I Uncertainty Principle.... Intent Behind It?

[Madi Araly]

Sure, I understand that POV and there's nothing wrong with it but you want to be careful about going to the point where basically you are assuming that all the thousands of very smart physicists who have worked on this stuff are idiots who got it all wrong.

As I said in a previous comment, that is clearly not my intent. I look up to these physicists more than anyone else, but I refuse to put clouded or blind faith in anything. After all, I wouldn't want to repeat Ancient Greece.

 

[anorlunda]

As I said in a previous comment, that is clearly not my intent. I look up to these physicists more than anyone else, but I refuse to put clouded or blind faith in anything. After all, I wouldn't want to repeat Ancient Greece.

Well then, don't be blind by choice. You should study QM yourself. May I recommend Leonard Susskind's video course on quantum mechanics? He is an excellent teacher. Here is a link to the first lecture.

 

[Madi Araly]

Well then, don't be blind by choice. You should study QM yourself. May I recommend Leonard Susskind's video course on quantum mechanics? He is an excellent teacher. Here is a link to the first lecture.

I recognize him, perhaps I've seen one of his videos before. I'll watch his lectures, thanks for the link!

 

[malawi_glenn]

The uncertainty property follows from Cauchy–Schwarz inequality

As some people have pointed out, the Heisenberg Uncertainty is a statistical relation, one can not thus ask what happens on an event-by-event basis

The standard deviation concerned in the Heisenberg uncertainty relation is INTRINSIC to the system/object itself, no matter how good detectors and devices we will invent - there will always be a non-zero standard deviation in our measurements

 

[Traruh Synred]

In a more fundamental you can think about this w/o explicit reference to commutation rules. A 'particle' is an quantized excitation of a field. It is more wave than particle. The particle like-behavior arises when the excitation is in the form of a wave-packet which is localized, but to get a localized excitation/wave you need to superimpose many wave lengths (inversely related to momenta via Planck's constant). The more localized the packet/'particle' the more different momenta (wavelengths) it needs to be constructed from, hence, the uncertainty principle. The inverse is also true. If you want to get an excitation that has a small range of momentum it need to will have to occupy a lot of space. You might look at wave packet on Wikipedia to see how it works.

-Traruh

 

[doctorshankar]

I've been pre-occupied with Heisenberg's uncertainty principle for around four years now, and I've come to fabricate a lot of questions.

The most pressing one, however, is as follows:
To me, the uncertainty principle seems to reference our (relatively) poorly controlled methods to measure a particle's momentum and position rather than being some special quantum phenomenon. Is this how it was intended?

If I measure a coffee mug's position using a crowbar, I change the coffee mug's momentum by measuring it. I do not, however, change its momentum simply by having knowledge of the coffee mug's position. This is how I think of the uncertainty principle, but was it meant this way? If it was, then doesn't that screw up multiple other concepts such as entanglement and electron configuration around nuclei?

Or did Heisenberg believe in some phenomenon that changed one of the particle's traits merely because we observed it?

Heisenberg's uncertainty principle is for subatomic particles/micro level. tea mug may be interpreted by Newtonian Physics. my understanding of the principle is that we can never be sure about anything at all. or for that matter truth can never be known to the human mind. this principle may have philosophical implications

 

[bhobba]

my understanding of the principle is that we can never be sure about anything at all. or for that matter truth can never be known to the human mind. this principle may have philosophical implications

That's reading more into it than what it says. Many do it, but IMHO it not particularly constructive.

Thanks
Bill

 

[GenesisAria]

This was an interesting video that gives a good view on the uncertainty issue along with other issues in quantum physics. Don't pay attention to the title, it doesn't reflect on the presentation itself. But it demonstrates a perspective you can possibly take on quantum physics that works around it's inherent issues, including uncertainty, the duality, entanglement and so on.


Funny enough this viewpoint actually complies with old electrical engineering descriptions of the universe.

 

[Markus Hanke]

Why is it that humans are so certain of this, though?

