Does realism imply locality or vice versa?

[stevendaryl]

I said: "I agree that Einstein and Bell (when deriving his theorem) assumed local realism (LR). But according to Wikipedia we have, "Local realism is a feature of classical mechanics and of classical electrodynamics" . (you will also find "realism, another principle which relates to the value of unmeasured quantities") And if you peruse the internet you find that almost all agree that classical mechanics is a deterministic theory.
Thus LR does not include nondeterministic theories. That is my position as well."

Fine, then where is is the fallacy in: Wikipedia ⇒ LR is CM ⇒ CM is deterministic?

Classical mechanics is an example of a locally realistic theory that happens to be deterministic. Locally realistic doesn't mean classical mechanics.

Your attempt at a nondeterministic LR theory with Brownian motion was wrong.

No, it wasn't. Brownian motion is not deterministic. It is a stochastic model.

I don't understand. QT predicts perfect anticorrelations when the measurement are at the same angle/axis. QT is not deterministic.

It's not locally realistic, either. That's the point Bell was making when he talked about determinism being a conclusion, rather than an assumption. A nondeterministic locally realistic theory cannot predict perfect anticorrelations in an EPR-type experiment. So in trying to come up with a locally realistic model of EPR, you may as well assume it is deterministic.

 

[stevendaryl]

Here's a paper that discusses Bell's notion of local realism: https://arxiv.org/pdf/0707.0401

It makes the same point that I've made:

Bell’s mathematical formulation of “local causality” – Eq. (1) – is stated in terms of probabilities...

In a (local) stochastic theory, however, even a complete specification of relevant beables in the past (e.g., those
in region 3 of Figure 2) may not determine the realized value of the beable in question (in region 1). Rather, the
theory specifies only probabilities for the various possible values that might be realized for that beable. Of course,
determinism is not really an alternative to, but is rather merely a special case of, stochasticity:

Bell was definitely not assuming determinism. He was assuming a broader class of models that included determinism as a special case.

 

[Zafa Pi]

Perhaps this is not worth arguing about, except that Bell made the correct point that determinism is a conclusion, not an assumption, and I have been trying to explain what I think he meant by that. An example of the type of theory that would count as a nondeterministic locally realistic theory would be a theory of nondeterministic cellular automata. Suppose that spacetime is actually discrete (maybe at the Planck level). Then you could describe the state of the universe in the following terms:

1. Divide up the universe into tiny little cubes of size $L^3$, where $L$ is some tiny length. Then a cell is specified by a triple of integers $(i,j,k)$, which is the cell where $i L \leq x \leq (i+1)L$, $j L \leq y \leq (j+1)L$, $k L \leq z \leq (k+1) L$.
   
2. Each cell can be in one of a finite number of possible states: $\sigma_1, \sigma_2, ..., \sigma_N$
3. Divide time up into discrete intervals, of length $L/c$.
4. Then describe the complete state of the world by a function $S(i,j,k)$, where $i, j, k$ range over integers, specifying the state of cell with label $(i,j,k)$
5. Describe the evolution of the state by a probability transition function: $P(n | n_0 , n_1, ..., n_8)$.

The meaning of the probability transition function is this: If at time $t$, you have a cell in state $\sigma_{n_0}$, and its 8 nearest neighbors are in states $\sigma_{n_1}, \sigma_{n_2}, ... \sigma_{n_3}$, then at time $t + L/c$, that cell will be in some state $\sigma_{n}$ with a probability given by $P(n | n_0 , n_1, ..., n_8)$.

This is a nondeterministic theory. But it is local in the sense that the next state of any cell depends only on the state of that cell and the states of cells that it is touching.

This edifice is a diversion. How about having the first n states be either H or T and let the n+1st state be H or T with probability ½ each.

 

[stevendaryl]

Bell went on to say (the figure is reproduced by me):

Full specification of what happens in 3 makes events in 2 irrelevant for predictions about 1 in a locally causal theory.

In the figure, regions 1 and 2 represent local observations or measurements. The triangles represent the backwards lightcones of those observations. (You can understand the diagram as a 2-D representation of spacetime, with the horizontal direction representing space and the vertical direction representing time.) Bell allows for the observations in regions 1 and 2 to be probabilistic, but that the probabilities themselves must depend on facts in the backward lightcone. More than that, even, the result of an observation in region 1 can depend only on facts about close-by regions of space (where "close-by" is defined in terms of light cones). So in the figure above, anything in the distant past that affects observations in region 1 must do so through more immediate regions of spacetime such as region 3.

