I A skeptic's view on Bohmian Mechanics

rubi

As usual, you use double standards. Classical mechanics also claims to be deterministic, hence not represented by an ensemble but by a single collection of classical particles. Nevertheless, you do not say that themodynamics is incompatible with classical mechanics.
This is why something like the ergodic hypothesis is needed. A classical mechanical system is definitely only in one pure state and all observables are determined by this state. In order to predict the correct thermodynamics, it must be the case that the particles of this one state are distributed in a chaotic way, such that averages turn out to be the same as if we had computed the ensemble average.

• dextercioby

A. Neumaier

As usual, you use double standards. Classical mechanics also claims to be deterministic, hence not represented by an ensemble but by a single collection of classical particles. Nevertheless, you do not say that themodynamics is incompatible with classical mechanics.
This is because statistical mechanics does not make the assumption of equilibrium to deduce results.

The equilibrium assumption is made in statistical mechanics only for special systems known to be in equilibrium
, such as crystals, water at rest, or a gas at uniform pressure and temperature. For all other cases (the majority of real world situations), there is nonequilibrium statistical mechanics which derives hydrodynamics and elasticity theory under much weaker assumptions, or kinetic theory for very dilute gases. The latter give a fairly realistic description of the universe at last, which is not in equilibrium.

On the other hand, Bohmian mechanics cannot even start without assuming Bohmian quantum equilibrium of the whole universe! This is quite a different sort of assumption. And it is assumed without any justification except that it is needed to reproduce Born's rule.

Thus the same standards result in a very different appraisal of the two theories.

Demystifier

2018 Award
No, it doesn't lead to Bell's inequalities. The joint assumption of both
(1) $A(\lambda,\vec a,\vec b) = A(\lambda,\vec a)$, $B(\lambda,\vec a,\vec b) = B(\lambda,\vec b)$
(2) $P(\lambda,\vec a,\vec b) = P(\lambda)$
leads to Bell's inequality. In a hidden variable theory, we can deny (1), which leads to non-locality or we can deny (2), which (according to Bell) leads to superdeterminism, or we can of course deny both. Apparently, Bohmian mechanics does not only violate (1), but it also violates (2), which would mean that it is not just non-local, but also superdeterministic (according to Bell).

By using an expression for the correlations, given by $\left<A(\vec a) B(\vec b)\right> = \int A(\lambda,\vec a) B(\lambda,\vec b) P(\lambda, \vec a, \vec b)\mathrm d\lambda$, we can in principle write down a local hidden variable theory. The usual argument is that because it violates (2), it must be superdeterministic. I don't understand, why this argument does not equally apply to Bohmian mechanics, if BM violates (2), which it does.
Ah, now I see the source of confusion. When I was referring to a quantity $P(\lambda)$ for the first time, it was not the same quantity as you defined it above. What you call $P(\lambda)$, Bell calls $\rho(\lambda)$.

So let me explain it all over again, now using your conventions. In BM we have
$$A(\lambda,\vec a,\vec b) \neq A(\lambda,\vec a)$$
$$B(\lambda,\vec a,\vec b) \neq B(\lambda,\vec b)$$
So when the measurement setup changes, then $A(\lambda,\vec a,\vec b)$ and $B(\lambda,\vec a,\vec b)$ change. Due to this, there is no need to change $P(\lambda)$ and introduce superdeterminism in BM.

• PeterDonis

Demystifier

2018 Award
This is because statistical mechanics does not make the assumption of equilibrium to deduce results.

The equilibrium assumption is made in statistical mechanics only for special systems known to be in equilibrium
, such as crystals, water at rest, or a gas at uniform pressure and temperature. For all other cases (the majority of real world situations), there is nonequilibrium statistical mechanics which derives hydrodynamics and elasticity theory under much weaker assumptions, or kinetic theory for very dilute gases. The latter give a fairly realistic description of the universe at last, which is not in equilibrium.

On the other hand, Bohmian mechanics cannot even start without assuming Bohmian quantum equilibrium of the whole universe! This is quite a different sort of assumption. And it is assumed without any justification except that it is needed to reproduce Born's rule.

