2-2

a_⊗b_:=KroneckerProduct[a,b]
CompConj[A_] := A /.Complex[0,n_]->-Complex[0,n]
Adj[x_]:=CompConj[Transpose[x]]
Brac[a_,b_]:=(Adj[a].b)[[1]][[1]]

𝕀=(({{1, 0}, {0, 1}}));
𝕆=(({{0, 0}, {0, 0}}));

Logic & Simulation of
Hardy's Nonlocality Argument

Nicholas Wheeler
Reed College Physics Department
1 May 2010

Hyperlink["Lucian Hardy","http://www.perimeterinstitute.ca/index.php?option=com_content&task=view&id=30&Itemid=72&pi=1078"]

Introduction

In 1993, Lucian Hardy (PhD 1992) devised a scheme for demonstrating quantum nonlocality ("Nonlocality for two particles without inequalities for almost all entangled states," Phys. Rev. Letters 71, 1665-1668,  (1993)) that Mermin has called "the best version of Bell's theorem." Hardy's scheme resembles Bell's in that it concerns a composite pair of qubits in an entangled state (taken by Bohm/Bell to be the singlet state, but by Hardy to be an adjustably entangled state) and two players—Alice and Bob, each of whom is equipped to examine the state in either of two ways. It differs, however, from Bell's in that in proceeds without reference to any analog of "Bell's inequality" and does not seek statistical evidence of correlation: it places one in position to (in principle) deduce nonlocality from the evidence of a single measurement. In the latter respect it has much in common with the GHZ argument, but is simpler, both conceptually and experimentally (involves only two qubits, not three).

Hardy chose to present his argument as a diabolically clever but seemly unmotivated hat trick (the motivation emerges only retrospectively, at the end of the argument), and to employ a cumbersome notation. For these reasons, I on first reading found his argument difficult to understand. And the efforts of more recent popular expositors—see, for example,

Hyperlink["Dan Styer's Account","http://www.oberlin.edu/physics/dstyer/StrangeQM/Hardy.pdf"]

—have (in my view) enjoyed only very limited success. In New Hardy.nb I provide a "computational version" of the material I present on pages 77-85 of Chapter 1 in my Quantum II Notes (2009), but that discussion can be criticized on the ground that it is so encumbered with distracting details (which I now recognize to be largely irrelevant) that it goes only a little way toward clarifying the simple essence of Hardy's argument.

Implications of Hardy's argument were put to experimental test by Paul Kwiat and co-workers (A. G. White, D. F. V. James, P. H. Eberhard & P. G. Kwiat, "Nonmaximally entangled states: production, characterization and utilization," Phys. Rev. Letters 83, 3103-3107 (1999)). A simplified version of Kwiat's experiment was devised by Mark Beck and co-workers (J. A. Carlson, M. D. Olmstead & M. Beck, "Quantum mysteries tested: an experiment implementing Hardy's test of local realism," AJP 74, 180-186 (2006)). Peter Wills has undertaken to reproduce Beck's experiment as a thesis project, and it was that effort that motivated me to revisit the subject, my new objective being to construct a computer simulation of such experiments. 

Yesterday I devised the elementary mathematical lemma that exposes quite clearly the essence of Hardy's argument, and that enables one to proceed efficiently to the results upon which the projected simulation must be based.

A mathematical lemma

Let p and q be real numbers subject to the contstraint (p^2)+(q^2)=1. Then

(Subscript[u, 1])=(({{p}, {q (E^(I ϕ))}}));
(Subscript[u, 2])=(({{-q}, {p (E^(I ϕ))}}));

are orthogonal unit vectors. Nothing essential to the present discussion will be lost if (for notational simplicity) we assume our orthonormal 2-vectors to be real, of which we will need two pairs:

(Subscript[u, 1])=(({{p}, {q }}));
(Subscript[u, 2])=(({{-q}, {p }}));
(Subscript[v, 1])=(({{P}, {Q }}));
(Subscript[v, 2])=(({{-Q}, {P }}));

Construct

(Subscript[uu, 1])=(Subscript[u, 1])⊗(Subscript[u, 1]);
(Subscript[uu, 2])=(Subscript[u, 1])⊗(Subscript[u, 2]);
(Subscript[uu, 3])=(Subscript[u, 2])⊗(Subscript[u, 1]);
(Subscript[uu, 4])=(Subscript[u, 2])⊗(Subscript[u, 2]);

(Subscript[uv, 1])=(Subscript[u, 1])⊗(Subscript[v, 1]);
(Subscript[uv, 2])=(Subscript[u, 1])⊗(Subscript[v, 2]);
(Subscript[uv, 3])=(Subscript[u, 2])⊗(Subscript[v, 1]);
(Subscript[uv, 4])=(Subscript[u, 2])⊗(Subscript[v, 2]);

(Subscript[vu, 1])=(Subscript[v, 1])⊗(Subscript[u, 1]);
(Subscript[vu, 2])=(Subscript[v, 1])⊗(Subscript[u, 2]);
(Subscript[vu, 3])=(Subscript[v, 2])⊗(Subscript[u, 1]);
(Subscript[vu, 4])=(Subscript[v, 2])⊗(Subscript[u, 2]);

(Subscript[vv, 1])=(Subscript[v, 1])⊗(Subscript[v, 1]);
(Subscript[vv, 2])=(Subscript[v, 1])⊗(Subscript[v, 2]);
(Subscript[vv, 3])=(Subscript[v, 2])⊗(Subscript[v, 1]);
(Subscript[vv, 4])=(Subscript[v, 2])⊗(Subscript[v, 2]);

LEMMA: 4-vectors constructed in this way are orthonormal:

Table[Assuming[{(p^2)+(q^2)==1,(P^2)+(Q^2)==1}, Simplify[Brac[(Subscript[uv, i]),(Subscript[uv, j])]]],{i,1,4},{j,1,4}]//TableForm

Similarly (if redundantly)

Table[Assuming[{(p^2)+(q^2)==1,(P^2)+(Q^2)==1}, Simplify[Brac[(Subscript[uu, i]),(Subscript[uu, j])]]],{i,1,4},{j,1,4}]//TableForm

So each of those four quartets provides an orthonormal basis in 4-space.

Let

Z=(({{(Subscript[z, 1])}, {(Subscript[z, 2])}, {(Subscript[z, 3])
(Subscript[z, 4])}}))

be an arbitrary 4-vector. Its coordinates with respect to (say) the uv-basis are

Table[Brac[(Subscript[uv, i]),Z],{i,1,4}]//TableForm

Assuming[{(p^2)+(q^2)==1,(P^2)+(Q^2)==1}, Simplify[(∑_(i=1)^4)(Subscript[uv, i])Brac[(Subscript[uv, i]),Z]==Z]]

The 4-vector Z—which is typically entangled—is displayed here as a superposition of separable/unentangled states.

Hardy's application of the LEMMA

With

in mind, Hardy introduces the unit 4-vector

Ψ=α (({{1}, {0}}))⊗(({{1}, {0}}))-β(({{0}, {1}}))⊗(({{0}, {1}}));
Ψ//MatrixForm

to describe a ξ-parameterized population of entangled states of a pair of qubits. The qubits are disentangled if αβ = 0 and maximally entangled (one of the Bell states)if α = β. The "almost all entangled states" to which Hardy alludes in the title of his paper arise when neither of those special conditions is satisfied. 

