# Witnessing Bell violations through probabilistic negativity

Benjamin Morris<EMAIL_ADDRESS>School of Physics and Astronomy and
Centre for the Mathematics and Theoretical Physics of Quantum Non-Equilibrium
Systems, University of Nottingham, University Park, Nottingham NG7 2RD, United
Kingdom Lukas J. Fiderer  Institute for Theoretical Physics, University of
Innsbruck, 6020 Innsbruck, Austria Ben Lang  School of Physics and Astronomy
and Centre for the Mathematics and Theoretical Physics of Quantum Non-
Equilibrium Systems, University of Nottingham, University Park, Nottingham NG7
2RD, United Kingdom Daniel Goldwater  School of Mathematical Sciences and
Centre for the Mathematics and Theoretical Physics of Quantum Non-Equilibrium
Systems, University of Nottingham, University Park, Nottingham NG7 2RD, United
Kingdom

###### Abstract

Bell’s theorem shows that no hidden-variable model can explain the measurement
statistics of a quantum system shared between two parties, thus ruling out a
classical (local) understanding of nature. In this work we demonstrate that by
relaxing the positivity restriction in the hidden-variable probability
distribution it is possible to derive quasiprobabilistic Bell inequalities
whose sharp upper bound is written in terms of a negativity witness of said
distribution. This provides an analytic solution for the amount of negativity
necessary to violate the CHSH inequality by an arbitrary amount, therefore
revealing the amount of negativity required to emulate the quantum statistics
in a Bell test.

##  1 Introduction

It has now been 60 years since John Stewart Bell wrote his famous paper on the
Einstein-Podolsky-Rosen (EPR) paradox bell1964einstein , and 50 years since
the first experimental Bell test freedman1972experimental . The majority of
physicists are perfectly happy to concede that in the lab we see experimental
results consistent with the postulates of quantum mechanics. However, the
implications of these mathematical postulates on the ‘reality’ of the
wavefunction is still very much up for debate caves2002quantum ;
harrigan2010einstein ; pusey2012reality ; colbeck2012system ;
mermin2014physics ; ringbauer2015measurements .

These Bell experiments remain as some of the most important demonstrations for
the reality of the quantum state and the death of a ‘local realism’ picture of
nature. In such an experiment a physical system is distributed between
spatially separated observers, and we allow these observers to perform
measurements on their local system. The emerging statistics prove that
physical systems are not bound to behave locally (in accordance to local
hidden-variable models). Rather, the statistics are consistent with the
postulates governing quantum mechanics.

In this work we remove the postulates of quantum mechanics and instead allow a
physical system to be distributed according to a quasiprobability (hidden-
variable) distribution that is allowed to take negative values. Although we
are perfectly content with real negative numbers in physics, negative
quasiprobabilities– despite receiving support from individuals such as Dirac
dirac1942bakerian  and Feynman feynman1987negative  and having a solid
mathematical foundation ruzsa1988algebraic ; khrennikov2007generalized – have
been a long debated issue in theoretical physics muckenheim1986review . See,
for example, the extensive discussion surrounding the interpretation of
negative values in the Wigner distribution ferrie2011quasi ;
veitch2012negative . In the majority of considerations, quasiprobability
distributions are used to describe states that are not directly observed; that
is, all observable measurement statistics must be governed by ordinary
probability distributions. As an example, a Wigner function may assign a
negative quasiprobability to a particle having a particular position/momentum
combination, but any physical measurement, constrained by Heisenberg
uncertainty, will have an all-positive outcome distribution. This feature
ensures that no outcome is ever predicted to be seen occurring a negative
number of times feynman1987negative , and similarly protects the
quasiprobability physicist from falling victim to ‘Dutch book’ arguments
(de2017theory, , Ch.3).

An important motivator for this work is the result of Al-Safi and Short
al2013simulating  (expanded upon by the authors of oas2014exploring ) which
showed that it is possible to simulate all non-signalling correlations, (those
which adhere to the principles of special relativity) popescu1994quantum ;
peres2004quantum , if one allows negative values in a probability
distribution. However, physical reality does not explore this full set of
correlations – but rather, is restricted to those achievable by quantum
correlations. Therefore the question that we pose in this work is:

> “What are the restrictions on the negativity in a hidden-variable
> probability distribution such that it can emulate the statistics seen in a
> physical Bell experiment?”

In order to answer this question we construct CHSH inequalities for two
parties clauser1969proposed  whose degree of violation is witnessed by the
amount of negativity present in the hidden-variable probability distribution.
Our witness yields a value of 0 for a quasiprobability distribution which is
entirely positive, such as that which would describe an ordinary classical
system.

We start by describing the setup necessary for the construction of these
nonlocal experiments, introducing the probability distributions admitted by
classical, quantum, and non-signalling theories, and giving the famous Bell
scores that these distributions respectively allow one to reach in such
nonlocal experiments. We then give the definition of a quasiprobability
distribution and motivate negativity witnesses as a quantitative method of
detecting negativity in said distributions. Our main result is that the
violation of the CHSH inequality (and $n$-measurement generalisations) can be
exactly characterised by a negativity witness of the hidden-variable
distribution defined over the local states, and that there exists
quasiprobability distributions which can saturate (up to the no-signalling
limit) any such violation, whilst still having well-defined local statistics.
This shows that it is possible to recapture the nonlocal features of Bell
experiments through having a finite amount of negativity allowed in a hidden-
variable distribution over scenarios which are, in themselves, entirely local
and classical.

Figure 1: A source $S$ distributes a system between two spatially separated
observers, Alice ($A$) and Bob ($B$). Alice and Bob choose to measure their
local part of the system with measurements $x_{A},x_{B}$, with possible
outcomes $y_{A},y_{B}\in\\{-1,+1\\}$. The statistics of their measurement
outcomes depending upon the physical system being distributed by the source.

##  2 Setup

Let us consider the following experimental setup. A source $S$ distributes a
system between $2$ observers, the $k$th observer can choose some measurement
$x_{k}\in\\{0,1,\dots,X_{k}\\}$ and record some outcome
$y_{k}\in\\{1,2,\dots,Y_{k}\\}$, the possible values of $k$ being $\\{A,B\\}$.
A specific experimental setup is characterised by the conditional probability,

$\displaystyle P\left(y_{A},y_{B}|x_{A},x_{B}\right).$ (1)

The physical theory governing the behaviour of the system and experiment
determines the achievability of certain conditional probability distributions
resulting from these experiments. We are interested in the following three
physical theories:

$\bullet$ Classical theory admits probability distributions of the following
form,

$\displaystyle
P\left(y_{A},y_{B}|x_{A},x_{B}\right)=\sum_{\lambda_{A},\lambda_{B}}P_{A}\left(y_{A}|x_{A},\lambda_{A}\right)P_{B}\left(y_{B}|x_{B},\lambda_{B}\right)P_{\Lambda}\left(\lambda_{A},\lambda_{B}\right),$
(2)

where $P_{\Lambda}\left(\lambda_{A},\lambda_{B}\right)$ is a joint probability
distribution defined over local hidden variables. With each choice of hidden
variables we associate a local scenario governed by ordinary local probability
distributions $P_{k}\left(y_{k}|x_{k},\lambda_{k}\right)$ for the observables
$y_{A},y_{B}$. The hidden-variable probability distribution $P_{\Lambda}$
determines how such local scenarios are mixed, and the probability
distributions $P_{k}\left(y_{k}|x_{k},\lambda_{k}\right)$ are called
“$\lambda_{k}$-local” because they belong to the scenario associated with a
particular value of $\lambda_{k}$, not to be confused with the (observable)
marginal probability distributions that are obtained by marginalising the
total probability distribution, equation (2).

