# How can a "state" in a contextuality scenario be contextual?

As also discussed in the answers to What do noncontextual scenarios with no quantum model represent? and Can the Peres-Mermin square be reframed as a statement on the associated conditional outcome probabilities?, a contextuality scenario, as defined in (Leifer and Duarte 2020), is a triple $$\mathfrak C=(X,\mathcal M,\mathcal N)$$ with $$X$$ a set of outcomes, $$\mathcal M$$ a set of measurement contexts, and $$\mathcal N$$ a set of maximal partial measurement contexts.

A typical example of this is the Specker triangle as the contextuality scenario with $$X=\{a,b,c\}$$, $$\mathcal M=\{\{a,b\},\{a,c\},\{b,c\}\}$$ and $$\mathcal N=\varnothing$$. The idea is to model a situation where there are three possible outcomes, with labels $$a,b,c$$, and three possible measurement settings (or measurement contexts, if you will), corresponding to the elements of $$\mathcal M$$. Given any measurement context $$M\in\mathcal M$$, one and only one outcome is possible.

One can then consider value functions and states on a contextuality scenario. A value function is some $$v:X\to\{0,1\}$$ which represents a deterministic outcome assignment. It prescribes for any possible outcome whether that outcome will be found or not. A state is the probabilistic counterpart of a value function: a map $$\omega:X\to[0,1]$$ assigning a probability to each outcome (both $$v$$ and $$\omega$$ are required to satisfy a few constraints to make sure the above interpretations are sensible). For example, the Specker triangle above has no value function, but it admits the state $$\omega(a)=\omega(b)=\omega(c)=\frac12$$.

Another function one can consider is a noncontextual state, which is defined (c.f. Def II.11 in Leifer and Duarte 2020) as a state $$\omega:X\to[0,1]$$ such that $$\omega(a) = \sum_{v\in V_{\mathfrak C} } p_v v(a),$$ with $$(p_v)_v$$ a probability distribution on $$V_{\mathfrak C}$$, and where $$V_{\frak C}$$ was defined previously as the set of all value functions. States that are not noncontextual are said to be contextual.

What confuses me about this is the fact that, looking at the fact that all states are functions exclusively of the outcomes, how can any state be contextual? The paper explicitly mentions that value function are always noncontextual, being defined on the set of outcomes regardless of the context. Why doesn't the same argument apply to states? Where is the information about the context entering the definition?

To be clear, I'm aware of the fact that one can have contextual quantum scenarios, in the sense of not being possible to determine probability outcomes without specifying the measurement setting. My doubt is specifically about the fact that, using this definition of a "state", I'm not seeing where information about the context is to be fed.

• Hi, just a fast comment. A state $\omega$ is called contextual when is not in $C_{\mathfrak{C}}$. So, any state, when exists, that is not inside the set described is said to be contextual. Importantly, not every state can be explained using the value functions statistics. It is subtle, but the noncontextual assumption, i.e., information about consistency among contexts, is being considered in the definition of the value functions -- note that they are defined over scenarios.
– R.W
Commented Sep 9, 2022 at 4:26
• @R.W but that's my point: states are also only defined over the set of outcomes $X$ (is that what you mean with "are defined over scenarios"? If not, I'm missing something; value functions are not defined over contexts, are they?). So what makes them "contextual"? Is there a simple example showing the "contextuality" of a state that is not in $C_{\frak C}$?
– glS
Commented Sep 9, 2022 at 6:23
• for example, considering the state that can be defined for the Specker triangle. This oughts to be contextual according to this definition, as there's no value functions at all. Still, we can assign probabilities for each outcomes regardless of contexts: $\omega(x)=1/2$ for all $x\in X=\{a,b,c\}$. So in what sense is this state "contextual"?
– glS
Commented Sep 9, 2022 at 6:27
• Check Def. 2.6; value functions are not just any $v:X \to [0,1]$, they must satisfy constraints over a scenario, otherwise they are not value functions. That's what I meant with 'defined over a scenario'.
– R.W
Commented Sep 10, 2022 at 6:49
• @R.W sure, you need to have exactly one $1$ in each context, which is analogous to the requirement for states that probabilities be normalised in each context. I still don't see what's "contextual" about eg the state above. It's defined on outcomes, like value functions, and it has to satisfy some requirements in each context, like value functions. For something to be "contextual" I'd expect that outcome probabilities depend on contexts, like we have in QM, no?
– glS
Commented Sep 10, 2022 at 8:56

So, I'll try to answer with the story of the question:

What confuses me about this is the fact that, looking at the fact that all states are functions exclusively of the outcomes, how can any state be contextual?

