Probabilistics of Hardy's Scheme

Nicholas Wheeler
Reed College Physics Department
11 May 2010

Introduction

Bohm, Bell and later Hardy all contemplate situations in which Alice & Bob do measurements on their respective components of a composite pair of 2-state systems. The scheme—drawn on the presumption that Alice makes the first measurement, Bob the second—is illustrated below:

Show[Graphics[{Thickness[0.008],Line[{{0,0},{1,0}}]}],
Graphics[{Red,Thickness[0.008],Line[{{1,-1},{1,+1}}]}],
Graphics[{Thickness[0.008],Line[{{1,+1},{2,+1}}]}],
Graphics[{Thickness[0.008],Line[{{1,-1},{2,-1}}]}],
Graphics[{Blue,Thickness[0.008],Line[{{2,1/2},{2,3/2}}]}],
Graphics[{Blue,Thickness[0.008],Line[{{2,-1/2},{2,-3/2}}]}],
Graphics[{Thickness[0.008],Line[{{2,3/2},{3,3/2}}]}],
Graphics[{Thickness[0.008],Line[{{2,1/2},{3,1/2}}]}],
Graphics[{Thickness[0.008],Line[{{2,-1/2},{3,-1/2}}]}],
Graphics[{Thickness[0.008],Line[{{2,-3/2},{3,-3/2}}]}],
Graphics[Text[Style["p", FontSize->12],{1.1,0.5}]],
Graphics[Text[Style["1-p", FontSize->12],{1.17,-0.5}]],
Graphics[Text[Style["q", FontSize->12],{2.1,1.25}]],
Graphics[Text[Style["1-q", FontSize->12],{2.17,.75}]],
Graphics[Text[Style["r", FontSize->12],{2.1,-.75}]],
Graphics[Text[Style["1-r", FontSize->12],{2.17,-1.25}]],Graphics[Text[Style["Aup Bup", FontSize->18],{3.4,1.50}]],
Graphics[Text[Style["Aup Bdn", FontSize->18],{3.4,0.50}]],
Graphics[Text[Style["Adn Bup", FontSize->18],{3.4,-0.50}]],
Graphics[Text[Style["Adn Bdn", FontSize->18],{3.4,-1.50}]],
Graphics[Text[Style["ρ", FontSize->18],{-0.13,0}]],
PlotLabel->Style["Bohm/Bell/Hardy Scheme—Alice First", Blue, FontFamily->"Helvetica", FontSize->14]]

Clear[p,q,r,P]

Alice's results are, in effect, marginal probabilities—independent of the results of Bob's subsequent results:

(Subscript[P, Aup])=p;
(Subscript[P, Adn])=1-p;

Bob's results are conditional probabilites, contingent upon the results of Alice's prior measurement:

(Subscript[P, Bup,Aup])=q;
(Subscript[P, Bdn,Aup])=1-q;
(Subscript[P, Bup,Adn])=r;
(Subscript[P, Bdn,Adn])=1-r;

REMARK: The preferred notation here would be (Subscript[P, Bup|Aup]), which Mathematica disallows. Whence the comma.

Joint probabilities are obtained by appeal to the formula

(Subscript[P, AupBup])=(Subscript[P, Bup,Aup])(Subscript[P, Aup])
(Subscript[P, AupBdn])=(Subscript[P, Bdn,Aup])(Subscript[P, Aup])
(Subscript[P, AdnBup])=(Subscript[P, Bup,Adn])(Subscript[P, Adn])
(Subscript[P, AdnBdn])=(Subscript[P, Bdn,Adn])(Subscript[P, Adn])

These sum to unity:

(Subscript[P, AupBup])+(Subscript[P, AupBdn])+(Subscript[P, AdnBup])+(Subscript[P, AdnBdn])//Simplify

From the joint probabilities we can recover Alice's marginal probabilities

(Subscript[P, Aup])=(Subscript[P, AupBup])+(Subscript[P, AupBdn])//Simplify
(Subscript[P, Adn])=(Subscript[P, AdnBup])+(Subscript[P, AdnBdn])//Simplify

and can—if we assume the joint probabilities are independent of the sequence in which Alice/Bob make their measurements (Bohm/Bell/Hardy contemplate situations in which it would be relativistically impossible to assign covariant meaning to such a sequence)—assign marginal probabilities also to Bob:

(Subscript[P, Bup])=(Subscript[P, AupBup])+(Subscript[P, AdnBup])
(Subscript[P, Bdn])=(Subscript[P, AupBdn])+(Subscript[P, AdnBdn])

(Subscript[P, Bdn])==1-(Subscript[P, Bup])//Simplify

(Subscript[P, Aup,Bup])=((Subscript[P, AupBup])/(Subscript[P, Bup]))
(Subscript[P, Aup,Bdn])=((Subscript[P, AupBdn])/(Subscript[P, Bdn]))
(Subscript[P, Adn,Bup])=((Subscript[P, AdnBup])/(Subscript[P, Bup]))
(Subscript[P, Adn,Bdn])=((Subscript[P, AdnBdn])/(Subscript[P, Bdn]))