$\bullet$ Quantum theory endows us with a Hilbert space structure for our
quantum states that admits probability distributions of the following form,

$\displaystyle
P\left(y_{A},y_{B}|x_{A},x_{B}\right)=\operatorname{Tr}\left[\left(M^{(A)}_{y_{A}|x_{A}}\otimes
M^{(B)}_{y_{B}|x_{B}}\right)\rho\right],$ (3)

where $\rho\succeq 0$ and $M^{(k)}_{y_{k}|x_{k}}$ are Positive Operator-Valued
Measures (POVMs) 2000quantum  for each $k$.

$\bullet$ No-signalling principle, our third physical theory, prohibits the
sending of information faster than the speed of light popescu1994quantum ;
peres2004quantum . Such a theory has the conditions on its conditional
probability distribution that for any $k\in\\{A,B\\}$,

$\displaystyle\sum_{y_{k}}P\left(y_{A},y_{B}|x_{A},x_{B}\right)$ (4)

is independent of $x_{k}$. These three physical theories ranging from the most
restrictive(classical), to the least restrictive (no-signalling), with quantum
theory existing somewhere between the two popescu1994quantum .

Representing the full set of correlations that the quantum conditional
probability distribution in equation (3) allows one to reach is a notorious
problem, and the set has recently been shown to be not closed slofstra2019set
. Therefore we instead restrict ourselves to studying the achievable bounds
that these conditional probabilities allow one to reach in nonlocal
experiments ; the original and most famous of which being the Bell inequality
bell1964einstein .

######  Definition 1 (Bell inequality).

Given observers $A$ and $B$, each with measurement choice
$x_{k}\in\\{0_{k},1_{k}\\}$ with outcomes $y_{k}\in\\{-1,+1\\}$, experiments
performed on the systems adhere to the bound,

$\displaystyle\left|E(0_{A},0_{B})-E(0_{A},1_{B})+E(1_{A},0_{B})+E(1_{A},1_{B})\right|\leq
X,$ (5)

where both the correlation measure
$E(x_{A},x_{B})=\sum_{y_{A},y_{B}}y_{A}y_{B}\,P(y_{A},y_{B}|x_{A},x_{B})$ and
bound $X\in\mathbb{R}^{+}$ are theory dependent. The left-hand side of that
inequality is often called the score of the experiment.

Each physical theory admits a different conditional probability distribution,
and hence a different achievable bound $X$. Classical theory has the CHSH
bound $X=2$ clauser1969proposed , quantum theory has the Tsirelson bound of
$X=2\sqrt{2}$ cirel1980quantum  and non-signalling distributions $X=4$
popescu1994quantum . We are interested in the achievable bounds of a classical
system’s probability distribution when the hidden-variable distribution in
said probability distribution can be negative.

##  3 Results

We now define an important object for this work, the quasiprobability
distribution.

######  Definition 2 (Quasiprobability distribution).

We define a quasiprobability distribution as
$\tilde{P}_{\Lambda}:\Lambda_{1}\times\dots\times\Lambda_{N}\rightarrow\mathbb{R}$
where $\Lambda_{i}\subset\mathbb{R}$ and $|\Lambda_{i}|<\infty\,\forall\,i$,
that is properly normalised, such that,

$\displaystyle\sum_{\lambda_{1},\dots,\lambda_{N}}\tilde{P}_{\Lambda}\left(\lambda_{1},\dots,\lambda_{N}\right)=1.$
(6)

It can be seen that the collection of functions adhering to the above
definition forms a convex set, which we will denote $\tilde{\mathcal{P}}$, a
super set of the convex set of positive probability distributions
$\mathcal{P}\subset\tilde{\mathcal{P}}$. We must now determine how to quantify
the presence of negativity in our quasiprobability distributions. To this end
we will use a well-known method for quantitatively detecting properties of a
quantum state, witnesses terhal2000bell ; lewenstein2000optimization ;
eisert2007quantitative ; chabaud2021witnessing . Let us therefore proceed by
defining a _negativity witness:_

######  Definition 3 (Negativity witness).

Given some properly normalised probability distribution $P$, a well-defined
negativity witness is one which,

$\displaystyle\mathcal{N}(P)=0\,\,\forall\,P\in{\mathcal{P}}.$ (7)

We may additionally require such a witness to ‘faithfully’ detect negativity,

$\displaystyle\mathcal{N}(P)>0\,\,\forall\,P\in\tilde{\mathcal{P}}{\backslash}\mathcal{P}.$
(8)

In the following, we consider classical local hidden-variable models as
defined in equation (2), but we replace the hidden-variable probability
distribution $P_{\Lambda}$ with a quasiprobability distribution
$\tilde{P}_{\Lambda}$,

$\displaystyle
P\left(y_{A},y_{B}|x_{A},x_{B}\right)=\sum_{\lambda_{A},\lambda_{B}}P_{1}\left(y_{A}|x_{A},\lambda_{A}\right)P_{2}\left(y_{B}|x_{B},\lambda_{B}\right)\tilde{P}_{\Lambda}\left(\lambda_{A},\lambda_{B}\right).$
(9)

This corresponds to a scenario where different local statistics of
observations, governed by $\lambda$-local (ordinary) probability distributions
$P_{k}\left(y_{k}|x_{k},\lambda_{k}\right)$, are mixed according to a
quasiprobability distribution $\tilde{P}_{\Lambda}$. However, when
$\tilde{P}_{\Lambda}$ takes negative values, we should no longer think of the
model as an ignorance mixture of valid local scenarios but rather as a
nonlocal model al2013simulating . Furthermore, when compared with ordinary
hidden-variable models not all combinations of hidden-variable and
$\lambda$-local probability distributions are valid; only those combinations
which lead to well-defined $P\left(y_{A},y_{B}|x_{A},x_{B}\right)$ are valid,
i.e., comprised of values between 0 and 1 (the normalisation condition is
always fulfilled).

In addition to the correlation function between two measurements $x_{A}$ and
$x_{B}$, ${E}(x_{A},x_{B})$, it will also be useful to define
$\lambda_{k}$-local expectation values corresponding to an imagined scenario
where observer $k$ is able to perform measurement $x_{k}$ in the local
scenario corresponding to $\lambda_{k}$,

$\displaystyle\langle
k\rangle_{\lambda_{k}}^{x_{k}}:=\sum_{y_{k}}y_{k}P_{k}(y_{k}|x_{k},\lambda_{k}).$
(10)

This $\lambda_{k}$-local expectation value will be useful to formulate our
results, but does not correspond to the actual observations which are
themselves governed by equation (9).

We are now in a position to state the main result of this work, the
quasiprobabilistic Bell inequality.

