Does the HUP alone ensure the randomness in QT?

Question

This answer of mine has been strongly criticized on the ground that it is no more than a philosophical blabbering. Well, it may well be. But people seem to be of the opinion that HUP alone does not ensure randomness and you need Bell's theorem and other features for the randomness in QM. However, I still believe it is the HUP which is all one needs to appreciate the probabilistic feature of QM. Bell's theorem or other such results reinforces this probabilistic view only. I am very much curious to know the right answer.

Answer 1

I'm not sure if an undergrads perspective would be useful here- but I'll give it a shot (at worst I'll learn something new.)

David Griffith's "Introduction to Quantum Mechanics" takes great care to motivate the uncertainty principle from more basic founding postulates of Q.M. First Hilbert space and the state vector, as the description of the particle, are defined. Next classical observables are formulated as operators on the state vector. Eigenvalues and the basis of the operators are explored and it is revealed that for certain (conjugate) operators, the state vector cannot be written in the same basis if a unique value for those operator's corresponding observable is desired. It is shown that such operators do not commute. It is finally shown that from non-commuting the uncertainty principle can be mathematically derived.

So the point of this summary (all of which I'm sure you already know well) is the order that things are done. Griffiths is so far my favorite text book author, and I'm sure there is a reason he laid things out so explicitly. He stresses the classical nature of the observables and how the state vector is truly fundamental. It always seemed to me (and thus how I understand it) that what he was getting at is that observables like position and momentum are classical and what we are doing is trying to perform classical observation on a quantum system. When we attempt to do this, we are putting limitations on the state vector that nature simply doesn't do on her own. The result of this is that we end up with non-comeasurable observables, simply because of our classical bias in "translating" the true state of the particle, which is simply not completely expressible solely in terms of classical observables. To me this, what Q.M. is actually doing, seems more fundamental than the HUP. Perhaps it borders on metaphysics- but it seems to be the logical conclusion of the math/algorithms.

And because Bell's Theorem is mentioned: the inputs for this theorem are already there in Q.M.- the theorem simply tells us how to properly combine them and then conclude the character of the correlations between observables. In a way (once again seeming to me) it "measures" what kind of probabilities we are expressing in our theory.

Answer 2

It's very strange for someone to say that "Bell's theorem ensures something in quantum mechanics". Bell's theorem is a theorem - something that can be mathematically proved to hold given the assumptions. It's valid in the same sense as $1+1=2$. Is $1+1=2$ needed for something in physics? Maybe - but the question clearly makes no sense. Mathematics is always valid in physics - and everywhere else.

However, even the assumptions of Bell's theorem surely can't be "necessary building blocks" for some results in quantum mechanics because Bell's theorem is not a theorem about quantum mechanics at all. It is a theorem (an inequality) about local realist theories - exactly the kind of theories that quantum mechanics is not. Whether someone needs $1+1=2$ doesn't matter because this fact is imposed upon him, anyway. Any proof may be modified so that $1+1=2$ is needed and any proof may be modified so that $1+1=2$ is not needed.

But even if one ignores the comment about "Bell's theorem and other such results" that can't possibly have anything to do with the question, it's nontrivial to make the question precise. The uncertainty principle is normally formulated as a part of quantum mechanics - we say that $\Delta x$ and $\Delta p$ can't have well-defined sharp values at the same moment. What it means for them not to have sharp values? Well, obviously, it means that one measures their values with an error margin, and the fluctuations or choice of the measured value from the allowed distribution has to be random.

If it were not random, there would have to be another quantity for which one should do the same discussion. Again, if the uncertainty principle applied to this hidden variables (and a complementary one), it would imply that its values have to be random. Do you allow me to assume that the HUP holds for whatever variables we have? If you do, obviously, there has to be random things in the Universe.

But even the term "random" is too ill-defined. Do you require some special vanishing of correlations etc.? If you do, shouldn't you describe what those requirements are?

So I don't think it's possible to fully answer vague questions of this kind. I would say a related comment that quantum mechanics - with its random character - is the only mathematically possible and self-consistent framework that is compatible with certain basic observations of the quantum phenomena. The outcomes in quantum mechanics take place randomly, with probabilities and probability distributions that can be calculated from the squared probability amplitudes, and all other attempts to modify the basic framework of quantum mechanics have been ruled out.

If it's so, and it is so, there's really no point in trying to decompose the postulates of quantum mechanics into pieces because the pieces only combine into a viable theoretical structure, able to explain the behavior of important worlds such as ours, when all these postulates are taken seriously at the same moment.

Answer 3

In my opinion HUP is not a "principle" but a consequence of the mathematical framework of QM - it is derived rather than "postulated".

Randomness or uncertainty in measuring some variable in some state is not strictly related to the uncertainty of its canonically conjugate variable. HUP establishes some limitation on them and that's it. What I want to underline is that, say, uncertainty in momentum is determined with the given QM state itself.

About randomness, it is easy to understand if we remember that the information is gathered with help of photons. When the number of photons in one "observation" is large, their average is well determined and it is what the classical physics deals with. If the number of photons is small, the uncertainty makes an impression of a strong randomness in measuring, say, position of a body. Even the Moon position is uncertain if based on few-photon measurements.

