(r) +

∫g(r − r0)V (r0)ψ(r0) dr0, g(r) = − m

2π~2e ikr



ψ = ψ0 +

∫gVψ

ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ


ψ = ψ0 +

∫gVψ0 +

∫ ∫gVgVψ0 +

∫ ∫ ∫gVgVgVψ0 + · · ·

+V

g + + + . . .ψ

0ψ

0


Quantum paradoxes and other fun stuff

“About your cat, Mr. Schrodinger – I have good news and bad news . . . ”C. Segre (IIT) PHYS 406 - Spring 2015 April 23, 2015 4 / 22

Einstein Podolsky Rosen paradox

“Can quantum-mechanical description of physical reality be consideredcomplete?,” Physical Review 47, 777-779 (1935)

“If, without in any way disturbing a system, we can predictwith certainty (i.e., with probability equal to unity) thevalue of a physical quantity, then there exists an elementof physical reality corresponding to this physical quantity.”

If ψ is an eigenfuction of an opera-tor A, then

for example take momentum as theoperator

Aψ = aψ

ψ = e ip0x/~

p =~i

∂

∂x

pψ =~i

∂

∂xe ip0x/~ =

~i

i

~p0e

ip0x/~ = p0ψ

thus the momentum in state ψ is said to be real







Aψ = aψ

ψ = e ip0x/~

p =~i

∂

∂x

pψ =~i

∂

∂xe ip0x/~ =

~i

i

~p0e

ip0x/~ = p0ψ








Aψ = aψ

ψ = e ip0x/~

p =~i

∂

∂x

pψ =~i

∂

∂xe ip0x/~ =

~i

i

~p0e

ip0x/~ = p0ψ








Aψ = aψ

ψ = e ip0x/~

p =~i

∂

∂x

pψ =~i

∂

∂xe ip0x/~ =

~i

i

~p0e

ip0x/~ = p0ψ








Aψ = aψ

ψ = e ip0x/~

p =~i

∂

∂x

pψ =~i

∂

∂xe ip0x/~ =

~i

i

~p0e

ip0x/~ = p0ψ








Aψ = aψ

ψ = e ip0x/~

p =~i

∂

∂x

pψ =~i

∂

∂xe ip0x/~ =

~i

i

~p0e

ip0x/~ = p0ψ








Aψ = aψ

ψ = e ip0x/~

p =~i

∂

∂x

pψ =~i

∂

∂xe ip0x/~ =

~i

i

~p0e

ip0x/~ = p0ψ








Aψ = aψ

ψ = e ip0x/~

p =~i

∂

∂x

pψ =~i

∂

∂xe ip0x/~

=~i

i

~p0e

ip0x/~ = p0ψ








Aψ = aψ

ψ = e ip0x/~

p =~i

∂

∂x

pψ =~i

∂

∂xe ip0x/~ =

~i

i

~p0e

ip0x/~

= p0ψ








Aψ = aψ

ψ = e ip0x/~

p =~i

∂

∂x

pψ =~i

∂

∂xe ip0x/~ =

~i

i

~p0e

ip0x/~ = p0ψ








Aψ = aψ

ψ = e ip0x/~

p =~i

∂

∂x

pψ =~i

∂

∂xe ip0x/~ =

~i

i

~p0e

ip0x/~ = p0ψ




If Aψ = aψ does not hold, how-ever, A cannot be said to have aparticular value as we know fromthe position operator q

qψ = xe ip0x/~ 6= aψ

P(a, b) =

∫ b

aψ∗ψ dx =

∫ b

adx = b − a

there is an equal probability of measuring any value of the position

this is a specific instance of the fact that if two operators do not commute[A,B] 6= 0 then precise knowledge of one precludes such knowledge of theother

the authors thus conclude that

1 the quantum-mechanical description of reality given by the wavefunction is not complete, or

2 when the operators corresponding to two physical quantities do notcommute the two quantities cannot have simultaneous reality




qψ = xe ip0x/~

6= aψ

P(a, b) =

∫ b

aψ∗ψ dx =

∫ b

adx = b − a










P(a, b) =

∫ b

aψ∗ψ dx =

∫ b

adx = b − a










P(a, b) =

∫ b

aψ∗ψ dx

=

∫ b

adx = b − a










P(a, b) =

∫ b

aψ∗ψ dx =

∫ b

adx

= b − a










P(a, b) =

∫ b

aψ∗ψ dx =

∫ b

adx = b − a










P(a, b) =

∫ b

aψ∗ψ dx =

∫ b

adx = b − a










P(a, b) =

∫ b

aψ∗ψ dx =

∫ b

adx = b − a










P(a, b) =

∫ b

aψ∗ψ dx =

∫ b

adx = b − a










P(a, b) =

∫ b

aψ∗ψ dx =

∫ b

adx = b − a










P(a, b) =

∫ b

aψ∗ψ dx =

∫ b

adx = b − a








Suppose that we have two systems, I and II which interact over a timeperiod 0 ≤ t ≤ T

if we know the state of each system before t = 0, then by using theSchrodinger equation, we can calculate the state of the combined systemI+II at any subsequent time, specifically t > T and we call it Ψ

suppose that a physical quantity Ahas eigenvalues and eigenfunctions

then the system-wide wavefunctionas a function of x1 is

where here ψn(x2) are the “coeffi-cients” of the expansion in un(x1)

a1, a2, a3, . . .

u1(x1), u2(x1), u3(x1), . . .