You see, you are thinking of position and momentum as two completely separated, isolated entities. But the thing is, they are not - you can describe any given system in terms of position, or in terms of momentum, and you will find that these two descriptions are related via an operation called a Fourier transform. Without going into the mathematical details, the consequence of this relationship is that, if you decrease the uncertainty in position ( i.e. you measure position more precisely ), you will at the same time increase the uncertainty in momentum, and vice versa. You focus on position, the momentum gets smeared out; you focus on momentum, the position gets smeared out. Therefore, there is a lower limit as to how accurately you can determine both simultaneously - that's just precisely the HUP. This is a fundamental fact of nature, and not due to any limitation of our instruments.

 

[Quandry]

You can know somethings position by a single measurement. To know its momentum you require two measurements. You can never know what happened between the two measurements.

 

[weezy]

Absolutely not. The POINT of the HUP is that it is not at all a measurement problem but rather a fundamental fact of nature.

Wave-particle duality?

 

[weezy]

You see, you are thinking of position and momentum as two completely separated, isolated entities. But the thing is, they are not - you can describe any given system in terms of position, or in terms of momentum, and you will find that these two descriptions are related via an operation called a Fourier transform. Without going into the mathematical details, the consequence of this relationship is that, if you decrease the uncertainty in position ( i.e. you measure position more precisely ), you will at the same time increase the uncertainty in momentum, and vice versa. You focus on position, the momentum gets smeared out; you focus on momentum, the position gets smeared out. Therefore, there is a lower limit as to how accurately you can determine both simultaneously - that's just precisely the HUP. This is a fundamental fact of nature, and not due to any limitation of our instruments.

Is wave-particle duality the reason for this?

 

[phinds]

Wave-particle duality?

"Wave particle duality" is has been a deprecated concept for something like 100 years now. Yeah, I know you hear about it in pop-sci but it's a waste of time. Quantum objects are not particles nor are they waves. They are quantum objects and there is no "duality" involved, just the fact that they will exhibit wave characteristics if you specifically measure for that and particle characteristics if you measure for that.

 

[weezy]

"Wave particle duality" is has been a deprecated concept for something like 100 years now. Yeah, I know you hear about it in pop-sci but it's a waste of time. Quantum objects are not particles nor are they waves. They are quantum objects and there is no "duality" involved, just the fact that they will exhibit wave characteristics if you specifically measure for that and particle characteristics if you measure for that.

From what I've learned so far the reason for existence of HUP is because a "wave packet" fulfills the criteria for not having a well defined position/momentum. Yes this may not be real answer but I find it hard to imagine the physical reality of HUP otherwise.

 

[bhobba]

Wave-particle duality?

Is wave-particle duality the reason for this?

The wave particle duality is one of those ideas introduced in beginning texts and popularizations but was really overthrown when Dirac came up with the transformation theory at the end of 1926:
http://www.lajpe.org/may08/09_Carlos_Madrid.pdf

The reason for the HUP is simply the math of non-commuting operators.

Its sometimes counterproductive looking for explanations other than the math - this is one case IMHO.

Thanks
Bill

 

[weezy]

The wave particle duality is one of those ideas introduced in beginning texts and popularizations but was really overthrown when Dirac came up with the transformation theory at the end of 1926:
http://www.lajpe.org/may08/09_Carlos_Madrid.pdf

The reason for the HUP is simply the math of non-commuting operators.

Its sometimes counterproductive looking for explanations other than the math - this is one case IMHO.

Thanks
Bill

Then why do most introductory textbooks explain HUP using the wave-packet picture? I am referring to Arthur Beiser's modern physics where he shows how the uncertainty relations come about from wave-particle picture of quantum objects and how Fourier transforms relate position and momentum. I am also familiar with Griffiths' explanation where he uses standard deviations to derive and commutator relations to arrive at HUP. But gives no explanation why the operators don't commute.

 

[Nugatory]

Then why do most introductory textbooks explain HUP using the wave-packet picture? I am referring to Arthur Beiser's modern physics where he shows how the uncertainty relations come about from wave-particle picture of quantum objects and how Fourier transforms relate position and momentum.

(You said "particle" once and "packet" once. Was that intended?)

