Clarifying the Meaning of "Random" in Quantum Physics

[zonde]

Yes you have. It is in the mathematical assumptions of the inequalities by stating that
causality can only act in a linear way , i.e. the outcomes A and B are in Linear combination with vectors a and b.

I am not talking about original Bell inequality but instead about very simple model to which I gave link in my post #14. In that model there are no such assumptions.

 

[carllooper]

In mathematics there is no such thing as pure randomness. In mathematics we can only approximate such an idea, with some help from statistics. For example, we can write an algorithm for a random number generator. But we'll call the results of such an algorithm "pseudo-random" in order to maintain the idea of pure randomness (which otherwise escapes us).

We might say that pure randomness might express something similar to what mathematics otherwise expresses by the concept of zero. Or even better (but not quite mathematical): the concept of null.

In terms of signal theory, pure randomness might be called noise. But noise is ambiguous because noise (so called) will often have some identifiable cause. If we call it "noise" it is often to distinguish it from some other signal in which we're interested. Such "noise" would really be a signal. One (if we understood the cause) we might filter out in order to concentrate on the other signal in which we're interested.

More interesting is the idea of a kind of fundamental noise. And by this we can suggest the meaning of such might be defined negatively: as "not a signal", not just a signal in which we're not interested, but not a signal at all. The absence of a signal. A zero signal or better: a null signal.

C

 

[MacroMeasurer]

This might be a silly question but when people say that something on the quantum level is completely "random," (except for general probability) does that mean, according to theory at least, if you were to go back in time and repeat an experiment exactly that the results could just as easily be different as the same, or that the results of a given experiment are unpredictable beforehand aside form general likeliness of many different possible events but the results would still be the same in the aforementioned scenario? Or, I suppose, do currently accepted theories not have an answer for this?

Let me know if this is a reasonable answer for a layperson:

There is an epistemological difference from classical physics: in quantum mechanics, there is no way for anyone to precisely predict the outcome of any particular measurement.

Normally when people say that a coin has a 50/50 chance of coming up heads, people take that to mean "for our purposes," or "as far as we can tell." The common intuition** is that a much smarter, better informed entity could know in advance how the coin or dice will fall. Whether we call them demons, psychics, or aliens, lots of discussions about determinism and randomness involve someone or something that does know before a coin is flipped whether it will land face up. Even we mortal humans can expect that as physicists get better theories and tools dice will seem less random, and supercomputers using superfast cameras will someday predict how a coin spinning in the air will land.

Quantum randomness does not have that caveat. When we know as much as there is to know, we will still end up saying 50/50 for "quantum coins." There is (some) disagreement about whether there is a specific predetermined outcome waiting for us to observe it, which is another way of discussing what would happen if we rewound the clock. But there is general agreement that no demon, alien, or supercomputer can know the outcome in advance, even if there is in principle a right answer. When the theory tells you that something is 50/50, like e.g. spin along an axis orthoganol to your last measurement, the idea is that nobody could give you a better guess than that.***
** Folk theory, pre-1950 understanding of classical physics, etc.--meaning, in particular, ignoring chaos.

***The theory does include exact values, but only for things we can never directly measure. That is equivalent to the fairness of a coin: you can't see that a given coin has an exactly "50/50" chance of coming up heads using a microscope, you can only test for it statistically by flipping the coin repeatedly.

 

[atyy]

But there is general agreement that no demon, alien, or supercomputer can know the outcome in advance, even if there is in principle a right answer. When the theory tells you that something is 50/50, like e.g. spin along an axis orthoganol to your last measurement, the idea is that nobody could give you a better guess than that.

This part is not agreed. If nonlocal hidden variables exist, and a demon or alien has control over them, then the quantum randomness is converted to classical randomness. At present, no human (we believe) has such control, so quantum randomness is believed for practical purposes to create "true" randomness that is useful for making a secure code.

Here is an example of using quantum mechanics to certify that the random number generator is "truly" random.

http://arxiv.org/abs/1009.1567
Secure device-independent quantum key distribution with causally independent measurement devices
Lluis Masanes, Stefano Pironio, Antonio Acin

 

[carllooper]

Yes, the issue pivots on whether there is, or is not, regardless of whether we can know such or not, hidden variables (otherwise known as demons) that create what we otherwise nominate as true randomness - in which case true randomness would be a fiction. It would be, instead, what we'd otherwise call "pseudo-random" (or classical randomness) in the sense that such would have some sort of structure or formula, for which only the demon had the keys.