Ψ(x1, x2) =∞∑n=1

ψn(x2)un(x1)

if A is now measured and is found to have value ak then the first systemmust be left in state uk(x1) and the second system must be, thereforefound in state ψk(x2) so that Ψ = ψk(x2)uk(x1)








a1, a2, a3, . . .

u1(x1), u2(x1), u3(x1), . . .

Ψ(x1, x2) =∞∑n=1

ψn(x2)un(x1)






suppose that a physical quantity Ahas eigenvalues

and eigenfunctions



a1, a2, a3, . . .

u1(x1), u2(x1), u3(x1), . . .

Ψ(x1, x2) =∞∑n=1

ψn(x2)un(x1)






suppose that a physical quantity Ahas eigenvalues

and eigenfunctions



a1, a2, a3, . . .

u1(x1), u2(x1), u3(x1), . . .

Ψ(x1, x2) =∞∑n=1

ψn(x2)un(x1)









a1, a2, a3, . . .

u1(x1), u2(x1), u3(x1), . . .

Ψ(x1, x2) =∞∑n=1

ψn(x2)un(x1)









a1, a2, a3, . . .

u1(x1), u2(x1), u3(x1), . . .

Ψ(x1, x2) =∞∑n=1

ψn(x2)un(x1)









a1, a2, a3, . . .

u1(x1), u2(x1), u3(x1), . . .

Ψ(x1, x2) =∞∑n=1

ψn(x2)un(x1)









a1, a2, a3, . . .

u1(x1), u2(x1), u3(x1), . . .

Ψ(x1, x2) =∞∑n=1

ψn(x2)un(x1)









a1, a2, a3, . . .

u1(x1), u2(x1), u3(x1), . . .

Ψ(x1, x2) =∞∑n=1

ψn(x2)un(x1)









a1, a2, a3, . . .

u1(x1), u2(x1), u3(x1), . . .

Ψ(x1, x2) =∞∑n=1

ψn(x2)un(x1)

if A is now measured and is found to have value ak then the first systemmust be left in state uk(x1) and the second system must be, thereforefound in state ψk(x2)

so that Ψ = ψk(x2)uk(x1)








a1, a2, a3, . . .

u1(x1), u2(x1), u3(x1), . . .

Ψ(x1, x2) =∞∑n=1

ψn(x2)un(x1)




If instead of operator A, we chooseto expand the overall wavefunctionin terms of the eigenfunctions of B,then we have

b1, b2, b3, . . .

v1(x1), v2(x1), v3(x1), . . .

Ψ(x1, x2) =∞∑s=1

ϕs(x2)vs(x1)

if B is measured and found to be br , then the system can be said to be inthe state Ψ = ϕr (x2)vr (x1) and the second system must be in state ϕr (x2)

Thus, as a consequence of two different measurements made on System I,System II can be left in states with two different wave functions, when it isfar away from, and not interacting with System I

Thus one can assign two different wave functions, ψk(x2) and ϕr (x2), tothe same reality (System II after interaction with System I)




b1, b2, b3, . . .

v1(x1), v2(x1), v3(x1), . . .

Ψ(x1, x2) =∞∑s=1

ϕs(x2)vs(x1)







b1, b2, b3, . . .

v1(x1), v2(x1), v3(x1), . . .

Ψ(x1, x2) =∞∑s=1

ϕs(x2)vs(x1)







b1, b2, b3, . . .

v1(x1), v2(x1), v3(x1), . . .

Ψ(x1, x2) =∞∑s=1

ϕs(x2)vs(x1)







b1, b2, b3, . . .

v1(x1), v2(x1), v3(x1), . . .

Ψ(x1, x2) =∞∑s=1

ϕs(x2)vs(x1)







b1, b2, b3, . . .

v1(x1), v2(x1), v3(x1), . . .

Ψ(x1, x2) =∞∑s=1

ϕs(x2)vs(x1)







b1, b2, b3, . . .

v1(x1), v2(x1), v3(x1), . . .

Ψ(x1, x2) =∞∑s=1

ϕs(x2)vs(x1)






Suppose that ψk(x2) and ϕr (x2) are eigenfunctions of two non-commutingoperators, P and Q with eigenvalues pk and qr

By measuring either A or B on System I, we are able to predict withcertainty, and without disturbing System II, either the value of P (pk) orQ (qr ) According to our criterion for reality, in the first case, P must bean element of reality and in the second case, Q must be an element ofreality but ψk(x2) and ϕr (x2) were shown to be part of the same realityand this leads to a contradiction of the postulate that “when the operatorscorresponding to two physical quantities do not commute the twoquantities cannot have simultaneous reality”

Thus, they conclude that “the quantum-mechanical description of realitygiven by the wave function is not complete”

If this “realist” version of quantum mechanics is correct, there must besome local hidden variables which specify the state of the systemcompletely




By measuring either A or B on System I, we are able to predict withcertainty, and without disturbing System II, either the value of P (pk) orQ (qr )

