Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
    
Asking a separate question instead of abusing the comments is a very good idea! –  Ebenezer Sklivvze Apr 9 '11 at 10:17
1  
@Sklivvz: except that this question is not interested in the past discussion (as per @sb1's comment to my answer) so the whole talk about Bell's theorem (which is just misunderstanding on sb1's part anyway) shouldn't be present in this question at all. –  Marek Apr 9 '11 at 10:29
    
I'm appalled that so many people piled on your previous Answer without leaving any comments. Qudos to Marek for having left a comment, however the part of his comment that I agree with is that your Answer was not much to the point. It may be that the other downvoters felt that you didn't Answer the Question, not that it was philosophical blathering. I didn't downvote your Answer, but nor did I upvote it. –  Peter Morgan Apr 9 '11 at 13:03
    
@Peter: No, there were comments exchanged which were not quite friendly. I guess the moderator has removed them all except the first comment. –  user1355 Apr 9 '11 at 13:26
    
I am totally clueless about the above comment made by @Marek. He seems to assume a lot of things which makes one quite surprised and detested! –  user1355 Apr 9 '11 at 13:30

6 Answers 6

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.

share|improve this answer
    
It's true that the uncertainty principle is derived, but what you say in your third paragraph doesn't make much sense. There's not really anything classical about observables. In fact, they act very non-classically since they have nontrivial commutation relations with other things. Observables are operators on the Hilbert space of states, and "project" out (in some sense) the information contained in the state vector you're looking for. The "classical" things are more related to expectation values, not the operators. I think Griffiths discuses this somewhere in the exercises. –  Mr X Apr 10 '11 at 14:48
    
@Jeremy Price what I was getting at is that things like "momentum" and "position" are not true quantum mechanical properties- rather they are classical measures that we apply to the quantum world. But it is a true interpretation about the expectation values, from what I've read. Which is the root of the uncertainty principle, is it not- as the uncertainty of an observable is expressed as a deviation from the expectation value? (In Griffith's derivation at least) –  jaskey13 Apr 10 '11 at 15:16
    
@jaskey13 I don't think it's right to say that about momentum and position. They're very real things quantum mechanically, we still have quantum mechanical analogues of, e.g., conservation of momentum and energy, despite the fact that they are not "well-defined" in a classical sense. In fact, if you look at how to derive the Schroedinger equation, you replace operators into E = p^2/2m and act this on a function, usually as a function of position, which is surely taking all of these properties very seriously and fundamentally! –  Mr X Apr 12 '11 at 16:55
    
@Jeremy Price Are these analogues those that come from an application of Ehrenfest's theorem? If not- could you please tell me what they are? –  jaskey13 Apr 12 '11 at 21:48
1  
My edition is surprisingly scant when it comes to conservation laws. Maybe it is time to move on to something more advanced –  jaskey13 Apr 12 '11 at 22:56

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.

share|improve this answer

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.

share|improve this answer

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.

share|improve this answer
    
You have misunderstood my point as well. I am well aware of all the fundamental postulates of QM. You need all those postulates for a fully functional Q.T. However, my point is UP is the postulate for the essential qualitative element of the randomness in the theory. –  user1355 Apr 9 '11 at 8:43
    
@sb1: that might be the case but you start your question with "people seem to be of the opinion that..." which is simply not the case. People talked about something completely different last time so I am not sure why you bring that in now if you only intend to give downvotes for people's replies. If you only want to talk about pure QM then I suggest you edit your question in order not to confuse people further. –  Marek Apr 9 '11 at 8:49
    
@sb1 I think there's a Useful Answer in here (my +1). –  Peter Morgan Apr 9 '11 at 13:37
    
@Peter: thank you. Well, I believe all I said is correct and relevant but whether I've read @sb1's mind correctly as to what his intents were with this question that's another story... –  Marek Apr 9 '11 at 14:14

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.

share|improve this answer
    
I have no confusion about the fundamentals of quantum mechanics more than any body else here. Your engineering example is cute but wrong in the sense that the error in measurement in that case can be made arbitrarily small by more accurate instruments. Any theory, comes with a HUP like principle where the uncertainty can't be made arbitrarily small has to have probabilistic features. That's my understanding. –  user1355 Apr 9 '11 at 16:02
2  
@sb1 : I think the moral of how this site (has to) work is that if a question appears to be asking for a Textbook explanation of something that is what will be provided by default. If one means to challenge, or extend, the textbooks on an apparently basic topic (which Fundamental ones are) then the question formation needs to be referenced and so on. Phrases like "People believe.." dont convey exactly what was intended. So yes there will be misunderstandings about what was intended here. I worked on the Question title itself this time as my source for your meaning. Try again though. –  Roy Simpson Apr 9 '11 at 21:18
    
@sb1, I think the "engineering example" Roy cites is relevant to this discussion, but I point out that it emerges in deterministic signal processing, that stochastic SP is not needed (which you both may know). I find that a good author on this issue in SP is Leon Cohen, who I think writes very clearly. His "Time-Frequency Distributions-A Review", PROCEEDINGS OF THE IEEE, VOL. 77, NO. 7, JULY 1989, Page 941, where he discusses the relationship between quantum and SP from 30 years experience, is a 40 pages that's well worth reading. –  Peter Morgan Apr 9 '11 at 22:32
1  
@sb1 The error in measurement in the SP case cannot be made arbitrarily small because the concept of measuring the amplitude of the signal at a given frequency requires measurement of the signal at all times, so that a perfect Fourier transform of the signal can be constructed. If we measure the signal for only a finite time, we can effectively only compute the Fourier components of the signal we want in convolution with a window function. –  Peter Morgan Apr 9 '11 at 22:41
    
@Peter, thanks for the link. Actually this Answer is only part of a larger Answer I had developed for this question, which developed the point about the time-frequency domain (and another example) much further. But when I checked with the OP question I found that my conclusions had nothing much to do with the original question, so I truncated the answer to what the OP seemed to be asking. This Answer is now being downvoted probably because it doesnt address an ambiguous question, so I will probably delete it and not answer any more ambiguous questions of this type. –  Roy Simpson Apr 10 '11 at 16:08

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.

share|improve this answer
    
UP can be derived from the Schrödinger equation and normally introductory text books derive it. But in advanced courses one learns that Schrödinger equation can be derived from the basic axioms of quantum theory. These axioms are held to be the most fundamental postulates about nature. These postulates directly leads us to the general uncertainty relationship. The catch here, imho, is UP encompasses the gist of the theory. In order to be a quantum theory all a theory needs is to be consistent with the UP. It is truly a fundamental principle of the QT in this sense. –  user1355 Apr 10 '11 at 15:10
    
-1 is not mine. –  user1355 Apr 10 '11 at 15:14
1  
@sb1 Downvoting was all too likely for my Answer. Downvotes are meaningless unless someone is wise enough to be able to say why, at least for other readers, if not for the Answerer. Your idea that HUP is truly a principle, and enough to make a theory a quantum theory, seems to me quite radical. I think I don't see that in quantum logic or axiomatic approaches? It's often done from the CCRs, which give CHSH, etc. Is there a proof that a theory that satisfies HUP (and what other conditions?) must violate the Bell inequalities? Otherwise, what you're proposing seems rather different from QM. –  Peter Morgan Apr 10 '11 at 17:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.