Hardy imposes two requirements to fix the design of the u and v vectors:

Hardy's first requirement: The coordinates of Ψ with respect to the uu basis are

Assuming[{(α^2)+(β^(2))==1,(p^2)+(q^2)==1}, Table[Brac[(Subscript[uu, i]),Ψ],{i,1,4}]]//TableForm

Hardy requires the leading coordinate to vanish:

Assuming[{(α^2)+(β^(2))==1,(p^2)+(q^2)==1}, Brac[(Subscript[uu, 1]),Ψ]]

Normalized values of p and q are now determined to within arbitrary ± signs. Setting

we have

Table[(Subscript[(OverBar[u]), i])=(Subscript[u, i])/.{p->((Sqrt[β])/(Sqrt[α + β])),q->-((Sqrt[α])/(Sqrt[α + β]))},{i,1,2}];

Table[(Subscript[(OverBar[uu]), i])=(Subscript[uu, i])/.{p->((Sqrt[β])/(Sqrt[α + β])),q->-((Sqrt[α])/(Sqrt[α + β]))},{i,1,4}]

Assuming[{(α^2)+(β^(2))==1},Simplify[ Table[Brac[(Subscript[(OverBar[uu]), i]),Ψ],{i,1,4}]]]//TableForm

Hardy's second requirement: The coordinates of Ψ with respect to the uv-basis are

Assuming[{(α^2)+(β^(2))==1,(p^2)+(q^2)==1,(P^2)+(Q^2)==1}, Table[Brac[(Subscript[uv, i]),Ψ],{i,1,4}]]/.{p->((Sqrt[β])/(Sqrt[α + β])),q->-((Sqrt[α])/(Sqrt[α + β]))}//TableForm

Hardy requires the final coordinate to vanish. Normalized values of P and Q are now determined to within an arbitrary ± signs. Setting

we have

Table[(Subscript[(OverBar[v]), i])=(Subscript[v, i])/.{P->((α^(3/2))/(Sqrt[(α^3)+(β^3)])),Q->-((β^(3/2))/(Sqrt[(α^3)+(β^3)]))},{i,1,2}];

Table[(Subscript[(OverBar[uv]), i])=(Subscript[uv, i])/.{p->((Sqrt[β])/(Sqrt[α + β])),q->-((Sqrt[α])/(Sqrt[α + β])),P->((α^(3/2))/(Sqrt[(α^3)+(β^3)])),Q->-((β^(3/2))/(Sqrt[(α^3)+(β^3)]))},{i,1,4}];

Assuming[{(α^2)+(β^(2))==1},Simplify[ Table[Brac[(Subscript[(OverBar[uv]), i]),Ψ],{i,1,4}]]]//TableForm

We have now used up our freedom to assign values to p, q, P and Q. But when we look to the coordinates of Ψ relative to the vu-basis, we find that they are identical to the preceding uv-coordinates, except for interchange of the middle two:

Table[(Subscript[(OverBar[vu]), i])=(Subscript[vu, i])/.{p->((Sqrt[β])/(Sqrt[α + β])),q->-((Sqrt[α])/(Sqrt[α + β])),P->((α^(3/2))/(Sqrt[(α^3)+(β^3)])),Q->-((β^(3/2))/(Sqrt[(α^3)+(β^3)]))},{i,1,4}];

Assuming[{(α^2)+(β^(2))==1},Simplify[ Table[Brac[(Subscript[(OverBar[vu]), i]),Ψ],{i,1,4}]]]//TableForm

Note in particular that the final coordinate again vanishes.

Looking finally to the (enforced) Ψ-coordinates relative to the vv-basis, we have

Table[(Subscript[(OverBar[vv]), i])=(Subscript[vv, i])/.{p->((Sqrt[β])/(Sqrt[α + β])),q->-((Sqrt[α])/(Sqrt[α + β])),P->((α^(3/2))/(Sqrt[(α^3)+(β^3)])),Q->-((β^(3/2))/(Sqrt[(α^3)+(β^3)]))},{i,1,4}];

giving

Assuming[{(α^2)+(β^2)==1},Simplify[ Table[Brac[(Subscript[(OverBar[vv]), i]),Ψ],{i,1,4}]]]//TableForm

Assuming[{(α^2)+(β^2)==1},Simplify[((β(-1+α β+(β^2)))/((α^2)-α β+(β^2)))==-((α(α-β)β)/(1-α β))]]

The vv-coordinates of Ψ are relatively complicated, but do entail

Assuming[{(α^2)+(β^2)==1},Simplify[(∑_(i=1)^4)((Brac[(Subscript[(OverBar[vv]), i]),Ψ])^2),{i,1,4}]]

as is required by the fact that Ψ is a unit vector.

Fundamental to Hardy's argument---as will emerge—is the fact that the (square of) the 4th coordinate (which will acquire the interpretation of a probability) almost always does not vanish. Defining

h[ξ_]:=(((-((α(α-β)β)/(1-α β))))^2)/.{α->Cos[ξ],β->Sin[ξ]}
h[ξ]

Plot[h[ξ],{ξ,0,π/2},Ticks->{{(1/4)π,(1/2)π},{}}]

Evidently h[ξ] vanishes at ξ=0 and ξ=π/2, where Ψ[ξ] is disentangled, and—curiously—at ξ=π/4, where

is maximally entangled (a Bell state). To locate the maxima of h[ξ] we proceed

hprime[ξ_]:=D[h[ξ],ξ]//Simplify
hprime[ξ]

Plot[((Sin[2 ξ] (-17 Cos[2 ξ]+Cos[6 ξ]+12 Sin[4 ξ]))/(2 (((-2+Sin[2 ξ]))^3))),{ξ,.0001,π/2-.0001}]

FindRoot[hprime[ξ]==0,{ξ,0.43}]
FindRoot[hprime[ξ]==0,{ξ,1.13}]

h[ξ]/.ξ->0.434692
h[ξ]/.ξ->1.1361

Plot[h[ξ],{ξ,0,π/2},Ticks->{{0.4347,(1/4)π,1.1361,(1/2)π},{0.0902}}, GridLines->{{0.4347,1.1361},{0.0902}}, AxesLabel->{"ξ"}, PlotStyle->Thick]

The intermediately entangled states

(Subscript[Ψ, Hardy1])=Cos[0.4347](({{1}, {0}}))⊗(({{1}, {0}}))-Sin[0.4347](({{0}, {1}}))⊗(({{0}, {1}}))//MatrixForm
(Subscript[Ψ, Hardy2])=Cos[1.1361](({{1}, {0}}))⊗(({{1}, {0}}))-Sin[1.1361](({{0}, {1}}))⊗(({{0}, {1}}))//MatrixForm

are called "Hardy states."