Bell says:

A theory will be said to be locally causal if the probabilities attached to values of local beables in a space-time region 1 are unaltered by specification of values of local beables in a space-like separated region 2, when what happens in the backward light cone of 1 is already sufficiently specified, for example by a full specification of local beables in a spacetime region 3...

There is nothing in this definition that assumes determinism.

 

[Zafa Pi]

Classical mechanics is an example of a locally realistic theory that happens to be deterministic. Locally realistic doesn't mean classical mechanics.

From the Wikipedia article I linked to:
Local realism[edit]
Einstein's principle of local realism is the combination of the principle of locality (limiting cause-and-effect to the speed of light) with the assumption that a particle must objectively have a pre-existing value (i.e. a real value) for any possible measurement, i.e. a value existing before that measurement is made.

Local realism is a feature of classical mechanics,

No, it wasn't. Brownian motion is not deterministic. It is a stochastic model.

Simplify. Coin flipping is a stochastic/probabilistic model. The word deterministic does not appear in probability theory. It is part of physical models, so I ask: is the outcome of flipping a coin determined? Einstein and other classical physicists would have said yes as they would have to Brownian motion, a classical theory.

 

[stevendaryl]

Simplify. Coin flipping is a stochastic/probabilistic model. The word deterministic does not appear in probability theory. It is part of physical models, so I ask: is the outcome of flipping a coin determined? Einstein and other classical physicists would have said yes as they would have to Brownian motion, a classical theory.

The issue is not what Einstein believed to be true. The issue is whether Bell's analysis assumes determinism. It doesn't, as Bell himself explained.

 

[Zafa Pi]

So perhaps we should say that "A theory will be said to be locally causal" [Bell] and LR have different definitions. All this is justifying my post #61.
I read Bell's proof and feel I understood it. From a classical physicist's point of view coin flipping and Brownian motion are deterministic, because nature is.
Arguing over definitions is tiring.

 

[stevendaryl]

So perhaps we should say that "A theory will be said to be locally causal" [Bell] and LR have different definitions. All this is justifying my post #61.

No, let's not say that. Deterministic theories are a special case, but Bell's notion is more general.

I read Bell's proof and feel I understood it. From a classical physicist's point of view coin flipping and Brownian motion are deterministic, because nature is. Arguing over definitions is tiring.

It's not an argument over definitions, it's an argument over whether Bell's proof assumed determinism. It doesn't. His proof applies to a larger class of theories than deterministic theories. It's a technical point which may or may not be interesting to anyone, except that people are often saying things like:

"Bell's argument applies to deterministic theories. So the failure of Bell's inequality only shows that our theory is nondeterministic. What's the big deal?"

That's fallacious reasoning, because determinism is not a necessary assumption in the derivation of Bell's inequalities. The inequality applies to local stochastic models, as well.

 

[stevendaryl]

It's not an argument over definitions, it's an argument over whether Bell's proof assumed determinism. It doesn't. His proof applies to a larger class of theories than deterministic theories.

What I should say is that Bell's factorizability condition applies to a larger class of theories. His actual proof assumed determinism, because you can't get perfect correlations/anti-correlations from a nondeterministic locally realistic theory.

 

[Jilang]

Just so I am clear on this. We are talking classical realism, right? With classical hidden variables? Are there such animals as quantum hidden variables?

 

[Zafa Pi]

Tho I gave you a deserved like I have two issues.

The answer to that is remarkable; yes we can, but if we want our model to consist of things which have some real objective existence (like position or momentum for example) then the only way to do it is to have these 'realistic' elements connected in some way that violates the bounds of special relativity.

What about ER = EPR? No violations or speeding tickets.

Bell's own masterful exposition of all this is still, for me, the best : https://cds.cern.ch/record/142461/files/198009299.pdf

Way too long winded and inelegant. Nielsen & Chuang on CHSH good, mine better.