But we can also entertain the equally valid idea of no such demon.

And we can formalise this non-demonic randomness to some extent, in the sense that we can characterise what it's statistical properties would be like. For example, we can say the sum of all random numbers between -1 and 1 would sum to exactly 0. It would require an infinite sum, but we can assume such, ie. we don't have to physically carry out the sum. We can play with the idea of the purely random and not find ourselves completely bereft of some intellectual ground on which to express such.

C

 

[A. Neumaier]

the paper suggests a strong laser as a way to implement the experiment which would demand nonlinear Maxwell equations in inhomogeneous media

The laser doesn't need to be so strong so that one would need these. Just strong enough so that the classical approximation is adequate. At this strength the Maxwell equations (which without qualification always mean the linear ones) are satisfied quite accurately in air, typically much more accurately than one can perform Bell-type experiments. And typical beam splitters do this very efficiently as modeled, too; otherwise it would be hardly possible to analyze optical quantum experiments.

 

[harrylin]

[..] a hidden-variable field theory is so nonlocal from the start that none of the assumptions used to justify Bell type inequalities are satisfied, hence the Bell inequalities cannot be derived.

"none of the assumptions" is pretty strong, only a single one suffices and I find it difficult to put my finger on that, so to say. ..
Could you please give an example of one assumption that clearly isn't valid for a hidden-variable field theory?

 

[A. Neumaier]

"none of the assumptions" is pretty strong, only a single one suffices and I find it difficult to put my finger on that, so to say. ..
Could you please give an example of one assumption that clearly isn't valid for a hidden-variable field theory?

I was exaggerating; I meant the assumption that a particle travels along exactly one path. This is clearly impossible for a solution of Maxwell's equation. A single solution can contain several rays whose neighborhoods have a significant energy density, and indeed, according to classical optics, a beam splitter produces such a solution.

 

[stevendaryl]

I misstated in #16 what I had meant to say; see the updated formulation there for the intended version. To derive Bell's inequality you need to make assumptions that are never satisfied when you start with Maxwell's equations. Thus Bell's inequality doesn't hold. But the quantum result is just based on properties of the Maxwell equation, hence the quantum field approach gives identical results with the experimental findings.

I'm not sure what you're saying here. It seems to me that the assumptions behind Bell's inequality can be formulated in a way that is independent of the distinctions between particles and fields.

Roughly speaking, let's assume SR (GR causes complications that seem irrelevant). Pick a rest frame. Pick a time interval $\delta t$. Now, relative to that rest frame, partition space into cubes of size $c \delta t$.

Let $\mathcal{R}$ be the set of all possible records, or histories, of a single cube during a single time interval. So classically, a record might consist of an identification of what particles were in the cube during that interval, what their positions and momenta were at times during the interval, what the values of various fields at different points within the cube were at times within that interval.

From classical probability and special relativity, we would expect that if $\mathcal{R}$ was a complete description of local conditions within a cube during an interval, then the behavior of the universe can be described by a grand transition function $\mathcal{T} : \mathcal{R}^{27} \times \mathcal{R} \rightarrow [0,1]$. The meaning of this is that $\mathcal{T}(\vec{r}, r')$ is the probability that a cube will have record $r'$ for the next interval, given that it and its neighboring cubes have record $\vec{r}$ during the current interval. So this is just saying that the behavior of a cube during the next interval is dependent on the behavior of its neighboring cubes during this interval. Note, I'm allowing the behavior to be nondeterministic, but I'm requiring the probabilities to be dependent only on local conditions. Also, I'm being sloppy about talking about probabilities, since in general there could be (and would be) uncountably many possible records. So I should be talking about probability distributions, instead of probabilities.

