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Shadow Theory

Sealed or Leaky Section 2

The source tetralemma

Section 3 of 17

2 The source tetralemma

2.1 Setting and premises

A Bell experiment realized in a source model is specified as follows. Let (Λ,A,ν)(\Lambda,\mathcal A,\nu) be a standard Borel probability space of complete initial source states of everything relevant to the run except the devices that generate the settings. Alice's and Bob's settings x,y∈{0,1}x,y\in\{0,1\} are drawn with probabilities q(x,y)>0q(x,y)>0. A Tier-1 readout is a measurable map τ:Λ→T\tau:\Lambda\to\mathcal T into a standard Borel space. Its value T=τ(λ)T=\tau(\lambda) is the nominated initial classical information. The Tier-1 readout is source-complete if λ=G(T)\lambda=G(T) almost surely for some measurable GG.

Status: Stated premises; unchanged from version 3.

The following premises are not derived here.

(SO) Single outcomes.

For each setting pair, Alice's and Bob's outcomes are ±1\pm1-valued random variables A,BA,B on the run. Each run has exactly one outcome per party, jointly distributed with λ\lambda.

(D) Determinism.

There are measurable FA,FB:Λ×{0,1}2→{±1}F_A,F_B:\Lambda\times\{0,1\}^2\to\{\pm1\} with A=FA(λ,x,y)A=F_A(\lambda,x,y) and B=FB(λ,x,y)B=F_B(\lambda,x,y) almost surely. Dependence on the distant setting is allowed.

(MI) Measurement independence.

The settings (x,y)(x,y) are independent of λ\lambda, and hence of TT.

(NS) Tier-1 no-signalling.

For almost every tt, the conditional box Pt(a,b∣x,y)=P(A=a,B=b∣x,y,T=t)P_t(a,b\mid x,y)=\Prb(A=a,B=b\mid x,y,T=t) is no-signalling: ∑bPt(a,b∣x,y)\sum_bP_t(a,b\mid x,y) does not depend on yy, and ∑aPt(a,b∣x,y)\sum_aP_t(a,b\mid x,y) does not depend on xx.

Realism here means existence of the stipulated source model, not an extra convex-linearity axiom. Write Exy=E[AB∣x,y]E_{xy}=\E[AB\mid x,y] and S=E00+E01+E10−E11S=E_{00}+E_{01}+E_{10}-E_{11}. Under (D), define the finite response variable

Z=(FA(λ,x,y),FB(λ,x,y))x,y∈{0,1}∈{±1}8. Z=\bigl(F_A(\lambda,x,y),F_B(\lambda,x,y)\bigr)_{x,y\in\{0,1\}} \in\{\pm1\}^8.

It settles every outcome at fixed settings, but need not contain every source distinction.

Remark 2.1 (The access quantifier)

Status: Scope.

(NS) is conditional no-signalling for the particular nominated TT, not merely no-signalling of the unconditioned laboratory marginals. The latter does not imply the former. Calling TT all information any observer could ever acquire requires an additional physical access claim, absent here. A conclusion for every admissible TT means a separate implication for each TT satisfying the premises, not proof that every conceivable observer or global source variable is admissible. An optional hypothesis (QT) requires every PtP_t to be quantum-realizable with a tensor-product structure; it does not replace classical TT by unrestricted quantum side information.

Lemma 2.2 (Outcome determinism already supplies single outcomes)

Status: Proved.

Within the stated probability model, (D) supplies the single-valued outcomes required by (SO). Consequently there is no model of the other three displayed premises that violates only (SO) while retaining (D) in its stated meaning.

Proof

For every setting pair, the two measurable functions in (D) take exactly one value in {±1}\{\pm1\} at each source state; their pushforward gives the joint outcome law with λ\lambda. Almost-sure qualifications can be imposed simultaneously for the four setting pairs by taking a finite union of null sets. A branching final vector is not either of these scalar outcome functions. Replacing (D) by deterministic evolution of that vector would change the premise.

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2.2 Guessing and hidden information

Lemma 2.3 (No-signalling guessing bound)

Status: Proved.

For a no-signalling ±1\pm1 box with CHSH value SS, every Alice or Bob marginal has guessing probability at most 3/2−S/43/2-S/4. For 2≤S≤42\le S\le4 this is a sharp bound.

