Does quantum physics imply the existence of randomness?

[Zafa Pi]

I'm curious to hear what the experts here think about the following thought experiment. You have a radioactive atom that was created in a nuclear reactor some years ago. You place the atom under a powerful detector that will signal you when the atom disintegrates. You know the half-life of the atom. Can you, at any time, predict with definiteness (up to that permitted by the uncertainty relation between time and energy) when that atom will decay? I say no, thinking as follows. Statistical properties like the half-life can give you definite information in the infinite limit of sample size, i.e. in this case, an infinite number of atoms, or infinite waiting time. I'm not sure right now, but I think that estimate of the probability of the particle decaying in the time interval dt>0 is the best you could do. Of course, you would know that the particle will decay if you wait eternally. Not only that, but you cannot discern the atom's history from your observation. You would have absolutely no idea when that batch of radioisotope was created (Well, only that it was more recent than 1941).
If, on the other hand the atom's nucleus possessed some internal machinery that determined the atom's fate, then ascertaining the values of parameters that govern the machinery's behavior might tell you when the atom will decay, and it might be a possible to learn how long the machinery has been ticking away. But, thanks to Bell and his theorem, we know that such an internal mechanism in a quantum particle cannot exist because that would entail the existence of forbidden "hidden variables".

Hidden variables is not forbidden if locality is violated. See post #14

 

[Zafa Pi]

Are the paths of molecules truly random, or deterministicly chaotic? I'm trying to think of examples of truly non-deterministic processes other than the quantum-mechanical.

Whether true randomness occurs in nature is not known and will likely never be known. The only physical theories/models that have intrinsic randomness involve QM.

 

[Zafa Pi]

Schrodinger equation is "describing the time-evolution of the system's wave function"
"The wave function is a complex-valued probability amplitude, and the probabilities for the possible results of measurements made on the system can be derived from it"
-Wiki
I think a logical conclusion would be to deduce the Schrodinger equation talks about how the particle is subject to probability. Where do you think is my line of reasonning wrong ?

A simple analogy is this: You have a ±1 valued random variable where the probability of 1 at time t ≥ 0 is t/(1+t). So the the probability distribution of the r.v. is totally determined by the time, but if you sample the r.v. at, say, time t = 1 then the probability of getting 1 = the probability of getting -1 = ½.

 

[naima]

(where "local" means that the response of a detector can be predicted using only the value of hidden variables in the past light cone of the detection event).

In Bertlmanns socks Bell writes the locality condition
$P(a,b,\lambda) = P(a,\lambda) P(b,\lambda)$ where a and b are the distant outcomes.
Is it equivalent to what you writes (with the past cones)?

 

[Nugatory]

In Bertlmanns socks Bell writes the locality condition
$P(a,b,\lambda) = P(a,\lambda) P(b,\lambda)$ where a and b are the distant outcomes.
Is it equivalent to what you writes (with the past cones)?

Yes, if I'm understanding your question properly.

 

[naima]

How can it be derived?

 

[Mark Harder]

While am am new to this stuff it seems that purely mathematically the Schrodinger equation for a free particle is the same as the Heat equation except with a complex constant coefficient. One would expect that it describes a diffusion process similar to a continuous time Brownian motion. In Feynmann's Lectures on Physics Book 3, he describes how this actually works. The Shroedinger equation for a free particle describes a continuous stochastic process similar to a Markov process except that instead of conditional probabilities, there are conditional complex amplitudes.

The Markov process analogy is an interesting one in that in a Markov process, the probability of transitioning from the current state to the next one is not conditional. At most, the transition probabilities depend only on properties of the current state. Randomness aside, MPs resemble classical mechanics in which the equations of motion ensure that knowing p and q at any time is sufficient to know their precise values in the next instant. The values of momentum and position at any previous time do not directly enter the equations of motion.

Following that thought, an important difference between classical and quantum mechanics is that the latter says that we don't even know exactly which state we are in at any given time, since the uncertainty relations place finite limits on how precisely p and q can be specified simultaneously. The equation that tells us how a QM system evolves in time is the Schroedinger eqn. Fortunately, I suppose, the SE is a linear differential eqn., so that the fuzziness of the future states will be remain within bounds; that is, the solutions of the SE won't 'blow up' the way they can in a highly nonlinear system. In the language of higher algebra, the SE respects the uncertainties of the present state when it maps it to a future state.