INTERPOLATED REMARK: Peter Wills, in the thesis which he defended yesterday (5 May), uses a different argument (source not cited) to obtain those same results. The argument is rather pretty, so I reproduce it: from

Assuming[(α^2)+(β^2)==1,Simplify[(((-((α(α-β)β)/(1-α β))))^2)]]

we are led to define

f[x_]:=(((x^2)(1-2x))/(((1-x))^2))

where

x=α β=Cos[ξ]Sin[ξ]

ranges on [0, 0.5] as ξ ranges on [0, π/2].  To locate (Subscript[x, max]) we proceed

Plot[{f[𝓍],f'[𝓍]},{𝓍,0,1/2}]

Solve[f'[x]==0,x]

%//N

The command

Solve[Cos[ξ]Sin[ξ]==(1/2) (3-(Sqrt[5])),ξ];

supplies four zeros, of which only the positive two are of interest. They are located at precisey the points obtained previously:

N[{ArcCos[(Sqrt[(1/2)+(1/2) (Sqrt[-13+6 (Sqrt[5])])])],ArcCos[(Sqrt[(1/2) (1-(Sqrt[-13+6 (Sqrt[5])]))])]}]

Here ends the INTERPOLATED REMARK.

So much for the mathematical underpinnings of Hardy's argument. I turn now to discussion of what this mathematics has to do with the foundations of quantum mechanics—specifically, the quantum mechanics of photons. I preface that discussion with a review of some aspects of the classical physics of (100% polarized monochromatic) light beams.

Classical light beams and polarizers

REMARK: Here I borrow from the review of Jones Calculus that appears in Quantum Eraser.nb (7 April 2010), where I was borrowing from the discussion that appears on pages 27-37 in my "Ellipsometry: Stokes' parameters & related constructs in optics & classical/quantum mechanics" (1999).

Erect an orthogonal frame in the plane normal to an on-rushing monochromatic optical planewave, and write

to describe the motion of the (transverse) E-field. With R. Clark Jones, fold that information into design of a "Jones vector"—a complex 2-vector

J=(({{(Subscript[ℰ, 1])(E^(I (Subscript[δ, 1])))}, {(Subscript[ℰ, 2])(E^(I (Subscript[δ, 2])))}}));

The time-dependent field can then be described

𝒲[t_]:=J*(E^(I ω t))
𝒲[t]//MatrixForm

Introduce hermitian matrices by a convention that departs slightly from Pauli's

(Subscript[𝕤, 0])=(({{1, 0}, {0, 1}}));
(Subscript[𝕤, 1])=(({{1, 0}, {0, -1}}));
(Subscript[𝕤, 2])=(({{0, 1}, {1, 0}}));
(Subscript[𝕤, 3])=(({{0, -I}, {I, 0}}));

Then these simple quadratic constructs

(Adj[J].(Subscript[𝕤, 0]).J)[[1]][[1]]
(Adj[J].(Subscript[𝕤, 1]).J)[[1]][[1]]
(Adj[J].(Subscript[𝕤, 2]).J)[[1]][[1]]/.(Subscript[δ, 2])->(Subscript[δ, 1])+δ//ExpToTrig
(Adj[J].(Subscript[𝕤, 3]).J)[[1]][[1]]/.(Subscript[δ, 2])->(Subscript[δ, 1])+δ//ExpToTrig

are recognized to be precisely the parameters (see "Ellipsometry," equation (20) page 12) that Stokes introduced to describe the polarization of such a beam. They possess the great merit that—since interpretable as "intensities"—they are susceptible to direct photometric measurement.

In the following figure I have set

so as to render (Subscript[ℰ, 1]^2)+(Subscript[ℰ, 2]^2)=1 automatic:

Manipulate[ParametricPlot[{Cos[θ]Cos[ t],Sin[θ]Cos[t+δ]},{t,0,2π}, PlotRange->{{-1,1},{-1,1}}, PlotLabel->"Ellipse Traced by Flying E-vector", LabelStyle->{Bold, Red}, Ticks->False, PlotStyle->Thick],{θ,0,(π/2)},{δ,((-π)/2),(π/2)}]

Beck restricts his attention—for reasons having to to do with experimental realities—to linearly polarized beams (photons), which result when the relative phase δ = 0 :

Manipulate[ParametricPlot[{Cos[θ]Cos[ t],Sin[θ]Cos[t]},{t,0,2π}, PlotRange->{{-1,1},{-1,1}}, PlotLabel->"Linear Polarization", LabelStyle->{Bold, Red}, Ticks->False, PlotStyle->Thick],{θ,0,(π/2)}]

The Stokes parameters of such beams assume the especially simple form

(Adj[(({{Cos[θ]}, {Sin[θ]}}))].(Subscript[𝕤, 0]).(({{Cos[θ]}, {Sin[θ]}})))[[1]][[1]]//TrigReduce
(Adj[(({{Cos[θ]}, {Sin[θ]}}))].(Subscript[𝕤, 1]).(({{Cos[θ]}, {Sin[θ]}})))[[1]][[1]]//TrigReduce
(Adj[(({{Cos[θ]}, {Sin[θ]}}))].(Subscript[𝕤, 2]).(({{Cos[θ]}, {Sin[θ]}})))[[1]][[1]]//TrigReduce
(Adj[(({{Cos[θ]}, {Sin[θ]}}))].(Subscript[𝕤, 3]).(({{Cos[θ]}, {Sin[θ]}})))[[1]][[1]]

Obviously orthogonal to (({{Cos[θ]}, {Sin[θ]}})) is the vector (({{-Sin[θ]}, {Cos[θ]}})). They are seen to refer to beams that are in an obvious sense "oppositely polarized"

Manipulate[ParametricPlot[{{Cos[θ]Cos[ t],Sin[θ]Cos[t]},{-Sin[θ]Cos[ t],Cos[θ]Cos[t]}},{t,0,2π}, PlotRange->{{-1,1},{-1,1}}, PlotLabel->"Opposite Linear Polarization", LabelStyle->{Bold, Red}, Ticks->False, PlotStyle->Thick],{θ,0,(π/2)}]

The adjustment (({{Cos[θ]}, {Sin[θ]}}))⟶(({{-Sin[θ]}, {Cos[θ]}})) causes the Stokes parameters (({{(Subscript[S, 1])}, {(Subscript[S, 2])}, {(Subscript[S, 3])}})) to change sign, which is to say: it sends one point on the (equator of) the Poincaré sphere to go into the diametrically opposite point:

(Adj[(({{-Sin[θ]}, {Cos[θ]}}))].(Subscript[𝕤, 0]).(({{-Sin[θ]}, {Cos[θ]}})))[[1]][[1]]//TrigReduce
(Adj[(({{-Sin[θ]}, {Cos[θ]}}))].(Subscript[𝕤, 1]).(({{-Sin[θ]}, {Cos[θ]}})))[[1]][[1]]//TrigReduce
(Adj[(({{-Sin[θ]}, {Cos[θ]}}))].(Subscript[𝕤, 2]).(({{-Sin[θ]}, {Cos[θ]}})))[[1]][[1]]//TrigReduce
(Adj[(({{-Sin[θ]}, {Cos[θ]}}))].(Subscript[𝕤, 3]).(({{-Sin[θ]}, {Cos[θ]}})))[[1]][[1]]

Let x and y be real numbers that conform to the normalization condition (x^2)+(y^2)=1. Then

X=(({{x}, {y (E^(I ϕ))}}));
Y=(({{-y}, {x (E^(I ϕ))}}));

are Jones vectors that refer to a pair of oppositely polarized beams:

(Adj[X].Y)[[1]][[1]]

