Toys can't play: physical agents in Spekkens' theory

Spekkens' toy theory is an epistemically-restricted theory [8]. Such theories distinguish two types of states: ontic states, which encode the physical state of a system, and epistemic states—the states of knowledge that an observer has about the system. In the toy theory, epistemic states are constrained by the knowledge balance principle, inspired by the Heisenberg uncertainty principle:

'If one has maximal knowledge, then for every system, at every time, the amount of knowledge one possesses about the ontic state of the system at that time must equal the amount of knowledge one lacks' [18].

In practice, suppose that a system can be in one of 2d possible ontic states, that is, specifying the value of $\log_2 (2d) = 1+ \log_2 d$ degrees of freedom is sufficient to identify each ontic state. Then an observer is only allowed to access at most $\log_2 d$ bits of information, and we say that the system has dimension d; this and other constraints will become clearer with the examples ahead.

Our results apply to the general case where a toy system can be decomposed into N subsystems of dimension d each. The continuous limit, analogous to a particle moving in space, corresponds to $d \to \infty$. For example, the elementary systems analogous to qubits (toy bits) have d = 2. In the following we stick to intuitive visualizations for small discrete dimensions (one or two toy bits); we review the general case in appendix A.

We can represent up to two discrete toy systems via simple grid diagrams. For higher dimensions, grid diagrams are not as useful; the interested reader may find a review of the necessary mathematical notation in the appendix. A single system with two degrees of freedom ($d = 2)$ has four different ontic states (labelled for example 1,2,3,4). To see this, note that answering two binary questions would be sufficient to identify the ontic state of the system, for example 'is the ontic state odd?' and 'is the ontic state smaller than 3?'. Each ontic state is represented by one of four boxes, . To represent an observer's knowledge about the ontic state—that is the epistemic state from their perspective—we colour in some of the boxes. For example, represents 'the observer knows that the system's ontic state is either o = 1 or o = 2.' For simplicity we usually omit the number labels. One can draw an analogy between toy states and quantum states:

Zoom In Zoom Out Reset image size

These are the only pure states at d = 2: they are states of maximal allowed information, according to the knowledge balance principle, for which the observer knows half of the degrees of freedom (for example represents 'the ontic state is odd'). The quantum analogy carries through to a stabilizer formulation of the toy theory [9]. The fully mixed state (a state of maximal ignorance) is represented as the mixture:

Zoom In Zoom Out Reset image size

Unlike quantum theory, this is the only physical mixed state at d = 1, that is the only mixed state that can emerge as a marginal of a globally pure state (like the entangled state which we will consider in a moment). The epistemic restriction implies that for a system composed of N elementary systems, an agent is only allowed to have access to exactly $0\leqslant J\leqslant N$ bits of information (corresponding to an epistemic state spanning $2^{2N-J}$ ontic states). For example, for N = 1 the valid epistemic states can span either 2 or 4 ontic states. The mixture:

Zoom In Zoom Out Reset image size

is not a valid epistemic state [18]. ²

When we consider two toy systems, we represent the ontic states with a $4\times4$ grid, where the rows determine the possible ontic states of system A and the columns those of system B, for example:

Zoom In Zoom Out Reset image size

We omit the subsystem labels when they are clear from context. There are also global toy states which are not product states, for example the classically correlated state:

Zoom In Zoom Out Reset image size

which is a mixed state, and the pure entangled state:

Zoom In Zoom Out Reset image size

Analogously to a Bell state in quantum theory, this latter state is used for toy teleportation and dense coding protocols [18]. Like their quantum analogues, entangled toy states are globally pure states with mixed marginals. In this example, the observer has maximal information about the correlations between A and B, and maximal ignorance about the reduced state of individual subsystems.

For discrete toy systems, taking the reduced state over one system corresponds to projecting the grid diagram into one axis. For example,

Zoom In Zoom Out Reset image size

More formally, if the global epistemic state is E_AB , the reduced (or marginal) state on system B is $E_B = \{o_B: \ (o_A, o_B) \in E_{AB} \}$. In the above example, $E_{AB} = \{(1,1), (1,2), (3,1), (3,2) \}$ so $E_B = \{1,2\}$.

For discrete dimensions, the allowed toy transformations are permutations of ontic states, constrained to mapping valid epistemic states to valid epistemic states (in all subsystems). For example consider the transformation that permutes the second and third ontic states,

Zoom In Zoom Out Reset image size

This permutation acts analogously to the Hadamard gate in quantum theory,

Zoom In Zoom Out Reset image size

We will see other examples (like a CNOT gate) further ahead; in particular we will see that a controlled Hadamard is not a valid operation, and we will explore the implications of this for agents' free choice.

In the toy theory, a measurement is a partition of the ontic state space into valid epistemic states, called the measurement basis. The outcome of the measurement is determined by the position of the ontic state. The observer can then update their knowledge; in the toy theory the measurement update rule leads to an ontic disturbance, which we will later see emerges from the physical implementation of a measurement. The consequence for the updated epistemic state is a bit cumbersome to explain, but will become clear from examples right ahead. In short, the updated description should guarantee that if the observer repeats the measurement they obtain the same outcome, and be compatible with their overall knowledge [18]. For example, consider the measurement: , where the numbered partitions correspond to the epistemic states of the measurement basis. This is analogous to a Pauli-Z measurement of a single qubit. Suppose that Bob measures a toy bit in state in this basis, and obtains outcome. This allows him to deduce that the ontic state previous to the measurement was 0; however

Zoom In Zoom Out Reset image size

is not a valid epistemic state. Rather, it is a pre- and post-selected state, a concept that has analogues in quantum theory. The smallest epistemic state compatible with his knowledge of the pre- and post-selection and with the requirement for repeated outcomes is . To see this, first note that there are four epistemic states compatible with Bob's knowledge,

Zoom In Zoom Out Reset image size

The requirement for repeated outcomes translates to 'the post-measurement description must be a subset of , so that if I apply the same measurement again, I am guaranteed to obtain the same outcome .' Of the four candidates, only the first epistemic state satisfies the condition,

Zoom In Zoom Out Reset image size

This is analogous to the quantum measurement update rule for projective outcomes: if Bob measured $\vert +\rangle$ in the Z basis and obtained outcome 0, he would henceforth describe the state as $\vert 0\rangle$. Now suppose that Bob is making the same measurement, but now on his half of the entangled toy Bell state (1.1). The global view of Bob's local measurement is:

Zoom In Zoom Out Reset image size

If Bob obtains outcome he can deduce that the global ontic state before the measurement was one of the two coloured squares in the bottom-left corner,

Zoom In Zoom Out Reset image size

This is the pre- and post-selected state that describes his knowledge of the ontic state of AB just before the measurement. There are two valid epistemic states that are compatible with this knowledge and live in the measurement partition ,

Zoom In Zoom Out Reset image size

Of the two candidates, Bob picks the leftmost (the smallest state), which is the description that makes the most use of his knowledge about the state before the measurement (picking the state on the right would be discarding what he knew of the correlations between the two systems). The quantum analogy is when Bob makes a local Z measurement on a Bell state $\propto \vert 00\rangle + \vert 11\rangle$, and upon seeing outcome 0, updates his global description of the state to $\vert 00\rangle$.

In quantum theory, the post-measurement state of a system S emerges from the physical picture by moving the Heisenberg cut one level up and thinking of a projective measurement on the agent's memory M after the physical entangling operation between S and M. For example, the (unnormalized) post-measurement state when obtaining outcome 1 on a Z measurement of a qubit in state $\vert \psi\rangle_S$ can be expressed as:

Zoom In Zoom Out Reset image size

For details on the general case we refer to appendix C.1.

In quantum theory, if Alice and Bob share a Bell state and measure their individual qubits in the computational basis (corresponding to the observables Z_A and Z_B ), they can make inferences about each other's outcomes (figure 4). For example, if the shared state is:

$$\begin{align*} \frac{1}{\sqrt{2}} \left( \vert 00\rangle_{AB} + \vert 11\rangle_{AB} \right), \end{align*}$$

and Bob obtains outcome b = 0, he can update his description of the global post-measurement state to $\vert 00\rangle_{AB}$ and infer with certainty that Alice must obtain outcome a = 0; analogously for b = 1.

Zoom In Zoom Out Reset image size

This scenario can be reenacted in the toy theory, with similar results: Bob can make a deterministic prediction about Alice's outcome (figure 5). Alice and Bob share the entangled state analogous to a Bell state,

Zoom In Zoom Out Reset image size

Zoom In Zoom Out Reset image size

Bob measures his system in the toy-Z basis, M_Z = . If he obtains outcome $B = 0 =$ , he can update his description of the shared state to

Zoom In Zoom Out Reset image size

From this description, Bob can infer that if Alice now measures her system in the same basis, she will obtain outcome $A = 0 =$ with certainty, that is, '$B = 0 \implies A = 0$' Analogously, he can conclude that '$B = 1 \implies A = 1$'. A formal proof can be found in appendix B.3.

Wigner's friend experiment was first proposed by Wigner [20]. The setting involves a quantum system R and an observer A (Alice) performing a measurement on this system in a closed laboratory, as well as an outside observer Wigner. For Alice in the lab, the outcome of the experiment is recorded in the device she is using to measure the system R, for example, as a position of a pointer (or an entry in her memory). However, Wigner does not have any information about Alice getting a particular outcome, and describes the evolution of the closed lab as a unitary (reversible) process, and assigns an entangled state to R and A. In the language of quantum mechanics, if we assume that the pointer is initially in the state $\vert 0\rangle_A$, and the state of the measured system is $\frac{1}{\sqrt{2}}\vert 0\rangle_R+\frac{1}{\sqrt{2}}\vert 1\rangle_R$ this process corresponds to:

$$\begin{align} \left(\frac{1}{\sqrt{2}}\vert 0\rangle_R+\frac{1}{\sqrt{2}}\vert 1\rangle_R\right)\vert 0\rangle_A \rightarrow \frac{1}{\sqrt{2}}\vert 0\rangle_R\vert 0\rangle_A+\frac{1}{\sqrt{2}}\vert 1\rangle_R\vert 1\rangle_A. \end{align} \tag{ 5.1 }$$

Alice and Wigner turn out to have descriptions of the same setting which are vastly different from each other. We will not discuss numerous conceptual implications of the original thought experiment here—a review can be found in [21]. However, we would still like to see how we can model this setting in the toy theory (here we will do so in the original epirestricted picture).

