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While investigating the EPR Paradox, it seems like only two options are given, when there could be a third that is not mentioned - Heisenberg's Uncertainty Principle being given up.

The setup is this (in the wikipedia article): given two entangled particles, separated by a large distance, if one is measured then some additional information is known about the other; the example is that Alice measures the z-axis and Bob measures the x-axis position, but to preserve the uncertainty principle it's thought that either information is transmitted instantaneously (faster than light, violating the special theory of relativity) or information is pre-determined in hidden variables, which looks to be not the case.

What I'm wondering is why the HUP is not questioned? Why don't we investigate whether a situation like this does indeed violate it, instead of no mention of its possibility? Has the HUP been verified experimentally to the point where it is foolish to question it (like gravity, perhaps)?

Edit

It seems that all the answers are not addressing my question, but addressing waveforms/commutative relations/fourier transforms. I am not arguing against commutative relations or fourier transforms. Is not QM the theory that particles can be represented as these fourier transforms/commutative relations? What I'm asking this: is it conceivable that QM is wrong about this in certain instances, for example a zero state energy, or at absolute zero, or in some area of the universe or under certain conditions we haven't explored? As in:

Is the claim then that if momentum and position of a particle were ever to be known somehow under any circumstance, Quantum Mechanics would have to be completely tossed out? Or could we say QM doesn't represent particles at {absolute zero or some other bizarre condition} the same way we say Newtonian Physics is pretty close but doesn't represent objects moving at a decent fraction of the speed of light?

Example

EPR Paradox: "It considered two entangled particles, referred to as A and B, and pointed out that measuring a quantity of a particle A will cause the conjugated quantity of particle B to become undetermined, even if there was no contact, no classical disturbance."

"According to EPR there were two possible explanations. Either there was some interaction between the particles, even though they were separated, or the information about the outcome of all possible measurements was already present in both particles."

These are from the wikipedia article on the EPR Paradox. This seems to me to be a false dichotomy; the third option being: we could measure the momentum of one entangled particle, the position of the other simultaneously, and just know both momentum and position and beat the HUP. However, this is just 'not an option,' apparently.

Clarification

I'm not disputing that two quantities that are fourier transforms of each other are commutative / both can be known simultaneously, as a mathematical construct. Nor am I arguing that the HUP is indeed false. I'm looking for justification not just that subatomic particles can be models at waveforms under certain conditions (Earth like ones, notably), but that a waveform is the only thing that can possibly represent them, and any other representation is wrong. You van verify the positive all day long, that still doesn't disprove the negative. It is POSSIBLE that waveforms do not correctly model particles in all cases at all times. This wouldn't automatically mean all of QM is false, either - just that QM isn't the best model under certain conditions. Why is this not discussed?

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I +1d to get rid of the downvote you had. It's the last line that did it for me. –  Olly Price Aug 13 '12 at 22:37
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Anyone who is downvoting care to elaborate on where my question is unclear, unuseful or shows no effort? I'd be glad to improve it if I can. –  Ehryk Aug 13 '12 at 23:40
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Try Bohmian mechanics. –  MBN Sep 6 '12 at 11:02
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@Ehryk: Not my downvote, but this question is a waste of time. You misunderstood what EPR is all about. The EPR effects have nothing to do with HUP, and you can show that they are inconsistent with local variables determining experimental outcomes without doing quantum mechanics, just from the experimental outcomes themselves. This means the weirdness is not due to the formalism, but really there in nature. –  Ron Maimon Sep 11 '12 at 6:15
    
So in a universe without the commutative relation/HUP, where the commutative relation was sometimes zero / position and momentum could both be known, where's the paradox with EPR? You could just determine the values of both entangled particles, no paradox necessary. –  Ehryk Sep 11 '12 at 7:24
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12 Answers 12

up vote 5 down vote accepted
+50

In precise terms, the Heisenberg uncertainty relation states that the product of the expected uncertainties in position and in momentum of the same object is bounded away from zero.

Your entanglement example at the end of your edit does not fit this, as you measure only once, hence have no means to evaluate expectations. You may claim to know something but you have no way to check it. In other entanglement experiments, you can compare statistics on both sides, and see that they conform to the predictions of QM. In your case, there is nothing to compare, so the alleged knowledge is void.