According to our criterion for reality, in the first case, P must bean element of reality and in the second case, Q must be an element ofreality but ψk(x2) and ϕr (x2) were shown to be part of the same realityand this leads to a contradiction of the postulate that “when the operatorscorresponding to two physical quantities do not commute the twoquantities cannot have simultaneous reality”






















Problem 12.1

Consider a two-level system, |φa〉 and |φb〉, with 〈φi |φj〉 = δij

The two-particle state

α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉

cannot be expressed as a product |ψr (1)〉 |ψs(2)〉 for any one-particlestates |ψr 〉 and |ψs〉

Start by assuming

α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉 = |ψr (1)〉 |ψs(2)〉

expand |ψr 〉 and |ψs〉 in terms ofthe two basis states

|ψr 〉 = A |φa〉+ B |φb〉|ψs〉 = C |φa〉+ D |φb〉


Problem 12.1



α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉


Start by assuming

α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉 = |ψr (1)〉 |ψs(2)〉




Problem 12.1



α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉


Start by assuming

α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉 = |ψr (1)〉 |ψs(2)〉




Problem 12.1



α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉


Start by assuming

α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉 = |ψr (1)〉 |ψs(2)〉




Problem 12.1



α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉


Start by assuming

α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉 = |ψr (1)〉 |ψs(2)〉


|ψr 〉 = A |φa〉+ B |φb〉

|ψs〉 = C |φa〉+ D |φb〉


Problem 12.1



α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉


Start by assuming

α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉 = |ψr (1)〉 |ψs(2)〉




Problem 12.1

α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉= [A |φa(1)〉+ B |φb(1)〉] [C |φa(2)〉+ D |φb(2)〉]

= AC |φa(1)〉 |φa(2)〉+ AD |φa(1)〉 |φb(2)〉+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉

taking the inner products:

〈φa(1)| 〈φb(2)| −→ α = AD

〈φa(1)| 〈φa(2)| −→ 0 = AC

〈φb(1)| 〈φa(2)| −→ β = BC

〈φb(1)| 〈φb(2)| −→ 0 = BD

if any of the coefficients (A, B, C , or D) are zero, then either α or β mustbe zero which is a condition excluded by assumption


Problem 12.1

α |φa(1)〉 |φb(2)〉+ β |φb(1)〉 |φa(2)〉= [A |φa(1)〉+ B |φb(1)〉] [C |φa(2)〉+ D |φb(2)〉]= AC |φa(1)〉 |φa(2)〉+ AD |φa(1)〉 |φb(2)〉

+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉


〈φa(1)| 〈φb(2)| −→ α = AD

〈φa(1)| 〈φa(2)| −→ 0 = AC

〈φb(1)| 〈φa(2)| −→ β = BC

〈φb(1)| 〈φb(2)| −→ 0 = BD



Problem 12.1


+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉


〈φa(1)| 〈φb(2)| −→ α = AD

〈φa(1)| 〈φa(2)| −→ 0 = AC

〈φb(1)| 〈φa(2)| −→ β = BC

〈φb(1)| 〈φb(2)| −→ 0 = BD



Problem 12.1


+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉


〈φa(1)| 〈φb(2)|

−→ α = AD

〈φa(1)| 〈φa(2)| −→ 0 = AC

〈φb(1)| 〈φa(2)| −→ β = BC

〈φb(1)| 〈φb(2)| −→ 0 = BD



Problem 12.1


+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉


〈φa(1)| 〈φb(2)| −→ α = AD

〈φa(1)| 〈φa(2)| −→ 0 = AC

〈φb(1)| 〈φa(2)| −→ β = BC

〈φb(1)| 〈φb(2)| −→ 0 = BD



Problem 12.1


+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉


〈φa(1)| 〈φb(2)| −→ α = AD

〈φa(1)| 〈φa(2)|

−→ 0 = AC

〈φb(1)| 〈φa(2)| −→ β = BC

〈φb(1)| 〈φb(2)| −→ 0 = BD



Problem 12.1


+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉


〈φa(1)| 〈φb(2)| −→ α = AD

〈φa(1)| 〈φa(2)| −→ 0 = AC

〈φb(1)| 〈φa(2)| −→ β = BC

〈φb(1)| 〈φb(2)| −→ 0 = BD



Problem 12.1


+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉


〈φa(1)| 〈φb(2)| −→ α = AD

〈φa(1)| 〈φa(2)| −→ 0 = AC

〈φb(1)| 〈φa(2)|

−→ β = BC

〈φb(1)| 〈φb(2)| −→ 0 = BD



Problem 12.1


+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉


〈φa(1)| 〈φb(2)| −→ α = AD

〈φa(1)| 〈φa(2)| −→ 0 = AC

〈φb(1)| 〈φa(2)| −→ β = BC

〈φb(1)| 〈φb(2)| −→ 0 = BD



Problem 12.1


+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉


〈φa(1)| 〈φb(2)| −→ α = AD

〈φa(1)| 〈φa(2)| −→ 0 = AC

〈φb(1)| 〈φa(2)| −→ β = BC

〈φb(1)| 〈φb(2)|

−→ 0 = BD



Problem 12.1


+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉


〈φa(1)| 〈φb(2)| −→ α = AD

〈φa(1)| 〈φa(2)| −→ 0 = AC

〈φb(1)| 〈φa(2)| −→ β = BC

〈φb(1)| 〈φb(2)| −→ 0 = BD



Problem 12.1


+ BC |φb(1)〉 |φa(2)〉+ BD |φb(1)〉 |φb(2)〉


〈φa(1)| 〈φb(2)| −→ α = AD

〈φa(1)| 〈φa(2)| −→ 0 = AC

〈φb(1)| 〈φa(2)| −→ β = BC

〈φb(1)| 〈φb(2)| −→ 0 = BD



Bell’s inequality

“On the Einstein Podolsky Rosen paradox,” J.S. Bell, Physics 1, 195-200(1964).