In the toy theory, we consider again two subsystems: the measured system R and Alice's memory register A. The individual states and the joint state of the systems R and A can be pictured as:

Zoom In Zoom Out Reset image size

Alice measures R in the toy-Z basis. She describes her measurement as M_Z = and sees a definite outcome a. Wigner sees Alice's measurement as a reversible transformation (the toy CNOT), and updates his description of the joint state of R and A as:

Zoom In Zoom Out Reset image size

Alice, on the other hand, has the subjective experience of seeing one outcome a and write it to her memory. We can describe the knowledge update from the perspective of the different agents as:

Zoom In Zoom Out Reset image size

In the framework of the toy theory, the difference between Alice's and Wigner's descriptions has a straightforward interpretation. Thanks to the knowledge balance principle, in the state of the maximal knowledge an agent can have maximal information either about an individual system or about how these individual systems are correlated. Hence, Alice's and Wigner's epistemic states do not contradict each other, and simply represent two different ways an agent can view a composite system (two states of knowledge about the same ontic state). In [19], it is shown that the Wigner and Alice's views discrepancy can be reproduced (and interpreted!) in any realistic toy model. Alice's collapsed state can simply be understood as more coarse-grained compared to Wigner's; the correlations of her state are beyond her description level. She can still get away with it, though, as the correlations are not used in later dynamics—which is not the case for the next scenario we present.

In the Frauchiger–Renner setting [17], four quantum agents (Alice, Bob, Ursula and Wigner) perform a series of measurements and reasoning steps, reaching a logical contradiction,

$$\begin{align} (w = \text{ok} \wedge u = \text{ok}) \implies b = 1 \implies a = 1 \implies w = \text{fail}, \end{align} \tag{ 5.2 }$$

that is, when both Ursula and Wigner obtain outcomes 'ok' in their measurements, they can reason based on their observation that that Bob predicted that Alice predicted that Wigner would obtain a different outcome, 'fail', with certainty. The experimental protocol consists of individual steps that we covered so in this manuscript: two qubits R and S are initially prepared in an entangled Hardy state [22]; Alice measures R and Bob measures S; then Ursula measures Alice's lab (including R and Alice's memory A), and finally Wigner measures Bob's lab (including S and Bob's memory B). The contradiction is found for a specific choice of initial state and measurement bases, which are described in appendix C.2. ⁵ For a pedagogical discussion of the original paradox and the assumptions behind it, we refer to our previous work [23]; for discussions of broader implications for abstract logic and interpretations of quantum theory see for example [5, 21, 24, 25]. The paradox has also been shown to arise in other physical theories, namely box world [7].

Our question is whether a similar multi-agent logical paradox can be found in Spekkens' toy theory; we will see that it cannot, partially because of the restrictions in the individual operations like conditional state preparation, and partly because this is an explicitly non-contextual epistemic theory. Since the toy theory does not allow Alice to perform non-orthognal conditional state preparation of S, we follow the version of the experiment where all relevant correlations are encoded in the initial state of RS. The global system of Alice and Bob's labs is composed of four subsystems: two systems R and S measured by Alice and Bob, and Alice's and Bob's memory registries A and B (figure 6). The order of measurements follows the original experiment, and the systems can be of an arbitrary dimension k. We show that in the toy theory, there is no choice of initial state and measurements by the four agents that can lead to a logical contradiction in this experimental scenario. A formal description of the setting and proof can be found in appendix B.4.

Theorem 5.1. There exists no valid epistemic state that can be used to reach a logical contradiction in the Frauchiger–Renner setting in the toy theory.

Zoom In Zoom Out Reset image size

To tackle the more general case of toy systems of arbitrary dimensions, we must review heavier formalism [10, 11]. Our main results are expressed in this language, but we add intuitive descriptions that convey the main message and do not require learning the formalism.

In the continuous case, we represent toy systems through observables q_i and p_i for each of the n subsystems, analogous to position and momentum: for example, a toy particle moving in 3D would have n = 3. For arbitrary dimensions, we represent a valid epistemic state $(\mathbf{V}, \mathbf{v})$ by the known observables $\mathbf{V} = \langle\, \mathbf{f}_1, \dots, \mathbf{f}_k\rangle$ and the valuation $\mathbf{v}$ of those observables. For example, if all we know about a 1D continuous system is that $\mathbf{f}_1 = 2q_1 - p_1 = 5$, then $\mathbf{V} = \Bigg\langle \begin{pmatrix} 2 \\ -1 \end{pmatrix}\Bigg\rangle$ and we can have for example $\mathbf{v} = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$, as $(2, -1) \begin{pmatrix} 3 \\ 1 \end{pmatrix} =$ $2\times3 - 1\times1 = 5$. The set of ontic states compatible with this epistemic state are all those that share the valuation $\mathbf{v}$ for the observables in $\mathbf{V}$; this includes $o_1 = \begin{pmatrix} 3 \\ 1 \end{pmatrix}$, but also for instance $o_2 = \begin{pmatrix} 2 \\ -1 \end{pmatrix}$, as $(2, -1) o_2 = 5$, $o_3 = \begin{pmatrix} 4 \\ 3 \end{pmatrix}$, and so on.

The complete epistemic restriction in then given by the principle of classical complementarity:

"The valid epistemic states are those wherein an agent knows the values of a set of quadrature variables that commute relative to the Poisson bracket, and is maximally ignorant otherwise." [10]

Here, "maximal ignorance" means that there is a uniform probability over all other values of variables. It can be shown that this complementarity principle requires observables to be linear, i.e. quadrature observables.

We represent quadrature observables as a vector $\mathbf{f} \in \mathbb{Z}^{2n}_d / \mathbb{R}^{2n}$ if we consider n systems of dimension d or n continuous systems. Then the Poisson bracket is defined for both the continuous and discrete case, where the sum has to be understood in mod d in the discrete case:

$$\begin{align} [\,f,g] = \sum_{i = 1}^n f_{2i - 1} g_{2i} - f_{2i} g_{2i-1} \end{align} \tag{ A.1 }$$

where f_j denotes the jth entry of $\mathbf{f}\;$ ^{7 , 8} [10].

To continue our example suppose that we bring in a second 1D system, and we know the local observable corresponding to the position of this system, for instance $\mathbf{f}_2 = q_2 = 10$. The global epistemic state is then specified by:

$$\begin{align*} \left(\mathbf{V}, \mathbf{v}^{^{\prime}}\right) = \left(\langle\, \mathbf{f}_1, \mathbf{f}_2\rangle, \mathbf{v}^{^{\prime}} \right) = \left(\Bigg\langle{\begin{pmatrix} 2 \\ -1 \\ 0 \\ 0 \end{pmatrix} , \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}}\Bigg\rangle, \begin{pmatrix} 3 \\ 1 \\ 10 \\ 0 \end{pmatrix} \right), \end{align*}$$

which is a product state. On the other hand, if instead of q₂ we knew a global property, like that the positions of the two systems were perfectly correlated, $q_1 = q_2$, we could represent this through a new observable $\mathbf{f}_3 = q_1 - q_2 = 0$, and so our global epistemic state would be:

$$\begin{align*} (\mathbf{V}^{^{\prime}}, \mathbf{v}^{^{\prime\prime}}) = (\langle\, \mathbf{f}_3\rangle, \mathbf{v}^{^{\prime\prime}} ) = \left(\left\langle \begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \end{pmatrix} \right\rangle, \begin{pmatrix} 3 \\ 1 \\ 3 \\ 0 \end{pmatrix} \right). \end{align*}$$

Valid reversible transformations are sympletic transformations, represented by a pair $(U, \mathbf{a})$, where $\mathbf{a}$ is an ontic state ⁹ and U is a sympletic matrix. Sympletic matrices are those that satisfy $U^T \,J U = J$, where

$$\begin{align} \mathbf{J} = \begin{pmatrix} 0 & 1 & 0 & 0 & \dots \\[6pt] -1 & 0 & 0 & 0 & \\[6pt] 0 & 0 & 0 & -1 & \\[6pt] 0 & 0 & 1 & 0 & \\ \vdots & & & &\ddots \\ \end{pmatrix}, \end{align} \tag{ A.2 }$$

and is used to write the symplectic inner product ¹⁰ . The transformation $(U, \mathbf{a})$ transforms an each ontic state from $\mathbf{o}$ to $U (\mathbf{o} + \mathbf{a})$. An epistemic state $(V,{\mathbf{v}})$ transforms under such a symplectic transformation as:

$$\begin{align} (V,{\mathbf{v}}) \to (\ (U^T)^{-1} V,\ U({\mathbf{v}}+\mathbf{a})\ ). \end{align} \tag{ A.3 }$$

The reason the transformation implements $(U^T)^{-1}$ on the vector space of known variables is that the transformation transforms an ontic state $\mathbf{o}$ to $U \mathbf{o}$, so if we know that an ontic state was compatible with the known variables before it must also be compatible afterwards.