 

[Zafa Pi]

What I should say is that Bell's factorizability condition applies to a larger class of theories. His actual proof assumed determinism, because you can't get perfect correlations/anti-correlations from a nondeterministic locally realistic theory.

I took some time away to stroll down internet lane. I peeked in various sites to get a better handle on determinism, even reading past threads of PF. I find it a mess, it reminds me when people try to distinguish innate random from random. I notice that in physics discussions nondeterministic or nonlocal do not necessarily mean not deterministic or not local as they would in math.

Along the way I found several folks with fine pedigrees that think Bell is wrong, but this is an irrelevant aside.

I mentioned before that I thought your automata model and Brownian motion are excessively baroque compared to simple coin flipping. Do you consider the actual act of flipping a "fair" coin nondeterministic, and/or would you take the theory of iid sequences of 0,1 valued r.v.s (with P(1) = P(0)) for your stochastic model?
I am admittedly confused about what you've been saying. Would you tell me what you mean by nondeterministic and provide an example from this wider class.

I can't think of anything that requires a measurement that is deterministic in the classical sense.

 

[stevendaryl]

Do you consider the actual act of flipping a "fair" coin nondeterministic,

Are you asking whether it's REALLY nondeterministic? You can't answer that question unless you know the actual laws of the universe. What you can answer is: Is it deterministic or nondeterministic according to this theory or that theory. According to Newtonian mechanics, it's deterministic. According to quantum mechanics, it might be nondeterministic.

I am admittedly confused about what you've been saying. Would you tell me what you mean by nondeterministic and provide an example from this wider class.

First, the definition of deterministic: A theory is deterministic if the state at one time uniquely determines the state at future times. A theory is nondeterministic if more than one future state is possible from a given present state. You want an example of the latter type of theory? That's what I described to you, and you dismissed it, saying that you preferred a coin flip. So a coin flip is an example. If my theory of coin flips is that every flip has a 50/50 chance of resulting in "heads" or "tails", then that's a nondeterministic theory.

 

[Zafa Pi]

you cannot predict perfect correlations and anticorrelations unless the theory is deterministic. - only CFD theories can predict perfect correlation/anti-correlation.

No, let's not say that. Deterministic theories are a special case, but Bell's notion is more general. It's not an argument over definitions, it's an argument over whether Bell's proof assumed determinism. It doesn't. His proof applies to a larger class of theories than deterministic theories.

So deterministic theories and CFD theories are the same for you? They are for me. Can't QM predict the perfect correlation?

In the wider class I mentioned above I was thinking of your larger class.

The 1964 Bell proof he assumed determinism as you said (hidden variables and CFD). Can you provide a Bell type theorem with proof that doesn't assume determinism?

I have several in mind that have coin flipping involved, but I would still call them deterministic. Maybe they're not??

I have a simple version of a Bell theorem and proof with a claim that experiments violate the inequality with no words that a freshman math or physics major hasn't seen before. Yet I am still curious about what you mean, you know stuff that I don't.

I am overwhelmed with the amount of disagreement and confusion I see in this area - in PF threads and in the published literature.

 

[stevendaryl]

So deterministic theories and CFD theories are the same for you? They are for me. Can't QM predict the perfect correlation?

I wish that people hadn't brought up CFD with respect to QM, because it seems like a complicated concept that doesn't clarify anything. But it seems to me that CFD is only with respect to certain types of changes in the actual history of the world: If I had taken a left turn at Albuquerque I would have ended up at El Paso. We imagine an alternate world that is exactly like ours, except that I made a different choice. So CFD only makes sense relative to certain localized turning points in the history of the world. To me, that implies that there are two different ways CFD might fail:

1. There might be no such turning points. In a superdeterministic theory, it doesn't make sense to ask what would have happened if Alice had done something different than she did, because Alice's choice was set in stone long ago, and there are no alternative histories.
2. There might be no unique answer to the question: What would have happened? In a nondeterministic theory, the outcomes of experiments are not determined, and so there is no answer to a question such as: What result would Alice have gotten if she had measured X?

CFD is not exactly the same as determinism, though. Superdeterministic theories don't obey CFD, because there are no alternatives to ask about. And a nondeterministic theory might obey CFD for certain turning points and certain consequences, even if they don't for all pairs of turning-point/consequence. For example, if I flip a coin to decide whether to go left or right, and I go right, there might be a counterfactual answer to the question: What would have happened to me if the coin had landed tails?