The reason why the Heisenberg uncertainty relation is undoubted is that it is a simple algebraic consequence of the formalism of quantum mechanics and the fundamental relation $[x,p]=i\hbar$ that stood at the beginning of an immensely successful development. Its invalidity would therefore imply the invalidity of most of current physics.

Bell inequalities are also a simple algebraic consequence of the formalism of quantum mechanics but already in a more complex set-up. They were tested experimentally mainly because they shed light on the problem of hidden variables, not because they are believed to be violated.

The Heisenberg uncertainty relation is mainly checked for consistency using Gedanken experiments, which show that it is very difficult to come up with a feasible way of defeating it. In the past, there have been numerous Gedanken experiments along various lines, including intuitive and less intuitive settings, and none could even come close to establishing a potential violation of the HUP.

Edit: One reaches experimental limitations long before the HUP requires it. Nobody has found a Gedankenexperiment for how to do defeat the HUP, even in principle. We don't know of any mechanism to stop an electron, thereby bringing it to rest. It is not enough to pretend such a mechanism exists; one must show a way how to achieve it in principle. For example, electron traps only confine an electron to a small region a few atoms wide, where it will roam with a large and unpredictable momentum, due to the confinement.
Thus until QM is proven false, the HUP is considered true. Any invalidation of the foundations of QM (and this includes the HUP) would shake the world of physicists, and nobody expects it to happen.

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Why wouldn't it just invalidate it under certain conditions? For example: by some means, we completely arrest an electron. Position = center of device, momentum = 0. Both known simultaneously. Couldn't we just say QM is 'not a valid model for arrested particles but works for moving ones' without invalidating most of current physics? –  Ehryk Sep 7 '12 at 21:43
    
Any invalidation of the foundations of QM would shake the world of physicists. - But the center of a device is usually poorly definable, and an electron cannot be arrested completely, neither in position nor in momentum. One reaches experimental limitations long before the HUP requires it. - In the past, there have been numerous Gedanken experiments along similar and many other lines, and none could even come close to establishing a violation of the HUP. –  Arnold Neumaier Sep 9 '12 at 13:12
    
In practice, I get this. I'm not claiming that we can make such a machine now, or soon. But the HUP seems to say such a machine cannot exist, now or ever, with more advanced races or technologies or anything. Are you saying an electron cannot be arrested completely - anywhere, ever, inside a black hole, at absolute zero - under no conditions ever? –  Ehryk Sep 9 '12 at 19:40
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until QM is proven false, the HUP is true. –  Arnold Neumaier Sep 10 '12 at 11:53
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@Ehryk: here's why you're seeming nonsensical to everyone here: at small length scales, an electron looks very, very much like a wave. You get interference patterns and everything. Now, you want to 'stop' it. Well, a "slower" electron has a longer wavelength than a "faster" one, but this longer wavelength is going to spread it out farther. By the time you get to your limit of a 'stopped' electron, the electron will be spread out over all of space. –  Jerry Schirmer Sep 14 '12 at 23:13
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In quantum mechanics, two observables that cannot be simultaneously determined are said to be non-commuting. This means that if you write down the commutation relation for them, it turns out to be non-zero. A commutation relation for any two operators $A$ and $B$ is just the following $$[A, B] = AB - BA$$ If they commute, it's equal to zero. For position and momentum, it is easy to calculate the commutation relation for the position and momentum operators. It turns out to be $$[\hat x ,\hat p] = \hat x \hat p - \hat p \hat x = i \hbar$$ As mentioned, it will always be some non-zero number for non-commuting observables. So, what does that mean physically? It means that no state can exist that has both a perfectly defined momentum and a perfectly defined position (since $ |\psi \rangle$ would be both a right eigenstate of momentum and of position, so the commutator would become zero. And we see that it isn't.).

So, if the uncertainty principle was false, so would the commutation relations. And therefore the rest of quantum mechanics. Considering the mountains of evidence for quantum mechanics, this isn't a possibility.