For example consider the state:

$$\begin{align*} \left(\mathbf{V} = \Bigg\langle \begin{pmatrix} 2 \\ -1 \end{pmatrix}\Bigg\rangle, \mathbf{v} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} \right), \end{align*}$$

and the transformation that swaps position and momentum:

$$\begin{align*} U = \begin{pmatrix} 0 & 1 \\[6pt] 1 & 0 \end{pmatrix}. \end{align*}$$

This transforms the above state to:

$$\begin{align*} \left(\mathbf{V}^{^{\prime}} = \Bigg\langle \begin{pmatrix} -1 \\ 2 \end{pmatrix}\Bigg\rangle, \mathbf{v}^{^{\prime}} = \begin{pmatrix} 1 \\ 3 \end{pmatrix} \right). \end{align*}$$

The ontic state $o_2 = \begin{pmatrix} 2\, \\ -1\,\end{pmatrix}$ which was compatible with the knowledge before the transformation is still valid after the transformation as $\begin{pmatrix} -1\, \\ 2\,\end{pmatrix}^T (U o_2) = 5$.

In the continuous case measurement consists of a vector space V_π of observables that which can have outcomes ${\mathbf{v}}_{\pi} \in \Omega$, all possible (inequivalent) outcomes also result in a partition of the ontic state space, like in the discrete case. The probability to get outcome ${\mathbf{v}}_{\pi}$ if the system is in an ontic state $\mathbf{m}$ is given by [10]:

$$\begin{align} \xi({\mathbf{v}}_{\pi}|\mathbf{m}) = \delta_{V_{\pi}^{\perp} + {\mathbf{v}}_{\pi}}(\mathbf{m}). \end{align} \tag{ A.4 }$$

Intuitively, this means that we can only obtain measurement outcomes that are compatible with the ontic state of the system. We can denote a measurement V_π and its outcome ${\mathbf{v}}_{\pi}$ by the pair $(V_{\pi}, {\mathbf{v}}_{\pi})$. With the conditional probability distribution $\xi({\mathbf{v}}_{\pi}|\mathbf{m})$ we can calculate the probability for a measurement outcome given the epistemic state $(V,{\mathbf{v}})$ [10],

$$\begin{align} P({\mathbf{v}}_{\pi}|(V,{\mathbf{v}})) = \sum_{\mathbf{m} \in \Omega} \xi({\mathbf{v}}_{\pi}|\mathbf{m}) \ \mu_{(V,{\mathbf{v}})}(\mathbf{m}), \end{align} \tag{ A.5 }$$

where $\mu_{(V,{\mathbf{v}})}(\mathbf{m})$ is the equal probability distribution over all ontic states compatible with the knowledge of the observables in V having valuations $\mathbf{v}$.

For example consider the state:

$$\begin{align*} \left(\mathbf{V} = \Bigg\langle \begin{pmatrix} 2 \\ 0 \end{pmatrix}\Bigg\rangle, \mathbf{v} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} \right), \end{align*}$$

and we want to measure the position:

$$\begin{align*} \Bigg\langle \begin{pmatrix} 1 \\ 0 \end{pmatrix}\Bigg\rangle. \end{align*}$$

The epistemic state is such that we know the position to be $2\,\cdot 3 = 6$, so the only compatible measurement outcome is:

$$\begin{align*} {\mathbf{v}}_{\pi} = \begin{pmatrix} 6 \\ 0 \end{pmatrix}. \end{align*}$$

The post-measurement state of the system (given the information about the outcome) is as follows.

Theorem A.1 (Measurement update rule [16]). When an epistemic state $(V,{\mathbf{v}})$ is subjected to a measurement V_π , and outcome ${\mathbf{v}}_{\pi}$ is obtained, the epistemic state is updated to $(V^{^{\prime}}, \mathbf{v}^{^{\prime}})$, where

$$\begin{align} V^{^{\prime}} &= V_{\pi} \oplus V_{\text{commute}}, \end{align} \tag{ A.6 }$$

$$\begin{align} {\mathbf{v}}^{^{\prime}} &\in (V_{\pi}^{\perp} + {\mathbf{v}}_{\pi}) \cap (V^{\perp}_{\text{commute}} + {\mathbf{v}}), \end{align} \tag{ A.7 }$$

$$\begin{align} V_{\text{commute}} &= \{\mathbf{f} \in V: \ [\mathbf{f}, \mathbf{f}_\pi] = 0, \ \forall \ \mathbf{f}_\pi \in V_\pi \} \subseteq V. \end{align} \tag{ A.8 }$$

We consider the case where both the measured system and the pointer are continuous 1D systems, characterized by the observables $q_S, p_S, q_M$ and p_M . Note that we cannot start the pointer in the analogous of a Gaussian state, as each toy observable can only be either fully known or completely unknown. We start instead with a pointer well-localized in position space, with $q_M = x_0$. Suppose that we want to measure the position of the first system, S. If S starts in a state of well-defined position q, then we expect to end up in a 'classically corelated' toy state analogous to $\vert q\rangle_S \vert x_0 +q\rangle_M$. If on the other hand S starts with well-defined momentum p and undefined position, we would expect the final global state to be somehow analogous to a superposition $\sum_q \alpha(q,x_0, p) \vert q\rangle_S\vert x_0 + q\rangle_P$. Theorem B.6 shows that in the toy theory there exists a transformation that produces a final state that is analogous to the quantum case.

Theorem B.5 (Coherent copy of position). Let the epistemic state of the memory system be initialized as $q_{\text{memory}} = 0$, and let the initial epistemic state of the measured system be $(V_{\text{information}} = \langle {\mathbf{v}}_1\,\rangle,{\mathbf{v}})$ with ${\mathbf{v}}_1,{\mathbf{v}} \in \mathbb{Z}_d^{2}$ or $\mathbb{R}^2$ so that the initial epistemic state of the composite system is:

$$\begin{align} \left(\left\langle \begin{pmatrix} v^1_1 \\[6pt] v^2_1 \\[6pt] 0 \\[6pt] 0 \end{pmatrix}, \begin{pmatrix} 0 \\[6pt] 0 \\[6pt] 1 \\[6pt] 0 \end{pmatrix} \right\rangle, \begin{pmatrix} v^1 \\[6pt] v^2 \\[6pt] 0 \\[6pt] 0 \end{pmatrix}\right). \end{align} \tag{ B.10 }$$

Then the transformation:

$$\begin{align} S = \begin{pmatrix} 1 & 0 & 0 & 0 \\[6pt] 0 & 1 & 0 & -1 \\[6pt] 1 & 0 & 1 & 0 \\[6pt] 0 & 0 & 0 & 1 \\[6pt] \end{pmatrix} \end{align} \tag{ B.11 }$$

correlates the position of the memory system and the information system. In prime and continuous dimensions tracing out the memory system results in a mixture of all possible measurement outcomes of the position of the information system.

Proof. We apply the transformation to the initial state:

$$\begin{align} S: \left(\left\langle \begin{pmatrix} v^1_1 \\[6pt] v^2_1 \\[6pt] 0 \\[6pt] 0 \end{pmatrix}, \begin{pmatrix} 0 \\[6pt] 0 \\[6pt] 1 \\[6pt] 0 \end{pmatrix} \right\rangle, \begin{pmatrix} v^1 \\[6pt] v^2 \\[6pt] 0 \\[6pt] 0 \end{pmatrix}\right) \to \left(\left\langle \begin{pmatrix} v^1_1 \\[6pt] v^2_1 \\[6pt] 0 \\[6pt] v^2_1 \\[6pt] \end{pmatrix}, \begin{pmatrix} -1 \\[6pt] 0 \\[6pt] 1 \\[6pt] 0 \end{pmatrix} \right\rangle, \begin{pmatrix} v^1 \\[6pt] v^2 \\[6pt] v^1 \\[6pt] 0 \end{pmatrix}\right). \end{align} \tag{ B.12 }$$

This transformation ensures that the position of the memory system is always equal to the position of the information system.

Furthermore, if we trace out the memory system, that is taking the marginal of the probability distribution of the ontic state over the memory system. The probability distribution before marginalisation is the uniform distribution over $V^{\perp}+{\mathbf{v}} = \text{span}((0,-1,0,1)^T,(v_1^{\perp},v_2^{\perp},v_{1}^{\perp},0)^T) +(v^1,v^2,v^1,0)^T$ where $(v_1^{\perp},v_2^{\perp})$ is the vector spanning $V_{information}^{\perp}$. After marginalisation this results in a uniform probability distribution over $V^{\perp}+{\mathbf{v}}$ with the last two entries removed $V^{^{\prime} \perp}+\mathbf{v^{^{\prime}}} = \text{span}((0,1)^T,(v_1^{\perp},v_2^{\perp})^T) +(v^1,v^2)$. Therefore, if $v_1^{2} \neq 0$ the traced out state is the maximally mixed state and therefore an equal mixture of all positions of the information system. In non prime dimensions, even if $v_1^{2} \neq 0$, the state does not need to be the maximally mixed state. On the other hand if $v_1^{2} = 0$, the state is the state where the information system has definite position v¹.