Consider an experiment wherein a pair of spin one-half particles areprepared in a singlet spin state and are moving freely in oppositedirections.

Suppose that two apparati are prepared to measure the spin state of eachparticle, ~σ1 and ~σ2 by means of Stern-Gerlach magnets which do notinterfere with the other particle.

If we measure the component ~σ1 · a and get +1 then we can presume that~σ2 · a = −1 without doing the measurement, the result is predetermined.

Assume that this predetermination is characterized by parameters λ.

The result, A, of measuring ~σ1 · a is then determined by a and λ and theresults, B, of measuring ~σ2 · b is then determined by b and λ.

A(a, λ) = ±1, B(b, λ) = ±1


Bell’s inequality







A(a, λ) = ±1, B(b, λ) = ±1


Bell’s inequality







A(a, λ) = ±1, B(b, λ) = ±1


Bell’s inequality




If we measure the component ~σ1 · a and get +1

then we can presume that~σ2 · a = −1 without doing the measurement, the result is predetermined.



A(a, λ) = ±1, B(b, λ) = ±1


Bell’s inequality







A(a, λ) = ±1, B(b, λ) = ±1


Bell’s inequality







A(a, λ) = ±1, B(b, λ) = ±1


Bell’s inequality






The result, A, of measuring ~σ1 · a is then determined by a and λ

and theresults, B, of measuring ~σ2 · b is then determined by b and λ.

A(a, λ) = ±1,

B(b, λ) = ±1


Bell’s inequality







A(a, λ) = ±1, B(b, λ) = ±1


Bell’s inequality

Result B for particle 2 is assumed not to depend on a and neither does Adepend on b.

If ρ(λ) is the probability distribu-tion of λ, then the expectationvalue of the product of ~σ1 · a and~σ2 · b is

and it should be equivalent to thequantum expectation value

P(a, b) =

∫ρ(λ)A(a, λ)B(b, λ) dλ

〈~σ1 · a |~σ2 · b〉 = −a · b

The essence of Bell’s paper is that this last statement is not possiblebecause of the presence of the hidden local variable(s) λ


Bell’s inequality




P(a, b) =


〈~σ1 · a |~σ2 · b〉 = −a · b



Bell’s inequality




P(a, b) =


〈~σ1 · a |~σ2 · b〉 = −a · b



Bell’s inequality




P(a, b) =


〈~σ1 · a |~σ2 · b〉 = −a · b



Bell’s inequality




P(a, b) =


〈~σ1 · a |~σ2 · b〉 = −a · b



Bell’s inequality




P(a, b) =


〈~σ1 · a |~σ2 · b〉 = −a · b



Aside: Problem 4.50

Suppose we have 2 spin 12 fermions in a singlet configuration. If Sa1 is the

component of the first particle’s spin angular momentum in the direction aand Sb2 is the component of the second particle’s spin angular momentumin the b direction. Compute 〈Sa1Sb2〉

The singlet state can be written as

if we choose a ≡ z and put b in thex-z plane

|0 0〉 =1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

Sa1 = Sz1

Sb2 = cos θSz2 + sin θSx2

Sa1Sb2 |0 0〉 =

1√2

[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

=1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉

+ sin θSx2 |↓2〉

)

− Sz1 |↓1〉 (cos θSz2 |↑2〉

+ sin θSx2 |↑2〉

)

]


Aside: Problem 4.50





|0 0〉 =1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

Sa1 = Sz1


Sa1Sb2 |0 0〉 =

1√2

[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

=1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉


)

− Sz1 |↓1〉 (cos θSz2 |↑2〉


)

]


Aside: Problem 4.50





|0 0〉 =1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

Sa1 = Sz1


Sa1Sb2 |0 0〉 =

1√2

[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

=1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉


)

− Sz1 |↓1〉 (cos θSz2 |↑2〉


)

]


Aside: Problem 4.50





|0 0〉 =1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

Sa1 = Sz1


Sa1Sb2 |0 0〉 =

1√2

[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

=1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉


)

− Sz1 |↓1〉 (cos θSz2 |↑2〉


)

]


Aside: Problem 4.50





|0 0〉 =1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

Sa1 = Sz1


Sa1Sb2 |0 0〉 =

1√2

[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

=1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉


)

− Sz1 |↓1〉 (cos θSz2 |↑2〉


)

]


Aside: Problem 4.50





|0 0〉 =1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

Sa1 = Sz1


Sa1Sb2 |0 0〉 =

1√2

[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

=1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉


)