The two assumptions that (1) $\mathcal{R}$ is a complete description of the conditions within a cube during an interval, and (2) Einstein causality, imply that knowledge of conditions in distant cubes can give you no more information about future possibilities for one cube than knowing about conditions within neighboring cubes. Putting that more coherently: I pick a particular cube. If I know about conditions in that cube and all its neighboring cubes during one interval, then I can make a probabilistic prediction about conditions in that cube during the next interval. Those probabilistic predictions cannot be changed by more knowledge of conditions in non-neighboring cubes. This basically makes the universe into a gigantic 3D cellular automaton (except for the fact that the records are pulled from an uncountably infinite set, while the theory of cellular automata require the records to be taken from a finite set; I'm not sure how important that distinction is for the purposes of this discussion).

Completeness is an important assumption here. Obviously, if my records are incomplete---that is, there are microscopic details that I failed to take into account--then it is possible that additional correlations between distant cubes could be implemented by those microscopic details, which would give the erroneous appearance of nonlocality. If I took those microscopic details into account, then local information would be sufficient to predict the future of a cube.

Bell's theorem applied to the EPR experiment shows that it can't be described this way, and that it can't be fixed by assuming microscopic details that we failed to take into account. In the EPR experiment, no matter how fine-grained your descriptions of the cubes, information about conditions in distant cubes can tell you something about the future behavior of this cube.

But getting back to your point about Maxwell's equations, it seems to me that there is nothing about my "cellular automaton" model of the universe that would prevent the use of the electromagnetic field as part of the record for a cube.

 

[A. Neumaier]

getting back to your point about Maxwell's equations, it seems to me that there is nothing about my "cellular automaton" model of the universe that would prevent the use of the electromagnetic field as part of the record for a cube.

A single solution of the classical Maxwell equations may after some time (e.g., after passing a beam splitter) have significant energy in many of your cells simultaneously, even if it is localized at time $t=0$ in a single one. This is contrary to your assumed particle behavior. You must assume that upon a split the particle remains in a single cell though with a certain probability only, while the wave remains a single object divided into two beams.

 

[zonde]

I meant the assumption that a particle travels along exactly one path. This is clearly impossible for a solution of Maxwell's equation. A single solution can contain several rays whose neighborhoods have a significant energy density, and indeed, according to classical optics, a beam splitter produces such a solution.

Multiple paths from the source does not add anything new to the argument. You can check this using the model I gave. This simple model that does not use any specific assumptions:
https://www.physicsforums.com/showthread.php?p=2817138#post2817138
If you say that you can get around action at a distance please point out how one should modify this model to get expected correlations without action at a distance.

 

[A. Neumaier]

This simple model that does not use any specific assumptions:
https://www.physicsforums.com/showthread.php?p=2817138#post2817138

The specific assumption the setting you linked to assumes is locality. But Maxwell's equations are not local in Bell's (or your) sense. They are exactly as nonlocal as quantum mechanics itself.

 

[zonde]

The specific assumption the setting you linked to assumes is locality. But Maxwell's equations are not local in Bell's (or your) sense. They are exactly as nonlocal as quantum mechanics itself.

You mean that two measurements happen at the same place? Basically that distance is an illusion? And you derived that philosophical crap from Maxwell's equations?
I probably misunderstood you. Right?

 

[TrickyDicky]

You mean that two measurements happen at the same place? Basically that distance is an illusion? And you derived that philosophical crap from Maxwell's equations?

Where did you get all that? Do you understand how a simple plane wave doesn't fulfill locality as defined in the model you linked(and that you claimed made no assumptions)?

 

[zonde]

Do you understand how a simple plane wave doesn't fulfill locality as defined in the model you linked

No I do not understand. This is not about simply correlated outcomes. It's about changes caused over the distance. From that post: "In other words, in a local world, any changes that occur in Miss A's coded message when she rotates her SPOT detector are caused by her actions alone."
Pay attention to word "changes". If by rotation of her detector Miss A can cause changes in coded message that comes out of Mr B's detector then it's not simple correlation between two places but clearly action at a distance. And I do not understand how you can get "action at a distance" type nonlocality out of simple plane wave (not given any time for it to propagate to the distant place).

 

[stevendaryl]

A single solution of the classical Maxwell equations may after some time (e.g., after passing a beam splitter) have significant energy in many of your cells simultaneously, even if it is localized at time $t=0$ in a single one. This is contrary to your assumed particle behavior. You must assume that upon a split the particle remains in a single cell though with a certain probability only, while the wave remains a single object divided into two beams.