To obtain some intuition, let us look at two examples. Recall that this sympletic transformation acts on an initial state as $(V, \mathbf{v}) \to ((S^T)^{-1}V, S\mathbf{v})$. Consider the initial state analogous to $\vert \phi\rangle_S\vert x_0\rangle_M$, that is the product state between an arbitrary state of S and a well-defined memory position $q_M = x_0$. This state is transformed as:

$$\begin{align} \left(\left\langle \begin{pmatrix} v_1 \\[6pt] v_2\\[6pt] 0 \\[6pt] 0 \end{pmatrix}, \begin{pmatrix} 0 \\[6pt] 0 \\[6pt] 1 \\[6pt] 0 \end{pmatrix} \right\rangle, \begin{pmatrix} w_1 \\[6pt] w_2 \\[6pt] x_0 \\[6pt] 0 \end{pmatrix}\right) \quad \longrightarrow \quad \left(\left\langle \begin{pmatrix} v_1 \\[6pt] v_2\\[6pt] 0 \\[6pt] -v_2 \end{pmatrix}, \begin{pmatrix} 1 \\[6pt] 0 \\[6pt] 1 \\[6pt] 0 \end{pmatrix} \right\rangle, \begin{pmatrix} w_1 \\[6pt] w_2 \\[6pt] x_0 + w_1\\[6pt] 0 \end{pmatrix}\right). \end{align} \tag{ B.13 }$$

In particular, after a quick simplification we can see that it transforms the initial state analogous to $\vert q\rangle_S \vert x_0\rangle_M$ as:

$$\begin{align} \left( \left\langle \begin{pmatrix} 1 \\[6pt] 0 \\[6pt] 0 \\[6pt] 0 \\[6pt] \end{pmatrix}, \begin{pmatrix} 0 \\[6pt] 0 \\[6pt] 1 \\[6pt] 0 \\[6pt] \end{pmatrix} \right\rangle, \begin{pmatrix} q \\[6pt] 0 \\[6pt] x_0 \\[6pt] 0 \\[6pt] \end{pmatrix}\right) \qquad \longrightarrow \qquad \left( \left\langle \begin{pmatrix} 1 \\[6pt] 0 \\[6pt] 0 \\[6pt] 0 \\[6pt] \end{pmatrix}, \begin{pmatrix} 0 \\[6pt] 0 \\[6pt] 1 \\[6pt] 0 \\[6pt] \end{pmatrix} \right\rangle, \begin{pmatrix} q \\[6pt] 0 \\[6pt] q+x_0 \\[6pt] 0 \\[6pt] \end{pmatrix}\right), \end{align} \tag{ B.14 }$$

that is, a state analogous to $\vert q\rangle_S\vert x_0+q\rangle_M$, as desired (it satisfies $q_S = q$ and $q_M = q+ x_0$). On the other hand, U transforms the state analogous to $\vert p\rangle_S \vert x_0\rangle_M$ as:

$$\begin{align} \left( \left\langle \begin{pmatrix} 0 \\[6pt] 1 \\[6pt] 0 \\[6pt] 0 \\[6pt] \end{pmatrix}, \begin{pmatrix} 0 \\[6pt] 0 \\[6pt] 1 \\[6pt] 0 \\[6pt] \end{pmatrix} \right\rangle, \begin{pmatrix} 0 \\[6pt] p \\[6pt] x_0 \\[6pt] 0 \\[6pt] \end{pmatrix}\right) \qquad \longrightarrow \qquad \left( \left\langle \begin{pmatrix} 0 \\[6pt] 1 \\[6pt] 0 \\[6pt] -1 \\[6pt] \end{pmatrix}, \begin{pmatrix} 1 \\[6pt] 0 \\[6pt] 1 \\[6pt] 0 \\[6pt] \end{pmatrix} \right\rangle, \begin{pmatrix} 0 \\[6pt] p \\[6pt] x_0 \\[6pt] 0 \\[6pt] \end{pmatrix}\right), \end{align} \tag{ B.15 }$$

that is, a state such that $p_S-p_M = p$ and $q_S + q_M = x_0$. Similarly to the quantum case, the reduced states of just S or just M are fully mixed.

Theorem B.6 (Coherent copy of an arbitrary observable). Let the state of the information system be $\left(W, {\mathbf{w}} \right)$, where W such that it defines a state on an arbitrary amount of systems. Let $\mathbf{f}$ be an observable, such that $\mathbf{f} = ((S^M)^T)^{-1} \mathbf{q_1}$ where $\mathbf{q_1}$ the position of the first system of the information system and S^M a symplectic transformation. Furthermore, let $\mathbf{v} = (T_{mem}^T)^{-1} \mathbf{q}$ be an observable with T a symplectic transformation and $\mathbf{q}$ the position of the memory system. Let the memory system be initially in the state where the value of $\mathbf{v}$ is 0, such that the total initial state is $(V_{tot},{\mathbf{v}}_{tot}) = \left(\langle (\mathbf{v}, 0,0, \dots)^T, (0,0,\mathbf{w_1})^T, \dots (0,0,\mathbf{w_k})^T\rangle, (0,0,{\mathbf{w}})^T \right)$, where $\mathbf{w_1},{\ldots},\mathbf{w_k}$ span W.

Then the transformation

$$\begin{align} (T_{mem} \otimes {\unicode{x1D7D9}}_{mes} ) ({\unicode{x1D7D9}}_{mem} \otimes S^M_{mes}) (S \otimes {\unicode{x1D7D9}}_{mes_2,..,mes_N} ) ({\unicode{x1D7D9}}_{mem} \otimes (S^M_{mes})^{-1}) (T_{mem}^{-1} \otimes {\unicode{x1D7D9}}_S ) ), \end{align} \tag{ B.16 }$$

correlates the value of $\mathbf{v}$ of the memory system and with the value of $\mathbf{f}$ of the information system. In prime and continuous dimensions tracing out the memory system results in a mixture of all possible measurement outcomes of $\mathbf{f}$ of the information system.

Proof. The first part of the transformation $({\unicode{x1D7D9}}_{mem} \otimes (S^M_{mes})^{-1}) (T_{mem}^{-1} \otimes {\unicode{x1D7D9}}_S ) )$ performs a basis change, such that the observable $\mathbf{f}$ of the information system in the old basis is transformed to $\mathbf{q_1}$ in the new basis and the observable $\mathbf{v}$ of the memory system is transformed to the position of the memory system. Therefore, the case where we copy $\mathbf{q_1}$ into the position of the memory system is related by a basis change to the case where we copy $\mathbf{f}$ into the value of v of the memory system. By an analogous calculation as in theorem B.5 the transformation $(S \otimes {\unicode{x1D7D9}}_{mes_2,..,mes_N} )$ correlates the position of the memory system with $\mathbf{q_1}$. This means that after the transformation with $(S \otimes {\unicode{x1D7D9}}_{mes_2,..,mes_N} )$ the state $(V_{tot}^{^{\prime}},{\mathbf{v}}^{^{\prime}})$ is such that $V_{tot}^{^{\prime}}$ contains a vector of the form $(1,0,-1,0,0,0,0, \dots, 0)^{T} \in V^{^{\prime}}_{tot}$ and the valuation of $(1,0,-1,0,0,0,0, \dots, 0)^{T}$ is zero. The last part of the transformation $({\unicode{x1D7D9}}_{mem} \otimes S^M_{mes}) (T_{mem} \otimes {\unicode{x1D7D9}}_S ) )$ transforms $(1,0,-1,0,0,0,0, \dots, 0)^{T} \to (\mathbf{v}, -\mathbf{f} )$ and the valuation of the new vector is still zero as the transformation is symplectic. Therefore, the above transformation correlates the value of $\mathbf{f}$ on the information system with the value of v on the memory system.

As the transformation $({\unicode{x1D7D9}}_{mem} \otimes (S^M_{mes})^{-1}) (T_{mem}^{-1} \otimes {\unicode{x1D7D9}}_S ) )$ acts only locally on the information system and the memory system, we can first transform back the joint memory and system state, then trace out the memory system and finally only transform the information system back and get the same result as if we would have just traced out the memory system. To determine the marginal we must consider the probability distribution over the ontic states induced by the epistemic state. This probability distribution is just the uniform distribution over the ontic states in $U^{\perp}+ \mathbf{u}$, where $(U,\mathbf{u})$ is the state after the measurement update. The vector space $(S_{M}^T \otimes T_{mem}^T U)^{\perp}$ is given by $V^{\perp}_M \oplus \langle({0,1,0,-1},\dots,0)^{T}\rangle$ where $V_{M}^{\perp}$ is spanned by the vectors $(v_i^{(1)},0, \mathbf{v}_i)$ where $\mathbf{v}_i$ are the vectors spanning $(S_M^T W)^{\perp}$ and $v_i^{(k)}$ is the kth entry of $\mathbf{v}_i$. Taking the marginal of the probability distribution over the memory system gives the uniform probability distribution over $(S_M^T W)^{\perp} \oplus \langle(0,1,0,\dots,0)^{T}\rangle + S_M^{-1}{\mathbf{w}}$, as marginalisation results in removing the the entries in the vectors of the systems we marginalized over. Therefore, we can conclude the state of the marginal system is $(M = (S_M^T W) \cap \langle(0,1,{\ldots},0)^{T}\rangle^{\perp}, S_M^{-1}{\mathbf{w}})$. Thus all vectors in M are of the form $(1,0,{\ldots})^{T}$ or $(0,0,{\ldots})^{T}$. This means that $M \oplus \langle (1,0,0,{\ldots},0)^{T} \rangle$ is a set of commuting observables and $(1,0,0,{\ldots},0)^{T}$ is linearly independent of M if and only if $(0,1,0,{\ldots},0)^{T}$ was not already contained in $(S_M^T W)$. In this case, for each value $q \in Z_{d}$ or $\mathbb{R}$ there exists a valuation vector $\mathbf{v}_q$ such that $(1,0,0,{\ldots},0)^{T}$ has the valuation q and the valuation of M is constant. This means that M is a mixture of states with all possible values of q₁ except if $\mathbf{q_1}$ was already known. After transforming the marginalized system back with S_M , the results we found for $\mathbf{q_1}$ before the transformation with S_M hold now for $\mathbf{f}$.