What makes a state contextual or not is not related to them being functions of the outcomes, but related to how these depend on the measurement contexts of the scenario.

The paper explicitly mentions that value function are always noncontextual, being defined on the set of outcomes regardless of the context. Why doesn't the same argument apply to states? Where is the information about the context entering the definition?

The value functions are noncontextual because any of the assignments will not really depend on the context information with which the other measurements were performed. This does translates to noncontextuality at the level of states because states are noncontextual if they can be written as probabilistic mixtures of value functions. The context information is entering in the definition of the value functions, so here is crucial we go back to the definition of them:

Def. 2.6: A value function $$v: X \to [0,1]$$ on a contextuality scenario $$\mathfrak{C} = (X,\mathcal{M},\mathcal{N})$$ is a function that assigns a value $$0$$ or $$1$$ to every outcome such that (1) For every $$M \in \mathcal{M}, v(a)=1$$ for exactly one $$a \in M$$, (2) For every $$N \in \mathcal{N}, v(a)=1$$ for at most one $$a\in N$$.

This definition is the crucial one where the information about the measurement contexts is relevant. Note that what it is saying is that for any measurement context, there can only be at at most one value assignment $$1$$, which implies that the global and local structure of the contexts are logically consistent.

My doubt is specifically about the fact that, using this definition of a "state", I'm not seeing where information about the context is to be fed.

This context information is indirect; the context information for classical states is described in terms of them being possibly described by a statistical mixture of value assignments.

States are also only defined over the set of outcomes X (is that what you mean with "are defined over scenarios"? If not, I'm missing something; value functions are not defined over contexts, are they?). So what makes them "contextual"? Is there a simple example showing the "contextuality" of a state that is not in C_{\mathfrak{C}}?

Here is the biggest source of confusion. Both states and value functions are defined with respect to a given scenario. Therefore, existence or not is dependent on the scenario one considers. So, yes, note that for the definition of every type of state they mention over a scenario and note also that the set of states related to a particular type are always with respect to a given scenario and this is why the notation $$C_{\mathfrak{C}}$$ or $$Q_{\mathfrak{C}}$$ etc. A state is not just a function $$\omega: X \to [0,1]$$.

you need to have exactly one 1 in each context, which is analogous to the requirement for states that probabilities be normalised in each context. I still don't see what's "contextual" about eg the state above. It's defined on outcomes, like value functions, and it has to satisfy some requirements in each context, like value functions. For something to be "contextual" I'd expect that outcome probabilities depend on contexts, like we have in QM, no?

Note that when they are contextual, they do depend on the contexts. Take for instance the Specker's triangle and the only state that this scenario has, given by $$\omega(a) = \omega(b)=\omega(c)=1/2$$. This state is contextual since you cannot simply say that the state of the system is given, even statistically, by some particular elements of the measurements. There is no local 'existence' of the state unless a complete 'global' picture over all contexts is given. If you observe $$a$$ you cannot say that 'the system was in state a' or 'with some $$p$$ the system is in state a and with $$1-p$$ the system is in c' etc.

• thanks for the answer. This helps. In synthesis, being a function of outcomes is a red herring. So, you're saying the Specker triangle example is contextual because there's no way to say what the outcome in each context is before the measurement. That makes sense, of course. But do you have a simple example of a contextual scenario with both a contextual and a non-contextual state? That would help to further highlight the relevant differences. I suppose I could build such an example based on some known non-contextual quantum state, but a simpler toy example might be even better in this context
– glS
Commented Sep 11, 2022 at 17:07
• I think that the simplest such scenario might be the KCBS; I can improve the answer on this line. Sorry if it has became too long.
– R.W
Commented Sep 12, 2022 at 5:53