The 1964 Bell proof he assumed determinism as you said (hidden variables and CFD). Can you provide a Bell type theorem with proof that doesn't assume determinism?

Bell's inequality is provable without assuming determinism. The usual proof (assuming determinism) goes like this:

Assume that there are functions $A(\alpha, \lambda)$ and $B(\beta, \lambda)$ giving Alice's and Bob's results, respectively as a function of Alice's setting, $\alpha$, Bob's setting, $\beta$ and the hidden variable $\lambda$. This framing assumes that the results are deterministic functions of $\alpha, \beta, \lambda$. That's not necessary to get the inequality.

Instead, let's assume that Alice's result is only probabilistically related to her setting and the hidden variable, and similarly for Bob's result. There are two possible results, which we'll say are +1 and -1. The quantity we're interested in is the correlation, $E(\alpha, \beta)$, which is the average value, over many trials, of the product of their results, for fixed $\alpha$ and $\beta$. We can compute this in the following way:

$E(\alpha, \beta) = P_{++}(\alpha, \beta) - P_{+-}(\alpha, \beta) - P_{-+}(\alpha, \beta) + P_{--}(\alpha, \beta)$

where $P_{++}(\alpha, \beta)$ is the probability Alice gets +1 and Bob gets +1, $P_{+-}(\alpha, \beta)$ is the probability Alice gets +1 and Bob gets -1, etc.

Now, a local explanation of the correlation in terms of hidden variables would involve the following three functions:

1. $P(\lambda)$: the probability density for $\lambda$
   
2. $P_A(\alpha, \lambda)$: the probability that Alice gets +1 (the probability that she gets -1 is just $1 - P_A(\alpha, \lambda)$)
   
3. $P_B(\beta, \lambda)$: the probability that Bob gets +1 (the probability that he gets -1 is just $1 - P_B(\beta, \lambda)$)

Determinism would imply that $P_A$ and $P_B$ are either 0 or 1 for each choice of $\alpha, \beta, \lambda$. But we're not assuming determinism. Then in terms of these functions:

• $P_{++}(\alpha, \beta) = \int d\lambda P(\lambda) P_A(\alpha, \lambda) P_B(\beta, \lambda)$
• $P_{+-}(\alpha, \beta) = \int d\lambda P(\lambda) P_A(\alpha, \lambda) (1 - P_B(\beta, \lambda))$
• $P_{-+}(\alpha, \beta) = \int d\lambda P(\lambda) (1 - P_A(\alpha, \lambda)) P_B(\beta, \lambda)$
• $P_{--}(\alpha, \beta) = \int d\lambda P(\lambda) (1 - P_A(\alpha, \lambda)) (1 - P_B(\beta, \lambda))$

So:

$E(\alpha, \beta) = \int d\lambda P(\lambda) (4 P_A(\alpha, \lambda) P_B(\beta, \lambda) - 2 P_A(\alpha, \lambda) - 2 P_B(\beta, \lambda) + 1)$

Now the trick is to rewrite $P_A(\alpha, \lambda) = \frac{1}{2}(X_A(\alpha, \lambda) + 1)$, $P_B(\beta, \lambda) = \frac{1}{2}(X_B(\beta, \lambda) + 1)$, where $X_A(\alpha, \lambda)$ and $X_B(\beta, \lambda)$ are both numbers between +1 and -1. In terms of these,

$E(\alpha, \beta) = \int d\lambda P(\lambda) X_A(\alpha, \lambda) X_B(\beta, \lambda)$

Now, following Bell (or maybe CHSH, who developed a related inequality) we compute the combination:

$C(\alpha, \alpha', \beta, \beta') \equiv E(\alpha, \beta) + E(\alpha', \beta) + E(\alpha', \beta) - E(\alpha', \beta')$

In terms of our hidden variable, $\lambda$, this becomes:

$C(\alpha, \alpha', \beta, \beta') = \int d\lambda P(\lambda) C_\lambda(\alpha, \alpha', \beta, \beta')$

where
$C_\lambda(\alpha, \alpha', \beta, \beta') \equiv X_A(\alpha, \lambda) X_B(\beta, \lambda) + X_A(\alpha', \lambda) X_B(\beta, \lambda) + X_A(\alpha, \lambda) X_B(\beta', \lambda) - X_A(\alpha', \lambda) X_B(\beta', \lambda)$