Thus the same standards result in a very different appraisal of the two theories.
As I already said, there is no point in arguing about BM as we (you and me) cannot even agree of the foundations of classical statistical mechanics, or even on the concept of probability. I wish you a good luck with your thermal interpretation inspired by your own interpretation of thermodynamics and statistical mechanics. rubi

Ah, now I see the source of confusion. When I was referring to a quantity $P(\lambda)$ for the first time, it was not the same quantity as you defined it above.
What other quantity $P(\lambda)$ did you refer to then?

Due to this, there is no need to change $P(\lambda)$ and introduce superdeterminism in BM.
It seems like $P(\lambda)$ depends on $\vec a$, $\vec b$ through $\lambda$, which determines the settings of the measurement apparata. In order to compute expectation values like $\left<A(\vec a)B(\vec b)\right>$, you need to tune $\lambda$ such that it reflects the particular settings.

DrChinese

Gold Member
I just wanted to say to the participants above (especially Arnold Neumaier, demystifier, rubi) that I very much enjoyed reading the back and forth. Although I doubt many minds were changed, the scope and intensity of the debate was very enlightening to me.

Thanks! • eloheim, rubi and Demystifier

Demystifier

2018 Award
What other quantity $P(\lambda)$ did you refer to then?
In BM $P=|\Psi|^2$, where $\Psi$ is the conditional wave function.

Demystifier

2018 Award
Although I doubt many minds were changed
Even if we haven't changed our minds, at least we sharpened our arguments. • DrChinese

DrChinese

Gold Member
Even if we haven't changed our minds, at least we sharpened our arguments. I couldn't agree more. And you responded with grace at each step. Not easy from those two, coming at you from every side. I give everyone kudos for making their strongest arguments. You had me following links to some material I had not seen before (there's a lot of that it seems).

rubi

In BM $P=|\Psi|^2$, where $\Psi$ is the conditional wave function.
Okay, I actually had this in mind, too. But isn't this also the distribution with respect to which expectation values are computed?

A. Neumaier

As I already said, there is no point in arguing about BM as we (you and me) cannot even agree of the foundations of classical statistical mechanics, or even on the concept of probability. I wish you a good luck with your thermal interpretation inspired by your own interpretation of thermodynamics and statistical mechanics. All my arguments here are solely about the shut-up-and-calculate part of probability, statistics, and Bohmian mechanics. Thus they are independent of the details how one interprets probability or thermodynamics or statistical mechanics.

Demystifier

2018 Award
All my arguments here are solely about the shut-up-and-calculate part of probability, statistics, and Bohmian mechanics.
No they are not. You neither shut up nor calculate. Last edited:

Demystifier

2018 Award
Okay, I actually had this in mind, too. But isn't this also the distribution with respect to which expectation values are computed?
To clarify those things completely, one would need to rewrite the equations of BM in the language of Bell's theorem. As far as I am aware, nobody, not even Bell himself, has done it explicitly.

The important question is: Does BM involve fine tuning of initial conditions? For if it does, then it is superdeterministic. Well, the catch is that "fine tuning" is not a precisely defined concept. BM assumes quantum equilibrium at the beginning of the experiment. Bohmians don't think of it as "fine tuning", but some critiques, like Neumaier, think of it as fine tuning. It is somewhat subjective, but if you want to call quantum equilibrium "fine tuning", then yes, BM is "superdeterministic". But even if one calls it "superdeterminism", it is a rather soft kind of superdeterminism, much softer than superdeterminism needed in local theories.

• eloheim

rubi

To clarify those things completely, one would need to rewrite the equations of BM in the language of Bell's theorem. As far as I am aware, nobody, not even Bell himself, has done it explicitly.
I don't understand why this is not immediate. Aren't the particle positions $\vec x$ supposed to be the hidden variables and aren't they supposed to be distributed according to $\left|\Psi\right|^2$? If that is the case, then the spin along some axis should be given by some functions $A,B(\vec x,\vec a,\vec b)$ and the correlations should be given by $\int A(\vec x,\vec a,\vec b) B(\vec x,\vec a,\vec b) \left|\Psi\right|^2\mathrm d\vec x$. Now the question is whether $\left|\Psi\right|^2$ depends on $\vec a$,$\vec b$, but it seems to me that it does, because BM must model the measurement apparatus and the environment in order supposedly reproduce QM and of course the angles $\vec a$, $\vec b$ are some functions of the hidden variables $\vec x$, because they are determined by the positions of the atoms that make up the apparatus. Hence, it seems to be that BM meets Bell's definition of superdeterminism. It wouldn't be that way if BM would work without taking the measurement theory into account, but we learned in the last thread that it won't reproduce QM this way.

