The relaxed Hardy test under no-signalling: at most 1/4 for four parties

Abstract

For four parties with two outcomes each, every no-signalling box that satisfies the relaxed Hardy conditions of S. S. Bhattacharya, A. Roy, A. Mukherjee and R. Rahaman (arXiv:1507.07327) has success probability q at most 1/4. This refutes their conjecture that the optimal success probability under no-signalling is 1/3 for any number of parties and any dimension, and corrects their observation of the value 1/3 for four parties.

Definition 1.1 (No-signalling boxes).

$$\forall N:\mathbb{N},\forall d:\mathbb{N},\forall P:(\mathrm{Fin}\left(N\right)\to \mathrm{Bool})\to \left((\mathrm{Fin}\left(N\right)\to \mathrm{Fin}\left(d\right))\to \mathbb{R}\right),\mathrm{IsNSBox}\left(N,d,P\right)\iff ((\forall s:\mathrm{Fin}\left(N\right)\to \mathrm{Bool},\forall x:\mathrm{Fin}\left(N\right)\to \mathrm{Fin}\left(d\right),0\le P\left(s,x\right))\wedge \left((\forall s:\mathrm{Fin}\left(N\right)\to \mathrm{Bool},\sum _{x}P\left(s,x\right)=1)\wedge \forall p:\mathrm{Fin}\left(N\right),\forall s:\mathrm{Fin}\left(N\right)\to \mathrm{Bool},\forall t:\mathrm{Fin}\left(N\right)\to \mathrm{Bool},(\forall k:\mathrm{Fin}\left(N\right),k\ne p\Rightarrow s_k=t_k)\Rightarrow \forall x:\mathrm{Fin}\left(N\right)\to \mathrm{Fin}\left(d\right),\sum _{a}P\left(s,x[p:=a]\right)=\sum _{a}P\left(t,x[p:=a]\right)\right))$$

Formalization. D5/S3/Quantum/Entanglement/RelaxedHardyNoSignallingOptimum.IsNSBox (✓ std3).

Citation. Some Sankar Bhattacharya; Arup Roy; Amit Mukherjee; Ramij Rahaman (2015). Witnessing Genuine Mutipartite Non-locality. DOI: 10.48550/arXiv.1507.07327. URL: https://arxiv.org/abs/1507.07327v1.

Commentary.

A box of N parties with d outcomes assigns to every input string s in {u, v}^N (in Lean a map Fin N -> Bool, with false for u and true for v) a probability distribution P(s, .) on the outcome strings x in (Fin d)^N. It is no-signalling when, for every party p and all input strings s, t that agree away from p, the marginal of the other parties is the same: summing P(s, x) over the outcome a of p, with x[p := a] the string x with a at p, gives the same value for s and t.

Definition 1.2 (The relaxed Hardy conditions).

$$\forall N:\mathbb{N},\forall d:\mathbb{N},\forall P:(\mathrm{Fin}\left(N\right)\to \mathrm{Bool})\to \left((\mathrm{Fin}\left(N\right)\to \mathrm{Fin}\left(d\right))\to \mathbb{R}\right),\forall q:\mathbb{R},\mathrm{RelaxedHardy}\left(N,d,P,q\right)\iff (((0<q)\wedge \forall x:\mathrm{Fin}\left(N\right)\to \mathrm{Fin}\left(d\right),(\forall k:\mathrm{Fin}\left(N\right),x_k=0)\Rightarrow P\left(\varnothing ,x\right)=q)\wedge \left((\forall r:\mathrm{Fin}\left(N\right),\forall x:\mathrm{Fin}\left(N\right)\to \mathrm{Fin}\left(d\right),(\forall k:\mathrm{Fin}\left(N\right),k\ne r\Rightarrow x_k=0)\Rightarrow \left(x_r\ne d-1\Rightarrow P\left(\{r\},x\right)=0\right))\wedge \exists j:\mathrm{Fin}\left(N\right),\forall i:\mathrm{Fin}\left(N\right),i\ne j\Rightarrow \forall x:\mathrm{Fin}\left(N\right)\to \mathrm{Fin}\left(d\right),((x_i=d-1)\wedge \left((x_j=d-1)\wedge \forall k:\mathrm{Fin}\left(N\right),((k\ne i)\wedge k\ne j)\Rightarrow x_k=0\right))\Rightarrow P\left(\{i,j\},x\right)=0\right))$$

Formalization. D5/S3/Quantum/Entanglement/RelaxedHardyNoSignallingOptimum.RelaxedHardy (✓ std3).