(Stokes remarked that such beams do not interfere.) Construct the associated projection matrices

𝕏=X.Adj[X];
𝕐=Y.Adj[Y];

which I demonstrate to be projective, orthogonal and complete:

Assuming[(x^2)+(y^2)==1,Simplify[𝕏.𝕏-𝕏==𝕆]]
Assuming[(x^2)+(y^2)==1,Simplify[𝕐.𝕐-𝕐==𝕆]]
Assuming[(x^2)+(y^2)==1,Simplify[𝕏.𝕐==𝕆]]
Assuming[(x^2)+(y^2)==1,Simplify[𝕏+𝕐==𝕀]]

If we define

jx=(Adj[X].J)[[1]][[1]]
jy=(Adj[Y].J)[[1]][[1]]

then we can display an arbitrary J-beam as a weighted superposition of X and Y-beams:

Assuming[(x^2)+(y^2)==1,Simplify[J==jx X+jy Y]]

The intensities of the X and Y-beams are given by expressions

Assuming[(x^2)+(y^2)==1,Simplify[ComplexExpand[CompConj[jx]jx]]]
Assuming[(x^2)+(y^2)==1,Simplify[ComplexExpand[CompConj[jy]jy]]]

that when added give back the intensity of the J-beam:

Assuming[(x^2)+(y^2)==1,Simplify[%+%%]]

If we assume that the J-beam is linearly polarized, and that so also are the X and Y-beams, then the preceding statements simplify markedly:

jx=(Adj[X].J)[[1]][[1]]/.{(Subscript[δ, 1])->0,(Subscript[δ, 2])->0,ϕ->0}
jy=(Adj[Y].J)[[1]][[1]]/.{(Subscript[δ, 1])->0,(Subscript[δ, 2])->0,ϕ->0}

Expand[Assuming[(x^2)+(y^2)==1,Simplify[ComplexExpand[CompConj[jx]jx]/.{(Subscript[δ, 1])->0,(Subscript[δ, 2])->0,ϕ->0}]]]
Expand[Assuming[(x^2)+(y^2)==1,Simplify[ComplexExpand[CompConj[jy]jy]/.{(Subscript[δ, 1])->0,(Subscript[δ, 2])->0,ϕ->0}]]]
Assuming[(x^2)+(y^2)==1,Simplify[%+%%]]

The projection matrices 𝕏 and 𝕐 serve individually to describe the action of crossed polarizers. When a J-beam is incident upon an X-polarizer, its X-component is transmitted and its Y-component is extinguished:

(Subscript[J, in])=J;
(Subscript[J, out])=𝕏.J;

(Subscript[J, in])//MatrixForm
(Subscript[J, out])//MatrixForm

The intensity attenuation factor is

Simplify[((ComplexExpand[Adj[(Subscript[J, out])].(Subscript[J, out])])/(ComplexExpand[Adj[(Subscript[J, in])].(Subscript[J, in])]))][[1]][[1]]/.(x^2)+(y^2)->1

which when the incident beam is linearly polarized and the polarizer is a linear polarizer (the case in which we will acquire primary interest) assumes the simple form.

%/.{(Subscript[δ, 1])->0,(Subscript[δ, 2])->0,ϕ->0}

REMARK: The action of polarizers can be described

(Subscript[(({{(Subscript[S, 0])}, {(Subscript[S, 1])}, {(Subscript[S, 2])}, {(Subscript[S, 3])}})), in])⟶(Subscript[(({{(Subscript[S, 0])}, {(Subscript[S, 1])}, {(Subscript[S, 2])}, {(Subscript[S, 3])}})), out])=𝕄.(Subscript[(({{(Subscript[S, 0])}, {(Subscript[S, 1])}, {(Subscript[S, 2])}, {(Subscript[S, 3])}})), in])

where 𝕄 is a 4×4 "Mueller matrix." But I will refrain from pursuit of that elegantly useful topic: for detailed discussion, see "Ellipsometry."

Photons and meters

The Jones vector

J=(({{(Subscript[ℰ, 1])}, {(Subscript[ℰ, 2])(E^(I δ))}}));

—a complex unit vector if, as will be understood,

—will be taken now to describe the quantum state of a photon, of which the Stokes parameters continue as before to describe the polarization. The linearly polarized photons to which we will restrict our attention are described by real unit 2-vectors (δ=0).

It is with regard to the action of "polarizers" that the classical and quantum theories differ most markedly. And the differences comes to this: the action of a classical polarizer, as determined by a downstream photometer, is deterministic, while the action of a quantum polarizer—a prototypical "meter"—as determined by a downstream counter, is probabilistic. 

Action of a classical polarizer: Reverting to a notation introduced at the outset, let p and q be a normalized pair of real numbers (marking a point on the unit circle). Then

(Subscript[u, 1])=(({{p}, {q }}));
(Subscript[u, 2])=(({{-q}, {p }}));

are orthonormal, and their associated projectors

(Subscript[𝕦, 1])=(Subscript[u, 1]).Adj[(Subscript[u, 1])];
(Subscript[𝕦, 2])=(Subscript[u, 2]).Adj[(Subscript[u, 2])];

are orthogonal and complete. Suppose the Jones vector of a linearly polarized optical beam J has been developed in the u-basis:

(Subscript[J, in])=(Subscript[j, 1])(Subscript[u, 1])+(Subscript[j, 2])(Subscript[u, 2]);

where (Subscript[j, 1]) and (Subscript[j, 2]) are a normalized pair of real numbers. Presentation of the beam to a (Subscript[𝕦, 1])-polarizer produces with certainty the beam

(Subscript[J, out1])=(Subscript[j, 1])(Subscript[u, 1]);

which has intensity

Assuming[(p^2)+(q^2)==1,Simplify[(Adj[(Subscript[J, out1])].(Subscript[J, out1]))[[1]][[1]]]]

while presentation to a (Subscript[𝕦, 2])-polarizer produces (with certainty)

(Subscript[J, out2])=(Subscript[j, 2])(Subscript[u, 2]);

which has intensity

Assuming[(p^2)+(q^2)==1,Simplify[(Adj[(Subscript[J, out2])].(Subscript[J, out2]))[[1]][[1]]]]

Action of a quantum polarizer: Turning to the analogous quantum mechanical sitution, we begin by incorporating the 𝕦-matrices into the design of matrix that represents the action of a "𝕌-meter"

𝕌=(Subscript[𝕦, 1])-(Subscript[𝕦, 2]);

Assuming[(p^2)+(q^2)==1,Simplify[Eigenvalues[𝕌]]]

When presented with a photon in state J the meter either announces +1 ("up"), signaling that it prepared a photon in state  (Subscript[u, 1]) (this it does with probability (Subscript[j, 1]^2)), else it announces "dn", signaling that it prepared a photon in state  (Subscript[u, 2]) (this it does with probability (Subscript[j, 2]^2)).