######  Theorem 1 (Quasiprobabilistic Bell inequality).

Given observers $A$ and $B$, each with measurement choice
$x_{k}\in\\{0_{k},1_{k}\\}$ with outcomes $y_{k}\in\\{-1,+1\\}$ whose systems
are distributed according to some quasiprobability distribution
$\tilde{P}_{\Lambda}$, then the quasiprobabilistic Bell inequality holds:

$\displaystyle\left|{E}(0_{A},0_{B})-{E}(0_{A},1_{B})+{E}(1_{A},0_{B})+{E}(1_{A},1_{B})\right|\leq\,2+\mathcal{N}(\tilde{P}_{\Lambda}),$
(11)

where

$\displaystyle\mathcal{N}(\tilde{P}_{\Lambda})\coloneq\begin{cases}\mathcal{N}_{+}(\tilde{P}_{\Lambda})\qquad\text{
if }{E}(1_{A},0_{B})+{E}(1_{A},1_{B})<0,\\\
\mathcal{N}_{-}(\tilde{P}_{\Lambda})\qquad\text{ else,}\end{cases}$ (12)

is a negativity witness, and

$\mathcal{N}_{\pm}(\tilde{P}_{\Lambda}):=\sum_{\lambda_{A},\lambda_{B}}\left[2\pm\left(\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle B\rangle_{\lambda_{B}}^{1_{B}}+\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{0_{B}}\right)\right]\left(\Big{|}\tilde{P}_{\Lambda}(\lambda_{A},\lambda_{B})\Big{|}-\tilde{P}_{\Lambda}(\lambda_{A},\lambda_{B})\right).$

The proof of this theorem begins analogously with Bell’s proof of the CHSH
bound bell2004speakable , but diverges when the assumption $P\in\mathcal{P}$
is made in Bell’s proof. The above result shows that if an arbitrary amount of
negativity is allowed in the hidden-variable probability distribution then the
upper bound of equation (11) can be arbitrarily large. However, it should be
noted that a natural limit of $4$ in the relevant Bell tests (i.e., for the
upper bound in the quasiprobabilistic Bell inequality) is imposed by the
requirement that $P\left(y_{A},y_{B}|x_{A},x_{B}\right)$ is a well-defined,
valid probability distribution 111This can be easily seen as for well-defined
experimental setups where $y_{k}=\\{-1,+1\\}$, i.e. for valid
$P\left(y_{A},y_{B}|x_{A},x_{B}\right)$, each term on the left hand side of
equation (11) can be at most $1$ and least $-1$. .

The previous result of Al-Safi and Short al2013simulating  showed that it was
possible to violate said inequality up to this no-signalling bound of $X=4$.
Therefore, in order to emulate the physical results seen in Bell tests
(Tsirelson bound) one needs a negative probability distribution whose witness
equals $\mathcal{N}(\tilde{P}_{\Lambda})=2(\sqrt{2}-1)$. In section 4 we show
that for any $\mathcal{N}({\tilde{P}_{\Lambda}}){\leq 2}$, there exist
quasiprobabilistic hidden-variable models with valid local measurement
statistics that saturate inequality (11). We would hope that if a physical
mechanism was discovered that allowed a joint hidden-variable probability
distribution to have the appearance of negativity, one would expect that said
physical mechanism was limited in such a way that it resulted in the Tsirelson
bound and more generally was able to reconstruct the limits on quantum
correlations.

It is also important to note that although said witness
$\mathcal{N}(\tilde{P}_{\Lambda})$ is a valid witness according to definition
3 it is not necessarily a ‘faithful’ one. However this can be rectified, at
the cost of loosening the bound, by redefining said witness. For example the
function
$\mathcal{N}^{\prime}({\tilde{P}_{\Lambda}}):=\sum_{\lambda_{A},\lambda_{B}}4\left(\Big{|}\tilde{P}_{\Lambda}(\lambda_{A},\lambda_{B})\Big{|}-\tilde{P}_{\Lambda}(\lambda_{A},\lambda_{B})\right)$,
is defined to be both a valid and ‘faithful’ witness.

There are numerous generalisations of the famous CHSH inequalities, such as
multiple parties perez2008unbounded , arbitrary outcomes cope2019bell , etc.
These would no doubt be interesting to study but we leave it to future work to
explore these other generalisations and instead focus on the scenario in which
Alice and Bob have access to an arbitrary number of measurement settings
wehner2006tsirelson .

######  Theorem 2.

Given observers $A$ and $B$, each with $n\geq 2$ measurements
$x_{k}\in\\{0_{k},1_{k},\dots,{n-1}_{k}\\}$ with outcomes
$y_{k}\in\\{-1,+1\\}$ whose systems are distributed according to some
quasiprobability distribution $\tilde{P}_{\Lambda}$,

$\displaystyle\left|\sum^{n-1}_{i=0}{E}(i_{A},i_{B})+\sum^{n-1}_{i=1}{E}(i_{A},i-1_{B})-{E}(0_{A},n-1_{B})\right|\leq\,2n-2+{\mathcal{N}_{n}({\tilde{P}_{\Lambda}})},$
(13)

where
$\mathcal{N}_{n}({\tilde{P}_{\Lambda}})=\sum^{n-1}_{i=1}\mathcal{N}^{(i)}(\tilde{P}_{\Lambda})$
is a negativity witness with

$\displaystyle\mathcal{N}^{(x)}(\tilde{P}_{\Lambda})\coloneq\begin{cases}\mathcal{N}^{(x)}_{+}(\tilde{P}_{\Lambda})\qquad\text{
if }{E}(0_{A},x_{B})+{E}(0_{A},x-1_{B})<0,\\\
\mathcal{N}^{(x)}_{-}(\tilde{P}_{\Lambda})\qquad\text{ else,}\end{cases}$ (14)

where
$\mathcal{N}^{(x)}_{\pm}({\tilde{P}_{\Lambda}})\coloneqq\sum_{\lambda_{A},\lambda_{B}}\left[2\pm\left(\langle
A\rangle_{\lambda_{A}}^{x_{A}}\langle B\rangle_{\lambda_{B}}^{x_{B}}+\langle
A\rangle_{\lambda_{A}}^{x_{A}}\langle
B\rangle_{\lambda_{B}}^{x-1_{B}}\right)\right]\left(\Big{|}{\tilde{P}_{\Lambda}}(\lambda_{A},\lambda_{B})\Big{|}-{\tilde{P}_{\Lambda}}(\lambda_{A},\lambda_{B})\right)$.

The proof of the above theorem can be found in appendix B, it utilises proof
by induction by chaining together the inequalities from theorem 1. In section
4 we show that the bound in theorem 2 can be saturated. Namely, for any
$\mathcal{N}_{n}(\tilde{P}_{\Lambda}){\leq 2}$, there exist well-defined
$P\left(y_{A},y_{B}|x_{A},x_{B}\right)$, characterised by a quasiprobability
hidden-variable distribution
$\tilde{P}_{\Lambda}(\lambda_{A},\dotsc,\lambda_{n})$, that saturate
inequality (13). In addition, analogously to the two measurement result, at
the cost of loosening the bound we can ensure that the above witness is also
‘faithful’ by choosing for all $x$,
$\mathcal{N}^{(x)}(\tilde{P}_{\Lambda})=\mathcal{N}^{\prime}({\tilde{P}_{\Lambda}})$.