− Sz1 |↓1〉 (cos θSz2 |↑2〉


)

]


Aside: Problem 4.50





|0 0〉 =1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

Sa1 = Sz1


Sa1Sb2 |0 0〉 =

1√2

[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

=1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉


)

− Sz1 |↓1〉 (cos θSz2 |↑2〉


)

]


Aside: Problem 4.50





|0 0〉 =1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

Sa1 = Sz1


Sa1Sb2 |0 0〉 =1√2

[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

=1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉


)

− Sz1 |↓1〉 (cos θSz2 |↑2〉


)

]


Aside: Problem 4.50





|0 0〉 =1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

Sa1 = Sz1


Sa1Sb2 |0 0〉 =1√2

[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

=1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉


)

− Sz1 |↓1〉 (cos θSz2 |↑2〉


)

]


Aside: Problem 4.50





|0 0〉 =1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

Sa1 = Sz1


Sa1Sb2 |0 0〉 =1√2

[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

=1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉 + sin θSx2 |↓2〉 )

− Sz1 |↓1〉 (cos θSz2 |↑2〉


)

]


Aside: Problem 4.50





|0 0〉 =1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

Sa1 = Sz1


Sa1Sb2 |0 0〉 =1√2

[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

=1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉 + sin θSx2 |↓2〉 )

− Sz1 |↓1〉 (cos θSz2 |↑2〉


) ]


Aside: Problem 4.50





|0 0〉 =1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

Sa1 = Sz1


Sa1Sb2 |0 0〉 =1√2

[Sz1(cos θSz2 + sin θSx2)] (|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

=1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉 + sin θSx2 |↓2〉 )

− Sz1 |↓1〉 (cos θSz2 |↑2〉 + sin θSx2 |↑2〉 ) ]


Aside: Problem 4.50

Sa1Sb2 |0 0〉 =1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉+ sin θSx2 |↓2〉)

− Sz1 |↓1〉 (cos θSz2 |↑2〉+ sin θSx2 |↑2〉) ]

=1√2

~2

4

{|↑1〉 [− cos θ |↓2〉+ sin θ |↑2〉]

+ |↓1〉 [cos θ |↑2〉+ sin θ |↓2〉]}

=~2

4

[− cos θ

1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

+ sin θ1√2

(|↑1〉 |↑2〉+ |↓1〉 |↓2〉)]

〈0 0|Sa1Sb2 |0 0〉 = −~2

4cos θ 〈0 0 | 0 0〉 = −~2

4cos θ = −~2

4a · b

putting this in terms of the Pauli spin matrices: 〈~σ1 · a |~σ2 · b〉 = −a · b


Aside: Problem 4.50

Sa1Sb2 |0 0〉 =1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉+ sin θSx2 |↓2〉)

− Sz1 |↓1〉 (cos θSz2 |↑2〉+ sin θSx2 |↑2〉) ]

=1√2

~2

4

{|↑1〉 [− cos θ |↓2〉+ sin θ |↑2〉]

+ |↓1〉 [cos θ |↑2〉+ sin θ |↓2〉]}

=~2

4

[− cos θ

1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

+ sin θ1√2

(|↑1〉 |↑2〉+ |↓1〉 |↓2〉)]

〈0 0|Sa1Sb2 |0 0〉 = −~2

4cos θ 〈0 0 | 0 0〉 = −~2

4cos θ = −~2

4a · b



Aside: Problem 4.50

Sa1Sb2 |0 0〉 =1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉+ sin θSx2 |↓2〉)

− Sz1 |↓1〉 (cos θSz2 |↑2〉+ sin θSx2 |↑2〉) ]

=1√2

~2

4

{|↑1〉 [− cos θ |↓2〉+ sin θ |↑2〉]

+ |↓1〉 [cos θ |↑2〉+ sin θ |↓2〉]}

=~2

4

[− cos θ

1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

+ sin θ1√2

(|↑1〉 |↑2〉+ |↓1〉 |↓2〉)]

〈0 0|Sa1Sb2 |0 0〉 = −~2

4cos θ 〈0 0 | 0 0〉 = −~2

4cos θ = −~2

4a · b



Aside: Problem 4.50

Sa1Sb2 |0 0〉 =1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉+ sin θSx2 |↓2〉)

− Sz1 |↓1〉 (cos θSz2 |↑2〉+ sin θSx2 |↑2〉) ]

=1√2

~2

4

{|↑1〉 [− cos θ |↓2〉+ sin θ |↑2〉]

+ |↓1〉 [cos θ |↑2〉+ sin θ |↓2〉]}

=~2

4

[− cos θ

1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

+ sin θ1√2

(|↑1〉 |↑2〉+ |↓1〉 |↓2〉)]

〈0 0| Sa1Sb2 |0 0〉 = −~2

4cos θ 〈0 0 | 0 0〉

= −~2

4cos θ = −~2

4a · b



Aside: Problem 4.50

Sa1Sb2 |0 0〉 =1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉+ sin θSx2 |↓2〉)

− Sz1 |↓1〉 (cos θSz2 |↑2〉+ sin θSx2 |↑2〉) ]

=1√2

~2

4

{|↑1〉 [− cos θ |↓2〉+ sin θ |↑2〉]