Much as in Brownian motion one would imagine continuous nowhere differentiable complex valued paths of states...

If the paths of QM states are not differentiable, could they be specified by the SE, which includes derivatives? I know that the diff eqs that describe Brownian motion are special, stochastic equations. Presumably these don't require derivatives in any way that I understand, but I wouldn't know.

 

[D2Bwrong]

So, I am not an expert in quantum physic, I just watched a lot of videos about it.

If I understand correctly, particles do not have a particular position as long as you don't observe them. With a certain equation, we can draw a cloud of probabilities which describes how likely the particle is to be at any location at any time. As I heard, this theory of quantum physics has proven itself to be extremely effective.

More than once, I had discussions with friends about whether or not our universe is purely deterministic or if it contains randomness. I am more on the deterministic side, and a argument that I often face is that quantum physics theory implies the existence of randomness.

On the surface, it seems to me like I can compare quantum physic's probabilistic nature to that of a coin toss. Probability theory is extremely effective to predict the distribution the multiple results of many throws will respect, even though these events have a deterministic nature.

Could it be that the same thing is happening with quantum physic?
Could it be that some deterministic processus is what generate the probabilistic distribution that lies within quantum physic?
Or is there some aspect of the theory I fail to understand?

See if this makes it simple.
Quantum mechanics is a probabilistic procedure. It is deterministic since you can predict an outcome. Once a starting point is known it has a random nature. For example of the coin toss: If we want the result of 1000 tosses starting with no previous tosses, then the odds of the result of each individual coin tosses are equivalent and so are the totals. If however, the results of the first 50 tosses is 35 heads and 15 tails, then the odds of the outcome of the total number of tosses being equivalent diminishes even though the odds of each individual toss remains constant. It may be easier to visualize if you consider the game of poker where the possibilities are more numerous and probabilities of winning change with each card dealt even though each player starts with the same odds. That is the randomness.

 

[naima]

Bell uses formulas in his proof.
he has a definition of locality where the speed of light does not appear.
How can we derive things like the pas light cones of the devices?

 

[Mark Harder]

See if this makes it simple.
Quantum mechanics is a probabilistic procedure. It is deterministic since you can predict an outcome. Once a starting point is known it has a random nature. For example of the coin toss: If we want the result of 1000 tosses starting with no previous tosses, then the odds of the result of each individual coin tosses are equivalent and so are the totals. If however, the results of the first 50 tosses is 35 heads and 15 tails, then the odds of the outcome of the total number of tosses being equivalent diminishes even though the odds of each individual toss remains constant. It may be easier to visualize if you consider the game of poker where the possibilities are more numerous and probabilities of winning change with each card dealt even though each player starts with the same odds. That is the randomness.

A deterministic process is one which, given exact initial and intermediate conditions, will arrive at an exactly known end-point. What, then, do you mean by 'predict an outcome' of a probabilistic procedure? You can only predict the probability of a specific outcome of a stochastic (random) process.

What do you mean by 'equivalent'. Does the word mean 'equal', i.e. 50% in the case of coin tosses? Or, do you intend to say something more than 'equal' when you use the word 'equivalent'. If I assume that you mean 'equal', then it seems like you are perilously close to saying that the probabilities of H and T for the remaining 950 tosses are somehow conditioned on the result of those first 50 tosses. Such is not the case, except for the case where the remaining tosses can produce no fewer than 35 heads and 15 tails. If the first 999 tosses come up heads, the probability that the last toss is tails is still 50%. The probability that all 1000 tosses turn up heads is extremely small, but not zero, and the probability of such a compound event can also be calculated. Yes, the probability of a compound event in poker, such as holding a straight flush or 2 pairs, is conditional. The probability of one of these compound events increases or decreases as the draw of cards proceeds, even though the probability of drawing anyone card is 1/52. Still, I'd like to hear more about how you relate these thoughts to QM.

 

[LaserMind]

The collapse of the wavefunction could not be *completely random* in practice because that would mean an exact point-position to an infinite number of decimal places.