Citation. Some Sankar Bhattacharya; Arup Roy; Amit Mukherjee; Ramij Rahaman (2015). Witnessing Genuine Mutipartite Non-locality. DOI: 10.48550/arXiv.1507.07327. URL: https://arxiv.org/abs/1507.07327v1.

Commentary.

The outcome value 0 stands for the paper’s outcome 1 and the value d - 1 for its outcome d. With all inputs u, all outcomes 1 have probability q > 0. With v at party r only, every outcome string with 1 at the other parties and a value other than d at r has probability 0. For one fixed party j and every i other than j, with v at i and j, the outcome string with d at i and j and 1 elsewhere has probability 0. In the display an input string is written as the set of parties with input v: the empty set for all inputs u, {r} for v at r only, {i, j} for v at i and j (in Lean the maps k |-> false, k |-> decide (k = r) and k |-> decide (k = i or k = j)).

Definition 1.3 (The conjecture).

$$claim\iff (\forall N:\mathbb{N},\forall d:\mathbb{N},3\le N\Rightarrow \left(2\le d\Rightarrow \forall \epsilon :\mathbb{R},0<\epsilon \Rightarrow \exists P:(\mathrm{Fin}\left(N\right)\to \mathrm{Bool})\to \left((\mathrm{Fin}\left(N\right)\to \mathrm{Fin}\left(d\right))\to \mathbb{R}\right),\exists q:\mathbb{R},(\mathrm{IsNSBox}\left(N,d,P\right))\wedge \left((\mathrm{RelaxedHardy}\left(N,d,P,q\right))\wedge \frac{1}{3}-\epsilon <q\right)\right))$$

Formalization. D5/S3/Quantum/Entanglement/RelaxedHardyNoSignallingOptimum.claim (✓ std3).

Citation. Some Sankar Bhattacharya; Arup Roy; Amit Mukherjee; Ramij Rahaman (2015). Witnessing Genuine Mutipartite Non-locality. DOI: 10.48550/arXiv.1507.07327. URL: https://arxiv.org/abs/1507.07327v1.

Commentary.

For every number N >= 3 of parties, every common number d >= 2 of outcomes and every epsilon > 0, some no-signalling box satisfies the relaxed Hardy conditions with success probability q > 1/3 - epsilon. The paper conjectures that the optimal success probability is 1/3 for any dimension and any number of parties, which implies this whether the optimum is read as a maximum or a supremum; requiring N >= 3 and a common d only weakens the claim, and the fixed party j may be chosen.

Theorem 1.4 (At most 1/4 for four parties).

$$\neg claim$$

Proof. Machine-checked in Lean as D5/S3/Quantum/Entanglement/RelaxedHardyNoSignallingOptimum.result (✓ std3). ∎

Resolves. Problems/bhattacharya-2015-relaxed-hardy-no-signalling-optimum (refuted) by D5/S3/Quantum/Entanglement/RelaxedHardyNoSignallingOptimum.result.

Source. Repository-derived.

Acknowledgement. Some Sankar Bhattacharya; Arup Roy; Amit Mukherjee; Ramij Rahaman (2015). Witnessing Genuine Mutipartite Non-locality. DOI: 10.48550/arXiv.1507.07327. URL: https://arxiv.org/abs/1507.07327v1.

Commentary.

Take N = 4, d = 2 and a box with success probability q. Let m_r be the probability of 1 at all parties other than r under the inputs u. No-signalling at r and the second condition give probability m_r to the outcome d at r and 1 elsewhere under S_r. For i other than j, no-signalling at j, the third condition and nonnegativity give the outcome d at i and 1 elsewhere probability at least m_i under S_ij, and symmetrically for j. No-signalling at i and at j moves the marginal of the remaining parties back to the inputs u, so the outcome d at i and j with 1 elsewhere has probability at least q under u. These three outcome strings and the all-1 string are distinct, so normalization gives 4 q <= 1, and q > 1/3 - 1/12 = 1/4 is impossible. In Lean the four choices of j are split, and in each case the instantiated equalities and inequalities are combined by linarith.

References

• Truth anchor: D5/S3/Quantum/Entanglement/RelaxedHardyNoSignallingOptimum.IsNSBox
• Truth anchor: D5/S3/Quantum/Entanglement/RelaxedHardyNoSignallingOptimum.RelaxedHardy
• Truth anchor: D5/S3/Quantum/Entanglement/RelaxedHardyNoSignallingOptimum.claim
• Truth anchor: D5/S3/Quantum/Entanglement/RelaxedHardyNoSignallingOptimum.result