##  4 Example

In order to understand how to saturate the Bell inequality from Theorem 1, we
rewrite the left-hand side of equation (11) as,

$\displaystyle\Bigg{|}\sum_{\lambda}\mathcal{M}(\lambda)\,\tilde{P}_{\Lambda}(\lambda)\Bigg{|}.$
(15)

We replaced the hidden variables $\lambda_{A}$ and $\lambda_{B}$ with a single
hidden variable $\lambda$ because our example only uses a single hidden
variable $\lambda$. Further, $\mathcal{M}(\lambda):=\langle
A\rangle_{\lambda}^{0_{A}}\langle B\rangle_{\lambda}^{0_{B}}-\langle
A\rangle_{\lambda}^{0_{A}}\langle B\rangle_{\lambda}^{1_{B}}+\langle
A\rangle_{\lambda}^{1_{A}}\langle B\rangle_{\lambda}^{0_{B}}+\langle
A\rangle_{\lambda}^{1_{A}}\langle B\rangle_{\lambda}^{1_{B}}$ are the scores
of each of the $\lambda$-local distributions , that is
$-2\leq\mathcal{M}(\lambda)\leq 2$ holds.

For a given value of the negativity witness, we exceed the local bound
maximally by the simple strategy of weighting classical distributions with
$\mathcal{M}(\lambda)=+2$ with positive quasiprobability, while simultaneously
taking a classical distribution with $\mathcal{M}(\lambda)=-2$ with negative
weight. To ensure that the total probability distribution
$P\left(y_{A},y_{B}|x_{A},x_{B}\right)$ is well-defined we make a choice of
three deterministic classical distributions with positive weight and a fourth
with negative weight. Our four deterministic classical distributions can be
denoted $[(-,-)_{A},(-,+)_{B}],[(+,-)_{A},(-,-)_{B}],[(+,+)_{A},(+,+)_{B}]$
and $[(+,-)_{A},(-,+)_{B}]$. Here, our notation means that the distributions
can be produced by assigning the first pair of symbols to Alice and the second
to Bob. Each party chooses to read either the first or second of the symbols
given to them (this choice reflects their measurement setting $x_{k}$) while
the outcome of their measurement is determined by the symbol itself; that is,
$y_{k}=+1$ ($y_{k}=-1$) for a plus (minus) sign. This experimental description
of distributing classical information makes clear that these distributions are
local, with our hidden variable $\lambda$ indicating which of these sets the
source actually produces.

The source produces each of the distributions according to the following
quasiprobability distribution,

$\displaystyle\tilde{P}_{\Lambda}(\lambda)=\begin{cases}\frac{4+\mathcal{N}}{12}\qquad\text{
for }\lambda=1,2,3,\\\ -\frac{\mathcal{N}}{4}\qquad\,\,\,\text{ for
}\lambda=4,\end{cases}$ (16)

where $\mathcal{M}(\lambda)=2$ if $\lambda=1,2,3$ and
$\mathcal{M}(\lambda)=-2$ if $\lambda=4$. We can use tables to represent
$\lambda$-local probability distributions, and the total probability
distribution is then given as the weighted sum of such tables:

$\displaystyle\scriptsize\frac{4+\mathcal{N}}{12}\begin{pmatrix}\begin{bmatrix}[c|cccc]&\lx@intercol\hfil
y_{A}y_{B}\hfil\lx@intercol\\\ x_{A}x_{B}&\--&-+&+-&++\\\ \hline\cr
00&1&0&0&0\\\ 01&0&1&0&0\\\ 10&1&0&0&0\\\ 11&0&1&0&0\\\
\end{bmatrix}&+\begin{bmatrix}[c|cccc]&\lx@intercol\hfil
y_{A}y_{B}\hfil\lx@intercol\\\ x_{A}x_{B}&\--&-+&+-&++\\\ \hline\cr
00&0&0&1&0\\\ 01&0&0&1&0\\\ 10&1&0&0&0\\\ 11&1&0&0&0\\\
\end{bmatrix}+\begin{bmatrix}[c|cccc]&\lx@intercol\hfil
y_{A}y_{B}\hfil\lx@intercol\\\ x_{A}x_{B}&\--&-+&+-&++\\\ \hline\cr
00&0&0&0&1\\\ 01&0&0&0&1\\\ 10&0&0&0&1\\\ 11&0&0&0&1\\\
\end{bmatrix}\end{pmatrix}\scriptsize-\frac{\mathcal{N}}{4}\begin{bmatrix}[c|cccc]&\lx@intercol\hfil
y_{A}y_{B}\hfil\lx@intercol\\\ x_{A}x_{B}&\--&-+&+-&++\\\ \hline\cr
00&0&0&1&0\\\ 01&0&0&0&1\\\ 10&1&0&0&0\\\ 11&0&1&0&0\\\ \end{bmatrix}$
$\displaystyle\scriptsize\qquad\quad\,\,=\frac{1}{12}\begin{bmatrix}[c|cccc]&\lx@intercol\hfil
y_{A}y_{B}\hfil\lx@intercol\\\ x_{A}x_{B}&\--&-+&+-&++\\\ \hline\cr
00&4+\mathcal{N}&0&4-2\mathcal{N}&4+\mathcal{N}\\\
01&0&4+\mathcal{N}&4+\mathcal{N}&4-2\mathcal{N}\\\
10&8-\mathcal{N}&0&0&4+\mathcal{N}\\\
11&4+\mathcal{N}&4-2\mathcal{N}&0&4+\mathcal{N}\\\ \end{bmatrix}.$ (17)

The requirement that the resulting total probability distribution must be
valid implies $\mathcal{N}\leq 2$ which corresponds to the no-signalling
limit. Furthermore, it is easy to check that said distribution indeed gives a
value of $\mathcal{N}$ for the negativity witness.

The quasiprobabilistic Bell inequality score for this experiment is
$2+\mathcal{N}$, which upon substituting equation (16) into the negativity
witness, can be seen to saturate the bound. In appendix C we discuss how one
can generalise the above to the $n$-measurement scenario.

##  5 Conclusion

We have shown that there exists a relationship between the amount of
negativity allowed in a joint hidden-variable distribution, and the degree to
which said distribution can demonstrate nonlocality in a Bell experiment. In
particular, theorem 2 introduces a quasiprobabilistic Bell inequality, which
gives us a sharp bound in the scenario of two parties with $n$ inputs
(corresponding to a choice between $n$ measurements) and can be used
straightforwardly to reconstruct quantum statistics using nothing more than
local, separable classical probability distributions and a quasiprobability
distribution over them – granted an appropriately well spent budget of
negativity.