But even if one calls it "superdeterminism", it is a rather soft kind of superdeterminism, much softer than superdeterminism needed in local theories.
What makes the BM superdeterminism softer than the superdeterminism required in local hidden variable theories?

Demystifier

2018 Award
I don't understand why this is not immediate. Aren't the particle positions $\vec x$ supposed to be the hidden variables and aren't they supposed to be distributed according to $\left|\Psi\right|^2$? If that is the case, then the spin along some axis should be given by some functions $A,B(\vec x,\vec a,\vec b)$ and the correlations should be given by $\int A(\vec x,\vec a,\vec b) B(\vec x,\vec a,\vec b) \left|\Psi\right|^2\mathrm d\vec x$. Now the question is whether $\left|\Psi\right|^2$ depends on $\vec a$,$\vec b$, but it seems to me that it does, because BM must model the measurement apparatus and the environment in order supposedly reproduce QM and of course the angles $\vec a$, $\vec b$ are some functions of the hidden variables $\vec x$, because they are determined by the positions of the atoms that make up the apparatus. Hence, it seems to be that BM meets Bell's definition of superdeterminism. It wouldn't be that way if BM would work without taking the measurement theory into account, but we learned in the last thread that it won't reproduce QM this way.

What makes the BM superdeterminism softer than the superdeterminism required in local hidden variable theories?
During the night I have thought about all this once again (thank you for asking good questions that force me to think more carefully), and now I have some new insights. Now my central claim is this: The fact that $P(\lambda, \vec a,\vec b)$ depends on $\vec a$, $\vec b$ does not automatically imply that the theory is superdeterministic.

To explain this claim, let me present a counterexample: 18th and 19th century physics! Supoose that the world is described by classical non-relativistic mechanics, in which all forces are either Newton gravitational forces or Coulomb electrostatic forces. These forces involve an action over distance, so the laws of physics are nonlocal. Let $\vec a$ and $\vec b$ be positions of two macroscopic charged balls. They produce electric field which, nonlocally, influence the motion of all other charged particles. Let $\lambda$ be the positions and momenta of all atoms in the air around the balls. The atoms are hidden variables of 18th and 19th century, so $\lambda$ is a hidden variable. Furthermore, let us introduce probability in this deterministic theory by taking elements of Boltzmann statistical mechanics. We assume that atoms of the air are in a thermal equilibrium (which is an analogue of quantum equilibrium in BM). The motion of atoms depends on the electric field produced by the balls, so the Hamiltonian describing the motion of $\lambda$ depends on $\vec a$ and $\vec b$. Hence the Boltzmann distribution is a function of the form
$$P(\lambda | \vec a,\vec b)$$
Note that $P(\lambda | \vec a,\vec b)$ is conditional probability, which should be distinguished from joint probability $P(\lambda, \vec a,\vec b)$. They are related by the Bayes formula
$$P(\lambda | \vec a,\vec b)=\frac{P(\lambda, \vec a,\vec b)}{P(\vec a,\vec b)}$$
where
$$P(\vec a,\vec b)=\int d\lambda \, P(\lambda, \vec a,\vec b)$$
is marginal probability.

Now what is superdeterminism? Superdeterminism means that $\vec a$ and $\vec b$ cannot be chosen independently. In other words, superdeterminism means that
$$P(\vec a,\vec b)\neq P_a(\vec a) P_b(\vec b)..........(Eq. 1)$$
where
$$P_a(\vec a)=\int d\lambda \int d^3b \, P(\lambda, \vec a,\vec b)$$
and similarly for $P_b$. So is non-local classical mechanics superdeterministic? It can be superdeterministic in principle! Nevertheless, it is not superdeterministic in FAPP (for all practical purposes) sense. Even though there are forces between the charged balls, which in principle can make them correlated, in practice we can choose the positions $\vec a$ and $\vec b$ of the balls independently. That's because the force falls off with the distance, so the mutual influence between $\vec a$ and $\vec b$ can be neglected when the balls are very far from each other. Therefore $\vec a$ and $\vec b$ are not correlated FAPP, so classical non-local mechanics is not superdeterministic FAPP.