Lemma B.7 (Existence of symplectic transformation). Consider the same setting as in theorem B.6. In continuous and prime dimensions for any observable $\mathbf{f}$ on the information system and any observable v on the memory system there exist symplectic transformations such that

$$\begin{align} \mathbf{f} &= ((S_M)^T)^{-1} \mathbf{q_1} \end{align} \tag{ B.17 }$$

$$\begin{align} \mathbf{v} &= (T^T)^{-1} \mathbf{q}. \end{align} \tag{ B.18 }$$

Proof. Both transformations exist if we can find for any amount of systems a symplectic transformation where the first column of the transformation is given by an arbitrary vector. Let us call this vector $\mathbf{w} = (w_1,\dots,w_{2n})^T$. The set of vectors:

$$\begin{align} C = \left\{(w_1,\dots,w_{2n})^T,(0,0,w_3,w_4,\dots,w_{2n})^T,\dots, (0,\dots,0,w_{2n-1},w_{2n})^T \right\} \end{align} \tag{ B.19 }$$

all commute with each other. Furthermore, the first vector listed in this set $(w_1,\dots,w_{2n})^T$ has symplectic inner product 1 with the vector $(-1/w_2,0,0,\dots,0)$ if $w_2\,\neq 0$ (otherwise choose $(0,1/w_1,0,\dots,0)$ and commutes with the rest of the set C. For the second vector the same holds for $(1/w_2,0,-1/w_4,0,\dots,0)$ if $w_4\,\neq 0$ (otherwise choose $(1/w_2,0,0,1/w_3,\dots,0)$ or $(0,-1/w_1,0,1/w_3,\dots,0)$ if $w_2 = 0$). In a similar way for all vectors $\mathbf{u}$ in the set C such a vector $\mathbf{u}^{^{\prime}}$ can be constructed such that it has symplectic inner product is 1 with $\mathbf{u}$ and commutes with all other vectors in C. A transformation is sympletic if its columns are such that, the first two columns have symplectic inner product 1 with each other but commute with the rest, the same for the third and fourth column and so on for the following pairs of columns. Therefore, the transformation where the odd columns are the vectors from the set C, starting with the vector $\mathbf{w}$ and the even rows are the vectors we constructed from the previous odd column in the manner as described above is by construction symplectic and has $\mathbf{w}$ as its first column.

Theorem B.8 (Restrictions on conditional transformations). Let $(V_S^{i},{\mathbf{v}}_s^{i})$ be the state of a system and $(V_T^{i},{\mathbf{v}}\,_T^{i})$ the initial state of the system to be prepared in a conditional preparation scenario. Additionally, let S_ST be a symplectic transformation. Then, for ${\mathbf{v}}_{s,1}^{i},\dots, {\mathbf{v}}_{s,k}^{i}$ generating a partition of the ontic state space, i.e. $((V_S^{i})^{\perp} + {\mathbf{v}}_{s,1}^{i})\cup \dots \cup ((V_S^{i})^{\perp} + {\mathbf{v}}_{s,k}^{i}) = \Omega$ the marginals of the target system the final state $(V\,_T^{f} ,{\mathbf{v}}\,_T^{f})$ must be either identical or orthogonal. The number of identical states is equal for each separate type of identical states.

Proof. After the transformation the state of the system is:

$$\begin{align} \left(\left(S_{ST}^{-1}\right)^T \left(V_S^{i}\oplus V_T^{i}\right) ,S_{ST} \left({\mathbf{v}}_s^{i} \oplus {\mathbf{v}}\,_T^{i}\right)\right). \end{align} \tag{ B.20 }$$

This state is allowed as $V_{S}^{i}$ only has support on the first $2\,\text{amount of systems in } S$ entries of the system and $V_{T}^{i}$ on the rest. The marginal of the target system after the transformation is:

$$\begin{align} \left(\left(R_{T} \left(S_{ST}^{-1}\right)^T \left(V_S^{i}\oplus V_T^{i}\right)\right) ,\Pi_{T} S_{ST} {\mathbf{v}}_s^{i} \oplus {\mathbf{v}}\,_T^{i}\right) \end{align} \tag{ B.21 }$$

where $\Pi_{T}$ the projection on the last $2\,\text{amount of systems in } T$ entries of the system and R_T removes the vectors that have support in S. As the transformation cannot depend on ${\mathbf{v}}_{S}^{i}$, $\mathbf{V}_{T}^{f}$ is independent of ${\mathbf{v}}_{S}^{i}$. Therefore, changing ${\mathbf{v}}_{S}^{i}$ can only change the valuation ${\mathbf{v}}_s^{f}$ of the final state. Furthermore, changing the valuation either result in orthogonal or identical states.

If the transformation is irreversible, then we can always see the transformation as a reversible transformation on a larger system and subsequently tracing out some systems. Taking the trace effectively removes vectors from $V_{T}^f$. Therefore it can make states that were orthogonal identical, but it cannot produce non orthogonal or identical states, as also in this case the transformation cannot depend on $\mathbf{v}_{S}^{i}$.

This establishes that the different marginalised states on the target system only differ in their valuations which depend only on $\mathbf{v}_{s}^{i}$. By equation (B.21) this dependence is a linear map. Let us call this map C. So, two target valuations $\mathbf{v}_{T,1}^{f}, \mathbf{v}_{T,2}^{f}$ are identical if and only if $\mathbf{v}_{T,1}^{f} - \mathbf{v}_{T,2}^{f} \in Ker(C)$. Therefore, the sets of identical marginal target states are given by $Ker(C)+\mathbf{v}\,_{T}^{f}$ with $\mathbf{v}\,_{T}^{f}$ some valuation for a target state. Therefore, each set of identical target marginal states has the same number of elements.

Note that all transformations in Spekkens' toy theory are given by a symplectic matrix S and a shift vector $\mathbf{a} \in \Omega$, the shift vector only changes the value of the different known observables by a constant, independent of the vector $\mathbf{v}_{s}^{i}$. Therefore, adding shifts to transformations also does not help to obtain non-orthogonal states.

Corollary 1 (Restrictions on conditional transformations: example). In Spekkens' toy theory, there are no reversible transformations U that implement the action:

Proof. This is a direct application of theorem B.8, which is the formal version of theorem 3.1. It also has a direct and simple proof in the stabilizer version of the toy theory; we present that proof here for pedagogical purposes for the readers familiar with the stabilizer formalism.

In the stabilizer formalization of the toy theory [9], for appropriate dimensions, each valid epistemic state can be isomorphically identified with a set $S = \langle g_1\,g_2,\dots \rangle$ of commuting or anti-commuting Pauli operators forming a valid stabilizer. Each allowed transformation on toy states corresponds to a permutation Π on the subset of stabilizers. This permutation can be represented by a (unitary or anti-unitary) permutation matrix $V_\Pi$ that acts on each stabilizer $g\in S$ as $V_\Pi g V_\Pi^T$. In this case, we have:

This transformation Π would have to map the stabilized states as:

$$\begin{align*} \Pi: \quad \operatorname{span}\{\mathcal{Z}_A,\mathcal{Z}_S\} &\to \operatorname{span}\{\mathcal{Z}_A,\mathcal{Z}_S\}, \\ \operatorname{span}\{-\mathcal{Z}_A,\mathcal{Z}_S\} &\to \operatorname{span}\{\mathcal{Z}_A,\mathcal{X}_S\}. \end{align*}$$

This means that permutation Π would depend on the sign of $\mathcal{Z}_{A}$. This dependence should be explicit in the corresponding permutation matrix $V_\Pi$ (which needs to be either unitary or anti-unitary [9]). However, permutation matrices cannot depend on the sign of stabilizers [9], due to the linear nature of quantum theory. To see this, note that the permutation matrix would have to act on each stabilizer $g\in\{\mathcal{Z}_{A} \otimes {\unicode{x1D7D9}}_S ,- \mathcal{Z}_{A}\otimes {\unicode{x1D7D9}}_S, {\unicode{x1D7D9}}_A\otimes \mathcal{Z}_{S}\}$ as $V_\Pi \ g\ V^\dagger_\Pi$, so if the transformation on the first state behaves as:

$$\begin{align*} \Pi: \quad \langle \mathcal{Z}_A,\mathcal{Z}_S \rangle \to & \langle V_\Pi (\mathcal{Z}_A \otimes {\unicode{x1D7D9}}_S) V^\dagger_\Pi,\ V_\Pi ({\unicode{x1D7D9}}_A \otimes \mathcal{Z}_S) V^\dagger_\Pi \rangle = \operatorname{span}\{\mathcal{Z}_A,\mathcal{Z}_S\} , \end{align*}$$

then it must act on the second state as:

$$\begin{align*} \Pi: \quad \operatorname{span}\{-\mathcal{Z}_A,\mathcal{Z}_S\} \to\quad & \operatorname{span}\{V_\Pi (- \mathcal{Z}_A \otimes {\unicode{x1D7D9}}_S) V^\dagger_\Pi,\ V_\Pi ({\unicode{x1D7D9}}_A \otimes \mathcal{Z}_S) V^\dagger_\Pi\} \\ =& \operatorname{span}\{- V_\Pi ( \mathcal{Z}_A \otimes {\unicode{x1D7D9}}_S) V^\dagger_\Pi,\ V_\Pi ({\unicode{x1D7D9}}_A \otimes \mathcal{Z}_S) V^\dagger_\Pi\} \\ =& \operatorname{span}\{-\mathcal{Z}_A,\mathcal{Z}_S\} \neq \operatorname{span}\{\mathcal{Z}_A,\mathcal{X}_S\}. \end{align*}$$

Zoom In Zoom Out Reset image size

Zoom In Zoom Out Reset image size

The example of the Bell scenario in the generalized formalism goes as follows. Alice and Bob's shared entangled state can be expressed as:

$$\begin{align} (V,{\mathbf{v}}) = \left( \left\langle \begin{pmatrix} 1\\[6pt] 0 \\[6pt] 1 \\[6pt] 0 \end{pmatrix}, \begin{pmatrix} 0\\[6pt] 1 \\[6pt] 0 \\[6pt] -1 \end{pmatrix} \right\rangle, \begin{pmatrix} 1\\[6pt] 0 \\[6pt] 0 \\[6pt] 0 \end{pmatrix} \right). \end{align} \tag{ B.32 }$$