Addition

I think I should clarify the difference between the HUP and the classical observer effect. In classical physics, you also can't determine the position and momentum of a particle. Firstly, knowing the position to perfect accuracy would require you to use a light of infinite frequency (I said wavelength in my comment, that's a mistake), which is impossible. See Heisenberg's microscope. Also, determining the position of a particle to better accuracy requires you use higher frequencies, which means higher energy photons. These will disturb the velocity of the particle. So, knowing the position better means knowing the momentum less.

The uncertainty principle is different than this. Not only does it say you can't determine both, but that the particle literally doesn't have a well defined momentum to be measured if you know the position to a high accuracy. This is a part of the more general fact in quantum mechanics that it is meaningless to speak of the physical properties of a particle before you take measurements on them. So, the EPR paradox is as follows - if the particles don't have well-defined properties (such as spin in the case of EPR), then observing them will 'collapse' the wavefunction to a more precise value. Since the two particles are entangled, this would seem to transfer information FTL, violating special relativity. However, it certainly doesn't. Even if you now know the state of the other particle, you need to use slower than light transfer of information to do anything with it.

Also, Bell's theorem, and Aspect's tests based off of it, show that quantum mechanics is correct, not local realism.

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So how do we know that all particles have a non-commuting relationship, always and forever, under all conditions, even the ones we aren't able to measure or with technology or knowledge we don't yet possess? –  Ehryk Sep 6 '12 at 10:47
    
What if you define position and momentum as the two real numbers that you measure at time t from experiment? (That's what most people consider "position" and "momentum" to be anyway.) What is this "new" definition of position and momentum? –  Nick Sep 8 '12 at 23:11
    
Let me add this: I've taken QM and done those calculations for the commutation plenty of times to figure out what sets of compatible observables there are. But I could give someone a random formula for some random integral and divide by 6.3 and say "look, this always comes out to a real value -- thus position and momentum can't be simultaneously well-defined!" and that makes no sense whatsoever. Yeah, I know the whole spiel about eigenvalues and eigenstates and identical preparations of quantum systems, but what kind of physical experiment demonstrates this limit? –  Nick Sep 8 '12 at 23:15
    
Noncommutativity of operators nicely explains emission spectra, which I believe were the subject of Heisenberg's (?) initial ponderings. There's a nice bit of this history explained at page 40 of this book by Alain Connes alainconnes.org/docs/book94bigpdf.pdf (there is probably a more focused reference for this history, but I don't know of one) –  Ryan Thorngren Sep 9 '12 at 0:51
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The Heisenberg's relation is not tied to quantum mechanics. It is a relation between the width of a function and the width of its fourier transform. The only way to get rid of it is to say that x and p are not a pair of fourier transform: ie to get rid of QM.

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So if by any means at all (entanglement, future machines, or divine powers) one could measure both position and momentum simultaneously, then all of quantum mechanics is false? There could be no QM in a universe in which this is possible? –  Ehryk Aug 14 '12 at 9:35
    
You necessarily need to change the relationship between position and momemtum. It is mathematically impossible if they just form a fourier transform pair. But considering the huge amount of datas validating QM, one can try to extend QM by adding a small term in the pair or by using a fractional commutator (with fractional derivative) for instance. –  Shaktyai Aug 14 '12 at 9:43
    
How about saying x and p are a pair of fourier transforms USUALLY, but not in certain circumstances such as {inside a black hole, at absolute zero, under certain entanglement experiments, in a zero rest energy universe, etc.} How do we know that because QM is right USUALLY or from what we can observe, that it is right ALWAYS and FOREVER? –  Ehryk Sep 6 '12 at 10:43
    
That is to say: QM as we know it is not valid in these cases. There is no possible objections to such a statement, but for it to get accepted by the physicists, you need to prove that you can explain things in a simpler way and that you can predict something measurable. –  Shaktyai Sep 6 '12 at 10:46
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Because there is no proof whatsoever that QM fails. The day it fails we shall reconsider the question. However, there are many theorists working on alternative theories, so you have your chances. –  Shaktyai Sep 6 '12 at 15:09
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The wave formulation has in its seed the uncertainty relation.