 

[DrChinese]

The collapse of the wavefunction could not be *completely random* in practice because that would mean an exact point-position to an infinite number of decimal places.

Not sure what you mean by this. There is nothing stopping this.

 

[Isaac0427]

Not sure what you mean by this. There is nothing stopping this.

I think he was saying that the particle is more likely to be observed in one place than another, so there is no "complete randomness", i.e. there can't be an equal probability for a particle to be in any position. I get the concept, but I don't know if "complete randomness" is the right term.

 

[entropy1]

He means that an exact value is a theoretical phenomenon, not a practical one. Theoretically, you can balance a pin on its tip; but it is practically infeasable.

 

[Grinkle]

he collapse of the wavefunction could not be *completely random* in practice because that would mean an exact point-position to an infinite number of decimal places.

He means that an exact value is a theoretical phenomenon, not a practical one. Theoretically, you can balance a pin on its tip; but it is practically unfeasable.

What law of physics prevents a measurable / observable property from actually having a specific infinitely precise value? I read your statements as though this is axiomatic or obvious from inspection, it is not obvious to me, at least.

 

[LaserMind]

What law of physics prevents a measurable / observable property from actually having a specific infinitely precise value? I read your statements as though this is axiomatic or obvious from inspection, it is not obvious to me, at least.

I thought a property could not have an infinitely precise value because infinity is never ending. Kinda obvious!

 

[DrChinese]

He means that an exact value is a theoretical phenomenon, not a practical one. Theoretically, you can balance a pin on its tip; but it is practically infeasable.

He might mean this (that's how I interpreted too). But the only practical limitation is in us making a measurement, which certainly does not constrain an observable itself in any way. It could be completely random, and in fact behaves as such. There is no particular reason to believe collapse is not completely random - other than by pure assumption.

 

[DrChinese]

I thought a property could not have an infinitely precise value because infinity is never ending. Kinda obvious!

So obvious that... it is not obvious.

Suppose I have an observable that can take on 1 of 2 values. Are you saying that observable cannot be completely random because the value does not have an infinite number of decimal places? Because I wouldn't agree with that.

And I wouldn't agree for an observable that is continuous either.

 

[Isaac0427]

Suppose I have an observable that can take on 1 of 2 values. Are you saying that observable cannot be completely random because the value does not have an infinite number of decimal places? Because I wouldn't agree with that.

I could be wrong, but I think he's saying that if you had an observable, such as position, that could take on one of an infinite number of values (I mean, there are an infinite amount of values between 0 and 1), the observable can't be completely random. What I got from his message, in a more mathematical form, the wavefunction $\psi=Ne^{ix}$ where N is a constant satisfying normalization is "completely random," but there is no possible value of N, and thus the wavefunction is not practical.

 

[DrChinese]

I could be wrong, but I think he's saying that if you had an observable, such as position, that could take on one of an infinite number of values (I mean, there are an infinite amount of values between 0 and 1), the observable can't be completely random. What I got from his message, in a more mathematical form, the wavefunction $\psi=Ne^{ix}$ where N is a constant satisfying normalization is "completely random," but there is no possible value of N, and thus the wavefunction is not practical.

If someone asserted that, I would challenge it. The number of possible outcomes does not change whether something is or is not random.

 

[Isaac0427]

If someone asserted that, I would challenge it. The number of possible outcomes does not change whether something is or is not random.

I think complete randomness would be defined as there are no values that the particle is more or less likely to be in. Again, not too relevant to this discussion, but it is true that by that definition, complete randomness is impossible in quantum mechanics for the mathematical reason in my previous post.

 

[entropy1]

What law of physics prevents a measurable / observable property from actually having a specific infinitely precise value? I read your statements as though this is axiomatic or obvious from inspection, it is not obvious to me, at least.

I was thinking about virtual particles for instance; they would bump into the balanced pin and push it over. Space is not exactly empty. Another example would be that a 'trapped particle' would gain infinite momentum and its position wouldn't be exactly measured. That sort of thing.

 

[Grinkle]

I think complete randomness would be defined as there are no values that the particle is more or less likely to be in.

I am not able to be precise with my language, so maybe I just can't make more progress here. I will try, and I appreciate any help.