+ |↓1〉 [cos θ |↑2〉+ sin θ |↓2〉]}

=~2

4

[− cos θ

1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

+ sin θ1√2

(|↑1〉 |↑2〉+ |↓1〉 |↓2〉)]

〈0 0| Sa1Sb2 |0 0〉 = −~2

4cos θ 〈0 0 | 0 0〉 = −~2

4cos θ

= −~2

4a · b



Aside: Problem 4.50

Sa1Sb2 |0 0〉 =1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉+ sin θSx2 |↓2〉)

− Sz1 |↓1〉 (cos θSz2 |↑2〉+ sin θSx2 |↑2〉) ]

=1√2

~2

4

{|↑1〉 [− cos θ |↓2〉+ sin θ |↑2〉]

+ |↓1〉 [cos θ |↑2〉+ sin θ |↓2〉]}

=~2

4

[− cos θ

1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

+ sin θ1√2

(|↑1〉 |↑2〉+ |↓1〉 |↓2〉)]

〈0 0| Sa1Sb2 |0 0〉 = −~2

4cos θ 〈0 0 | 0 0〉 = −~2

4cos θ = −~2

4a · b



Aside: Problem 4.50

Sa1Sb2 |0 0〉 =1√2

[Sz1 |↑1〉 (cos θSz2 |↓2〉+ sin θSx2 |↓2〉)

− Sz1 |↓1〉 (cos θSz2 |↑2〉+ sin θSx2 |↑2〉) ]

=1√2

~2

4

{|↑1〉 [− cos θ |↓2〉+ sin θ |↑2〉]

+ |↓1〉 [cos θ |↑2〉+ sin θ |↓2〉]}

=~2

4

[− cos θ

1√2

(|↑1〉 |↓2〉 − |↓1〉 |↑2〉)

+ sin θ1√2

(|↑1〉 |↑2〉+ |↓1〉 |↓2〉)]

〈0 0| Sa1Sb2 |0 0〉 = −~2

4cos θ 〈0 0 | 0 0〉 = −~2

4cos θ = −~2

4a · b



Bell’s inequality

Assuming the probability distribu-tion ρ(λ) is normalized

we note that becauseA(a, λ) = ±1, B(b, λ) = ±1

∫ρ(λ) dλ = 1

P(a, b) ≥ −1

P(a, b) =

∫ρ(λ)A(a, λ)B(b, λ) dλ = −

∫ρ(λ)A(a, λ)A(b, λ) dλ

P(a, b) can only reach -1 when

defining c as a third unit vectora = b

A(a, λ) = −B(a, λ)

P(a, b)− P(a, c) = −∫ρ(λ)

[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)

]dλ

= −∫ρ(λ)

[A(a, λ)A(b, λ)− A(b, λ)A(b, λ)A(a, λ)A(c , λ)

]dλ

= −∫ρ(λ)A(a, λ)A(b, λ)

[1− A(b, λ)A(c , λ)

]dλ


Bell’s inequality



∫ρ(λ) dλ = 1

P(a, b) ≥ −1

P(a, b) =

∫ρ(λ)A(a, λ)B(b, λ) dλ = −




A(a, λ) = −B(a, λ)

P(a, b)− P(a, c) = −∫ρ(λ)

[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)

]dλ

= −∫ρ(λ)


]dλ

= −∫ρ(λ)A(a, λ)A(b, λ)

[1− A(b, λ)A(c , λ)

]dλ


Bell’s inequality



∫ρ(λ) dλ = 1

P(a, b) ≥ −1

P(a, b) =

∫ρ(λ)A(a, λ)B(b, λ) dλ = −




A(a, λ) = −B(a, λ)

P(a, b)− P(a, c) = −∫ρ(λ)

[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)

]dλ

= −∫ρ(λ)


]dλ

= −∫ρ(λ)A(a, λ)A(b, λ)

[1− A(b, λ)A(c , λ)

]dλ


Bell’s inequality



∫ρ(λ) dλ = 1

P(a, b) ≥ −1

P(a, b) =

∫ρ(λ)A(a, λ)B(b, λ) dλ = −




A(a, λ) = −B(a, λ)

P(a, b)− P(a, c) = −∫ρ(λ)

[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)

]dλ

= −∫ρ(λ)


]dλ

= −∫ρ(λ)A(a, λ)A(b, λ)

[1− A(b, λ)A(c , λ)

]dλ


Bell’s inequality



∫ρ(λ) dλ = 1

P(a, b) ≥ −1

P(a, b) =


= −∫ρ(λ)A(a, λ)A(b, λ) dλ



A(a, λ) = −B(a, λ)

P(a, b)− P(a, c) = −∫ρ(λ)

[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)

]dλ

= −∫ρ(λ)


]dλ

= −∫ρ(λ)A(a, λ)A(b, λ)

[1− A(b, λ)A(c , λ)

]dλ


Bell’s inequality



∫ρ(λ) dλ = 1

P(a, b) ≥ −1

P(a, b) =





A(a, λ) = −B(a, λ)

P(a, b)− P(a, c) = −∫ρ(λ)

[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)

]dλ

= −∫ρ(λ)


]dλ

= −∫ρ(λ)A(a, λ)A(b, λ)