Now Bohmian mechanics. I hope the analogy between the example above and BM is quite clear. There is only one important difference. In BM, the non-local force does not fall off with distance. Nevertheless, the mutual influence between $\vec a$ and $\vec b$ can also be neglected. That's because $\vec a$ and $\vec b$ are orientations of macroscopic apparatuses, and decoherence destroys entanglement (in a FAPP sense) between macroscopic degrees of freedom. Without entanglement (in the FAPP sense) there is no correlation between $\vec a$ and $\vec b$. And without correlation, that is when (Eq. 1) becomes equality, there is no superdeterminism.

The only thing I didn't explain is the following. If $P(\lambda|\vec a,\vec b)$ does not imply superdeterminism (in FAPP sense), what is the justification for using $P(\lambda)$ in the Bell theorem? I don't know at the moment, but let me tell that there are proofs of non-locality which do not rest on explicit introducing of $\lambda$, and personally I like such proofs much more. An example is the Hardy proof of non-locality, reviewed e.g. in my
https://arxiv.org/abs/quant-ph/0609163

stevendaryl

Staff Emeritus
Now my central claim is this: The fact that $P(\lambda, \vec a,\vec b)$ depends on $\vec a$, $\vec b$ does not automatically imply that the theory is superdeterministic.
The way I understand it is that it is the assumption of locality, together with the Bell's inequalities, that suggests superdeterminism. In deriving the Bell inequalities, we can write:

$P(A,B | \alpha, \beta) = \sum_\lambda P(\lambda | \alpha, \beta) P(A | \alpha, \beta, \lambda) P(B | \alpha, \beta, \lambda, A)$

(where, as usual, $A$ is Alice's result, $B$ is Bob's result, $\alpha$ is Alice's setting, $\beta$ is Bob's setting, and $\lambda$ is some hidden variable that is a potential common influence to Alice's and Bob's measurements.)

This form is perfectly general (I think); it is not hard to come up with a model of this sort that satisfies the quantum predictions for EPR. But assuming locality and no-superdeterminism further constrains the probabilities:
• $P(\lambda | \alpha, \beta) = P(\lambda)$
If $\lambda$ is a common influence to both Alice's and Bob's measurements, then it must be determined in the intersection of their backwards lightcones. But (assuming no-superdeterminism), the choice of $\alpha$ and $\beta$ can be made at the last moment, too late to influence $\lambda$.

We can similarly argue from locality and no-superdeterminism that
• $P(A|\alpha, \beta, \lambda) = P(A|\alpha, \lambda)$
• $P(B|\alpha, \beta, \lambda, A) = P(B|\beta, \lambda)$
So the final assumed form of the probability distributions is determined by locality and "free will" (no-superdeterminism):
• $P(A, B|\alpha, \beta) = \sum_\lambda P(\lambda) P(A|\alpha, \lambda) P(B|\beta, \lambda)$
And of course, Bell shows that no such model can reproduce the predictions of QM for EPR.

If you aren't assuming locality, then there is no problem in letting $\lambda$ depend on $\alpha$ and $\beta$, even if they are chosen at the last minute, as long as the choices are made before either measurement.

In the case where, for example, Bob's choice is made after Alice's measurement, then the above analysis doesn't quite apply, since in that case, $\lambda$ cannot depend on $\beta$. But if we don't assume locality, then Bob's result can depend on Alice's result, so again, there is no problem with Bell's inequality.

zonde

Gold Member
The fact that $P(\lambda, \vec a,\vec b)$ depends on $\vec a$, $\vec b$ does not automatically imply that the theory is superdeterministic.
And if we look at the situation at some earlier time when $\vec a$ and $\vec b$ is not yet set but instead we have only (yet to be determined) outputs or random number generators $rnd_a$ and $rnd_b$? These random numbers don't have the type of influence as in your classical analogue. And that is the type of dependency that we call superdeterminism (dependency between $\lambda$ and $rnd_a$, $rnd_b$).
On the other hand if, at the moment when we set $\vec a$ or $\vec b$, distribution of $\lambda$ changes and becomes dependent on $\vec a$ or $\vec b$ it would be non-locality instead of superdeterminism.

