# Knowledge of Quantum Hidden Variables Enables Backwards-In-Time Signaling
## 2 The Protocol  
Consider the Peres-Mermin square [6; 24] composed of the operators  
\[ \begin{bmatrix}\sigma_{1x}&\sigma_{2x}&\sigma_{1x}\sigma_{2x}\\ \sigma_{2y}&\sigma_{1y}&\sigma_{1y}\sigma_{2y}\\ \sigma_{1x}\sigma_{2y}&\sigma_{1y}\sigma_{2x}&\sigma_{1z}\sigma_{2z}\end{bmatrix}  \tag{1}\]where \(\sigma_{1j}\) are the operators that refer to the spin components of a first spin-1/2 particle and \(\sigma_{2j}\) of a second spin-1/2 particle, both located in [PERSON]'s lab. In [PERSON] words: \"The three operators in each row and each column commute, and their product is \(+1\), except those of the last column, whose product is \(-1\). There is obviously no way of assigning numerical values \(\pm 1\) to these nine operators with this multiplicative property\" [6]. This implies that we cannot assign definite values to the spin components of the two particles that are independent of what is (was, will be) measured on the other particle: a very simple consequence of quantum contextuality of hidden variables. Since non-contextual hidden variables are impossible [5; 6; 24], one must conclude that the values of the hidden variables of the spins must be fixed by the context, which is our assumption here. The idea of our protocol is to delay the choice of the context to a time successive to when the values of the hidden variables have been used to determine some \"signal\". This can be achieved through the protocol composed by the following steps (see Figure 1):  
1. [PERSON] finds out the (contextual) hidden values of \(\sigma_{1x}\), \(\sigma_{2x}\), \(\sigma_{1y}\), \(\sigma_{2y}\), \(\sigma_{1z}\sigma_{2z}\). Of course, quantum mechanics prevents us from knowing these values because they refer to non-commuting (complementary) properties. However, the explicit assumption here is that [PERSON] can know the hidden variables. As proven by [PERSON], the hidden variables are contextual, but the context will be established by [PERSON] in the future. Clearly, we suppose that the values of the hidden variables reflect the values that are (or were or will be) uncovered by the experiments that are (or were or will be) performed. As specified above, this is the minimal requirement for \"hidden variables\". We emphasize the distinction between attaining knowledge of hidden variables and performing a quantum measurement. If one knew the hidden variable, one could predict the measurement outcome. Here we assume that one could somehow know the hidden variables _without_ performing a quantum measurement. All interpretations of quantum mechanics prevent the knowledge of hidden variables (hence the name), but here we explore what would happen if it were possible to know them: if one assumes they exist, it is natural to ask what would happen if they could be known. 2. Consider the products in (1): take the product of the hidden values of \(\sigma_{1x}\), \(\sigma_{2x}\), \(\sigma_{1y}\), \(\sigma_{2y}\) and \(\sigma_{1z}\sigma_{2z}\). If this product is equal to \(+1\), then set a bit \(b=0\), if the product is equal to \(-1\), then set \(b=1\). The first choice is consistent with looking at the last line of the Peres-Mermin square, the second choice with looking at the last column. 3. Then in the (distant) future, [PERSON] can retroactively decide the value of \(b\) by deciding what to measure. If she measures \(\sigma_{1x}\sigma_{2x}\) and \(\sigma_{1y}\sigma_{2y}\) (which commute), then she sets  
Figure 1: Depiction of the protocol. First, [PERSON] finds out the values of the (contextual) hidden variables and uses them to calculate the value of a bit \(b\). Later she establishes the context by choosing which set of observables to measure. Her later choice determines the value of \(b\), sending it back in time. \(b=1\). Since the hidden variables, by assumption, reflect the preexisting values of the observables, this means that by setting \(b=1\) she sends back to her past self \(b=1\). Instead, if she measures \(\sigma_{1x}\sigma_{2y}\) and \(\sigma_{1y}\sigma_{2x}\) (which also commute), then she sends back to herself \(b=0\).