[1− A(b, λ)A(c , λ)

]dλ


Bell’s inequality



∫ρ(λ) dλ = 1

P(a, b) ≥ −1

P(a, b) =




defining c as a third unit vector

a = b

A(a, λ) = −B(a, λ)

P(a, b)− P(a, c) = −∫ρ(λ)

[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)

]dλ

= −∫ρ(λ)


]dλ

= −∫ρ(λ)A(a, λ)A(b, λ)

[1− A(b, λ)A(c , λ)

]dλ


Bell’s inequality



∫ρ(λ) dλ = 1

P(a, b) ≥ −1

P(a, b) =





a = b

A(a, λ) = −B(a, λ)

P(a, b)− P(a, c) = −∫ρ(λ)

[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)

]dλ

= −∫ρ(λ)


]dλ

= −∫ρ(λ)A(a, λ)A(b, λ)

[1− A(b, λ)A(c , λ)

]dλ


Bell’s inequality



∫ρ(λ) dλ = 1

P(a, b) ≥ −1

P(a, b) =

∫ρ(λ)A(a, λ)B(b, λ) dλ = −




a = b

A(a, λ) = −B(a, λ)

P(a, b)− P(a, c) = −∫ρ(λ)

[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)

]dλ

= −∫ρ(λ)


]dλ

= −∫ρ(λ)A(a, λ)A(b, λ)

[1− A(b, λ)A(c , λ)

]dλ


Bell’s inequality



∫ρ(λ) dλ = 1

P(a, b) ≥ −1

P(a, b) =

∫ρ(λ)A(a, λ)B(b, λ) dλ = −




A(a, λ) = −B(a, λ)

P(a, b)− P(a, c) = −∫ρ(λ)

[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)

]dλ

= −∫ρ(λ)


]dλ

= −∫ρ(λ)A(a, λ)A(b, λ)

[1− A(b, λ)A(c , λ)

]dλ


Bell’s inequality



∫ρ(λ) dλ = 1

P(a, b) ≥ −1

P(a, b) =

∫ρ(λ)A(a, λ)B(b, λ) dλ = −




A(a, λ) = −B(a, λ)

P(a, b)− P(a, c) = −∫ρ(λ)

[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)

]dλ

= −∫ρ(λ)


]dλ

= −∫ρ(λ)A(a, λ)A(b, λ)

[1− A(b, λ)A(c , λ)

]dλ


Bell’s inequality



∫ρ(λ) dλ = 1

P(a, b) ≥ −1

P(a, b) =

∫ρ(λ)A(a, λ)B(b, λ) dλ = −




A(a, λ) = −B(a, λ)

P(a, b)− P(a, c) = −∫ρ(λ)

[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)

]dλ

= −∫ρ(λ)


]dλ

= −∫ρ(λ)A(a, λ)A(b, λ)

[1− A(b, λ)A(c , λ)

]dλ


Bell’s inequality



∫ρ(λ) dλ = 1

P(a, b) ≥ −1

P(a, b) =

∫ρ(λ)A(a, λ)B(b, λ) dλ = −




A(a, λ) = −B(a, λ)

P(a, b)− P(a, c) = −∫ρ(λ)

[A(a, λ)A(b, λ)− A(a, λ)A(c , λ)

]dλ

= −∫ρ(λ)


]dλ

= −∫ρ(λ)A(a, λ)A(b, λ)

[1− A(b, λ)A(c , λ)

]dλ


Bell’s inequality

P(a, b)− P(a, c) =

∫ρ(λ){−A(a, λ)A(b λ)}

[1− A(b, λ)A(c , λ)

]dλ

The quantity in the square brackets is always either 0 or 2 and the rest,being an integral over a probability density will vary between positive andnegative. Thus, by taking the absolute value of this quantity we can writean overestimate of the quantity by setting the variable part to 1.

|P(a, b)− P(a, c)| ≤∫ρ(λ)

[1− A(b, λ)A(c , λ)

]dλ

≤∫ρ(λ) dλ−

∫ρ(λ)A(b, λ)A(c, λ) dλ

|P(a, b)− P(a, c)| ≤ 1 + P(b, c)


Bell’s inequality

P(a, b)− P(a, c) =

∫ρ(λ){−A(a, λ)A(b λ)}

[1− A(b, λ)A(c , λ)

]dλ

The quantity in the square brackets is always either 0 or 2

and the rest,being an integral over a probability density will vary between positive andnegative. Thus, by taking the absolute value of this quantity we can writean overestimate of the quantity by setting the variable part to 1.

|P(a, b)− P(a, c)| ≤∫ρ(λ)

[1− A(b, λ)A(c , λ)

]dλ

≤∫ρ(λ) dλ−


|P(a, b)− P(a, c)| ≤ 1 + P(b, c)


Bell’s inequality

P(a, b)− P(a, c) =

∫ρ(λ){−A(a, λ)A(b λ)}

[1− A(b, λ)A(c , λ)

]dλ

The quantity in the square brackets is always either 0 or 2 and the rest,being an integral over a probability density will vary between positive andnegative.