Let me be precise what is meant by the wave formulation: the amplitude over space points will give information about localization on space, while amplitude over momenta will give information about localization in momentum space. But for a function, the amplitude over momenta is nothing else but the Fourier transform of the space amplitude.

The following is jut a mathematical fact, not up to physical discussion: the standard deviation, or the spread of the space amplitude, multiplied by the spread of the momenta amplitude (given by the Fourier transform of the former) will be bounded from below by one.

So, it should be pretty clear that, as long as we stick to a wave formulation for matter fields, we are bound mathematically by the uncertainty relation. No work around over that.

Why we stick to a wave formulation? because it works pretty nicely. The only way someone is going to seriously doubt that is the right description is to either:

1) find an alternate description that at least explains everything that the wave formulation describes, and hopefully some extra phenomena not predicted by wave formulation alone.

2) find an inconsistency in the wave formulation. In fact, if someone ever manages to measure both momenta and position for some electron below the Planck factor, it would be definitely an inconsistency in the wave formulation. It would mean we would have to tweak the De Broglie ansatz or something equally fundamental about it. Needless to say, nothing like that has happened

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It's a mathematical fact IF the particle can indeed be wholly represented by that specific function, right? So in the entanglement experiment, perhaps that function does not represent the state of TWO entangled particles? Maybe we have entanglement wrong, or maybe that function does not represent particles in certain conditions? Why are these possibilities not even discussed? –  Ehryk Sep 6 '12 at 18:05
    
@Ehryk, because scientists, as all humans, tend to do the least amount of effort that will get the job done, it really does not make economical sense to do otherwise. As i said, there would be something to discuss if something in the experiment would not turn out as expected, but it does. If you want to do your life's mission to prove false the wave representation, then you need to build an experiment that will either confirm it or disprove it. then, people will likely start seriously discussing other possibilities. –  lurscher Sep 6 '12 at 18:13
    
We can't prove Zeus doesn't exist, yet we don't accept his existence because of this. An idea shouldn't have to be 'debunked' to have a healthy amount of doubt in it, yet the wave formulation representing all particles, everywhere, at all times and locations seems to be presented 'beyond doubt' - so why is it stated with such certainty about unknowability and when challenged, the opposition gives in without so much as a mention? –  Ehryk Sep 6 '12 at 18:26
    
(I'm not trying to prove it wrong, or stating that it is, I'm asking if it can be false and if so, why it's not treated as such) –  Ehryk Sep 6 '12 at 18:28
    
@Ehryk, suppose someone starts asking why physicists assume that we only have one time dimension, and why we don't try to debunk that. We would reply the same thing; we have no reason to devote resources to debunk something that seems to fit so nicely with existing phenomena, so the ball is in the court of the person that insist that, say, two-dimensional time makes great deal of sense for X or Y experiment. Then, if the experiment sounds like something that has not been tested, and is under budget to implement, maybe some experimentalists will try to do it. That is how science works –  lurscher Sep 6 '12 at 18:30
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If we want the position and the momentum to be well-defined at each moment of time, the particle has to be classical. We inherited these notions from classical mechanics, where they apply successfully. Also they apply at macroscopic level. So, it is a natural question to ask if we can keep their good behavior in QM. Frankly, there is nothing to stop us to do this. We can conceive a world in which the particles are pointlike all the time, and move along definite trajectories, and this will "beat HUP". This was the first solution to be looked for. Einstein and de Broglie tried it, and not only them. Even Bohr, in his model, envisioned electrons as moving along definite trajectories in the atom (before QM). David Bohm was able to develop a model which has at a sub-quantum level this property, and in the meantime behaves like QM at quantum level. The price to be paid is to allow interactions which "beat the speed of light", and to adjust the model whenever something incompatible with QM was found. IMHO, this process of adjustments still continues today, and this looks very much like adding epicycles in the heliocentric model. But I don't want to be unfair with Bohm and others: it is possible to emulate QM like this, and if we learn QM facts which contradict it, it will always be possible to find such a model which behaves like QM, but also has a subquantum level which consists of classical-like point particles with definite positions and momenta. At this time, these examples prove that what you want is possible. One may argue that they are unaesthetic, because they are indeed more complicated than QM. But this doesn't mean that they are not true. Also, at this time they don't offer anything testable which QM can't offer. So, while QM describes what we observe, the additional features of hidden variable theories are not observable, more complicated, and violate special relativity. Or, if they don't violate special relativity, they contradict what QM predicts and we observed in experiments of entanglement like that of Alan Aspect. If EPR presents us with two alternatives, (1) spooky action at distance, (2) QM is incomplete, and that you propose, (3) HUP is false, let's not forget that Aspect's experiment and many others confirmed the alternative (1).