Proof

Put αx=E[A∣x]\alpha_x=\E[A\mid x] and βy=E[B∣y]\beta_y=\E[B\mid y]. Positivity gives 2P(A≠B∣xy)≥∣αx−βy∣2P(A\ne B\mid xy)\ge|\alpha_x-\beta_y| and 2P(A=B∣xy)≥∣αx+βy∣2P(A=B\mid xy)\ge|\alpha_x+\beta_y|. Let L=P(A≠B∣00)+P(A≠B∣01)+P(A≠B∣10)+P(A=B∣11)L=P(A\ne B\mid00)+P(A\ne B\mid01)+P(A\ne B\mid10)+P(A=B\mid11), so S=4−2LS=4-2L. Writing u=α0−β0u=\alpha_0-\beta_0, v=α0−β1v=\alpha_0-\beta_1, w=α1−β0w=\alpha_1-\beta_0, z=α1+β1z=\alpha_1+\beta_1 gives 2L≥∣u∣+∣v∣+∣w∣+∣z∣2L\ge|u|+|v|+|w|+|z|. The combinations u+v−w+z=2α0u+v-w+z=2\alpha_0, −u+v+w+z=2α1-u+v+w+z=2\alpha_1, −u+v−w+z=2β0-u+v-w+z=2\beta_0 and u−v−w+z=2β1u-v-w+z=2\beta_1 bound all four biases by LL. Thus the marginal guessing probability is at most (1+L)/2=3/2−S/4(1+L)/2=3/2-S/4. Mixing the all-plus local box with a PR box attains this value. This is also the bound in [22, Eq. (14)].

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Theorem 2.4 (Hidden-information theorem)

Status: Proved conditional on (SO), (D), (MI), (NS).

Suppose S>2S>2. For each xx and yy,

pguess(A∣X=x,T)≤32−S4<1,Hmin⁡(Z∣T)≥fNS(S)>0,H(Z∣T)≥fNS(S)>0.\begin{align}p_{\rm guess}(A\mid X=x,T)&\le\tfrac32-\tfrac S4<1,\tag{1}\\ H_{\min}(Z\mid T)&\ge\fNS(S)>0,\tag{2}\\ H(Z\mid T)&\ge\fNS(S)>0. \tag{3}\end{align}

The same marginal bound holds for Bob. The readout is not source-complete: on a set of readout values of positive probability, its conditional source law assigns positive probability to different response values. Under (QT), for 2<S≤222<S\le2\sqrt2, replace the guessing bound by

GQ(S)=12(1+2−S2/4) G_Q(S)=\tfrac12\bigl(1+\sqrt{2-S^2/4}\bigr)

and the entropy bounds by fQ(S)=1−log⁡2(1+2−S2/4)\fQ(S)=1-\log_2(1+\sqrt{2-S^2/4}).

Proof

(MI) makes the law of TT independent of the settings, so P=∫Pt dP(t)P=\int P_t\,d\Prb(t) and S=∫St dP(t)S=\int S_t\,d\Prb(t). By (NS) the Alice marginal given x,tx,t is independent of yy. Lemma 2.3 and averaging prove (1). For fixed x,yx,y, the random variable FA(λ,x,y)F_A(\lambda,x,y) is a coordinate of ZZ. Hence, pointwise in tt, the largest atom of Z∣tZ\mid t is no larger than the largest atom of that coordinate. (MI) identifies this coordinate's conditional law with the observed law at x,y,tx,y,t. Averaging gives (2); Shannon entropy dominates average conditional min-entropy, by its pointwise version and Jensen's inequality.

If λ=G(T)\lambda=G(T) almost surely, then ZZ is a function of TT and its conditional entropy is zero, a contradiction. Regular conditional source laws are concentrated on τ−1(t)\tau^{-1}(t) almost everywhere. A nondegenerate finite conditional response distribution therefore supplies different responses in those fibres, as in Theorem 6.1.

For (QT), the single-box bound GQG_Q is imported from [22, Eq. (6)]. Extend it by 11 for s≤2s\le2 on the quantum CHSH domain. The extension is concave: at s=2s=2 its right derivative is −1/4-1/4, no larger than the left derivative 00. Jensen gives EGQ(St)≤GQ(S)\E G_Q(S_t)\le G_Q(S), and the preceding coordinate argument applies unchanged.

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Lemma 2.5 (Sharp Shannon refinement and simultaneous sharpness)

Status: Proved conditional on (SO), (D), (MI), (NS).