Our work sits within the long-established tradition of trying to understand
quantum theory through interpretative lenses which remove some particular
aspect from a classical worldview. Such approaches are wide and varied,
including superdeterminism hossenfelder2020rethinking ; adlam2018quantum ;
retro-causality price2012does ; invoking an irreducible role for subjectivity
in physics mermin2014physics ; fuchs2014introduction ; mueller2020law ; taking
physical reality to consist of interacting, separate realms
goldstein2001bohmian ; esfeld2014ontology ; allowing the relativity of pre and
post-selection bacciagaluppi2020reverse ; taking Hilbert space to be literal
carroll2021reality , and so on. Here we add to this list, in that we present
an additional way to re-capture the nonlocal features of quantum theory:
through having a finite amount of negativity allowed in a hidden-variable
distribution over scenarios which are, in themselves, entirely local and
classical. We are not claiming that such quasi distributions are ‘real’ –
only, more modestly, that such a perspective could not be ruled out at this
stage.

Pursuing this line of reasoning, we would hope that our results may help to
determine the fundamental restrictions on a system’s quasiprobability hidden-
variable distribution such that it captures the full character of physical
correlations. Put another way; we know that zero negativity can capture the
set of classical correlations, whilst un-bounded negativity can capture the
non-signalling set. Given that the set of quantum correlations lies between
these two – what are the restrictions on the quasiprobability hidden-variable
distribution which would suffice to identify the full set of quantum
correlations? We hope to explore this question in further work.

###### Acknowledgements.

We are grateful to Benjamin Yadin, Richard Moles and Thomas Veness for helpful
discussions and the Physical Institute for Theoretical Hierarchy (PITH) for
encouraging an investigation into this topic. BM acknowledges financial
support from the Engineering and Physical Sciences Research Council (EPSRC)
under the Doctoral Prize Grant (Grant No. EP/T517902/1). LF acknowledges
financial support from the Austrian Science Fund (FWF) through SFB BeyondC
(Grant No. F7102). BL is supported by Leverhulme Trust Research Project Grant
(RPG-2018-213). D.G. acknowledges support from FQXI (grant no. RFP-IPW-1907)

##  Appendix A proof of theorem 1

######  Theorem.

Given observers $A$ and $B$, each with measurement choice
$x_{k}\in\\{0_{k},1_{k}\\}$ with outcomes $y_{k}\in\\{-1,+1\\}$ whose systems
are distributed according to some quasiprobability distribution
$\tilde{P}_{\Lambda}$, then the quasiprobabilistic Bell inequality holds:

$\displaystyle\left|{E}(0_{A},0_{B})-{E}(0_{A},1_{B})+{E}(1_{A},0_{B})+{E}(1_{A},1_{B})\right|\leq\,2+\mathcal{N}(\tilde{P}_{\Lambda}),$
(18)

where

$\displaystyle\mathcal{N}(\tilde{P}_{\Lambda})\coloneq\begin{cases}\mathcal{N}_{+}(\tilde{P}_{\Lambda})\qquad\text{
if }{E}(1_{A},0_{B})+{E}(1_{A},1_{B})<0,\\\
\mathcal{N}_{-}(\tilde{P}_{\Lambda})\qquad\text{ else,}\end{cases}$ (19)

is a negativity witness, and

$\mathcal{N}_{\pm}(\tilde{P}_{\Lambda}):=\sum_{\lambda_{A},\lambda_{B}}\left[2\pm\left(\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle B\rangle_{\lambda_{B}}^{1_{B}}+\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{0_{B}}\right)\right]\left(\Big{|}\tilde{P}_{\Lambda}(\lambda_{A},\lambda_{B})\Big{|}-\tilde{P}_{\Lambda}(\lambda_{A},\lambda_{B})\right).$

###### Proof.

The first part of the proof follows Bell’s 1971 derivation of the CHSH
inequality bell2004speakable . For brevity in the proof we will just write
$\tilde{P}_{\Lambda}$ as $P$.

We start by rewriting the correlation function,

$\displaystyle{E}(x_{A},x_{B}):=$
$\displaystyle\sum_{y_{A},y_{B}}y_{A}y_{B}\sum_{\lambda_{A},\lambda_{B}}P_{A}\left(y_{A}|x_{A},\lambda_{A}\right)P_{B}\left(y_{B}|x_{B},\lambda_{B}\right){P}\left(\lambda_{A},\lambda_{B}\right)$
$\displaystyle=$ $\displaystyle\sum_{\lambda_{A},\lambda_{B}}\langle
A\rangle_{\lambda_{A}}^{x_{A}}\langle
B\rangle_{\lambda_{B}}^{x_{B}}\,{P}(\lambda_{A},\lambda_{B}),$ (20)

where $\langle k\rangle_{\lambda_{k}}^{x_{k}}$, defined in equation (10), is
the $\lambda_{k}$-local expectation value for observer $k$ performing
measurement $x_{k}$. Starting with the following difference between
correlation functions,

$\displaystyle{E}$ $\displaystyle(0_{A},0_{B})-{E}(0_{A},1_{B})$
$\displaystyle=$ $\displaystyle\sum_{\lambda_{A},\lambda_{B}}\left(\langle
A\rangle_{\lambda_{A}}^{0_{A}}\langle B\rangle_{\lambda_{B}}^{0_{B}}-\langle
A\rangle_{\lambda_{A}}^{0_{A}}\langle
B\rangle_{\lambda_{B}}^{1_{B}}\right){P}(\lambda_{A},\lambda_{B})$
$\displaystyle=$ $\displaystyle\sum_{\lambda_{A},\lambda_{B}}\left(\langle
A\rangle_{\lambda_{A}}^{0_{A}}\langle B\rangle_{\lambda_{B}}^{0_{B}}-\langle
A\rangle_{\lambda_{A}}^{0_{A}}\langle B\rangle_{\lambda_{B}}^{1_{B}}\pm\langle
A\rangle_{\lambda_{A}}^{0_{A}}\langle B\rangle_{\lambda_{B}}^{0_{B}}\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle B\rangle_{\lambda_{B}}^{1_{B}}\mp\langle
A\rangle_{\lambda_{A}}^{0_{A}}\langle B\rangle_{\lambda_{B}}^{0_{B}}\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{1_{B}}\right){P}(\lambda_{A},\lambda_{B})$
$\displaystyle=$ $\displaystyle\sum_{\lambda_{A},\lambda_{B}}\langle
A\rangle_{\lambda_{A}}^{0_{A}}\langle
B\rangle_{\lambda_{B}}^{0_{B}}\left(1\pm\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{1_{B}}\right){P}(\lambda_{A},\lambda_{B})-\sum_{\lambda_{A},\lambda_{B}}\langle
A\rangle_{\lambda_{A}}^{0_{A}}\langle
B\rangle_{\lambda_{B}}^{1_{B}}\left(1\pm\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{0_{B}}\right){P}(\lambda_{A},\lambda_{B}),$ (21)

where the “$\pm$” in equation (21) is to be understood as either “$+$” in all
terms or “$-$” in all terms. Taking the absolute value of both sides and using
the triangular inequality,

$\displaystyle\big{|}{E}$ $\displaystyle(0_{A},0_{B})-{E}(0_{A},1_{B})\big{|}$
$\displaystyle\leq$ $\displaystyle\Big{|}\sum_{\lambda_{A},\lambda_{B}}\langle
A\rangle_{\lambda_{A}}^{0_{A}}\langle
B\rangle_{\lambda_{B}}^{0_{B}}\left(1\pm\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{1_{B}}\right){P}(\lambda_{A},\lambda_{B})\Big{|}+\Big{|}\sum_{\lambda_{A},\lambda_{B}}\langle
A\rangle_{\lambda_{A}}^{0_{A}}\langle
B\rangle_{\lambda_{B}}^{1_{B}}\left(1\pm\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{0_{B}}\right){P}(\lambda_{A},\lambda_{B})\Big{|}.$
(22)