We are in position now to construct the scheme that arises when it is Bob who makes the first measurement:

Show[Graphics[{Thickness[0.008],Line[{{0,0},{1,0}}]}],
Graphics[{Blue,Thickness[0.008],Line[{{1,-1},{1,+1}}]}],
Graphics[{Thickness[0.008],Line[{{1,+1},{2,+1}}]}],
Graphics[{Thickness[0.008],Line[{{1,-1},{2,-1}}]}],
Graphics[{Red,Thickness[0.008],Line[{{2,1/2},{2,3/2}}]}],
Graphics[{Red,Thickness[0.008],Line[{{2,-1/2},{2,-3/2}}]}],
Graphics[{Thickness[0.008],Line[{{2,3/2},{3,3/2}}]}],
Graphics[{Thickness[0.008],Line[{{2,1/2},{3,1/2}}]}],
Graphics[{Thickness[0.008],Line[{{2,-1/2},{3,-1/2}}]}],
Graphics[{Thickness[0.008],Line[{{2,-3/2},{3,-3/2}}]}],
Graphics[Text[Style["𝒫", FontSize->12],{1.1,0.5}]],
Graphics[Text[Style["1-𝒫", FontSize->12],{1.17,-0.5}]],
Graphics[Text[Style["𝒬", FontSize->12],{2.1,1.25}]],
Graphics[Text[Style["1-𝒬", FontSize->12],{2.17,.75}]],
Graphics[Text[Style["ℛ", FontSize->12],{2.1,-.75}]],
Graphics[Text[Style["1-ℛ", FontSize->12],{2.17,-1.25}]],Graphics[Text[Style["Bup Aup", FontSize->18],{3.4,1.50}]],
Graphics[Text[Style["Bup Adn", FontSize->18],{3.4,0.50}]],
Graphics[Text[Style["Bdn Aup", FontSize->18],{3.4,-0.50}]],
Graphics[Text[Style["Bdn Adn", FontSize->18],{3.4,-1.50}]],
Graphics[Text[Style["ρ", FontSize->18],{-0.13,0}]],
PlotLabel->Style["Bohm/Bell/Hardy Scheme—Bob First", Blue, FontFamily->"Helvetica", FontSize->14]]

NOTE that the logic of the diagram has led us to place BupAdn=AdnBup where AupBdn formerly resided; ditto BdnAup=AupBdn  where AdnBup formerly resided.

We can now describe Bob's {𝒫, 𝒬, ℛ} in terms of Alice's {p, q, r}

𝒫=p q+(1-p) r;
𝒬=((p q)/(p q+(1-p) r));
ℛ=((p (1-q))/(p (1-q)+(1-p) (1-r)));

and could similarly describe Alice's {p, q, r} in terms of Bob's {𝒫, 𝒬, ℛ}.

The joint probabilities that result from Bob's scheme

𝒫 𝒬
𝒫(1-𝒬)//Simplify
(1-𝒫)ℛ//Simplify
(1-𝒫)(1-ℛ)//Simplify

differ only by the previously noted permutation from those that have been found to result from Alice's scheme:

(Subscript[P, AupBup])
(Subscript[P, AupBdn])//Simplify
(Subscript[P, AdnBup])//Simplify
(Subscript[P, AdnBdn])//Simplify

Hardy's 𝕌𝕌 arrangement

Hardy imposes the condition (Subscript[P, AupBup])= 0. To achieve that without at the same time forcing (Subscript[P, AupBdn])= 0 we set

q=0;

The joint probabilities become

(Subscript[P, AupBup])
(Subscript[P, AupBdn])
(Subscript[P, AdnBup])
(Subscript[P, AdnBdn])

More interesting are the marginal probabilities

(Subscript[P, Bup,Aup])
(Subscript[P, Bdn,Aup])
(Subscript[P, Bup,Adn])
(Subscript[P, Bdn,Adn])

(Subscript[P, Bup])=(Subscript[P, AupBup])+(Subscript[P, AdnBup]);
(Subscript[P, Bdn])=(Subscript[P, AupBdn])+(Subscript[P, AdnBdn]);

(Subscript[P, Aup,Bup])=((Subscript[P, AupBup])/(Subscript[P, Bup]))
(Subscript[P, Aup,Bdn])=((Subscript[P, AupBdn])/(Subscript[P, Bdn]))
(Subscript[P, Adn,Bup])=((Subscript[P, AdnBup])/(Subscript[P, Bup]))
(Subscript[P, Adn,Bdn])=((Subscript[P, AdnBdn])/(Subscript[P, Bdn]))

Specifically, from

(Subscript[P, Bup,Aup])
(Subscript[P, Bdn,Aup])

(Subscript[P, Aup,Bup])
(Subscript[P, Adn,Bup])

we see that if one gets "up" the other gets "dn"