Bob's Z measurement corresponds to observable $\left\langle \begin{pmatrix} 0\\[6pt] 0 \\[6pt] 0 \\[6pt] 1\,\end{pmatrix} \right\rangle$. If he obtains outcome p, he updates his description of the shared state to

$$\begin{align} \left( \left\langle \begin{pmatrix} 0\\[6pt] 0 \\[6pt] 0 \\[6pt] 1 \end{pmatrix}, \begin{pmatrix} 0\\[6pt] 1 \\[6pt] 0 \\[6pt] -1 \end{pmatrix} \right\rangle, \begin{pmatrix} 0\\[6pt] p \\[6pt] 0 \\[6pt] p \end{pmatrix} \right) = \left( \left\langle \begin{pmatrix} 0\\[6pt] 0 \\[6pt] 0 \\[6pt] 1 \end{pmatrix}, \begin{pmatrix} 0\\[6pt] 1 \\[6pt] 0 \\[6pt] 0 \end{pmatrix} \right\rangle, \begin{pmatrix} 0\\[6pt] p \\[6pt] 0 \\[6pt] p \end{pmatrix} \right). \end{align} \tag{ B.33 }$$

Now, if Alice measures her system in $\left\langle \begin{pmatrix} 0\\[6pt] 1 \\[6pt] 0 \\[6pt] 0\,\end{pmatrix} \right\rangle$, she will get the outcome p with certainty. Thus, if Alice and Bob share the state $(V,{\mathbf{v}})$, the conclusion '$B = p \implies A = p$' can be made with certainty. We can also check that the conditions of lemma B.11 are fulfilled:

• (a)  
  $\left\langle \begin{pmatrix} 0\\[6pt] 1 \\[6pt] 0 \\[6pt] 0\,\end{pmatrix} \right\rangle \subset \left\langle \begin{pmatrix} 0\\[6pt] 0 \\[6pt] 0 \\[6pt] 1 \end{pmatrix}\right\rangle \oplus \left\langle\begin{pmatrix} 0\\[6pt] 1 \\[6pt] 0 \\[6pt] -1 \end{pmatrix} \right\rangle$,
• (b)  
  $\begin{pmatrix} 0\\[6pt] p \\[6pt] 0 \\[6pt] p \end{pmatrix}$ $\in \left({\left\langle {\begin{pmatrix} 0\\[6pt] 0 \\[6pt] 0 \\[6pt] 1 \end{pmatrix}}\right\rangle}^{\perp} + \begin{pmatrix} 0\\[6pt] p \\[6pt] 0 \\[6pt] p \end{pmatrix} \right)$ $\cap \left({\left\langle {\begin{pmatrix} 0\\[6pt] 0 \\[6pt] 0 \\[6pt] 1 \end{pmatrix}} \right\rangle}^{\perp} + \begin{pmatrix} 0\\[6pt] p \\[6pt] 0 \\[6pt] p \end{pmatrix} \right)$ $\cap \left({\left\langle {\begin{pmatrix} 0\\[6pt] 1 \\[6pt] 0 \\[6pt] 0 \end{pmatrix}}\right\rangle}^{\perp} + \begin{pmatrix} 0\\[6pt] p \\[6pt] 0 \\[6pt] p \end{pmatrix} \right)$.

We can write an arbitrary global ontic state of Alice and Bob's labs before the measurements start as:

$$\begin{align} \mathbf{o} = \begin{pmatrix} \mathbf{o}_R \\[6pt] \mathbf{o}_A \\[6pt] \mathbf{o}_S \\[6pt] \mathbf{o}_B \end{pmatrix}, \end{align} \tag{ B.34 }$$

with $\mathbf{o}_R \in \mathbb{Z}_d^{n_R}$ or $\mathbb{R}^{n_R}$, $\mathbf{o}_A \in \mathbb{Z}_d^{n_A}$ or $\mathbb{R}^{n_A}$, $\mathbf{o}_S \in \mathbb{Z}_d^{n_S}$ or $\mathbb{R}^{n_S}$, and $\mathbf{o}_B \in \mathbb{Z}_d^{n_B}$ or $\mathbb{R}^{n_B}$. We allow for arbitrary measurements by the different agents, fixing only the range of systems measured and the order of measurements:

• Alice at t = 1  
  performs the following measurement

  $$\begin{align} V_{A} = \left\langle {\begin{pmatrix} {\mathbf{v}}_{A,1} \\[6pt] \mathbf{0} \end{pmatrix} , \dots , \begin{pmatrix} {\mathbf{v}}_{A,{k_A}} \\[6pt] \mathbf{0} \end{pmatrix}} \right\rangle \end{align} \tag{ B.35 }$$

  where ${\mathbf{v}}_{A} \in {\mathbb{Z}}_d^{n_R}$ or ${\mathbb{R}}^{n_R}$. Alice says she got the outcome '1' if she got the result ${\mathbf{v}}_{A = 1}.$
• Bob at t = 2  
  performs the following measurement

  $$\begin{align} V_{B} = \left\langle \begin{pmatrix} \mathbf{0} \\[6pt]{\mathbf{v}}_{B,{1}} \\[6pt] \mathbf{0} \\[6pt] \end{pmatrix}, \dots , \begin{pmatrix} \mathbf{0} \\[6pt]{\mathbf{v}}_{B,{k_B}} \\[6pt] \mathbf{0} \\[6pt] \end{pmatrix} \right\rangle, \end{align} \tag{ B.36 }$$

  where ${\mathbf{v}}_{B} \in \mathbb{Z}_d^{n_S}$ or $\mathbb{R}^{n_S}$. Bob says he got the outcome 1 if he got the result ${\mathbf{v}}_{B = 1}$.
• Ursula at t = 3  
  performs the following measurement:

  $$\begin{align} V_{U} = \left\langle \begin{pmatrix} {\mathbf{v}}_{U,1} \\[6pt] \mathbf{0} \end{pmatrix} , \dots , \begin{pmatrix} {\mathbf{v}}_{U,k_U} \\[6pt] \mathbf{0} \end{pmatrix} \right\rangle, \end{align} \tag{ B.37 }$$

  where ${\mathbf{v}}_{U,1}, {\mathbf{v}}_{U,2} \in \mathbb{Z}_d^{n_A+n_R}$ or $\mathbb{R}^{n_A+n_R}$. Ursula says she measured 'ok' if she got the outcome ${\mathbf{v}}_{U,ok}$ and 'fail' if she got the outcome ${\mathbf{v}}_{U,fail}$. These two vectors need to correspond to distinct outcomes, i.e. ${\mathbf{v}}_{U,ok} -{\mathbf{v}}_{U,fail} \notin V_{U}^{\perp}$.
• Wigner at t = 4  
  performs the following measurement:

  $$\begin{align} V_{W} = \left\langle \begin{pmatrix} \mathbf{0} \\[6pt] {\mathbf{v}}_{W,1} \\[6pt] \end{pmatrix}, \dots ,\begin{pmatrix}\mathbf{0} \\[6pt] {\mathbf{v}}_{W,k_W} \\[6pt] \end{pmatrix} \right\rangle, \end{align} \tag{ B.38 }$$

  where ${\mathbf{v}}_{W,1}, {\mathbf{v}}_{W,2} \in \mathbb{Z}_d^{n_B+n_S}$ or $\mathbb{R}^{n_W+n_s}$. Wigner says he measured 'ok' if he got the outcome ${\mathbf{v}}_{W,ok}$ and 'fail' if he got the outcome ${\mathbf{v}}_{W,fail}$. These two vectors need to correspond to distinct outcomes, i.e. ${\mathbf{v}}_{W,ok} -{\mathbf{v}}_{W,fail} \notin V_{W}^{\perp}$.

The linear algebraic statements used in the proof are formally justified in appendix B.1.

Theorem 5.1. There exists no valid epistemic state that can be used to reach a logical contradiction in the Frauchiger–Renner setting in the toy theory.

Proof. We separate the proof in four parts. First, we find necessary conditions on the epistemic state such that predictions can be made with certainty. Second, we find conditions such that the probability for Wigner and Ursula both get 'ok' is non zero. Finally, we show that two measurement outcomes which would lead to a contradiction must be the same and, thus, do not lead to a contradiction.

We follow the chain of reasoning and determine what V needs to fulfill to allow for predictions with certainty:

• 'U = ok$\boldsymbol{\implies}$ B = 1'  
  The first condition for predictions with certainty requires that $V_{B} \subset V_{\text{commute},U} \oplus V_U$ where $V_{\text{commute},U}$ denotes subset of V that commutes with all vectors in V_U . This means for all vectors ${\mathbf{v}}_B \in V_B$ these exist vectors ${\mathbf{v}} \in V_{\text{commute},U}, {\mathbf{v}}_U \in V_U$ such that ${\mathbf{v}}_B = {\mathbf{v}} + {\mathbf{v}}_U$.
• 'B = 1 $\boldsymbol{\implies}$ A = 1'  
  The first condition for predictions with certainty requires that $V_{A} \subset V_{\text{commute},B} \oplus V_B$ where $V_{\text{commute},B}$ denotes subset of V that commutes with all vectors in V_B . This means for all vectors ${\mathbf{v}}_A \in V_A$ these exist vectors ${\mathbf{v}} \in V_{\text{commute},B},{\mathbf{v}}_B \in V_B$ such that ${\mathbf{v}}_A = {\mathbf{v}} + {\mathbf{v}}_B$.
• 'A = 1 $\boldsymbol{\implies}$ W = fail'  
  The first condition for predictions with certainty requires that $V_{W} \subset V_{\text{commute},A} \oplus V_A$ where $V_{\text{commute},A}$ denotes subset of V that commutes with all vectors in V_A . This means for all vectors ${\mathbf{v}}_W \in V_W$ these exist vectors ${\mathbf{v}} \in V_{\text{commute},A},{\mathbf{v}}_A \in V_A$ such that ${\mathbf{v}}_W = {\mathbf{v}} + {\mathbf{v}}_A$.