Starting with the first term on the right-hand side of inequality (22), we
again apply the triangular inequality,

$\displaystyle\Big{|}\sum_{\lambda_{A},\lambda_{B}}\langle
A\rangle_{\lambda_{A}}^{0_{A}}\langle
B\rangle_{\lambda_{B}}^{0_{B}}\left(1\pm\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{1_{B}}\right){P}(\lambda_{A},\lambda_{B})\Big{|}\leq$
$\displaystyle\sum_{\lambda_{A},\lambda_{B}}\Big{|}\langle
A\rangle_{\lambda_{A}}^{0_{A}}\langle
B\rangle_{\lambda_{B}}^{0_{B}}\left(1\pm\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{1_{B}}\right){P}(\lambda_{A},\lambda_{B})\Big{|}$
$\displaystyle=$ $\displaystyle\sum_{\lambda_{A},\lambda_{B}}\Big{|}\langle
A\rangle_{\lambda_{A}}^{0_{A}}\langle
B\rangle_{\lambda_{B}}^{0_{B}}\Big{|}\Big{|}\left(1\pm\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{1_{B}}\right){P}(\lambda_{A},\lambda_{B})\Big{|}.$
(23)

As $y_{k}\in\\{-1,+1\\}$ we can say $\Big{|}\langle
k\rangle_{\lambda_{k}}^{x_{k}}\Big{|}\leq 1\,\,\forall\,k$, we can write,

$\displaystyle\Big{|}\sum_{\lambda_{A},\lambda_{B}}\langle
A\rangle_{\lambda_{A}}^{0_{A}}\langle
B\rangle_{\lambda_{B}}^{0_{B}}\left(1\pm\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{1_{B}}\right){P}(\lambda_{A},\lambda_{B})\Big{|}\leq$
$\displaystyle\sum_{\lambda_{A},\lambda_{B}}\Big{|}\left(1\pm\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{1_{B}}\right){P}(\lambda_{A},\lambda_{B})\Big{|}$
$\displaystyle=$
$\displaystyle\sum_{\lambda_{A},\lambda_{B}}\Big{|}\left(1\pm\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{1_{B}}\right)\Big{|}\Big{|}{P}(\lambda_{A},\lambda_{B})\Big{|}$
$\displaystyle=$ $\displaystyle\sum_{\lambda_{A},\lambda_{B}}\left(1\pm\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{1_{B}}\right)\Big{|}{P}(\lambda_{A},\lambda_{B})\Big{|}$
(24)

where we have used the fact that $\left(1\pm\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle B\rangle_{\lambda_{B}}^{1_{B}}\right)$
is necessarily non-negative because of the choice of eigenvalues,
$y_{k}\in\\{-1,+1\\}$.

Similarly, we find for the second term on the right-hand side of inequality
(22)

$\displaystyle\Big{|}\sum_{\lambda_{A},\lambda_{B}}\langle
A\rangle_{\lambda_{A}}^{0_{A}}\langle
B\rangle_{\lambda_{B}}^{1_{B}}\left(1\pm\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{0_{B}}\right){P}(\lambda_{A},\lambda_{B})\Big{|}\leq$
$\displaystyle\sum_{\lambda_{A},\lambda_{B}}\left(1\pm\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{0_{B}}\right)\Big{|}{P}(\lambda_{A},\lambda_{B})\Big{|}.$
(25)

By adding inequalities (24) and (25) we find the following upper bound for the
left-hand side of inequality (22),

$\displaystyle\big{|}{E}(0_{A},0_{B})-{E}(0_{A},1_{B})\big{|}\leq$
$\displaystyle\sum_{\lambda_{A},\lambda_{B}}\left[2\pm\left(\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle B\rangle_{\lambda_{B}}^{1_{B}}+\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{0_{B}}\right)\right]\Big{|}{P}(\lambda_{A},\lambda_{B})\Big{|}.$
(26)

So far the proof followed Bell’s 1971 derivation bell2004speakable  of the
CHSH inequality. In Bell’s derivation, one assumes that the joint probability
distribution is positive, $P(\lambda_{A},\lambda_{B})\geq 0$, which, using the
definition of the correlation function and the triangle inequality, leads to
the well-known CHSH inequality,
$\left|{E}(0_{A},0_{B})-{E}(0_{A},1_{B})+{E}(1_{A},0_{B})+{E}(1_{A},1_{B})\right|\leq\,2$.

We have to take another approach because here $P(\lambda_{A},\lambda_{B})$ can
be a quasiprobability distribution and thus take negative values. For each of
the two inequalities (26) (corresponding to the choice for “$\pm$”), we define
a negativity witness $\mathcal{N}_{\pm}(P)$ for some normalised distribution
$P\in\tilde{\mathcal{P}}$ as the difference obtained by replacing
$\left|P(\lambda_{A},\lambda_{B})\right|$ with $P(\lambda_{A},\lambda_{B})$ in
the right-hand side of inequality (26),

$\displaystyle\mathcal{N}_{\pm}(P):=\sum_{\lambda_{A},\lambda_{B}}\left[2\pm\left(\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle B\rangle_{\lambda_{B}}^{1_{B}}+\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{0_{B}}\right)\right]\left[\left|P\left(\lambda_{A},\lambda_{B}\right)\right|-P\left(\lambda_{A},\lambda_{B}\right)\right].$
(27)

Note that although this negativity witness is perfectly valid according to
definition 3, it is not faithful because $2\pm\left(\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle B\rangle_{\lambda_{B}}^{1_{B}}+\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle B\rangle_{\lambda_{B}}^{0_{B}}\right)$
may be zero for $P\in\tilde{\mathcal{P}}/\mathcal{P}$, i.e.,
$\mathcal{N}_{\pm}(P)$ may be zero for a quasiprobability distribution.
Nevertheless we can now write inequality (26) as,

$\displaystyle\big{|}{E}(0_{A},0_{B})-{E}(0_{A},1_{B})\big{|}\leq$
$\displaystyle\,\sum_{\lambda_{A},\lambda_{B}}\left[2\pm\left(\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle B\rangle_{\lambda_{B}}^{1_{B}}+\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{0_{B}}\right)\right]{P}(\lambda_{A},\lambda_{B})+\mathcal{N}_{\pm}(P).$
(28)

The first term on the right-hand side of inequality (28) can then be
simplified using the definition of the correlation function (20) and that
${P}(\lambda_{A},\lambda_{B})$ is normalized,

$\displaystyle\sum_{\lambda_{A},\lambda_{B}}\left[2\pm\left(\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle B\rangle_{\lambda_{B}}^{1_{B}}+\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{0_{B}}\right)\right]{P}(\lambda_{A},\lambda_{B})$
$\displaystyle=\sum_{\lambda_{A},\lambda_{B}}2{P}(\lambda_{A},\lambda_{B})\pm\sum_{\lambda_{A},\lambda_{B}}\left(\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle B\rangle_{\lambda_{B}}^{1_{B}}+\langle
A\rangle_{\lambda_{A}}^{1_{A}}\langle
B\rangle_{\lambda_{B}}^{0_{B}}\right){P}(\lambda_{A},\lambda_{B})$
$\displaystyle=2\pm\left[{E}(1_{A},0_{B})+{E}(1_{A},1_{B})\right].$ (29)