In generic quantum contexts it often proves advantageous to employ the density matrix formalism when analyzing the action of meters (when the quantum system is in a mixed state there is no alternative). Look, for example, to the case of a linearly polarized photon: the density matrix

(Subscript[ρ, in])=(({{(Subscript[ℰ, 1])}, {(Subscript[ℰ, 2])}})).Adj[(({{(Subscript[ℰ, 1])}, {(Subscript[ℰ, 2])}}))];
(Subscript[ρ, in])//MatrixForm

possesses the properties one expects of a pure-state density matrix:

Assuming[(Subscript[ℰ, 1]^2)+(Subscript[ℰ, 2]^2)==1,Simplify[Tr[(Subscript[ρ, in])]]]
Assuming[(Subscript[ℰ, 1]^2)+(Subscript[ℰ, 2]^2)==1,Simplify[Tr[(Subscript[ρ, in]).(Subscript[ρ, in])]]]

The meter announce "up/dn" with probabilities (that yield unity when added)

(Subscript[Prob, up])=Assuming[{(Subscript[ℰ, 1]^2)+(Subscript[ℰ, 2]^2)==1,(p^2)+(q^2)==1},Simplify[Tr[(Subscript[ρ, in]).(Subscript[𝕦, 1])]]]
(Subscript[Prob, dn])=Assuming[{(Subscript[ℰ, 1]^2)+(Subscript[ℰ, 2]^2)==1,(p^2)+(q^2)==1},Simplify[Tr[(Subscript[ρ, in]).(Subscript[𝕦, 2])]]]
Assuming[{(Subscript[ℰ, 1]^2)+(Subscript[ℰ, 2]^2)==1,(p^2)+(q^2)==1},Simplify[(Subscript[Prob, up])+(Subscript[Prob, dn])]]

The meter-created states are given respectively by

(Subscript[ρ, outup])=Assuming[{(Subscript[ℰ, 1]^2)+(Subscript[ℰ, 2]^2)==1,(p^2)+(q^2)==1},Simplify[(((Subscript[𝕦, 1]).(Subscript[ρ, in]).(Subscript[𝕦, 1]))/(Tr[(Subscript[ρ, in]).(Subscript[𝕦, 1])]))==(Subscript[𝕦, 1])]]
(Subscript[ρ, outdn])=Assuming[{(p^2)+(q^2)==1,((Subscript[j, 1])^2)+((Subscript[j, 2])^2)==1},Simplify[(((Subscript[𝕦, 2]).(Subscript[ρ, in]).(Subscript[𝕦, 2]))/(Tr[(Subscript[ρ, in]).(Subscript[𝕦, 2])]))==(Subscript[𝕦, 2])]]

Such computational labor can, however, be avoided if the input state is presented in "meter pre-adapted form" (as Hardy was at pains to do)

(Subscript[J, in])//MatrixForm

(Subscript[ρ, in])=Assuming[{(Subscript[ℰ, 1]^2)+(Subscript[ℰ, 2]^2)==1,(Subscript[j, 1]^2)+(Subscript[j, 2]^2)==1},Simplify[(Subscript[J, in]).Adj[(Subscript[J, in])]]];
(Subscript[ρ, in])//MatrixForm

To illustrate the point, I turn to a generic instance of the physical situation contemplated by Hardy:

Generic entangled pair of photons, meters of two generic designs

Alice has a 𝕦-meter, with eigenstates/projectors

(Subscript[u, 1])=(({{p}, {q }}));
(Subscript[u, 2])=(({{-q}, {p }}));
(Subscript[𝕦, 1])=(Subscript[u, 1]).Adj[(Subscript[u, 1])];
(Subscript[𝕦, 2])=(Subscript[u, 2]).Adj[(Subscript[u, 2])];

Bob has a 𝕧-meter, with

(Subscript[v, 1])=(({{P}, {Q }}));
(Subscript[v, 2])=(({{-Q}, {P }}));
(Subscript[𝕧, 1])=(Subscript[v, 1]).Adj[(Subscript[v, 1])];
(Subscript[𝕧, 2])=(Subscript[v, 2]).Adj[(Subscript[v, 2])];

Alice/Bob's meters are (in God's spectral resolutions) described by the traceless hermitian 4×4 matrices

𝕌=(Subscript[𝕦, 1])⊗𝕀-(Subscript[𝕦, 2])⊗𝕀;
𝕍=𝕀⊗(Subscript[𝕧, 1])-𝕀⊗(Subscript[𝕧, 2]);

𝕌//MatrixForm
𝕍//MatrixForm

The state vector Ψ of a composite 2-photon system lives in a 4-space, where the 4-vectors

(Subscript[uv, 1])=(Subscript[u, 1])⊗(Subscript[v, 1]);
(Subscript[uv, 2])=(Subscript[u, 1])⊗(Subscript[v, 2]);
(Subscript[uv, 3])=(Subscript[u, 2])⊗(Subscript[v, 1]);
(Subscript[uv, 4])=(Subscript[u, 2])⊗(Subscript[v, 2]);

provide an orthonormal basis. The associated projectors are

(Subscript[𝕦𝕧, 1])=(Subscript[uv, 1]).Adj[(Subscript[uv, 1])];
(Subscript[𝕦𝕧, 2])=(Subscript[uv, 2]).Adj[(Subscript[uv, 2])];
(Subscript[𝕦𝕧, 3])=(Subscript[uv, 3]).Adj[(Subscript[uv, 3])];
(Subscript[𝕦𝕧, 4])=(Subscript[uv, 4]).Adj[(Subscript[uv, 4])];

The meter pre-adapted representation of Ψ reads

Ψ=(Subscript[ψ, 1])(Subscript[uv, 1])+(Subscript[ψ, 2])(Subscript[uv, 2])+(Subscript[ψ, 3])(Subscript[uv, 3])+(Subscript[ψ, 4])(Subscript[uv, 4]);
Ψ//MatrixForm

where I will, for notational simplicity, assume the coordinates ψ are real. The associated density matrix reads

ρρ=Ψ.Adj[Ψ];
ρρ//MatrixForm

Alice makes a measurement, gets "up/dn" with probabilities that I use density matrix formalism to evaluate

(Subscript[Prob, (Subscript[A, up])])=Assuming[{(p^2)+(q^2)==1,(P^2)+(Q^2)==1,(Subscript[ψ, 1]^2)+(Subscript[ψ, 2]^2)+(Subscript[ψ, 3]^2)+(Subscript[ψ, 4]^2)==1},Simplify[Tr[ρρ.((Subscript[𝕦, 1])⊗𝕀)]]];
(Subscript[Prob, (Subscript[A, dn])])=Assuming[{(p^2)+(q^2)==1,(P^2)+(Q^2)==1,(Subscript[ψ, 1]^2)+(Subscript[ψ, 2]^2)+(Subscript[ψ, 3]^2)+(Subscript[ψ, 4]^2)==1},Simplify[Tr[ρρ.((Subscript[𝕦, 2])⊗𝕀)]]];

but which, as I demonstrate, can be obtained directly from the "meter pre-adapted" representation of Ψ:

Assuming[{(p^2)+(q^2)==1,(P^2)+(Q^2)==1,(Subscript[ψ, 1]^2)+(Subscript[ψ, 2]^2)+(Subscript[ψ, 3]^2)+(Subscript[ψ, 4]^2)==1},Simplify[(Subscript[Prob, (Subscript[A, up])])==(Subscript[ψ, 1]^2)+(Subscript[ψ, 2]^2)]]
Assuming[{(p^2)+(q^2)==1,(P^2)+(Q^2)==1,(Subscript[ψ, 1]^2)+(Subscript[ψ, 2]^2)+(Subscript[ψ, 3]^2)+(Subscript[ψ, 4]^2)==1},Simplify[(Subscript[Prob, (Subscript[A, dn])])==(Subscript[ψ, 3]^2)+(Subscript[ψ, 4]^2)]]