In total, this means for each ${\mathbf{v}}_W \in V_W$ there exist vectors ${\mathbf{v}}_1\in V_{\text{commute},A}$,${\mathbf{v}}_2\,\in V_{\text{commute},B}$,${\mathbf{v}}_3 \in V_{\text{commute},U}$, ${\mathbf{v}}_A \in V_A$,${\mathbf{v}}_B \in V_B$, and ${\mathbf{v}}_U \in V_U$ such that

$$\begin{align} \begin{split} {\mathbf{v}}_W &= {\mathbf{v}}_1 + {\mathbf{v}}_A \\ &= {\mathbf{v}}_1 + {\mathbf{v}}_2 + {\mathbf{v}}_B \\ &= {\mathbf{v}}_1 + {\mathbf{v}}_2 + {\mathbf{v}}_3 + {\mathbf{v}}_U. \end{split} \end{align} \tag{ B.39 }$$

Even if the above chain of reasoning holds, the paradox only occurs if, additionally, the probability for Ursula and Wigner to both get 'ok' $P(ok,ok)$ is not zero. As they perform a joint measurement their measurement is given by $V_{U} \oplus V_{W}$. Because Ursula and Wigner measure different subsystems we can choose equivalent measurement outcome vectors such that ${\mathbf{v}}_{U,ok} \in V_{W}^{\perp}$ and ${\mathbf{v}}_{W,ok} \in V_{U}^{\perp}$. Thus, the measurement outcome $W = ok, U = ok$ has has a shift vector ${\mathbf{v}}_{U,ok} + {\mathbf{v}}_{W,ok}$. For $P(ok,ok) \neq 0$ to hold the state $(V,{\mathbf{v}})$ has to fulfill

$$\begin{align} ((V_{U} \oplus V_{W})^{\perp} + {\mathbf{v}}_{U,ok} + {\mathbf{v}}_{W,ok}) \cap (V^{\perp} + {\mathbf{v}}) \neq \emptyset. \end{align} \tag{ B.40 }$$

Lemma B.2 allows us to conclude that this condition is equivalent to:

$$\begin{align} {\mathbf{v}}_{U,ok} + {\mathbf{v}}_{W,ok} - {\mathbf{v}} \in (V_{U} \oplus V_{W})^{\perp} \oplus V^{\perp}. \end{align} \tag{ B.41 }$$

Due to $(V^{\perp})^{\perp} = V$ for a submodule or subset V (lemma B.4), we can rewrite the expression $(V_{U}\oplus V_{W})^{\perp} \oplus V^{\perp}$:

$$\begin{align} \begin{split} \left(V_{U} \oplus V_{W}\right)^{\perp} \oplus V^{\perp} &= \left( \left( \left(V_{U} \oplus V_{W}\right)^{\perp} \oplus V^{\perp}\right)^{\perp} \right)^{\perp} \\ &= \left( \left(\left(V_{U} \oplus V_{W}\right)^{\perp}\right)^{\perp} \cap\left( V^{\perp}\right)^{\perp}\right)^{\perp} \\ &= \left( \left(V_{U} \oplus V_{W}\right) \cap V\right)^{\perp}. \end{split} \end{align} \tag{ B.42 }$$

Thus, equation (B.39) is equivalent to:

$$\begin{align} {\mathbf{v}}_{U,ok} + {\mathbf{v}}_{W,ok} - {\mathbf{v}} \in \left( \left(V_{U} \oplus V_{W}\right) \cap V\right)^{\perp}. \end{align} \tag{ B.43 }$$

We want to ensure that in the reasoning chain we cannot only conclude with certainty but we can also make the desired conclusions. For example, the previous conditions ensure that when Bob measures he gets an outcome with certainty (which can be calculated from the epistemic state), but the condition we consider here ensures that this outcome is B = 1:

• 'U = ok $\boldsymbol{\implies}$ B = 1'  
  We apply the second condition for predictions with certainty. Thus, it has to hold that

  $$\begin{align} (V_{B}^{\perp} + {\mathbf{v}}_{B = 1}) \cap (V_{\text{commute}, U}^{\perp} + {\mathbf{v}}) \cap (V_U^{\perp} + {\mathbf{v}}_{U,ok}) \neq \emptyset. \end{align} \tag{ B.44 }$$

  The two measurements are defined on two different subsystems. Therefore, we can choose, without loss of generality, ${\mathbf{v}}_{B = 1} \in V_U^{\perp}$ and ${\mathbf{v}}_{U,ok} \in V_B^{\perp}$. Additionally, it does not matter which intersection is calculated first. Therefore, we can first calculate the intersection $(V_{B}^{\perp} + {\mathbf{v}}_{B = 1}) \cap (V_U^{\perp} + {\mathbf{v}}_{U,ok})$ using lemmas B.1 and B.3

  $$\begin{align} & (V_{B}^{\perp} + {\mathbf{v}}_{B= 1}) \cap (V_U^{\perp} + {\mathbf{v}}_{U,ok}) \nonumber\\ & \quad = (V_{B}^{\perp} + {\mathbf{v}}_{U,ok} +{\mathbf{v}}_{B= 1}) \cap (V_U^{\perp} + {\mathbf{v}}_{U,ok} + {\mathbf{v}}_{B= 1}) \nonumber\\ & \quad = (V_{B}^{\perp} \cap V_U^{\perp}) + {\mathbf{v}}_{B = 1} +{\mathbf{v}}_{U,ok} \nonumber\\ & \quad = (V_{B} \oplus V_U)^{\perp} + {\mathbf{v}}_{B= 1} +{\mathbf{v}}_{U,ok}. \end{align} \tag{ B.45 }$$

  Plugging this result into equation (B.44) we find the condition

  $$\begin{align} ((V_{B} \oplus V_U)^{\perp} + {\mathbf{v}}_{B= 1} +{\mathbf{v}}_{U,ok}) \cap (V_{\text{commute}, U}^{\perp} + {\mathbf{v}}) \neq \emptyset. \end{align} \tag{ B.46 }$$

  Due to lemma B.1 this condition is equivalent to the condition:

  $$\begin{align} & {\mathbf{v}}_{B= 1} +{\mathbf{v}}_{U,ok} - {\mathbf{v}} \in (V_{B} \oplus V_U)^{\perp} \oplus V_{\text{commute}, U}^{\perp}\nonumber\\ & \quad = \left((V_{B} \oplus V_U) \cap V_{\text{commute}, U}\right)^{\perp}. \end{align} \tag{ B.47 }$$
• 'B = 1 $\boldsymbol{\implies}$ A = 1'  
  With the same reasoning as above we find

  $$\begin{align} {\mathbf{v}}_{A= 1} +{\mathbf{v}}_{B = 1} - {\mathbf{v}} \in \left((V_{B} \oplus V_A) \cap V_{\text{commute}, B}\right)^{\perp}. \end{align} \tag{ B.48 }$$
• 'A = 1 $\boldsymbol{\implies}$ W = fail'  
  With the same reasoning as above we find

  $$\begin{align} {\mathbf{v}}_{A= 1} +{\mathbf{v}}_{W,fail} - {\mathbf{v}} \in \left((V_{A} \oplus V_W) \cap V_{\text{commute}, A}\right)^{\perp}. \end{align} \tag{ B.49 }$$

In total, the epistemic state $(V,{\mathbf{v}})$ and the measurements of Alice, Bob, Ursula, and Wigner need to fulfil:

• (a)  
  $V_{B} \subset V_{\text{commute},U} \oplus V_U$
• (b)  
  $V_{A} \subset V_{\text{commute},B} \oplus V_B$
• (c)  
  $V_{W} \subset V_{\text{commute},A} \oplus V_A$
• (d)  
  ${\mathbf{v}}_{U,ok} + {\mathbf{v}}_{W,ok} - {\mathbf{v}} \in \left( \left(V_{U} \oplus V_{W}\right) \cap V\right)^{\perp}$
• (e)  
  ${\mathbf{v}}_{B = 1} +{\mathbf{v}}_{U,ok} - {\mathbf{v}} \in \left((V_{B} \oplus V_U) \cap V_{\text{commute}, U}\right)^{\perp}$
• (f)  
  ${\mathbf{v}}_{A = 1} +{\mathbf{v}}_{B = 1} - {\mathbf{v}} \in \left((V_{B} \oplus V_A) \cap V_{\text{commute}, B}\right)^{\perp}$
• (g)  
  ${\mathbf{v}}_{A = 1} +{\mathbf{v}}_{W,fail} - {\mathbf{v}} \in \left((V_{A} \oplus V_W) \cap V_{\text{commute}, A}\right)^{\perp}$

Let us assume we have found a state $(V,{\mathbf{v}})$ and measurements such that the above chain of reasoning holds, and $P(ok,ok) \neq 0$. Such a state would lead to a paradox. In the following, we show that such a state and measurements cannot exist.