Complete randomness might mean that given a set of possible values, there are no values in that set which the particle is more or less likely to be in. There may be other values not in the set of possibilities that the particle has zero chance of being in.

Is that a bad / unuseful definition of complete randomness? I would use the term "even probability distribution" instead of complete randomness to describe what I am saying.

I think you are saying that complete randomness means there is no value whatsoever excluded from the possible value set, and in addition there is a perfectly even probability distribution.

I never studied statistics. There must be some math to show whether or not a set containing infinitely many members can have an even probability distribution that sums to 1? Is that the mathematical issue, or am I way off base?

 

[Nugatory]

I think he's saying that if you had an observable, such as position, that could take on one of an infinite number of values (I mean, there are an infinite amount of values between 0 and 1), the observable can't be completely random.

If that's what we're talking about, it's not right. There is a perfectly satisfactory theory of continuous probability distributions and what "random" means in that context. You may not encounter it until a few years into college because, unlike the simpler discrete cases, you need a moderate amount of calculus just to get started, but it's there.

 

[Grinkle]

I was thinking about virtual particles for instance; they would bump into the balanced pin and push it over. Space is not exactly empty. Another example would be that a 'trapped particle' would gain infinite momentum and its position wouldn't be exactly measured. That sort of thing.

Those are measurement issues, not existence issues, aren't they?

I thought you might talking about something equivalent to asking if space-time is discrete or continuous (the answer is not obvious to me, I don't have any leaning one way or the other on which is more likely true) but maybe that is not what you are getting at.

 

[entropy1]

Those are measurement issues, not existence issues, aren't they?

Indeed, things won't let themselves measure exactly, in practice. Everything at least has a little vibration and/or uncertainty. (Recall I was responding to Lasermind with respect to this)

 

[DrChinese]

Complete randomness might mean that given a set of possible values, there are no values in that set which the particle is more or less likely to be in. There may be other values not in the set of possibilities that the particle has zero chance of being in.

Coming close to yours is the following, from Wikipedia: "In situations where a population consists of items that are distinguishable, a random selection mechanism requires equal probabilities for any item to be chosen."

However, that is not the case for QM, where there is randomness but some values may be more likely than others (same source): "According to several standard interpretations of quantum mechanics, microscopic phenomena are objectively random.[6] That is, in an experiment that controls all causally relevant parameters, some aspects of the outcome still vary randomly. For example, if you place a single unstable atom in a controlled environment, you cannot predict how long it will take for the atom to decay—only the probability of decay in a given time.[7] Thus, quantum mechanics does not specify the outcome of individual experiments but only the probabilities."

 

[Mark Harder]

The collapse of the wavefunction could not be *completely random* in practice because that would mean an exact point-position to an infinite number of decimal places.

Certainly an exact location of a quantum particle is impossible. The UP guarantees that. But I don't understand why a "completely random" collapse would imply exactness. In fact, it sounds to me that just the opposite is true: if the collapse is random (completely or not), then any future evolution of the system could not be determined. With random initial conditions, how could a system trajectory be determined exactly? Since the wavefunction is linear, we are guaranteed that any 2 solutions cannot diverge without limit. That's the best we can do when we attempt to specify future states from imprecise initial states.

 

[write4u]

Certainly an exact location of a quantum particle is impossible. The UP guarantees that. But I don't understand why a "completely random" collapse would imply exactness. In fact, it sounds to me that just the opposite is true: if the collapse is random (completely or not), then any future evolution of the system could not be determined. With random initial conditions, how could a system trajectory be determined exactly? Since the wavefunction is linear, we are guaranteed that any 2 solutions cannot diverge without limit. That's the best we can do when we attempt to specify future states from imprecise initial states.

Question: If the observer has a fixed position, would that not play a part in any equation of the quantum function?

IOW, the observer *determines* the final position of the particle even as the observer cannot predict the exact position of the particle while it is in motion.

Question: How could we construct precisely focused lasers if total randomness existed.

As I understood one of the questions, once the wave function has collapsed and the particle is manifest it has a precise location, relative to the observer. The problem lies in the *uncertainty* of the particle's position while in motion, but apparently we can control this randomness (to an extend).