For 2≤S≤42\le S\le4,

H(Z∣T)≥gNS(S):=(S−2)/2≥fNS(S). H(Z\mid T)\ge g_{\rm NS}(S):=(S-2)/2\ge\fNS(S). (4)

Both the first bound and Hmin⁡(Z∣T)≥fNS(S)H_{\min}(Z\mid T)\ge\fNS(S) are simultaneously sharp over this no-signalling class with classical TT.

Proof

Concavity of binary entropy on each half of [0,1][0,1] gives h2(p)≥2min⁡(p,1−p)h_2(p)\ge2\min(p,1-p). Lemma 2.3 implies, at each tt,

H(A∣x,T=t)≥(St−2)+/2. H(A\mid x,T=t)\ge (S_t-2)_+/2.

Since u↦(u−2)+u\mapsto(u-2)_+ is convex, averaging and the deterministic-coordinate inequality give H(Z∣T)≥(S−2)/2H(Z\mid T)\ge(S-2)/2. The comparison with fNS\fNS follows because a convex function lies below its endpoint chord: fNS(2)=0\fNS(2)=0, fNS(4)=1\fNS(4)=1.

For sharpness let r=(S−2)/2r=(S-2)/2, let C0C_0 be a Bernoulli flag of mean rr, and let R0R_0 be zero when C0=0C_0=0 and a fair bit when C0=1C_0=1. Take λ=(C0,R0)\lambda=(C_0,R_0), T=C0T=C_0, independent settings, and all-plus responses in the first case. In the second case use A=(−1)R0A=(-1)^{R_0} and B=(−1)R0+xyB=(-1)^{R_0+xy}. The conditional boxes are local deterministic and PR, respectively, hence both no-signalling. Their average CHSH value is 2+2r2+2r. The response tables in the PR branch are distinct and equiprobable, so H(Z∣T)=rH(Z\mid T)=r and pguess(Z∣T)=1−r/2=3/2−S/4p_{\rm guess}(Z\mid T)=1-r/2=3/2-S/4. This is an extremal mathematical construction, not a physical PR-box claim or quantum-class sharpness assertion.

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Remark 2.6 (Reading the entropies)

Status: Scope.

The bounds concern source information not determined by the nominated initial classical readout, given the displayed access premise. They are not observer-independent statements of permanent inaccessibility. They refer to the finite response ZZ, not differential entropy of a continuous source state. Under (QT), H(Z∣T)≥max⁡{gNS(S),fQ(S)}H(Z\mid T)\ge\max\{g_{\rm NS}(S),\fQ(S)\} follows, but the two available bounds do not establish an optimal quantum Shannon curve. A postselected finite run needs the separate analysis of Section 3.

2.3 The four explicit escape models and their exact logical status

Theorem 2.7 (Source tetralemma: exclusion and corrected necessity)

Status: Proved.

A source-complete model with S>2S>2 cannot jointly satisfy (SO), (D), (MI) and (NS). Within the single-outcome model class, each of (D), (MI) and (NS) is separately necessary for this exclusion: models (C2)–(C4) below satisfy the other premises, have identity readout, and reproduce

PQ(a,b∣x,y)=14(1+ab cxy),c00=c01=c10=−c11=1/2. P_Q(a,b\mid x,y)=\tfrac14(1+ab\,c_{xy}),\qquad c_{00}=c_{01}=c_{10}=-c_{11}=1/\sqrt2.

Model (C1) is a source-complete branching realization of the same branch-weight table, outside the single-outcome formalism. It is not a counterexample to Lemma 2.2. Thus four descriptions of escape remain, but the claim of four logically independent premises is false.

Proof

The exclusion is Theorem 2.4. The following four models all take the readout to be the complete specified initial state.

(C1) Branching: outside (SO) and the stated (D). Let the source be a unit ray of the two-qubit-plus-apparatus Hilbert space, prepared in Φ+⊗∣ready⟩\Phi^+\otimes\ket{\mathrm{ready}}. Independent settings select unitary measurement interactions with final vector

∑a,b(Pax⊗Pby)Φ+⊗∣a,b⟩. \sum_{a,b}(P_a^x\otimes P_b^y)\Phi^+\otimes\ket{a,b}.