Now, it would be much better for such models if they would stop adjusting themselves to mimic QM, and predict something new, like a violation of HUP. This would really be something.

In conclusion, yes, you are right and in principle it is possible to beat HUP. The reason why most physicists don't care too much about this, is that the known ways to beat HUP are ugly, have hidden elements, violate other principles. But others consider them beautiful and useful, and if you are interested, start with Bohm's theory and the more recent developments of this.


Update

Synopsis: The Certainty of Uncertainty

Violation of Heisenberg’s Measurement-Disturbance Relationship by Weak Measurements (arXiv link)

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This was rather helpful, so I appreciate it. I'm still just having difficulty wrestling with unknowability in relation to this; for example if we ever found a way to arrest a particle completely; we'd know it's position and momentum (0) both at the same time, and while it violated HUP, it could just be said 'this particle cannot be represented by a wavefunction.' The reach of the HUP seems to include this though, with no provisions, and just be accepted so OBVIOUSLY you can't stop a particle. Would we just say the particle is classical in that instance? –  Ehryk Sep 6 '12 at 18:20
    
@Cristi I see (and generally have no objections to) your argument, but that conclusion seems misleading. Yes, it's possible to beat HUP (by discarding quantum mechanics) in the same sort of sense that it's possible to create a macroscopic stable wormhole: not strictly ruled out, but there is no evidence to support it. So I think it's misleading to be saying that this is possible. –  David Z Sep 6 '12 at 18:45
    
@David Zaslavsky: Thanks. To make clear my conclusion, and less misleading, I wrote the first, rather lengthy, paragraph. This contains for instance the statement "while QM describes what we observe, the additional features of hidden variable theories are not observable, more complicated, and violate special relativity." Anyway, I considered it would be more misleading to claim that one knows HUP can't be violated no matter what. –  Cristi Stoica Sep 6 '12 at 19:31
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@Ehryk: "What happened to particle-wave duality?". Particles are represented as wavefunctions. They are defined on the full space, but may have a small support (bump functions). At limit, when concentrated at a point, bump becomes Dirac's $\delta$ function. Then, it has definite position $x$, but indefinite wave vector, so it spreads immediately (this corresponds to HUP). Its "dual" is a pure wave (with definite wave vector $k_x$, hence momentum $p_x$). The "particle-wave" duality refers to these two extreme cases. But most of the times the "wavicles" are somewhere between these two extremes. –  Cristi Stoica Sep 6 '12 at 20:41
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@Ehryk: "How does a wave have mass?" They have momentum and energy: multiply wave 4-vector with $\hbar$ and obtain $4$-momentum, so yes, they have mass. Interesting thing: the rest mass $m_0$ is the same, even though in general the wave 4-vector is undetermined. By "undetermined" you can understand that the wavefunction is a superposition of a large (usually infinite) number of pure wavefunctions. Pure wavefunctions have definite wave vector (hence momentum), but totally undetermined position. –  Cristi Stoica Sep 6 '12 at 20:50
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You are asking if a more complete theory might show that HUP is wrong and that position and momentum do exist simultaneously. But a more complete theory has to explain all the observations that QM already explains, and those observations already show that position and momentum cannot have definite values simultaneously. This is known because when particles such as photon, electrons, or even molecules are sent through a pair of slits one at a time, an interference pattern on the detector plate appears that shows that the probability of the measured location and time follows a specific mathematical relationship. The fact that certain regions have zero probability shows that before measurement, the particles exist in a superposition of possible states, such that the wave function for those states can cancel out with other states resulting in areas of low probability of observation. The observed relationships through increasingly complex experiments rules out possibilities other than what is described by QM. The only way that QM could be superseded by a new theory is for new observations to be made that violate QM, but the new theory would still result in the same predictions as QM in the circumstances that QM has already been tested. Since HUP results directly from QM, HUP would also follow from a new theory with the only possible exception in conditions such as super high energy conditions such when a single particle is nearly a black hole.

