# Realism and counterfactual definiteness in Bell's theorem

1. Dec 13, 2015

### ShayanJ

Usually its said that the violation of Bell's inequality means that any theory that contains the assumptions of locality and realism doesn't agree with QM and observations. But sometimes I hear people talk about counter-factual definiteness instead of realism(or maybe the presence of both!) as the underlying assumption of Bell's inequality. But I really have problem understanding the difference between realism and counter-factual definiteness, actually it seems to me that they're the same and we only have local realism that is rejected by the violation of Bell's inequality.
I'll appreciate any discussion on these issues.
Thanks

2. Dec 13, 2015

### zonde

Realism is well established term in philosophy and all of the science assumes that realism holds in order for science to be meaningful. But in QM people by "realism" usually mean something different, like counterfactual definiteness. But just the same it sounds quite weird to hear something like "QM violates realism". So I prefer to use "counterfactual definiteness" and leave "realism" for it's proper philosophical meaning.

3. Dec 13, 2015

### stevendaryl

Staff Emeritus
I agree that "realism" is a fuzzy, philosophical concept. But the notion of a "local realistic model" is pretty clear. Basically, a local realistic model has a notion of the "state" of the universe at a particular time (there's probably some generalization to make the notion covariant), and the state factors into a local state for each small region of space. The local state for a small region might (for instance) be described by giving the values of various fields within that region, and the momenta and spins of particles within that region. In a local realistic model, a measurement that takes place at a particular time, confined to a small region of space, simply reveals information about the local state.

Quantum mechanics is definitely not a local realistic theory, in this sense. It does have a notion of a "state", but that notion of state does not factor into local states for each small region of space. Particularly because of entanglement, the QM state contains nonlocal information that can't be factored. "If Alice measures spin-up along axis $\vec{\alpha}$, then Bob will measure spin-down along axis $\vec{\alpha}$" is a fact about Alice and Bob that doesn't factor into a fact about Alice plus a fact about Bob. So even though there might be some sense in which the word "realism" applies to QM, and maybe there's a sense in which the word "local" applies to QM, but it's definitely not a local realistic model, in the sense of Bell. Before Bell's theorem, some people held out hope that maybe a new theory might be developed that would be a local realistic model, but would make the same predictions as QM. But Bell proved this impossible.

To me, Counterfactual Definiteness is not the same as local realism. CFD as it relates to EPR is the claim that in the alternate history in which Alice chose to measure spin along a different axis than $\alpha$, she would have gotten a definite result. I think this goes beyond the assumption of local realism, in that it is assuming that the outcomes of measurements are deterministic functions of the local state. I don't think local realism needs to assume that. You could imagine that the outcome is nondeterministic. That would still be locally realistic, I think, as long as the probabilities for the various outcomes depended only on the local state.

4. Dec 13, 2015

### billschnieder

The situation with the Bell stuff is an utter mess if you ask me.
Realism means objects have properties even when they have not been measured. Counterfactual definiteness means unmeasured (counterfactual) properties have single definite values. Counterfactuals deal with statements of the type "If x then A". It means such arguments statements are still valid even if x is false. For example, if Alice measured along "a" and got -1, then if Bob measured along "a" he must get +1. According to QM, that statement is true, even if Bob never measured along "a", it is still meaningful to talk about a definite value Bob would have obtained by measuring along "a". As you can appreciate, these kinds of statements form the cornerstone of most of logic and can't be easily rejected.

Another example, consider a stream of entangled spin-1/2 particle pairs heading toward Alice and Bob. They've picked in advance a pair of settings from the set {(a,b), (a,d), (c,b), (c,d)}. All the particles will be measured at just one of those pairs. You are tasked with predicting what the expected value of the product of their results would be based on what they picked. You can calculate predictions of E(a,b), E(a,d), E(c,b) and E(c,d) using QM. Note that they will measure just one of those. The question then is, which one is correct?According to CFD, they are all correct. If they pick (a,b), they will measure E(a,b), and if they picked (a,d) they would have measured E(a,d), etc. This is Counterfactual definiteness. The theory predicts a single definite result even if it is not measured. It is not as easy to reject CFD and be consistent in your application of QM.

Bell used CFD in two ways in his original derivation of the inequalities.
1) He used the prediction of QM that if Alice measured along "a" and got -1, then if Bob measured along "a" he must get +1. This allowed him to substitute B(a, λ) = - A(a,λ)
2) He used 3 measurement angles ("a", "b", "c") when only 2 stations were present. ie The inequality involves E(a,b), E(a,c), E(b,c) -- even if Alice and Bob only picked (a,b) for all their measurements, it is still true that they would have obtained E(a,c) had they picked (a,c) instead, OR they would have obtained E(b,c) had they picked (b,c) instead.

5. Dec 13, 2015

### ddd123

Noob question: I thought CFD wasn't really about determinism. Suppose a statistical distribution holds for observables in EPR. Then "what if" questions apply statistically, say an expected statistical correlation is broken by another choice of observable.

6. Dec 13, 2015

### billschnieder

Correct. See my second example above.