Use A0op=σzA_0^{\rm op}=\sigma_z, A1op=σxA_1^{\rm op}=\sigma_x, B0op=(σz+σx)/2B_0^{\rm op}=(\sigma_z+\sigma_x)/\sqrt2, B1op=(σz−σx)/2B_1^{\rm op}=(\sigma_z-\sigma_x)/\sqrt2. Orthogonal pointer branches have squared norms PQ(a,b∣x,y)P_Q(a,b\mid x,y), a no-signalling table. The complete vector is a deterministic function of source and settings, but no single actual pair A,BA,B has been selected. This is complete-state determinism, not (D). The branch-weight reading is the relative-state alternative [30, 31, 32]; derivation of subjective probabilities from branch weights is not a premise or result here.

(C2) Fundamental chance: violates only (D). Take Λ\Lambda to be pure rays with ν=δΦ+\nu=\delta_{\Phi^+}, τ=id\tau=\mathrm{id}, and independent settings. Sample one pair from PQP_Q using fundamental chance not encoded in the initial source state. Then (SO), (MI) and (NS) hold, while (D) fails. Completeness here is completeness of the initial physical state, not a hidden initial encoding of subsequent chance. Pure-ray chance ontologies supply the relevant contrast [2, 33].

(C3) Measurement dependence: violates only (MI). Let λ=(a0,a1,b0,b1)∈{±1}4\lambda=(a_0,a_1,b_0,b_1)\in\{\pm1\}^4, A=axA=a_x, B=byB=b_y, and take any positive setting law q(x,y)q(x,y) with

ν(λ∣x,y)=14PQ(ax,by∣x,y). \nu(\lambda\mid x,y)=\tfrac14 P_Q(a_x,b_y\mid x,y).

The two unused entries are uniform. Summing over them reproduces PQP_Q and normalizes each conditional source law. All its entries are positive, so each setting pair remains in the support at every λ\lambda. Conditional on T=λT=\lambda, the responses are local deterministic and satisfy (NS). They satisfy (SO) and (D), but the source law depends on the settings. This is a measurement-dependent construction of the Brans type [35].

(C4) Accessible nonlocality: violates only (NS). Take λ=(u,v)\lambda=(u,v) uniform on [0,1]2[0,1]^2, independently of the settings, with T=λT=\lambda, and define

A=sgn(1/2−u),B=A sgn((1+cxy)/2−v), A=\mathrm{sgn}(1/2-u),\qquad B=A\,\mathrm{sgn}((1+c_{xy})/2-v),

assigning sgn(0)=+1\mathrm{sgn}(0)=+1 on the irrelevant threshold sets. Its averaged joint probabilities are PQP_Q. It satisfies (SO), (D) and (MI). For y=1y=1, on the positive-measure interval between (1−1/2)/2(1-1/\sqrt2)/2 and (1+1/2)/2(1+1/\sqrt2)/2, Bob's definite output changes with xx. Its box conditional on the readable λ\lambda therefore violates (NS), although its unconditioned marginals do not signal. No claim that such complete access is physically available is required for the countermodel.

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Proposition 2.8 (The three-branch form fails)

Status: Proved.

The assertion that every source-complete realization of quantum Bell statistics is branching, source-indeterministic, or an operational surrogate violating convex-linear realism is false. Neither convex-linearity nor its failure is needed in the exclusion theorem.

Proof

(C3) and (C4) are single-outcome, source-deterministic and source-complete. Neither is forced into the proposed surrogate branch. With a single preparation, convex-linearity imposes no nontrivial requirement; random mixtures of admitted preparation laws can moreover be represented by their mixtures. A density-matrix-only stochastic surrogate is already in the chance branch for the nondegenerate Tsirelson marginals. None of the arguments of Theorem 2.4 uses convex-linearity. These observations refute the proposed universal three-way claim; they do not classify every possible surrogate.

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Proposition 2.9 (Where convex-linearity bites: the chance branch)

Status: Proved conditional on a convex-linear ontological model reproducing the trine table and tomographically complete qubit statistics.

A pure-state-complete chance model is possible, as in (C2). Nevertheless its preparation-law readout cannot in general be identified with only the density matrix if convex-linearity is imposed. The fibre over I/2I/2 contains ontic preparation laws at pairwise TV distance at least 1/41/4, and at summed distance at least 1/21/2 from the trine mixture. This is a statement about preparation laws, not a proof that each pure ray is an incomplete individual state.

Proof

Apply Corollary 8.4 and Theorem 8.3. The Beltrametti–Bugajski model distinguishes proper-mixture laws while retaining rays as individual states [2].

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