Basically you have to get used to the idea that particles are really quantized fluctuations in a field and that the field exists in a superposition of states. Any better theory will simply provide additional details about why the field behaves in that way.

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"Accept it as true until it's debunked" is not scientific. "When a particle can be perfectly represented by a waveform and ONLY a waveform, then it cannot have definite momentum and position" is acceptable. Asserting the "When" is "Always and Forever" is not. –  Ehryk Sep 15 '12 at 0:05
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if can help

Open timelike curves violate Heisenberg's uncertainty principle

http://arxiv.org/pdf/1206.5485v1.pdf

...and show that the Heisenberg uncertainty principle between canonical variables, such as position and momentum, can be violated in the presence of interaction-free CTCs....

Foundations of Physics, March 2012, Volume 42, Issue 3, pp 341-361

http://arxiv.org/pdf/1008.0433v2.pdf

...considering that a D-CTC-assisted quantum computer can violate both the uncertainty principle...

Phys Rev Lett, 102(21):210402, May 2009.

arxiv 0811.1209

...show how a party with access to CTCs, or a "CTC-assisted" party, can perfectly distin- guish among a set of non-orthogonal quantum states....

Phys. Rev. A 82, 062330 2010.

arxiv 1003.1987v2

...and can be interacted with in the way described by this simple model, our results confirm those of Brun et al that non-orthogonal states can be discriminated...

...Our work supports the conclusions of Brun et al that an observer can use interactions with a CTC to allow them to discriminate unknown, non-orthogonal quantum states – in contradiction of the uncertainty principle...

.

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The only way to make Heisenberg's principle irrelevant is to measure the speed and the position (to make it simple) of a fundamental particle.

In other words, you would have to observe a particle, without having it collide with a photon or reacting to a magnetic force, or without interacting with it.

There might be an other way, which would be to find a very general law (but not statistical) which describes the characteristics (spin, speed, position etc) of an elemental particle in an absolute way....

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I think that's just the observer effect, described in another answer, and I can beat that by hypothesyzing a future race that has developed a gravitational particle-position-and-momentum sensor machine, which does not use photons or interact with the particle in any way that would change the position or momentum (a read only sensor). Even in this case, the HUP says they CANNOT be known simultaneously. –  Ehryk Sep 6 '12 at 10:56
    
I want to know what evidence there is to support this, even in the case of such a hypothetical machine. –  Ehryk Sep 6 '12 at 10:57
    
In this case you interact using the gravitationnal interaction, so that's almost the same. –  Yves Sep 6 '12 at 11:21
    
Not really. Bombarding it with photons are distinct events; surrounding it by a machine that is sensitive to the gravitation inside of it would only exert the same gravity that any other matter around it would, and if done as stated in my hypothetical, would not alter the position or momentum in any way once the particle has settled inside the machine. –  Ehryk Sep 6 '12 at 11:24
    
Very interesting, and it would be possible if such a machine existed (my first point). But how would you measure something else than a change in the surronding gravitationnal field (which would imply an interaction with the particule) and how would you measure a spin ? It sounds like your method is equivalent to trying to measure an absolute quantity of energy, or to "forcing" the position or momentum of your particle, a case which doesn't fall under Heisenberg's principle. This reasoning might end up as a Ouroboros.. –  Yves Sep 6 '12 at 11:33
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"Heisenberg uncertainty principle" is a school term that is used in popular literature. It simply does not matter. What matters is the wavefunction and Schroedinger equation.

The EPR paradox experiment never used any explicit "uncertainty principle" in the proof.