For all ${\mathbf{v}}_W$ there exist ${\mathbf{v}}_A,{\mathbf{v}}_B$, and ${\mathbf{v}}_U$ be as in equation (B.39). Without loss of generality, we can choose equivalent measurement outcomes such that ${\mathbf{v}}_{B = 1} \in V_U^{\perp}$ and ${\mathbf{v}}_{U,ok} \in V_B^{\perp}$. Then it holds that ${\mathbf{v}}_U - {\mathbf{v}}_B \in V_{\text{commute}, U}$ and ${\mathbf{v}}_U - {\mathbf{v}}_B \in V_B \oplus V_U$. Thus, equation (B.47) implies that:

$$\begin{align} -{\mathbf{v}}_B^T {\mathbf{v}}_{B = 1} + {\mathbf{v}}_U^T {\mathbf{v}}_{U,ok} + {\mathbf{v}}_B^T {\mathbf{v}} -{\mathbf{v}}_{U}^T {\mathbf{v}} = 0. \end{align} \tag{ B.50 }$$

With the same argument we can find the following conditions:

$$\begin{align} -{\mathbf{v}}_B^T {\mathbf{v}}_{B = 1} + {\mathbf{v}}_A^T {\mathbf{v}}_{A = 1} + {\mathbf{v}}_B^T {\mathbf{v}} -{\mathbf{v}}_A^T {\mathbf{v}} &= 0, \end{align} \tag{ B.51 }$$

$$\begin{align} -{\mathbf{v}}_W^T {\mathbf{v}}_{W,fail} + {\mathbf{v}}_A^T {\mathbf{v}}_{A = 1} - {\mathbf{v}}_A^T {\mathbf{v}} +{\mathbf{v}}_W^T {\mathbf{v}} &= 0. \end{align} \tag{ B.52 }$$

Subtracting equation (B.51) from equation (B.52) find the condition:

$$\begin{align} {\mathbf{v}}_W^T {\mathbf{v}}_{W,fail} -{\mathbf{v}}_W^T {\mathbf{v}} + {\mathbf{v}}_B^T {\mathbf{v}}_{B = 1} - {\mathbf{v}}_B^T {\mathbf{v}} = 0. \end{align} \tag{ B.53 }$$

Adding up equations (B.53) and (B.50) we find the condition

$$\begin{align} -{\mathbf{v}}_W^T {\mathbf{v}}_{W,fail} +{\mathbf{v}}_W^T {\mathbf{v}} + {\mathbf{v}}_U^T {\mathbf{v}}_{U,ok} - {\mathbf{v}}_U^T {\mathbf{v}} = 0. \end{align} \tag{ B.54 }$$

If $P(ok,ok) \neq 0$ it holds that for all ${\mathbf{w}} \in \left(V_{U} \oplus V_{W}\right) \cap V$

$$\begin{eqnarray} &&{\mathbf{w}}^T({\mathbf{v}}_{U,ok} + {\mathbf{v}}_{W,ok} - {\mathbf{v}}) = 0. \end{eqnarray} \tag{ B.55 }$$

In particular, it holds that ${\mathbf{w}} = {\mathbf{v}}_U - {\mathbf{v}}_W \in \left(V_{U} \oplus V_{W}\right) \cap V$. Thus we can conclude that the following condition holds

$$\begin{align} -{\mathbf{v}}_W^T {\mathbf{v}}_{W,ok} + {\mathbf{v}}_U^T {\mathbf{v}}_{U,ok} -{\mathbf{v}}_U^T {\mathbf{v}} +{\mathbf{v}}_W^T {\mathbf{v}} = 0. \end{align} \tag{ B.56 }$$

We can subtract equation (B.54) from equation (B.56) and find

$$\begin{align} {\mathbf{v}}_W^T ({\mathbf{v}}_{W,ok} - {\mathbf{v}}_{W,fail}) = 0. \end{align} \tag{ B.57 }$$

Because ${\mathbf{v}}_U,{\mathbf{v}}_A,{\mathbf{v}}_B$ as in equation (B.39) can be found for all ${\mathbf{v}}_W$ it holds that ${\mathbf{v}}_{W,ok}- {\mathbf{v}}_{W,fail} \in V_W^{\perp}$. Thus, ${\mathbf{v}}_{W,ok}- {\mathbf{v}}_{W,fail}$ correspond to the same measurement outcome, as they have same valuation for any ${\mathbf{v}}_W \in V_W$. In summary, if there is a state and measurements such that the paradoxical chain of reasoning would hold, we have shown that then the two measurement outcomes that would lead to a paradox have to be the same outcome. Therefore, no such paradoxical chain of reasoning is possible.

We have two qubits R and S and two participants Alice and Bob, whose memory registers A and B are also modelled as one qubit each. There are two external agents Ursula and Wigner, whose quantum memories do not need to be explicitly modelled at this stage. The initial state of $R, S, A, B$ is a Hardy state of R and S [22], and erased memories:

$$\begin{align} \vert \psi_0\rangle_{RSAB} &= \frac1{\sqrt3} \left(\vert 0\rangle_R\vert 0\rangle_S + \vert 1\rangle_R \vert 0\rangle_S + \vert 1\rangle_R\vert 0\rangle_S \right) \otimes \vert 0\rangle_A \vert 0\rangle_B. \end{align} \tag{ C.7 }$$

We will describe the protocol and the evolution of the state of RSAB from the perspective of Ursula and Wigner, who put Alice and Bob's labs below the Heisenberg cut in a neo-Copenhagen interpretation (that is, they model Alice and Bob's measurements as reversible physical evolutions of quantum systems).

• t = 1  
  Alice measures R in the Z basis and stores the result in her memory. From Wigner's perspective, she is now entangled with R:

  $$\begin{align} \vert \psi_1\rangle_{RASB} &= \frac1{\sqrt3} \left(\vert 0\rangle_R\vert 0\rangle_A\vert 0\rangle_S + \vert 1\rangle_R\vert 1\rangle_A \vert 0\rangle_S + \vert 1\rangle_R\vert 1\rangle_A\vert 0\rangle_S \right) \otimes \vert 0\rangle_B. \end{align} \tag{ C.8 }$$

  Notation: we changed the order of the subsystems because this will be easier later.
• t = 2  
  Bob measures S in the Z basis and stores the result in his memory. The global state is now a Hardy state between the two labs,

  $$\begin{align} \vert \psi_2\rangle_{RASB} &= \frac1{\sqrt3} \left(\vert 0\rangle_R\vert 0\rangle_A\vert 0\rangle_S \vert 0\rangle_B + \vert 1\rangle_R\vert 1\rangle_A \vert 0\rangle_S \vert 0\rangle_B + \vert 1\rangle_R\vert 1\rangle_A\vert 0\rangle_S \vert 1\rangle_B \right). \end{align} \tag{ C.9 }$$
• t = 3  
  Ursula measures Alice's lab (RA) in the Bell basis; in particular we care about outcomes with non-zero probability, which correspond to the eigenstates:

  $$\begin{align} \vert \text{ok}\rangle_{RA} &= \frac{\vert 0\rangle_R \vert 0\rangle_A - \vert 1\rangle_R\vert 1\rangle_A}{\sqrt2}, \end{align} \tag{ C.10 }$$

  $$\begin{align} \vert \text{fail}\rangle_{RA} &= \frac{\vert 0\rangle_R \vert 0\rangle_A + \vert 1\rangle_R\vert 1\rangle_A}{\sqrt2}. \end{align} \tag{ C.11 }$$
• t = 4  
  Wigner measures Bob's lab (SB) in the Bell basis; again we label the two eigenstates with finite probability as:

  $$\begin{align} \vert \text{ok}\rangle_{SB} &= \frac{\vert 0\rangle_S \vert 0\rangle_B - \vert 1\rangle_S\vert 1\rangle_B}{\sqrt2}, \end{align} \tag{ C.12 }$$

  $$\begin{align} \vert \text{fail}\rangle_{SB} &= \frac{\vert 0\rangle_S \vert 0\rangle_B + \vert 1\rangle_S\vert 1\rangle_B}{\sqrt2}. \end{align} \tag{ C.13 }$$

The possibility of both Ursula and Wigner getting the outcome 'ok' is non-zero:

$$\begin{align*} P[u=w=ok]=|(\langle \text{ok}\vert_{RA}\langle \text{ok}\vert_{SB})\vert \psi_2\rangle_{RASB}|^2=\frac{1}{12}. \end{align*}$$

From now on, we post-select on this event. At time t = 3, Ursula reasons about the outcome that Bob observed at t = 2. Since we can regroup the global state before her measurement as:

$$\begin{align} \vert \psi_2\rangle_{RASB} = \sqrt{\frac23} \vert \text{fail}\rangle_{RA}\vert 0\rangle_S \vert 0\rangle_B + \frac{1}{\sqrt{3}}\vert 1\rangle_R\vert 1\rangle_A\vert 1\rangle_S\vert 1\rangle_B. \end{align} \tag{ C.14 }$$

Ursula concludes that the only possibility with non-zero overlap with her observation of $\vert \text{ok}\rangle_{RA}$ is that Bob measured $\vert 1\rangle_S$. We can write this inference as '$u = \text{ok} \implies b = 1$'. She can further reason about what Bob, at time t = 2 thought about Alice's outcome at time t = 1. Whenever Bob observes $\vert 1\rangle_S$, he can use the same form of $\vert \psi_2\rangle_{RASB}$ to conclude that Alice must have measured $\vert 1\rangle_R$. We can write this as '$b = 1\,\implies a = 1$'. Finally, we can think about Alice's deduction about Wigner's outcome. Using the rewriting of the global state:

$$\begin{align} \vert \psi_2\rangle_{RASB} = \frac{1}{\sqrt{3}}\vert 0\rangle_R\vert 0\rangle_A \vert 0\rangle_S\vert 0\rangle_B + \sqrt{\frac{2}{3}} \vert 1\rangle_R\vert 1\rangle_A \vert \text{fail}\rangle_{SB}, \end{align} \tag{ C.15 }$$

we see that Alice reasons that, whenever she finds R in state $\vert 1\rangle_R$, then Wigner will obtain outcome 'fail' when he measures Bob's lab. That is, '$a = 1\,\implies w = \text{fail}$'. Thus, chaining together the statements (the same reasoning that allowed the reader to solve the three hats problem), we reach an apparent contradiction:

$$\begin{align*} w = u = \text{ok} \implies b = 1\implies a = 1\implies w = \text{fail}. \end{align*}$$

That is, when the experiments stops with $u = w = \text{ok}$, the agents can make deterministic statements about each other's reasoning and measurement results, concluding that Alice had predicted $w = \text{fail}$, hence arriving to a logical contradiction.