The measurement-created states that Alice ships to Bob would, in density matrix formalism, be computed

(Subscript[ρρ, Aup])=Assuming[{(p^2)+(q^2)==1,(P^2)+(Q^2)==1,(Subscript[ψ, 1]^2)+(Subscript[ψ, 2]^2)+(Subscript[ψ, 3]^2)+(Subscript[ψ, 4]^2)==1},Simplify[((((Subscript[𝕦, 1])⊗𝕀).ρρ.((Subscript[𝕦, 1])⊗𝕀))/(Tr[ρρ.((Subscript[𝕦, 1])⊗𝕀)]))]];
(Subscript[ρρ, Adn])=Assuming[{(p^2)+(q^2)==1,(P^2)+(Q^2)==1,(Subscript[ψ, 1]^2)+(Subscript[ψ, 2]^2)+(Subscript[ψ, 3]^2)+(Subscript[ψ, 4]^2)==1},Simplify[((((Subscript[𝕦, 2])⊗𝕀).ρρ.((Subscript[𝕦, 2])⊗𝕀))/(Tr[ρρ.((Subscript[𝕦, 2])⊗𝕀)]))]];

but become immediately evident if one works from the "meter pre-adapted" representation of Ψ:

Assuming[{(p^2)+(q^2)==1,(P^2)+(Q^2)==1,(Subscript[ψ, 1]^2)+(Subscript[ψ, 2]^2)+(Subscript[ψ, 3]^2)+(Subscript[ψ, 4]^2)==1},Simplify[(Subscript[ρρ, Aup])==((((Subscript[ψ, 1])(Subscript[uv, 1])+(Subscript[ψ, 2])(Subscript[uv, 2])).Adj[((Subscript[ψ, 1])(Subscript[uv, 1])+(Subscript[ψ, 2])(Subscript[uv, 2]))])/((Subscript[ψ, 1]^2)+(Subscript[ψ, 2]^2)))]]
Assuming[{(p^2)+(q^2)==1,(P^2)+(Q^2)==1,(Subscript[ψ, 1]^2)+(Subscript[ψ, 2]^2)+(Subscript[ψ, 3]^2)+(Subscript[ψ, 4]^2)==1},Simplify[(Subscript[ρρ, Adn])==((((Subscript[ψ, 3])(Subscript[uv, 3])+(Subscript[ψ, 4])(Subscript[uv, 4])).Adj[((Subscript[ψ, 3])(Subscript[uv, 3])+(Subscript[ψ, 4])(Subscript[uv, 4]))])/((Subscript[ψ, 3]^2)+(Subscript[ψ, 4]^2)))]]

Bob will, with his 𝕧-meter, get "up/dn" with conditional probabilities

(Subscript[Prob, AupBup])=Assuming[{(p^2)+(q^2)==1,(P^2)+(Q^2)==1,(Subscript[ψ, 1]^2)+(Subscript[ψ, 2]^2)+(Subscript[ψ, 3]^2)+(Subscript[ψ, 4]^2)==1},Simplify[Tr[(Subscript[ρρ, Aup]).(𝕀⊗(Subscript[𝕧, 1]))]==((Subscript[ψ, 1]^2)/((Subscript[ψ, 1]^2)+(Subscript[ψ, 2]^2)))]]
(Subscript[Prob, AupBdn])=Assuming[{(p^2)+(q^2)==1,(P^2)+(Q^2)==1,(Subscript[ψ, 1]^2)+(Subscript[ψ, 2]^2)+(Subscript[ψ, 3]^2)+(Subscript[ψ, 4]^2)==1},Simplify[Tr[(Subscript[ρρ, Aup]).(𝕀⊗(Subscript[𝕧, 2]))]==((Subscript[ψ, 2]^2)/((Subscript[ψ, 1]^2)+(Subscript[ψ, 2]^2)))]]
(Subscript[Prob, AdnBup])=Assuming[{(p^2)+(q^2)==1,(P^2)+(Q^2)==1,(Subscript[ψ, 1]^2)+(Subscript[ψ, 2]^2)+(Subscript[ψ, 3]^2)+(Subscript[ψ, 4]^2)==1},Simplify[Tr[(Subscript[ρρ, Adn]).(𝕀⊗(Subscript[𝕧, 1]))]==((Subscript[ψ, 3]^2)/((Subscript[ψ, 3]^2)+(Subscript[ψ, 4]^2)))]]
(Subscript[Prob, AdnBdn])=Assuming[{(p^2)+(q^2)==1,(P^2)+(Q^2)==1,(Subscript[ψ, 1]^2)+(Subscript[ψ, 2]^2)+(Subscript[ψ, 3]^2)+(Subscript[ψ, 4]^2)==1},Simplify[Tr[(Subscript[ρρ, Adn]).(𝕀⊗(Subscript[𝕧, 2]))]==((Subscript[ψ, 4]^2)/((Subscript[ψ, 3]^2)+(Subscript[ψ, 4]^2)))]]

If, on the other hand, Bob makes the first measurement, his expected results are described by marginal probabilities

(Subscript[Prob, Bup])=(Subscript[ψ, 1]^2)+(Subscript[ψ, 3]^2)
(Subscript[Prob, Bdn])=(Subscript[ψ, 2]^2)+(Subscript[ψ, 4]^2)

and Alice's expected results are described by the conditional probablities

(Subscript[Prob, BupAup])=((Subscript[ψ, 1]^2)/((Subscript[ψ, 1]^2)+(Subscript[ψ, 3]^2)))
(Subscript[Prob, BupAdn])=((Subscript[ψ, 3]^2)/((Subscript[ψ, 1]^2)+(Subscript[ψ, 3]^2)))
(Subscript[Prob, BdnAup])=((Subscript[ψ, 2]^2)/((Subscript[ψ, 2]^2)+(Subscript[ψ, 4]^2)))
(Subscript[Prob, BdnAdn])=((Subscript[ψ, 4]^2)/((Subscript[ψ, 2]^2)+(Subscript[ψ, 4]^2)))

The joint probabilities—computed by appeal to the general formula

—are independent of who makes the first measurement, and given very simply by

(Subscript[Prob, upup])=(Subscript[ψ, 1]^2)
(Subscript[Prob, updn])=(Subscript[ψ, 2]^2)
(Subscript[Prob, dnup])=(Subscript[ψ, 3]^2)
(Subscript[Prob, dndn])=(Subscript[ψ, 4]^2)

These results make very clear the advantage gained by working from the meter pre-adapted representation of the initial composite state. Which is precisely what Hardy has prepared us to do.