Sorry, I still don't see that. In terms of my cellular model (which is basically Bell's "local beable") the classical record within a cell would include both the amplitude and direction of the E and B fields inside the cell at every moment in time, the presence or absence of particles, and for each particle, the position at each moment of time. I think that it would be true (classically) that the record for the next time interval would depend only on the records for neighboring cells in this time interval. The classical version would only give particle-like properties to electrons and would only give wavelike properties to the electromagnetic field, but both types of phenomena can be modeled by a cellular model, it seems to me.

 

[stevendaryl]

Where did you get all that? Do you understand how a simple plane wave doesn't fulfill locality as defined in the model you linked(and that you claimed made no assumptions)?

I'm with zonde. I don't know what you mean by saying that a plane wave is nonlocal.

The sense of nonlocal that is relevant for Bell's inequality is this: If you confine your attention to a small region of space, then what happens to that region at time $t+\delta t$ depends only on facts about what's going on at points less than $c \delta t$ away from that region, at time $t$. I don't see how that is violated by plane waves.

 

[TrickyDicky]

I'm with zonde. I don't know what you mean by saying that a plane wave is nonlocal.

The sense of nonlocal that is relevant for Bell's inequality is this: If you confine your attention to a small region of space, then what happens to that region at time $t+\delta t$ depends only on facts about what's going on at points less than $c \delta t$ away from that region, at time $t$. I don't see how that is violated by plane waves.

There is no finite region of a plane wave where you can apply that sense of locality. That's how it is violated.

 

[A. Neumaier]

Bell's theorem applied to the EPR experiment shows that it can't be described this way

I hadn't noticed that you were talking about a myriad of tiny cubes with arbitrary neighboring interaction; indeed, in this case you can approximate a field.

But then your just cited claim is no longer justified; Bell's assumptions are very different. You'd have to give a new proof.

 

[stevendaryl]

There is no finite region of a plane wave where you can apply that sense of locality. That's how it is violated.

I don't understand what you mean. Maxwell's equations are local, regardless of whether your solution is a plane wave or not.

You have a region, say the set of points $(x,y,z)$ such that $x^2 + y^2 + z^2 < R^2$
You're trying to predict the behavior of the electromagnetic field in that region at time $t+\delta t$.
Then it is sufficient to know the values of the electromagnetic field in the region $x^2 + y^2 + z^2 < (R+c \delta t)^2$ at time $t$

[edit:] What I said is not true if you have charges within the region. So for simplicity, let's assume that we also know that there are no charges in the region $x^2 + y^2 + z^2 < (R+c \delta t)^2$ at time $t$.

I don't understand the sense in which plane waves make any difference.

 

[TrickyDicky]

I don't understand what you mean. Maxwell's equations are local, regardless of whether your solution is a plane wave or not.

You have a region, say the set of points $(x,y,z)$ such that $x^2 + y^2 + z^2 < R^2$
You're trying to predict the behavior of the electromagnetic field in that region at time $t+\delta t$.
Then it is sufficient to know the values of the electromagnetic field in the region $x^2 + y^2 + z^2 < (R+c \delta t)^2$ at time $t$

[edit:] What I said is not true if you have charges within the region. So for simplicity, let's assume that we also know that there are no charges in the region $x^2 + y^2 + z^2 < (R+c \delta t)^2$ at time $t$.

I don't understand the sense in which plane waves make any difference.

That's propagation of signals by any field. The definition of locality usually (in
Einstein definition in Wikipedia or the
one linked by zonde and use in Bell's
theorem) talks about objects, and for a
plane wave in vacuum the only way to
define an object includes infinite
extension, for al its properties as a wave.

 

[TrickyDicky]

I don't understand what you mean. Maxwell's equations are local, regardless of whether your solution is a plane wave or not.

They are local in the operational sense of finite signal propagation speed. But that sense is respected by quantum violations of Bell's inequalities.
Mathematically it is clear how monochromatic plane waves violate the (special relativistic) Minkowskian sense of locality(Einstein locality). Such solutions of Maxwell's equation in vacuum are singular in Minkowski space hyperplanes.