Intro quantum mechanics almost always starts with Schrodinger's equation in the position basis, because the math required (elementary differential equations, a microskosh of complex analysis) is accessible to a second-year undergrad. The Fourier transform relationship between position and momentum then gets you to position-momentum uncertainty pretty directly. The upsides of this approach are that it's a good start for building an intuition about how QM behaves; you can solve some real and interesting problems that way (the hydrogen atom, for example); and the computational techniques are very important. The downside is that if you don't go further, you may be misled into thinking that this one particular case is the whole story; and if you do go further you have to go back over the same ground using a more sophisticated text, something like Ballentine.

I am also familiar with Griffiths' explanation where he uses standard deviations to derive and commutator relations to arrive at HUP. But gives no explanation why the operators don't commute.

For the specific case of position and momentum, it's easy to calculate the commutator directly, for example here: http://quantummechanics.ucsd.edu/ph130a/130_notes/node109.html

 

[bhobba]

Then why do most introductory textbooks explain HUP using the wave-packet picture?

You must start somewhere even if its not quite correct.

Unfortunately you see this a bit in physics, especially QM, and its really bad in QFT. You need to unlearn things as you go along. Feynman, for example, worried about it but saw no other way:
http://arxiv.org/abs/quant-ph/0609163

You need to see a 'correct' explanation of the double slit that does not depend on the wave-particle duality:
https://arxiv.org/ftp/quant-ph/papers/0703/0703126.pdf

The above is not understandable by the beginning student, which is why its not done that way.

But to make matters worse even that isn't correct as you get even more advanced:
http://arxiv.org/pdf/1009.2408.pdf

Its insidious - unfortunately.

Thanks
Bill

 

[bhobba]

But gives no explanation why the operators don't commute.

The deep answer is symmetry.

See Chapter 3 Ballentine.

And no it can't be explained simply - the proof is deep and tricky - but the result is - well - beauty incarnate.

Thanks
Bill

 

[weezy]

(You said "particle" once and "packet" once. Was that intended?)For the specific case of position and momentum, it's easy to calculate the commutator directly, for example here: http://quantummechanics.ucsd.edu/ph130a/130_notes/node109.html

I'm aware of the calculation for commutator b/w position and momentum. What I'm seeking is the reason for this? Experimentally it's quite evident but why does it have to be fundamental?

 

[weezy]

The deep answer is symmetry.

See Chapter 3 Ballentine.

And no it can't be explained simply - the proof is deep and tricky - but the result is - well - beauty incarnate.

Thanks
Bill

I will be very grateful if you are able to provide me with a link for this.

 

[bhobba]

I will be very grateful if you are able to provide me with a link for this.

Sorry - you need to get the textbook:
https://www.amazon.com/dp/9814578584/?tag=pfamazon01-20

It derives the actual operators from which the commutation relations trivially follows.

Thanks
Bill

 

[smilodont]

If you measure any fundamental observable very precisely - say position - any non-commuting partner observable moves into a superposition of states (momentum, for example). When you later measure that, you will find that the new outcome (for momentum) is random and uncorrelated to any prior measurement of that observable.

Keep in mind that for all practical purposes, particles do not have simultaneous (well-defined) values for both position and momentum. Experiments on entangled particle pairs demonstrate this very convincingly.

You said, 'When you later measure that, you will find that the new outcome (for momentum) is random and uncorrelated to any prior measurement of that observable.'
Why is this? I get that it is not a matter of us having messed anything up due to the act of measurement. Is it a matter of inconsistent timing? I.e., you measured two things at different times and they MAY have changed between times? Not sure, so I ask.

 

[bhobba]

Why is this?

Its because of the commutation relationship between position and momentum operators:
http://physics.stackexchange.com/questions/10362/how-does-non-commutativity-lead-to-uncertainty

As I mentioned the deep reason they have the form they do is symmetry, but, although very beautiful, its an advanced topic.