Thus, inequality (28) becomes

$\displaystyle\big{|}{E}(0_{A},0_{B})-{E}(0_{A},1_{B})\big{|}\leq$
$\displaystyle\,2\pm\left[{E}(1_{A},0_{B})+{E}(1_{A},1_{B})\right]+\mathcal{N}_{\pm}(P).$
(30)

Now, we choose the inequality corresponding to “$+$” if
$\left[{E}(1_{A},0_{B})+{E}(1_{A},1_{B})\right]$ is negative, and the
inequality corresponding to “$-$” else. This allows us to write

$\displaystyle\big{|}{E}(0_{A},0_{B})-{E}(0_{A},1_{B})\big{|}\leq$
$\displaystyle\,2-\left|{E}(1_{A},0_{B})+{E}(1_{A},1_{B})\right|+\mathcal{N}(P),$
(31)

where we defined

$\displaystyle\mathcal{N}(P)\coloneq\begin{cases}\mathcal{N}_{+}(P)\qquad\text{
if }{E}(1_{A},0_{B})+{E}(1_{A},1_{B})<0,\\\ \mathcal{N}_{-}(P)\qquad\text{
else.}\end{cases}$ (32)

From inequality (31), we obtain

$\displaystyle\big{|}{E}(0_{A},0_{B})-{E}(0_{A},1_{B})\big{|}+\left|{E}(1_{A},0_{B})+{E}(1_{A},1_{B})\right|\leq$
$\displaystyle\,2+\mathcal{N}(P),$ (33)

and with one final use of the triangular inequality we find a CHSH-type
inequality for arbitrary $P\in\tilde{P}$,

$\displaystyle\left|{E}(0_{A},0_{B})-{E}(0_{A},1_{B})+{E}(1_{A},0_{B})+{E}(1_{A},1_{B})\right|\leq$
$\displaystyle\,2+\mathcal{N}(P),$ (34)

completing the proof.

∎

##  Appendix B proof of theorem 2

######  Theorem.

Given observers $A$ and $B$, each with $n\geq 2$ measurements
$x_{k}\in\\{0_{k},1_{k},\dots,{n-1}_{k}\\}$ with outcomes
$y_{k}\in\\{-1,+1\\}$ whose systems are distributed according to some
quasiprobability distribution $\tilde{P}_{\Lambda}$,

$\displaystyle\left|\sum^{n-1}_{i=0}{E}(i_{A},i_{B})+\sum^{n-1}_{i=1}{E}(i_{A},i-1_{B})-{E}(0_{A},n-1_{B})\right|\leq\,2n-2+{\mathcal{N}_{n}({\tilde{P}_{\Lambda}})},$
(35)

where
$\mathcal{N}_{n}({\tilde{P}_{\Lambda}})=\sum^{n-1}_{i=1}\mathcal{N}^{(i)}(\tilde{P}_{\Lambda})$
is a negativity witness with

$\displaystyle\mathcal{N}^{(x)}(\tilde{P}_{\Lambda})\coloneq\begin{cases}\mathcal{N}^{(x)}_{+}(\tilde{P}_{\Lambda})\qquad\text{
if }{E}(0_{A},x_{B})+{E}(0_{A},x-1_{B})<0,\\\
\mathcal{N}^{(x)}_{-}(\tilde{P}_{\Lambda})\qquad\text{ else,}\end{cases}$ (36)

where
$\mathcal{N}^{(x)}_{\pm}({\tilde{P}_{\Lambda}})\coloneqq\sum_{\lambda_{A},\lambda_{B}}\left[2\pm\left(\langle
A\rangle_{\lambda_{A}}^{x_{A}}\langle B\rangle_{\lambda_{B}}^{x_{B}}+\langle
A\rangle_{\lambda_{A}}^{x_{A}}\langle
B\rangle_{\lambda_{B}}^{x-1_{B}}\right)\right]\left(\Big{|}{\tilde{P}_{\Lambda}}(\lambda_{A},\lambda_{B})\Big{|}-{\tilde{P}_{\Lambda}}(\lambda_{A},\lambda_{B})\right)$.

###### Proof.

The proof is similar to the creation of chained CHSH inequalities, see
braunstein1990wringing  for an intuitive description, and works by induction
in $n$.
Anchor step $n=2$: see theorem 1.
Inductive step: Suppose that theorem 2 holds for $n=k$. We will prove the
theorem for $n=k+1$. Starting from the left-hand side of equation (35) for
$n=k+1$, we find

$\displaystyle\left|\sum^{k}_{i=0}{E}(i_{A},i_{B})+\sum^{k}_{i=1}{E}(i_{A},i-1_{B})-{E}(0_{A},k_{B})\right|$
(37)
$\displaystyle=\left|\sum^{k-1}_{i=0}{E}(i_{A},i_{B})+\sum^{k-1}_{i=1}{E}(i_{A},i-1_{B})-{E}(0_{A},k_{B})+{E}(k_{A},k_{B})+{E}(k_{A},k-1_{B})\right|$
(38)
$\displaystyle=\left|\sum^{k-1}_{i=0}{E}(i_{A},i_{B})+\sum^{k-1}_{i=1}{E}(i_{A},i-1_{B})-{E}(0_{A},k_{B})+{E}(k_{A},k_{B})+{E}(k_{A},k-1_{B})\right.$
$\displaystyle\qquad+{E}(0_{A},k-1_{B})-{E}(0_{A},k-1_{B})\Biggr{|}$ (39)
$\displaystyle\leq\left|\sum^{k-1}_{i=0}{E}(i_{A},i_{B})+\sum^{k-1}_{i=1}{E}(i_{A},i-1_{B})-{E}(0_{A},k-1_{B})\right|+$
$\displaystyle\qquad\left|{E}(k_{A},k_{B})+{E}(k_{A},k-1_{B})+{E}(0_{A},k-1_{B})-{E}(0_{A},k_{B})\right|$
(40) $\displaystyle\leq
2k-2+\sum^{k-1}_{i=1}\mathcal{N}^{(i)}(\tilde{P}_{\Lambda})+2+\mathcal{N}^{(k)}(\tilde{P}_{\Lambda})$
(41) $\displaystyle=2(k+1)-2+\mathcal{N}_{k+1}(\tilde{P}_{\Lambda}),$ (42)

which concludes the induction. The inequality in line (40) is the triangle
inequality, and we proceed from that line by using the induction hypothesis
and theorem 1 for measurements $0_{A}$, $k_{A}$ for Alice, and $k-1_{B}$, and
$k_{B}$ for Bob. ∎

##  Appendix C Saturation of the $n$-measurement quasiprobabilistic Bell
inequality

We can generalise the $2$-measurement example from the main text to $n$
measurements in the following way. Using equation (20), we rewrite the left-
hand side of equation (13) as,

$\displaystyle\Bigg{|}\sum_{\lambda}\mathcal{M}(\lambda)\,\tilde{P}_{\Lambda}(\lambda)\Bigg{|},$
(43)

where we use only a single hidden variable $\lambda$, and

$\displaystyle\mathcal{M}(\lambda):=\sum^{n-1}_{i=0}\langle
A\rangle_{\lambda}^{i_{A}}\langle
A\rangle_{\lambda}^{i_{B}}+\sum^{n-1}_{i=1}\langle
A\rangle_{\lambda}^{i_{A}}\langle A\rangle_{\lambda}^{i-1_{B}}-\langle
A\rangle_{\lambda}^{0_{A}}\langle A\rangle_{\lambda}^{n-1_{B}}$ (44)

are the scores of each of the $\lambda$-local distributions, that is
$-(2n-2)\leq\mathcal{M}(\lambda)\leq 2n-2$ holds.