7. Dec 13, 2015

### wle

From what I've seen, a lot of the confusion about Bell's theorem seems to come from implicit (and illogical) beliefs, e.g.:
• There can only be one set of starting assumptions that let you derive the Bell inequalities. The proof I saw is "the" way to do it.
• I read a derivation of a Bell inequality that assumes X, Y, and Z, therefore all of X, Y, and Z are necessary to derive a Bell inequality.
• If there are different starting assumptions that let you derive a Bell inequality, then I can choose which set of assumptions is refuted by a Bell inequality violation.
Off the top of my head, I know at least four sets of starting assumptions that you can make that will let you derive Bell inequalities*:
1. Locality + determinism (inferred from QM's prediction of perfect correlations or anticorrelations for certain measurements).
2. Locality, in the sense explicitly formulated by Bell in his 1976 essay "The Theory of Local Beables" and in later works.
3. No FTL signalling + determinism.
4. A "noncontextuality" assumption, which says that it should be possible to attribute an underlying joint probability distribution for all the possible results of a Bell experiment, including the measurements that aren't performed.
These are conceptually different assumptions to start with, but they turn out to be equivalent at the mathematical level:
• You can make any Bell-local model deterministic by adding more hidden variables, so there is no difference between 1 and 2.
• The difference between "locality" (in the sense defined by Bell) and "no FTL signalling" collapses if you assume determinism, so 3 is the same as 1.
• You can formally reexpress any Bell-local model in the form of a noncontextual model and vice versa, so 2 and 4 are equivalent in this sense.
In particular, they imply exactly the same Bell inequalities, and all of them are refuted by a Bell violation.

*For 1, 2, and 3, in addition to the usual "no superdeterminism" assumption. (It's implicit in 4)

Last edited: Dec 13, 2015
8. Dec 13, 2015

### ShayanJ

So...realism means properties have values even if not measured. CFD means properties have single definite values even if not measured. Right?
But this still doesn't clear the difference between them for me. It seems to me that you guys are saying a realistic theory can be probabilistic and non-deterministic but a CFD theory can't. But this doesn't seem reasonable. I can't think of a theory that is non-deterministic but assigns values to variables even if not measured. In fact I think my main problem is that I don't understand how you can assign values to unmeasured properties, but don't assign them single definite values!

I think it would make things more clear if you explain what realism is in this example.

Thanks guys

9. Dec 14, 2015

### stevendaryl

Staff Emeritus
I wouldn't say that. CFD means that measurements have definite results even if not performed. So a question of the form: "What result would I have gotten if I had measured such-and-such?" has a definite answer, even if you never measured such-and-such. CFD is about the road not traveled, it's not about hidden values.

I guess it's not worth worrying about the difference, since any local realistic model with perfect correlations must obey CFD. And as wle points out, if any model is locally realistic, then there is a model with the same predictions that obeys CFD.

But to me, they are different. Suppose you have a device with a light bulb on it. There is a button on the device. When you push the button, the light either glows red, or it glows blue. That device would not obey CFD, because there is no answer to the question: "What color would I have seen if I had pushed the button?" unless you actually did push the button.

I consider that the device can be described by a perfectly good locally realistic theory, though. That theory is just nondeterministic: The future state of the device is not determined.

Mathematically, you can describe a class of "locally realistic devices" by a process model, which is defined by a set of states and a set of events, and a set of transitions of the form

$S \otimes e \rightarrow S'$

meaning that if the device is in state $S$ and event $e$ happens, then the device can change into state $S'$.

CFD in these terms would be the claim that for every pair $S, e$ there is exactly one $S'$ such that $S \otimes e \rightarrow S'$.

It's not that important, the distinction between local realism and CFD, except for the fact that if you tell people that Bell's theorem assumes determinism, then people's first reaction: Oh, then that just means that QM must be nondeterministic. But nondeterminism doesn't make any difference. It's the "local realism" part that is ruled out by QM.

10. Dec 14, 2015

### ShayanJ

So QM without wave-function collapse is a local realistic theory, right?(But it seems to me Bohmian mechanics is a counter-example for this! EDIT: If Bohmian mechanics contains entanglement, then its not local so its not local realistic which means its not a counter-example to what I said. Does it contain entanglement?)
I think I understand it now. Thanks.

Last edited: Dec 14, 2015
11. Dec 14, 2015

### stevendaryl

Staff Emeritus
No, I wouldn't say so. A local realistic theory has a local notion of state. Every point in spacetime has some collection of properties that are true at that point: Maybe the values of fields, maybe the presence or absence of various types of particles, etc. Then the local realistic theory evolves locally: The state at one point in spacetime can depend only on the state at points in the backward lightcone.

Quantum mechanics is not a local realistic theory in this sense, because there is no notion of the state at a particular point in spacetime. There is a quantum mechanical state, but it's nonlocal. In the case where Alice is measuring one particle's spin and, far away, Bob is measuring another particle's spin, the spin state for the two particles might be described by $\frac{1}{\sqrt{2}} (|up\rangle |down\rangle - |down\rangle|up\rangle)$. That's nonlocal, because it doesn't say what Bob's particle state is, nor what Alice's particle state is, it only says something about the composite state of the two particles.