Another reason, although not quite as strikingly deep and beautiful, is the connection between commutation and Poisson brackets:
http://bolvan.ph.utexas.edu/~vadim/Classes/2017s/brackets.pdf

Thanks
Bill

 

[weezy]

Its because of the commutation relationship between position and momentum operators:
http://physics.stackexchange.com/questions/10362/how-does-non-commutativity-lead-to-uncertainty

http://physics.stackexchange.com/questions/10362/how-does-non-commutativity-lead-to-uncertainty

http://physics.stackexchange.com/questions/10362/how-does-non-commutativity-lead-to-uncertainty

The first answer was brilliant. Although I have a doubt with the statement that not many operators exist that can transform a Ket to a null vector. Is it really so?

 

[bhobba]

The first answer was brilliant. Although I have a doubt with the statement that not many operators exist that can transform a Ket to a null vector. Is it really so?

Yes its so. It means linear operators with an inverse. If AX = 0 A(-1)AX = 0 ie X = 0.

The usual operators in QM are invertable because they are considered diagonalizable ie of the form A = ∑ yi |xi><xi| Let A' = ∑ 1/yi |xi><xi|. A'A = I.

If the above is goobly gook you need to study linear algebra in bra ket notation:
http://quantum.phys.cmu.edu/CQT/chaps/cqt03.pdf

Thanks
Bill

 

[weezy]

Yes its so. It means linear operators with an inverse. If AX = 0 A(-1)AX = 0 ie X = 0.

The usual operators in QM are invertable because they are considered diagonalizable ie of the form A = ∑ yi |xi><xi| Let A' = ∑ 1/yi |xi><xi|. A'A = I.

If the above is goobly gook you need to study linear algebra in bra ket notation:
http://quantum.phys.cmu.edu/CQT/chaps/cqt03.pdf

Thanks
Bill

Thank you for clearing my doubts.

 

[smilodont]

Then why do most introductory textbooks explain HUP using the wave-packet picture? I am referring to Arthur Beiser's modern physics where he shows how the uncertainty relations come about from wave-particle picture of quantum objects and how Fourier transforms relate position and momentum. I am also familiar with Griffiths' explanation where he uses standard deviations to derive and commutator relations to arrive at HUP. But gives no explanation why the operators don't commute.

You said, 'The reason for the HUP is simply the math of non-commuting operators.' By this, do you mean that the way this math works forces you to be 'uncertain'? If so, this seems like an artifact of the math, rather than a result of correct modelling. Or, to put it another way, if we modeled it some other way, it wouldn't be 'uncertain'? Correct?

 

[Nugatory]

You said, 'The reason for the HUP is simply the math of non-commuting operators.' By this, do you mean that the way this math works forces you to be 'uncertain'? If so, this seems like an artifact of the math, rather than a result of correct modelling. Or, to put it another way, if we modeled it some other way, it wouldn't be 'uncertain'? Correct?

Yes, but only helpful if that hypothetical other model also produces predictions that match the results of the countless experiments that agree with the model that we're using - and such a thing doesn't seem to be on offer anywhere. We don't use the mathematical structure of quantum mechanics because we revel in perverse and counterintuitive results, we use it because out of all the countless mathematical models out there, it's the one that best describes the universe we live in.

That doesn't mean that you have to like it, of course. Einstein went to his grave convinced that there had to be something better - and didn't live to see the wave of experimental results that have proven that any alternative model would have to retain all the features that he found most distasteful in QM.

 

[smilodont]

Yes, but only helpful if that hypothetical other model also produces predictions that match the results of the countless experiments that agree with the model that we're using - and such a thing doesn't seem to be on offer anywhere. We don't use the mathematical structure of quantum mechanics because we revel in perverse and counterintuitive results, we use it because out of all the countless mathematical models out there, it's the one that best describes the universe we live in.

That doesn't mean that you have to like it, of course. Einstein went to his grave convinced that there had to be something better - and didn't live to see the wave of experimental results that have proven that any alternative model would have to retain all the features that he found most distasteful in QM.

Then Einstein was right, if we notice the uncertainty caused by the mathematics. This is a sign of something that is an issue. No insult meant here to anyone.

But, I appreciate the answer. I have gotten vague answers before and this helps. Thanks.