B Does quantum physics imply the existence of randomness?

[Nugatory]

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".

Bell's inequality does not preclude all internal machinery of the type that you're describing. It does preclude any mechanism in which the theory governing the behavior of the hypothetical hidden variables is local (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).

Thus, Bell's theorem leaves room for deterministic theories (as well as hidden variable theories that are not deterministic) as long as they are non-local.

 

[vanhees71]

In quantum theory a certain kind of states, socalled "pure states", are represented by a vector $|\psi \rangle$ in an infinite-dimensional vector space called Hilbert space. One realization of the Hilbert space is the space of square integrable functions. That refers to the socalled position representation,
$$\psi(\vec{x})=\langle \vec{x}|\psi \rangle.$$
The physical meaning is that
$$P(\vec{\psi})=|\psi(\vec{x})|^2$$
is the probability density for finding the particle, prepared in the state described by this particular square-integrable wave function, at position $\vec{x}$. Square integrable means that the integral over $P(\vec{x})$ exists, and you can normalize it properly,
$$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} P(\vec{x})=1,$$
i.e. the particle is present somewhere in space with certainty.

Now, of course you cannot only measure the position of the particle but also, e.g., its momentum. Then the quantum theoretical formalism tells you that the corresponding momentum-wave function is given as the Fourier transform of the position wave function,
$$\tilde{\psi}(\vec{p})=\left (\frac{1}{2 \pi \hbar} \right)^3 \int_{\mathbb{R}} \mathrm{d}^3 \vec{x} \psi(\vec{x}) \exp \left (-\frac{\mathrm{i} \vec{x} \cdot \vec{p}}{\hbar} \right ).$$
Then the probability distribution for the momentum of the particle is given by
$$\tilde{P}(\vec{p}) = |\tilde{\psi}(\vec{p})|^2.$$
From the math of the Fourier integral it follows that also this probability distribution is properly normalized too.

 

[Chris1983]

Picture a six-sided dice. There are only six possible values when the dice is rolled. If you want to roll a specific value, then if that number is a whole integer with a value in the range of one through 6 you might say the dice gives a random result. The universe is random. If your desired result is not within that range, and the range was determined by another factor, then you might say that the result is predetermined to not give you your desired value. The universe is deterministic.

Do you consider a dice to be random because, in theory, it could return any of a range of values on any given roll? Or is it deterministic because it can only give value from a range that was determined when the dice was first made a cube?

What if you don't need a specific value, just any value within the range available on a dice? Then any roll will always return your desired value. Or the same if your want any value that is not in the dice range.

Deterministic vs random could be seen as a question of how accurate your measurement needs to be; or, they are not mutually exclusive. The universe proceeds along a course determined by its earliest state, but it has an unknowable amount of variation, at random.

Perhaps "god doesn't play dice" just rolls off of the tongue better than "god doesn't make Rube-Goldberg machines."

 

[Mark Harder]

A perfect example of apparent randomness occurs often in weather forecasting. While each molecule in the local system behaves deterministically, there are just too many mathematical forces in play for us to make precise predictions, except in the most general terms such as the probable direction of the storm front.

Yes, the key word being 'apparent'. When considering chaotic systems that are deterministic, random behavior is a model of the true dynamics. Probability theory is applied as if the system is truly random. I still wonder if there if there is a component of molecular dynamics that is truly random. Is heat random? Given a dose of thermal energy, are molecular motions deterministic while thermal energy itself is randomly partitioned among them? In performing computer simulations of molecular motion for example, the software applies to each atom the mean energy specified by statistical dynamics, then let's the simulation run according to the appropriate deterministic differential equations of the atoms to yield their not-quite-predictable behavior. But, in nature, isn't the thermal energy apportioned apparently at random, in which case the initial conditions for those equations should have random components? Obviously, I'm a little confused and the discussion is a little off-topic, having little to do with QM..

 

[Zafa Pi]

The postulates themselves don't imply that randomness is intrinsic: such a conclusion will depend on which interpretation you buy into. For example, in Bohmian mechanics and the many-worlds interpretation the randomness is only apparent: the state of the system always evolves deterministically.

Not true. It depends on the postulates. Copenhagen with collapse has intrinsic randomness with regard to measurements.

 

[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.