 

[TheMeInTeam]

You bring up what seems to me to be the deepest question in this topic. It seems that when physical processes are deterministically chaotic, probably theory is only a model of something that, though deterministic, is so complex that in practice is practically impossible to calculate. In such cases - like tossing a coin or casting dice, - choosing probability as a model for the process is a kind of "fudging". Here, probability describes non-stochastic systems for which we can have incomplete knowledge only. On the other hand, assigning probabilities to a quantum event is an appropriate description of a truly random process that exists in nature. The physical meaning of stochastic variables is not the same for all cases, in other words.

Is it really certain that the dice example isn't comparable to the quantum event one? From what I can tell reading up on this there's a good chance we're "fudging" in both cases.

 

[Grinkle]

Is it really certain that the dice example isn't comparable to the quantum event one?

That is re-stating the title of the thread, imo.

 

[TheMeInTeam]

That is re-stating the title of the thread, imo.

True, but as far as I can tell the answer is "no", because we don't know, don't even have evidence to prefer a conclusion. Pick your interpretation and the answer changes, right? We don't know which interpretation is correct, if any, but that's good enough reject a conclusion of "implied existence of randomness".

 

[Grinkle]

don't even have evidence to prefer a conclusion

See post 31. To me, non-locality is also weird. If quantum determinism exists, then non-locality must also exist. I think.

Its not evidence to prefer one or the other, but it is certainly not so easy to dismiss quantum randomness if that implies a rational person must then be accepting non-locality as the explanation for the Bell theorem experiments.

 

[TheMeInTeam]

See post 31. To me, non-locality is also weird. The existence of both quantum determinism and non-locality is contradicted by experiment. I think.

Its not evidence to prefer one or the other, but it is certainly not so easy to dismiss quantum randomness if that implies a rational person must then be accepting non-locality as the explanation for the Bell theorem experiments.

Any of it can "seem weird" when thinking in terms of every day experiences of what we're used to thinking.

Why do we necessarily need hidden variables to get to determinism? Going back to the dice example, the eye is not enough instrumentation to detect that the outcome a roll is, depending exactly on how the dice is thrown and how it interacts with the landing surface, deterministic.

Just as non-locality seems weird to you (and I'm not sure it's required in principle, but maybe I'm missing something), I find it strange that this point is the only thing in physics I've heard of that lacks clear causality and is somehow accepted by many regardless...even as we don't completely understand it yet. This one topic is exceptional with how we know things work otherwise, posited because we don't understand it! Quite a bit of the discussion here just assumes a wave function collapse outright...which might or very well might not be an accurate framework to use.

To me the best answer is "we don't know yet."

 

[carllooper]

Why is red light red? It's silly to say "we don't know yet" because the very thing we do know, if only for ourselves, is that red light gives us a particular sensation we can call red. What we don't know is how to formulate that in mathematical terms. I might see red light as cyan, and you otherwise see it as red, but we will both still call it "red". And in the same way, the mathematics of colour, based on the frequency of light, and otherwise used in technologies of colour (such as colour photography) is not in any way altered by this ambiguity in how we otherwise perceive colour. The same model works for both of us, whether you or I perceive red light as red or as cyan. A mathematical model, as much as words such as "red", are unable to make any distinction between sensations as otherwise personally experienced, even if we can otherwise entertain just such distinctions.

It becomes a useful thing that a model can work regardless of personal perception, but it also demonstrates how a physical/mathematical model may not ever be able to encode the full perceptual situation. There may very well be a limit to the scope of what physics and mathematics alone, can describe.

Randomness is particularly resistant to mathematical description. Indeed one might say, in the context of mathematics, there is no such thing as randomness. It may very well be that mathematics is inherently faulty with respect to randomness. In statistics, on the other hand, randomness is at least an operating assumption, regardless of whether there is a mathematical solution (a pseudo-random solution) or not. What matters in statistics is the aggregate effects regardless of the the precise nature of the causes (or lack thereof).

An absence of a causal model for wave function collapse doesn't mean we fail to see, in a given experiment, what is otherwise meant by wave function collapse. The collapse still happens, so to speak. It doesn't need us and our models, or lack thereof, in order to take place. We still see individual particle detections, and we still see the pattern of those detections, and we still see the mathematical correlation between the pattern and a wave function as derived from the geometry of the setup.