Uncertainty in measurements is a fundamental feature of states in physics. Determinism is possible only for "well-averaged" measurements. Look at the Ehrenfest equations - they involve average (expectation) values. It implies many-many measurements. In other words, the classical determinism is due to its inclusive character.

Answer 4

Well, you misinterpreted what I (and others) said at least in two important ways.

1. Bell's theorem surely isn't responsible for randomness in QM. That's because it doesn't actually tell you anything about QM itself, only about other theories trying to reproduce the same results that QM (and nature) produces. The reason I mentioned it is that it (severely) restricts the class of non-random theories that can describe the nature. Without such a theorem one might hope (and people still do) that it is possible to construct a deterministic framework that could be compatible with observations. So HUP certainly doesn't imply intrinsic randomness. You need further work to establish that no viable theory (and not just QM) is deterministic. Measurement of violation of Bell's inequalities is what does it (at least if one assumes locality).

2. QM is based on lots of principles. HUP is fundamental (and is built-in by including non-commutative operators into the framework) but no less fundamental than other postulates. Trying to isolate one particular feature of a theory doesn't always make sense. You could try to obtain deterministic QM by removing HUP but that essentially means letting $\hbar \to 0$ and obtaining classical physics, thereby losing all the other special effects of QM.

In other words, your statement "HUP which is all one needs to appreciate the probabilistic feature of QM" couldn't be more far removed from reality. To appreciate this probabilistic aspect, one needs to master the mathematical formalism of QM, the way it connects to experiment and the way measurements are interpreted. HUP is only a small part in it and actually, the one thing you almost never care for as it is built-in into the theory from the start.

Answer 5

The title question is

Does the HUP alone ensure the randomness of QM?

I claim that the answer to this question is:No.

The HUP has the basic forms:

$$\Delta E.\Delta t \ge \hbar$$

$$\Delta x.\Delta p \ge \hbar$$

Furthermore quantum mechanics books prove that for non-commuting observables:

$$[P,Q] \neq 0$$

$$\Delta P.\Delta Q \ge \hbar$$

So the HUP is proven generally as a consequence of the non-commutativity of the observables. In order to understand why there are non-commuting observables in QM takes us to the rest of the postulates of QM and so explains why the other answers say that HUP is a consequence of QM in toto.

However there is more to the topic of "QM randomness" than this, and we have not yet responded to your remarks about Bell's Theorem.

The first point to note is that in classical engineering, there is a concept of time-domain and frequency domain (for a wave) and the associated law:

$$\Delta \omega.\Delta t \ge 1$$

This law is a consequence of the Fourier transform between these domains and therefore:

$$e^{-i\omega t}$$

So the HUP formula is more widespread than just quantum mechanics. Of course if one puts

$$E=\hbar \omega$$

then one obtains

$$\Delta E.\Delta t \ge \hbar$$

once again!

So where does quantum randomness (assuming for the moment, that that is the correct term) come from?

One published book that makes this point explicitly is Roger Penrose "The Emperor's New Mind", p297

[In quantum collapse..] these real numbers play a role as actual probabilities for the alternatives in question. Only one of the alternatives survives into the actuality of physical experience.. It is here, and only here, that the non-determinism of quantum theory makes its entry.

The italics is mine (and this is where Penrose introduces his R definition for describing quantum wave function reduction). Thus if you are familiar with quantum mechanics then this is the reduction postulate (in words).

So we have several different concepts in play here: HUP, QM Postulates, Bell's Theorem, randomness.

Answer 6

I think I'm largely going to repeat what Roy, Vladimir, and Jaskey13 have already said, but perhaps, I hope, not so totally that this won't be Useful.

I take it that HUP, despite its grandiose title, is not a principle; it's derived as a consequence of the various mathematical structures of QM. As such, HUP is a part of a characterization of the properties of QM. HUP is, however, something of a lesser part of that characterization because it is not enough to characterize all the differences between classical stochastic physics and QM. It is possible, as Roy says, to construct local classical models for which, under a reasonable physical interpretation of the mathematics, HUP is true.

I'm not completely sure what you mean by "HUP alone does not ensure randomness"? I suppose the interpretation of QM is all probability all the time. In various comments you protest, and I believe, that you know the axioms of QM and their basic interpretation well enough. What I take you to mean is that "HUP alone does not ensure intrinsic randomness". This qualification, which is fairly commonly used, makes sense, to me, of your following comment, with my qualification inserted, that "you need Bell's theorem and other features for the [intrinsic] randomness in QM", whereas the relevance of Bell inequalities to your Question seems to have troubled other people here.

I take “intrinsic” to be a rather coded way to say that a classical probability theory is not isomorphic to quantum probability theory. I've previously cited on Physics SE the presentations of Bell-CHSH inequalities that I think best make this clear, due to Landau and to de Muynck, here, where I note that you also left a notably Useful(8) Answer. Their derivations use the CCRs in a way that is not significantly more obscure than does the derivation of the HUP. I take the Bell-CHSH inequalities to be a reasonable lowest-order characterization of the difference. There is of course confusion concerning the relevance of locality to the Bell inequalities, which I think could get in the way of my discussion here, but I see that you have a relatively sophisticated view of that confusion.