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As @MarkM pointed out above, what I meant but wasn't able to espouse was a 'non-commutation' property (a term I've not heard of in this context), or the claim that the exact position and momentum of a particle cannot be known simultaneously. I thought this was semantically equivalent to the Heisenberg Uncertainty Principle, which I guess it is not. –  Ehryk Aug 13 '12 at 23:30
    
Also, from wikipedia: "The uncertainty principle is a fundamental concept in quantum physics." (from the disambiguation page, main article here: en.wikipedia.org/wiki/Uncertainty_principle ). Could you explain or give sources for it 'not matter'ing? Further, the wiki article on the EPR Paradox explicitly uses the Heisenberg Uncertainty Principle - I'm not claiming WP is any authority, but it would be the source of my confusion. –  Ehryk Aug 13 '12 at 23:38
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@Annix This isn't true. Firstly, Heisenberg's matrix mechanics is an equally valid formulation of QM as wave mechanics, see Zettlli page 3. Second, the uncertainty principle is a part of wave mechanics. As you say, you can easily derive it from the Schrodinger equation. I find it odd that you say that this somehow makes the uncertainty principle irrelevant. You can't simultaneously know position and momentum to perfect accuracy, since localizing the position of the particle involves adding plane waves, which then makes the momentum uncertain. –  Mark M Aug 13 '12 at 23:58
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@Anixx If you claim that you may derive the HUP from the Schrödinger equation, you should show it. I actually think it is not possible, but I'm curious. One usually derive the HUP from the commutation relations and later one shows it is preserved by the unitary evolution. The Schr. equation tells us how the states evolve in time, while the HUP must be verified even in the initial state so I'm very skeptical about your derivation. In any case, the HUP is at least as fundamental as the Schr. equation and it is a term very often used in technical papers and seminars. –  drake Aug 14 '12 at 0:28
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@drake You can't derive it from the SE, but from the wave mechanics formulation (which is what I guess Annix means). See the'Proof of Kennard Inequality using Wave Mechanics' sub-section here: en.wikipedia.org/wiki/… However, I agree with you that the HUP is fundamental (see my above post.). –  Mark M Aug 14 '12 at 1:32
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Without gravity: The uncertainty principle is not really a principle because it is a derivable statement, it is not postulated. It is derivable and proven mathematically. Once you prove something you cannot unprove it. That means it cannot turn out to be false. For experimental verifications, see for example this article by Zeilinger et al and the references inside. Zeilinger is a world expert on quantum phenomena and it is expected that he will get Nobel prize in the future.

With gravity, (and that matters only at extremely high energy, as high as the Planck scale): Intuitively you can use the uncertainty principle to give an estimate about the energy needed to resolve tiny region of space. For sufficiently small region in space you will create a black hole. So there is a limit on the spacial resolution one can achieve, because of gravity. If you try to use higher energy you will create a bigger black hole. Bottom line is, uncertainty principle does not make sense in this case because space loses its meaning and it cannot be defined operationally.

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Things can be unproven if one of the axioms or postulates they are based on is proven false. HUP may be true if <x, y and z> are true, but it certainly is based on foundations (waveforms representing matter, for one) that are not infallible. –  Ehryk Aug 14 '12 at 11:37
    
@Ehryk You cannot unprove something by changing the postulates, because then you are talking about totally different problem. You can compare only 2 situations giving the same postulates/axioms. The axioms are true and not false in the sense that the coherent structure coming out of those postulates leads to predictions that are consistent with experimental observations. The world is quantum mechanical. –  Revo Aug 14 '12 at 16:03
    
You cannot unprove it as a model of how things could work, no, but you could show that it is just not the most accurate model of the world we live it - just like we can theorize about hyperbolic geometry as a model, though it's unlikely to be the model of reality. Is it the case that you could not have a variant of something like QM that produces similar results while in some instances allowing precise position and momentum values, in the same way newton's laws were 'good enough' for the values we had measured at non relativistic speeds up until that point? –  Ehryk Aug 15 '12 at 1:43
    
@Ehryk No. You could not have had something similar to Newtonian meachanics that underlies Quantum Mechanics. What you are thinking of has been thought of for long time ago, it is unknown as hidden variables theories. It has been proven experimentally that something like Newtonian mechanics or any deterministic theory cannot be the basis of Quantum Mechanics. May be you should also keep in mind the following main point: QM is more general than CM, hence it is more fundamental. Since QM is more general than CM, one should understand how CM emerges from QM, not the other way around. –  Revo Aug 15 '12 at 1:50
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@Ehryk One should understand CM in terms of QM not QM in terms of CM. –  Revo Aug 15 '12 at 1:52
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The way I see it, HUP cannot be disproven "at absolute zero", because absolute zero cannot be physically reached, er... due to HUP... is circular reasoning good enough? Let's try something else.