We again consider $4$ classical scenarios, $3$ of which achieve a score of
$2n-2$ but now the last achieving a score of $2n-6$. The source produces each
of the distributions according to the following quasiprobability distribution,

$\displaystyle\tilde{P}_{\Lambda}(\lambda)=\begin{cases}\frac{4+\mathcal{N}_{n}}{12}\qquad\text{
for }\lambda=1,2,3,\\\ -\frac{\mathcal{N}_{n}}{4}\qquad\,\,\,\text{ for
}\lambda=4,\end{cases}$ (45)

where $\lambda=1,2,3$ corresponds to classical distributions with score
$2n-2$, and $\lambda=4$ to $2n-6$. We can see that this distribution saturates
the $n$-measurement quasiprobabilistic Bell inequality from theorem 2,

$\displaystyle(2n-2)\tilde{P}_{\Lambda}(1)+(2n-2)\tilde{P}_{\Lambda}(2)+(2n-2)\tilde{P}_{\Lambda}(3)+(2n-6)\tilde{P}_{\Lambda}(4)=2n-2+\mathcal{N}_{n}.$
(46)

We now need to come up with the $\lambda$-local probability distributions
which result in a well-defined $P\left(y_{A},y_{B}|x_{A},x_{B}\right)$ and
gives the correct value for the witness $\mathcal{N}_{n}$. To do this we can
generalise the classical distributions from the main text for $n$
measurements, using the same notation as previously, such classical
distributions are,

$\displaystyle[(\overbrace{-,\dots,-}^{n})_{A},(\overbrace{-,\dots,-}^{n-1},+)_{B}]^{\lambda=1},$
$\displaystyle[(+,\overbrace{-,\dots,-}^{n-1})_{A},(\overbrace{-,\dots,-}^{n})_{B}]^{\lambda=2},[(\overbrace{+,\dots,+}^{n})_{A},(\overbrace{+,\dots,+}^{n})_{B}]^{\lambda=3},$
$\displaystyle[(+,\overbrace{-,\dots,-}^{n-1})_{A},(\overbrace{-,\dots,-}^{n-1},+)_{B}]^{\lambda=4}.$
(47)

It is easy to check that all such distributions achieve for $\lambda=1,2,3$ a
score $2n-2$, and for $\lambda=4$, $2n-6$. Since the distribution for
$\lambda=4$ enters into the total probability distribution,
$P\left(y_{A},y_{B}|x_{A},x_{B}\right)$, with negative weight, the other
$\lambda$-local distributions (with $\lambda=1,2,3$) must compensate for that
negativity to ensure that the total probability distribution is valid.

It is easy to see that this is indeed the case by observing that for each
combination of Alice and Bob’s signs for $\lambda=4$ that same combination of
symbols appear in the same places for at least one of the other distributions.
We also find that requiring positivity of the total probability distribution
also gives us the no-signalling condition:

$\displaystyle\tilde{P}_{\Lambda}(\lambda)+\tilde{P}_{\Lambda}(4)\geq 0\text{
for }\lambda=1,2,3\implies\mathcal{N}_{n}\leq 2\quad\forall n\geq 2.$ (48)

The final thing to check is that said distributions in equation (47) coupled
with the quasiprobability distribution in equation (45) gives the required
value $\mathcal{N}_{n}$ for the negativity witness
$\mathcal{N}_{n}({\tilde{P}_{\Lambda}})=\sum^{n-1}_{i=1}\mathcal{N}^{(i)}(\tilde{P}_{\Lambda})$
with

$\displaystyle\mathcal{N}^{(x)}(\tilde{P}_{\Lambda})\coloneq\begin{cases}\mathcal{N}^{(x)}_{+}(\tilde{P}_{\Lambda})\qquad\text{
if }{E}(0_{A},x_{B})+{E}(0_{A},x-1_{B})<0,\\\
\mathcal{N}^{(x)}_{-}(\tilde{P}_{\Lambda})\qquad\text{ else,}\end{cases}$ (49)

where
$\mathcal{N}^{(x)}_{\pm}({\tilde{P}_{\Lambda}})\coloneqq\sum_{\lambda_{A},\lambda_{B}}\left[2\pm\left(\langle
A\rangle_{\lambda_{A}}^{x_{A}}\langle B\rangle_{\lambda_{B}}^{x_{B}}+\langle
A\rangle_{\lambda_{A}}^{x_{A}}\langle
B\rangle_{\lambda_{B}}^{x-1_{B}}\right)\right]\left(\Big{|}{\tilde{P}_{\Lambda}}(\lambda_{A},\lambda_{B})\Big{|}-{\tilde{P}_{\Lambda}}(\lambda_{A},\lambda_{B})\right)$.

Firstly, we can see by going through the distributions in equation (47) that
for all measurement choices $x$, ${E}(0_{A},x_{B})+{E}(0_{A},x-1_{B})>0$
meaning that the witness we calculate for all $x$ in the sum of
$\mathcal{N}_{n}({\tilde{P}_{\Lambda}})$ is
$\mathcal{N}^{(x)}_{-}(\tilde{P}_{\Lambda})$. We then go through the
expectation values in the definition of
$\mathcal{N}^{(x)}_{-}(\tilde{P}_{\Lambda})$ for all $x$ for the $\lambda=4$
distribution given in equation (47), from which we can see,

$\displaystyle 2-\left(\langle A\rangle_{4}^{x_{A}}\langle
B\rangle_{4}^{x_{B}}+\langle A\rangle_{4}^{x_{A}}\langle
B\rangle_{4}^{x-1_{B}}\right)=\begin{cases}0\qquad\text{ for }x=1,\dots,n-2\\\
2\qquad\,\,\,\text{ for }x=n-1.\end{cases}$ (50)

meaning that upon calculating
$\mathcal{N}_{n}({\tilde{P}_{\Lambda}})\sum^{n-1}_{i=1}\mathcal{N}^{(i)}(\tilde{P}_{\Lambda})$,
we get

$\displaystyle\mathcal{N}_{n}({\tilde{P}_{\Lambda}})=$ $\displaystyle
2\left(\Big{|}{\tilde{P}_{\Lambda}}(4)\Big{|}-{\tilde{P}_{\Lambda}}(4)\right)$
$\displaystyle=$ $\displaystyle\mathcal{N}_{n},$ (51)

as required.

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