Thus, by taking the absolute value of this quantity we can writean overestimate of the quantity by setting the variable part to 1.

|P(a, b)− P(a, c)| ≤∫ρ(λ)

[1− A(b, λ)A(c , λ)

]dλ

≤∫ρ(λ) dλ−


|P(a, b)− P(a, c)| ≤ 1 + P(b, c)


Bell’s inequality

P(a, b)− P(a, c) =

∫ρ(λ){−A(a, λ)A(b λ)}

[1− A(b, λ)A(c , λ)

]dλ

The quantity in the square brackets is always either 0 or 2 and the rest,being an integral over a probability density will vary between positive andnegative. Thus, by taking the absolute value of this quantity we can write

an overestimate of the quantity by setting the variable part to 1.

|P(a, b)− P(a, c)| ≤∫ρ(λ)

[1− A(b, λ)A(c , λ)

]dλ

≤∫ρ(λ) dλ−


|P(a, b)− P(a, c)| ≤ 1 + P(b, c)


Bell’s inequality

P(a, b)− P(a, c) =

∫ρ(λ){−A(a, λ)A(b λ)}

[1− A(b, λ)A(c , λ)

]dλ

The quantity in the square brackets is always either 0 or 2 and the rest,being an integral over a probability density will vary between positive andnegative. Thus, by taking the absolute value of this quantity we can write

an overestimate of the quantity by setting the variable part to 1.

|P(a, b)− P(a, c)| ≤∫ρ(λ)

[1− A(b, λ)A(c , λ)

]dλ

≤∫ρ(λ) dλ−


|P(a, b)− P(a, c)| ≤ 1 + P(b, c)


Bell’s inequality

P(a, b)− P(a, c) =

∫ρ(λ){−A(a, λ)A(b λ)}

[1− A(b, λ)A(c , λ)

]dλ


|P(a, b)− P(a, c)| ≤∫ρ(λ)

[1− A(b, λ)A(c , λ)

]dλ

≤∫ρ(λ) dλ−


|P(a, b)− P(a, c)| ≤ 1 + P(b, c)


Bell’s inequality

P(a, b)− P(a, c) =

∫ρ(λ){−A(a, λ)A(b λ)}

[1− A(b, λ)A(c , λ)

]dλ


|P(a, b)− P(a, c)| ≤∫ρ(λ)

[1− A(b, λ)A(c , λ)

]dλ

≤∫ρ(λ) dλ−


|P(a, b)− P(a, c)| ≤ 1 + P(b, c)


Bell’s inequality

P(a, b)− P(a, c) =

∫ρ(λ){−A(a, λ)A(b λ)}

[1− A(b, λ)A(c , λ)

]dλ


|P(a, b)− P(a, c)| ≤∫ρ(λ)

[1− A(b, λ)A(c , λ)

]dλ

≤∫ρ(λ) dλ−


|P(a, b)− P(a, c)| ≤ 1 + P(b, c)


Bell’s inequality

Take the particular case of a ⊥ b

and c at an angle γ to a.

P(a, b) = 0

P(a, c) = − cos γ

P(b, c) = − cos(π

2− γ)

= − sin γ

Bell’s inequality becomes

| cos γ| ≤ 1− sin γ

a

b

cγ

Clearly this inequality does not hold for all cases meaning that anexperimental result predicted by quantum mechanics woud rule out anyhidden variables


Bell’s inequality

Take the particular case of a ⊥ b

and c at an angle γ to a.

P(a, b) = 0



2− γ)

= − sin γ



a

b

cγ



Bell’s inequality

Take the particular case of a ⊥ band c at an angle γ to a.

P(a, b) = 0



2− γ)

= − sin γ



a

b

cγ



Bell’s inequality


P(a, b) = 0



2− γ)

= − sin γ



a

b

cγ



Bell’s inequality


P(a, b) = 0



2− γ)

= − sin γ



a

b

cγ



Bell’s inequality


P(a, b) = 0



2− γ)

= − sin γ



a

b

cγ



Bell’s inequality


P(a, b) = 0



2− γ)

= − sin γ



a

b

cγ



Bell’s inequality


P(a, b) = 0



2− γ)

= − sin γ



a

b

cγ



Bell’s inequality


P(a, b) = 0



2− γ)

= − sin γ


| cos γ| ≤ 1− sin γ 0

0.5

1

1.5

2

0 π 2π

γ

a ⊥ b

|cos γ| 1-sin γ



Bell’s inequality


P(a, b) = 0



2− γ)

= − sin γ


| cos γ| ≤ 1− sin γ 0

0.5

1

1.5

2

0 π 2π

γ

a ⊥ b

|cos γ| 1-sin γ

Clearly this inequality does not hold for all cases

meaning that anexperimental result predicted by quantum mechanics woud rule out anyhidden variables


Bell’s inequality


P(a, b) = 0



2− γ)

= − sin γ


| cos γ| ≤ 1− sin γ 0

0.5

1

1.5

2

0 π 2π

γ

a ⊥ b

|cos γ| 1-sin γ



The EPR experiment

“Experimental test of Bell’s inequalities usint time-varying analyzers,”, A. Aspect, J.

Dalibard, and G. Roger, Phys. Rev. Lett. 49, 1804-1807 (1982).


The EPR experiment




The EPR experiment




The EPR experiment




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