Demystifier

2018 Award
And if we look at the situation at some earlier time when $\vec a$ and $\vec b$ is not yet set but instead we have only (yet to be determined) outputs or random number generators $rnd_a$ and $rnd_b$? These random numbers don't have the type of influence as in your classical analogue.
I guess you mean psudo-random, not trully random, right? If so, then it has a classical analogue because thermal fluctuations of classical atoms=$\lambda$ can serve as a pseudo-random generator.

And that is the type of dependency that we call superdeterminism (dependency between $\lambda$ and $rnd_a$, $rnd_b$).
Superdeterminism is not when $rnd_a$ and $rnd_b$ depend on $\lambda$. Superdeterminism is when this dependence is such that $rnd_a$ is correlated with $rnd_b$. As I have explained, such a correlation is FAPP absent in both classical and Bohmian mechanics.

rubi

Even though there are forces between the charged balls, which in principle can make them correlated, in practice we can choose the positions $\vec a$ and $\vec b$ of the balls independently. That's because the force falls off with the distance, so the mutual influence between $\vec a$ and $\vec b$ can be neglected when the balls are very far from each other. Therefore $\vec a$ and $\vec b$ are not correlated FAPP, so classical non-local mechanics is not superdeterministic FAPP.
That doesn't follow. If you have a statistical description of the situation, you can integrate out the short scales ("renormalize") and thereby obtain an effective model that describes only the long range effects. In general, you will obtain extra terms that correspond to a long range correlation, even though your initial model only contained short-range interactions. (See the Ising model for example.)

Now Bohmian mechanics. I hope the analogy between the example above and BM is quite clear. There is only one important difference. In BM, the non-local force does not fall off with distance. Nevertheless, the mutual influence between $\vec a$ and $\vec b$ can also be neglected. That's because $\vec a$ and $\vec b$ are orientations of macroscopic apparatuses, and decoherence destroys entanglement (in a FAPP sense) between macroscopic degrees of freedom. Without entanglement (in the FAPP sense) there is no correlation between $\vec a$ and $\vec b$. And without correlation, that is when (Eq. 1) becomes equality, there is no superdeterminism.
I don't think that decoherence will destroy correlations. It makes certain interference terms go away, which results in expectation values peaked on classical trajectories. However, classical trajectories needn't loose their correlations.

but let me tell that there are proofs of non-locality which do not rest on explicit introducing of $\lambda$, and personally I like such proofs much more. An example is the Hardy proof of non-locality, reviewed e.g. in my
https://arxiv.org/abs/quant-ph/0609163
Proofs like Hardy or GHZ aren't proofs of non-locality. They rather show that hidden variable models must be contextual, which can't be deduced from Bell's inequality alone. And they don't exclude superdeterministic models either.

But assuming locality and no-superdeterminism further constrains the probabilities:
• $P(\lambda | \alpha, \beta) = P(\lambda)$
If $\lambda$ is a common influence to both Alice's and Bob's measurements, then it must be determined in the intersection of their backwards lightcones. But (assuming no-superdeterminism), the choice of $\alpha$ and $\beta$ can be made at the last moment, too late to influence $\lambda$.
This is not a locality assumption. It just enforces that the angles $\alpha$, $\beta$ aren't correlated with $\lambda$. If that condition is violated, it is still possible that the reason for the correlation of $\alpha$ and $\beta$ lies in the common past. Thus the condition is not related to locality. It merely enforces that the angles can be freely choosen instead of being determined (possibly by the common past).

Demystifier

2018 Award
That doesn't follow.
So you think that classical mechanics is superdeterministic?

I don't think that decoherence will destroy correlations.
I do (in FAPP sense, of course).

However, classical trajectories needn't loose their correlations.
Nobody ever observed them, which confirms my claim that they are FAPP absent. To repeat, they are possible in principle, but impossible FAPP.

Proofs like Hardy or GHZ aren't proofs of non-locality.
Well, Hardy takes it as a proof of nonlocality.

They rather show that hidden variable models must be contextual,
Sure, they show that too. See my paper I mentioned above where I discuss it in more detail.