At last: Hardy's nonlocality argument

A composite pair of photons (one susceptible to measurement by Alice, the other by Bob) are assumed to be initially in the composite state

Ψ=α (({{1}, {0}}))⊗(({{1}, {0}}))-β(({{0}, {1}}))⊗(({{0}, {1}}));
Ψ//MatrixForm

Alice (ditto Bob) possesses both a 𝕦-meter and a 𝕧-meter. Prior to each experiment, Alice and Bob, acting independently, decide which of their respective meters they will use. Four meter-pairs are possible, so four meter pre-adapted representrations of Ψ acquire interest.  If the eigenvectors of  𝕦 and 𝕧 are denoted {(Subscript[u, 1]),(Subscript[u, 2])} and {(Subscript[v, 1]),(Subscript[v, 2])} then the elements of those pre-adapted bases are

in terms of which we have

plus two additional such equations that I refrain for the moment from writing out. Hardy fixes the Ψ-dependent design of 𝕦 (i.e., of  {(Subscript[u, 1]),(Subscript[u, 2])}) by imposing the requirement (Subscript[A, 1])=0. This places him in position to fix the design of 𝕧 (i.e., of  {(Subscript[v, 1]),(Subscript[v, 2])}) by imposing the further requirement that (Subscript[B, 4])=0. At this point he has

from which

follow as forced corollaries. (Note that the anticipated C-coefficients are actually a permutation of the B-coefficients—this in consequence of the special structure of Ψ.) Though only the 0s and the fact that (Subscript[D, 1])≠0 are of relevance to Hardy's argument, they are by now known functions of α and β. We have established that squared coefficients can be read as joint probabilities, and that these are independent of whether it is Alice or Bob who makes the first measurement.  I display those joint probabilities below as functions of ξ:

AProbabilities={0,((((Sqrt[α]) (Sqrt[β])))^2),((((Sqrt[α]) (Sqrt[β])))^2)+0.015,(((α-β))^2)}/.{α->Cos[ξ],β->Sin[ξ]};
BProbabilities={((((((Sqrt[α]) (Sqrt[β]) ((α^2)-(β^2)))/((Sqrt[α+β]) (Sqrt[(α^3)+(β^3)])))))^2),(((((α β (Sqrt[α+β]))/(Sqrt[(α^3)+(β^3)]))))^2),(((((Sqrt[(α^3)+(β^3)])/(Sqrt[α+β]))))^2),0}/.{α->Cos[ξ],β->Sin[ξ]};
CProbabilities={((((((Sqrt[α]) (Sqrt[β]) ((α^2)-(β^2)))/((Sqrt[α+β]) (Sqrt[(α^3)+(β^3)])))))^2),(((((Sqrt[(α^3)+(β^3)])/(Sqrt[α+β]))))^2),(((((α β (Sqrt[α+β]))/(Sqrt[(α^3)+(β^3)]))))^2),0}/.{α->Cos[ξ],β->Sin[ξ]};DProbabilities={(((((-α+β)/(-1+α β))))^2),((((((α^(3/2)) (β^(3/2)))/(1-α β))))^2),((((((α^(3/2)) (β^(3/2)))/(1-α β))))^2)+0.015,(((((β (-1+α β+(β^2)))/(1-α β))))^2)}/.{α->Cos[ξ],β->Sin[ξ]};

Plot[AProbabilities,{ξ,0,π/2}, Ticks->{{(1/4)π,(1/2)π},{-1,1}},AxesLabel->{"ξ"}, PlotLabel->"𝕦/𝕦 Probabilities", PlotStyle->{{Red,Thick},{Green,Thick},{Blue,Thick},{Black,Thick}}]
Plot[BProbabilities,{ξ,0,π/2}, Ticks->{{(1/4)π,(1/2)π},{-1,1}},AxesLabel->{"ξ"}, PlotLabel->"𝕦/𝕧 Probabilities", PlotStyle->{{Red,Thick},{Green,Thick},{Blue,Thick},{Black,Thick}}]
Plot[CProbabilities,{ξ,0,π/2}, Ticks->{{(1/4)π,(1/2)π},{-1,1}},AxesLabel->{"ξ"}, PlotLabel->"𝕧/𝕦 Probabilities", PlotStyle->{{Red,Thick},{Green,Thick},{Blue,Thick},{Black,Thick}}]
Plot[DProbabilities,{ξ,0,π/2}, Ticks->{{(1/4)π,(1/2)π},{-1,1}},AxesLabel->{"ξ"}, PlotLabel->"𝕧/𝕧 Probabilities", PlotStyle->{{Red,Thick},{Green,Thick},{Blue,Thick},{Black,Thick}}]

Except for those that vanish identically, all are non-zero except (in some instances) at
ξ = {0, π/4, π/2}. The 0s and 1s that appear at ξ=0 (where α=1 and β=0) can be understood as follows:

A note concerning the edge values of those curves: Those occur at ξ = 0 and ξ = 1, where Ψ is disentangled, so are susceptible to direct analysis. I do so as a check on the accuracy of the results in hand.

At ξ=0 we have

Ψ=up⊗up

while from

(Subscript[u, 1])[α_,β_]:=(({{((Sqrt[β])/(Sqrt[α+β]))}, {-((Sqrt[α])/(Sqrt[α+β]))}}))
(Subscript[u, 2])[α_,β_]:=(({{((Sqrt[α])/(Sqrt[α+β]))}, {((Sqrt[β])/(Sqrt[α+β]))}}))
(Subscript[v, 1])[α_,β_]:=(({{((-(α^(3/2))+(Sqrt[α]) β-(Sqrt[β]) (Sqrt[α β]))/((Sqrt[α+β]) (Sqrt[1-α β])))}, {((-α (Sqrt[β])+(β^(3/2))+(Sqrt[α]) (Sqrt[α β]))/((Sqrt[α+β]) (Sqrt[1-α β])))}}))
(Subscript[v, 2])[α_,β_]:=(({{((α (Sqrt[β])-(β^(3/2))-(Sqrt[α]) (Sqrt[α β]))/((Sqrt[α+β]) (Sqrt[1-α β])))}, {((-(α^(3/2))+(Sqrt[α]) β-(Sqrt[β]) (Sqrt[α β]))/((Sqrt[α+β]) (Sqrt[1-α β])))}}))

we have

(Subscript[u, 1])[1,0]==-dn
(Subscript[u, 2])[1,0]==up
(Subscript[v, 1])[1,0]==-up
(Subscript[v, 2])[1,0]==-dn

The equations

therefore become

which entail

and force all other coefficients to vanish, in precise agreement with the following quotations from the preceding figures:

A4B3C2D1={(((α-β))^2),(((((Sqrt[(α^3)+(β^3)])/(Sqrt[α+β]))))^2),(((((Sqrt[(α^3)+(β^3)])/(Sqrt[α+β]))))^2)+0.015,(((((-α+β)/(-1+α β))))^2)}/.{α->Cos[ξ],β->Sin[ξ]};

Plot[A4B3C2D1,{ξ,0,π/2}, Ticks->{{(1/4)π,(1/2)π},{-1,1}},AxesLabel->{"ξ"}, PlotLabel->"(Subscript[A, 4]), (Subscript[B, 2]), (Subscript[C, 2]) & (Subscript[D, 1]) Probabilities", PlotStyle->{{Red,Thick},{Green,Thick},{Blue,Thick},{Black,Thick}}]

A similar argument accounts for the facts at ξ = 1, where Ψ = -dn⊗dn.