Maybe try to imagine what would happen if HUP was to be violated? For one, I guess the proton - electron charges would cause one or two electrons to fall down into the nucleus, as HUP normally prevents that (if the electron fell down on nucleus we'd know it's position with great precision, requiring it to have indeterminate but large momentum, so it kind of settles for orbiting around nucleus).

If you know more about the stuff than I do, try to imagine what else would happen, and how likely is that effect. For example, if HUP violation would imply violation of 2nd law of thermodynamics, this would render HUP violation pretty unlikely.

That much from a layman.

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But then why can't we just say 'HUP is only for particles not at absolute zero'? It seems like violating it is 'not an option', even as above - so an electron falls into the nucleus. It has a measurable position and momentum. Why does HUP have to hold so strongly that we instead are comfortable with 'that particle must always have energy'? –  Ehryk Sep 6 '12 at 18:31
    
The way I see it "absolute zero" is purely theoretical concept. Look up Bose-Einstein condensate, get a feeling for what happens at extremely low temperatures and then try to project that further to zero. Doesn't click. So saying "HUP is only for particles at absolute zero" is like saying "HUP is for all particles", for absolute zero can't be reached. –  pafau k. Sep 6 '12 at 18:54
    
Do you have evidence or citations that nothing can be absolute zero? Or are you just asserting it? Note that saying 'we can't get to absolute zero' is different than 'no particle anywhere, at any time, can be at absolute zero.' –  Ehryk Sep 6 '12 at 19:10
    
Let me quote the beginning of Wikipedia entry on absolute zero :) "Absolute zero is the theoretical temperature at which entropy reaches its minimum value", note the word theoretical. Temperature always flows from + to -, so the simple explanation is: you'd have to have something below absolute zero to cool something else to absolute zero. (this would violate laws of thermodynamics). –  pafau k. Sep 6 '12 at 19:59
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Transfer heat from hot to hotter? Decrease the volume of the container. Cool matter? Increase the volume of the container. In both cases, heat is not 'transferred', but temperature (average kinetic energy) has been changed without the interaction of other matter, either hotter or colder. –  Ehryk Sep 10 '12 at 11:18
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The Heisenberg uncertainty principle forms one of the most important pillars in physics. It can't be proven wrong because too many experimentally determined phenomena are a result of the uncertainty principle. However, something may be discovered in the future that can make a modification to the uncertainty principle - in a similar way that Newton's laws were modified by Einstein's special relativity. Saying that the uncertainty principle is wrong is like saying that Newton's law is wrong.

In reply to the comments,

I'm not saying that it can be falsified. It can't. In a classical sense, it will always be correct, in a similar way that Newton's law will always be correct.However, it can be modified. Until the day that all the open questions in physics have been resolved, how can you claim that the uncertainty principle can't be modified further? Do we know everything about extra dimensions? Do we know everything about string theory and physics at the Planck scale?

By the way, it has already been modified.

Please check this link.

http://arxiv.org/abs/1106.0715

The uncertainty principle will always be correct. However, it can and has been modified. In its current formalism and interpretation, it could represent a special case of a larger underlying theory.

The claim that the current formalism and limitations to the uncertainty principle are absolute and can never be modified under any circumstance in the universe, is a claim that does not obey the uncertainty principle itself.

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The uncertainty principle is a lot closer to uncertainty law than your answer lets on. It's not really about measurement so much as it's about a Fourier Transform. –  Brandon Enright Jan 26 at 23:47
    
The Heisenberg Uncertainty Principle is an unfalsifiable claim? All of (good) science is falsifiable. See the first paragraph: en.wikipedia.org/wiki/Falsifiability –  Ehryk Jan 28 at 6:12
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protected by Qmechanic Jan 26 at 23:37

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