Last edited:

stevendaryl

Staff Emeritus
This is not a locality assumption. It just enforces that the angles $\alpha$, $\beta$ aren't correlated with $\lambda$. If that condition is violated, it is still possible that the reason for the correlation of $\alpha$ and $\beta$ lies in the common past.
But for $\alpha$ and $\beta$ to be correlated would be superdeterminism. $\alpha$ is the measurement choice made by Alice, and $\beta$ is the measurement choice made by Bob. They are free to use whatever means they like to make that choice. For the choices to be correlated is superdeterminism.

Demystifier

2018 Award
Because they are the same in the EPR definition of realism as deterministic.
You are probably right that EPR criterion is the reason why many think that non-realism and non-determinism are the same. Thank you for that insight!

But it is a misinterpretation of the EPR criterion. The EPR criterion says something like "If we can predict with certainty ... then there is an element of reality ..." So according to EPR, determinism implies reality. However, the converse is not true; reality does not imply determinism. Therefore determinism and reality are not equivalent, and not the same. The GRW theory is a well-known example of a theory (compatible with QM) which has elements of non-deterministic reality.

rubi

So you think that classical mechanics is superdeterministic?
Depends on the particular model. I'm just telling you that the short-rangedness of interactions in general leads to long-range correlations, which you denied. Also, in CM, the state isn't fine-tuned and the absence of correlations can then be justified on the basis of molecular chaos. However, in BM, the hidden variables must be distributed according to $\left|\Psi\right|^2$, which is a restriction on the allowed distributions that can in principle introduce correlations.

I do (in FAPP sense, of course).
However, that's just an opinion. I demand proof. If the wave-function evolves unitarily, correlations shouldn't go away, but rather propagate to finer parts of the system.

Nobody ever observed them.
That doesn't invalidate the argument.

Well, Hardy takes it as a proof of nonlocality.
He's wrong. Consistent histories is local despite allowing the Hardy state, so locality can't be ruled out by the argument.

But for $\alpha$ and $\beta$ to be correlated would be superdeterminism. $\alpha$ is the measurement choice made by Alice, and $\beta$ is the measurement choice made by Bob. They are free to use whatever means they like to make that choice. For the choices to be correlated is superdeterminism.
Which is what I'm saying. The $P(\lambda|\vec a,\vec b)=P(\lambda)$ condition prohibits superdeterminism, but it is unrelated to locality. The question remains: How can BM be non-superdeterministic despite violating this condition?

zonde

Gold Member
I guess you mean psudo-random, not trully random, right? If so, then it has a classical analogue because thermal fluctuations of classical atoms=$\lambda$ can serve as a pseudo-random generator.
Let it be pseudo-random. Why does it matters?

Superdeterminism is not when $rnd_a$ and $rnd_b$ depend on $\lambda$. Superdeterminism is when this dependence is such that $rnd_a$ is correlated with $rnd_b$. As I have explained, such a correlation is FAPP absent in both classical and Bohmian mechanics.
Look, if we can set hidden variable so that spin of entangled particles is always opposite in either measurement basis of Alice or Bob then we can violate Bell inequalities just as QM does.
I will use photons. Let's say entangled photons have opposite linear polarization in Alice's measurement base. Alice's measurement will give certain outcome based on two possible options HV or VH. Bob's measurement will give probabilistic outcome according to Malus law $P=\sin^2(\beta-\alpha)$ or $P=\cos^2(\beta-\alpha)$. And that is all we need to replicate predicted correlations of QM.

This is how superdeterminism can violate Bell's inequality. And there is no need for $\alpha$ to be correlated with $\beta$.

stevendaryl

Staff Emeritus
Which is what I'm saying. The $P(\lambda|\vec a,\vec b)=P(\lambda)$ condition prohibits superdeterminism, but it is unrelated to locality.
I disagree; it's the combination of no-superdeterminism and locality that prevents $\lambda$ from depending on $\vec{a}$ and $\vec{b}$. If the choice of $\lambda$ is only made after Alice and Bob make their choices, then it doesn't imply superdeterminism, but it does imply nonlocality, since Alice's choice influences the $\lambda$ that in turn influences Bob's result.

"A skeptic's view on Bohmian Mechanics"

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