We are in position now to construct the joint, marginal and conditional probabilities that arise in each of the four possible cases:

CASE UU:

From the table of joint probabilities

(Subscript[PUU, up,up])=0
(Subscript[PUU, up,dn])=(Subscript[A, 2]^2)
(Subscript[PUU, dn,up])=(Subscript[A, 3]^2)
(Subscript[PUU, dn,dn])=(Subscript[A, 4]^2)

(here the first subscript refers to Alice's meter reading, the second to Bob's) we conclude that if Alice/Bob both use 𝕦-meters then if one gets "up" the other necessarily gets "dn" (though it remains a possibility that both get "dn"). Another way to expose this same basic fact: construct the marginal probabilities

(Subscript[PUU, Aup])=(Subscript[A, 2]^2)
(Subscript[PUU, Adn])=(Subscript[A, 3]^2)+(Subscript[A, 4]^2)

(Subscript[PUU, Bup])=(Subscript[A, 3]^2)
(Subscript[PUU, Bdn])=(Subscript[A, 2]^2)+(Subscript[A, 4]^2)

and use

to construct tables of marginal probabilities

(Subscript[PUU, Aup|Bup])=(0/(Subscript[A, 3]^2))=0
(Subscript[PUU, Aup|Bdn])=((Subscript[A, 2]^2)/((Subscript[A, 2]^2)+(Subscript[A, 4]^2)))
(Subscript[PUU, Adn|Bup])=((Subscript[A, 3]^2)/(Subscript[A, 3]^2))=1
(Subscript[PUU, Adn|Bdn])=((Subscript[A, 4]^2)/((Subscript[A, 2]^2)+(Subscript[A, 4]^2)))

(Subscript[PUU, Bup|Aup])=(0/(Subscript[A, 2]^2))=0
(Subscript[PUU, Bup|Adn])=((Subscript[A, 3]^2)/((Subscript[A, 3]^2)+(Subscript[A, 4]^2)))
(Subscript[PUU, Bdn|Aup])=((Subscript[A, 2]^2)/(Subscript[A, 2]^2))=1
(Subscript[PUU, Bdn|Adn])=((Subscript[A, 4]^2)/((Subscript[A, 3]^2)+(Subscript[A, 4]^2)))

These state quite explicitly if Alice/Bob both use 𝕌-meters, and one gets "up" then the other gets "dn," and vice versa. We are not surprised to discover that in fact

(Subscript[PUU, up,dn])=(Subscript[A, 2]^2)=(Subscript[A, 3]^2)=(Subscript[PUU, dn,up])

CASE UV:

From the table of joint probabilities

(Subscript[PUV, up,up])=(Subscript[B, 1]^2)
(Subscript[PUV, up,dn])=(Subscript[B, 2]^2)
(Subscript[PUV, dn,up])=(Subscript[B, 3]^2)
(Subscript[PUV, dn,dn])=0

we obtain the marginal probabilities

(Subscript[PUV, Aup])=(Subscript[B, 1]^2)+(Subscript[B, 2]^2)
(Subscript[PUV, Adn])=(Subscript[B, 3]^2)

(Subscript[PUV, Bup])=(Subscript[B, 1]^2)+(Subscript[B, 3]^2)
(Subscript[PUV, Bdn])=(Subscript[B, 2]^2)

whence the conditional probabilities

(Subscript[PUV, Aup|Bup])=((Subscript[B, 1]^2)/((Subscript[B, 1]^2)+(Subscript[B, 3]^2)))
(Subscript[PUV, Aup|Bdn])=((Subscript[B, 2]^2)/(Subscript[B, 2]^2))=1
(Subscript[PUV, Adn|Bup])=((Subscript[B, 3]^2)/((Subscript[B, 1]^2)+(Subscript[B, 3]^2)))
(Subscript[PUV, Adn|Bdn])=(0/(Subscript[B, 2]^2))=0

(Subscript[PUV, Bup|Aup])=((Subscript[B, 1]^2)/((Subscript[B, 1]^2)+(Subscript[B, 2]^2)))
(Subscript[PUV, Bup|Adn])=((Subscript[B, 3]^2)/(Subscript[B, 3]^2))=1
(Subscript[PUV, Bdn|Aup])=((Subscript[B, 2]^2)/((Subscript[B, 1]^2)+(Subscript[B, 2]^2)))
(Subscript[PUV, Bdn|Adn])=(0/(Subscript[B, 3]^2))=0

If Alice uses her 𝕦 meter, and Bob uses his 𝕧 meter, then if Alice gets "dn" then Bob necessarily gets/got "up," and if Bob gets "dn" then Alice necessarily gets/got "up"

CASE VU:

The probabilities here follow from those of CASE UV by (Subscript[B, 2]^2)⇄(Subscript[B, 3]^2) interchange. Thus

(Subscript[PUV, Aup|Bup])=((Subscript[B, 1]^2)/((Subscript[B, 1]^2)+(Subscript[B, 2]^2)))
(Subscript[PUV, Aup|Bdn])=((Subscript[B, 3]^2)/(Subscript[B, 3]^2))=1
(Subscript[PUV, Adn|Bup])=((Subscript[B, 2]^2)/((Subscript[B, 1]^2)+(Subscript[B, 2]^2)))
(Subscript[PUV, Adn|Bdn])=(0/(Subscript[B, 2]^2))=0

(Subscript[PUV, Bup|Aup])=((Subscript[B, 1]^2)/((Subscript[B, 1]^2)+(Subscript[B, 3]^2)))
(Subscript[PUV, Bup|Adn])=((Subscript[B, 2]^2)/(Subscript[B, 32]^2))=1
(Subscript[PUV, Bdn|Aup])=((Subscript[B, 3]^2)/((Subscript[B, 1]^2)+(Subscript[B, 3]^2)))
(Subscript[PUV, Bdn|Adn])=(0/(Subscript[B, 2]^2))=0

Once again, Alice/Bob necessarily get opposite meter readings.

The situation thus far can be summarized as follows:

({{U, U}, {up, dn}, {dn, up}})⟹({{U, V}, {up, dn}, {dn, up}}) & ({{V, U}, {up, dn}, {dn, up}})⟹({{V, V}, {up, dn}, {dn, up}})

All of which could be attributed to hidden instruction sets that tell the respective photons how to respond when they encounter a 𝕦/𝕧 meter. Note particularly that the preceding line of argment leads to the conclusion that

But quantum mechanics assigns non-vanishing probabilities to those cases:

CASE VV:

(Subscript[PVV, up,up])=(Subscript[D, 1]^2)
(Subscript[PVV, up,dn])=(Subscript[D, 2]^2)
(Subscript[PVV, dn,up])=(Subscript[D, 3]^2)
(Subscript[PVV, dn,dn])=(Subscript[D, 4]^2)

Plot[{(((((α-β)/(1-α β))))^2),(((((α(α-β)β)/(1-α β))))^2)}/.{α->Cos[ξ],β->Sin[ξ]},{ξ,0,π/2}, Ticks->{{(1/4)π,(1/2)π},{-1,1}},AxesLabel->{"ξ"}, PlotLabel->"D-coefficients", PlotStyle->Thick]

Assuming[(α^2)+(β^2)==1, Simplify[-((β (-1+α β+(β^2)))/(1-α β))==((α(α-β)β)/(1-α β))]]