It might not be obvious, but since each of the Xs are between +1 and -1, it follows that $C_\lambda(\alpha, \alpha', \beta, \beta')$ must lie in the range $[-2, +2]$. So we have:

$\int d\lambda P(\lambda) (-2) \leq C(\alpha, \alpha', \beta, \beta') \leq \int d\lambda P(\lambda) (+2)$

And since the integral over all $\lambda$ of $P(\lambda)$ is just 1, we conclude:

$-2 \leq C(\alpha, \alpha', \beta, \beta') \leq +2$

So that's the same inequality as CHSH derived (a slight variant of Bell's original inequality), and it did not assume that the results were deterministic.

I have several in mind that have coin flipping involved, but I would still call them deterministic. Maybe they're not??

As I said, we can't know whether coin flips are really deterministic, or not. We can only say that they are deterministic or nondeterministic according to this or that model.

 

[stevendaryl]

where
$C_\lambda(\alpha, \alpha', \beta, \beta') \equiv X_A(\alpha, \lambda) X_B(\beta, \lambda) + X_A(\alpha', \lambda) X_B(\beta, \lambda) + X_A(\alpha, \lambda) X_B(\beta', \lambda) - X_A(\alpha', \lambda) X_B(\beta', \lambda)$

It might not be obvious, but since each of the Xs are between +1 and -1, it follows that $C_\lambda(\alpha, \alpha', \beta, \beta')$ must lie in the range $[-2, +2]$.

Letting

• $A \equiv X_A(\alpha, \lambda)$
• $A' \equiv X_A(\alpha', \lambda)$
• $B \equiv X_B(\beta, \lambda)$
• $B' \equiv X_B(\beta', \lambda)$
• $F(A,A', B, B') \equiv AB + A'B + AB' -A'B'$

We want to show that $-2 \leq F(A,A',B,B') \leq +2$, under the assumption that all 4 variables lie in the range $[-1, +1]$.

Let $A_{max}, B_{max}, A'_{max}, B'_{max}$ be a choice of the variables that maximizes $F$. Then:

• Either $A_{max} = \pm 1$ or $\frac{\partial F}{\partial A}|_{A = A_{max}, B=B_{max}, A'=A'_{max}, B' = B'_{max}} = 0$
• Either $A'_{max} = \pm 1$ or $\frac{\partial F}{\partial A'}|_{A = A_{max}, B=B_{max}, A'=A'_{max}, B' = B'_{max}} = 0$

Suppose $\frac{\partial F}{\partial A} = 0$. That implies that $B_{max} + B'_{max} = 0$. So in this case: $F(A_{max}, A'_{max}, B_{max}, B'_{max}) = A'_{max} (B_{max} - B'_{max}) \leq 2$

Suppose $\frac{\partial F}{\partial A'} = 0$. That implies that $B_{max} - B'_{max} = 0$. So in this case: $F(A_{max}, A'_{max}, B_{max}, B'_{max}) = A_{max} (B_{max} + B'_{max}) \leq 2$

So we conclude that if $F > 2$, it must be when $A_{max} = \pm 1$ and $A'_{max} = \pm 1$. We have two cases: they have the same sign, or they have opposite signs.

Suppose $A_{max} = \pm 1$ and $A'_{max} = +A_{max}$. Then $F(A_{max}, A'_{max}, B_{max}, B'_{max}) = \pm (B_{max} + B'_{max}) \pm (B_{max} - B'_{max}) = \pm 2 B_{max} \leq 2$

Suppose $A_{max} = \pm 1$ and $A'_{max} = -A_{max}$. Then $F(A_{max}, A'_{max}, B_{max}, B'_{max}) = \pm (B_{max} + B'_{max}) \mp (B_{max} - B'_{max}) = \pm 2 B'_{max} \leq 2$

So in all cases, $F(A,A',B, B') \leq 2$. We can similarly prove $-2 \leq F(A,A',B,B')$. So we conclude:

$-2 \leq F(A,A',B,B') \leq +2$