Misconception of the Heisenberg Uncertainty Principle - Comments

In summary, ZapperZ submitted a new PF Insights post discussing the common misconception of the Heisenberg Uncertainty Principle. He explains that this principle is not a limitation on our ability to measure both the position and momentum of a particle in a single measurement, but rather a mathematical consequence of the commutator between two operators. HUP is best understood in the concept of phase space, where classical physics defines a point by location and speed, while quantum mechanics has a different state space and quantum state.
  • #1
ZapperZ
Staff Emeritus
Science Advisor
Education Advisor
Insights Author
32,820
4,715
ZapperZ submitted a new PF Insights post

Misconception of the Heisenberg Uncertainty Principle

hupmisconceptions1-80x80.png


Continue reading the Original PF Insights Post.
 
Last edited by a moderator:
  • Like
Likes bhobba
Physics news on Phys.org
  • #2
There is a theorem on Fourier analysis which I think make clear this thing about HUP. The position and the momentum representations are connected by a Fourier transform, so I think that highlights the theoretical nature of the HUP.
 
  • #3
Very good explanation. I had read that it was related to the wave nature of photons, but I never understood it this clearly. Thanks.
 
  • #4
"Take note that the measurement uncertainty in a single is still the same as in the classical case"

What does this mean?
 
  • #5
bubblewrap said:
"Take note that the measurement uncertainty in a single is still the same as in the classical case"

What does this mean?
"Take note that the measurement uncertainty in a single is still the same as in the classical case. If I shoot the particle one at a time, I still see a distinct, accurate “dot” on the screen to tell me that this is where the particle hits the detector. "

What does this mean? ; couldn't edit the previous one, making a reply instead;;
 
  • #6
bubblewrap said:
"Take note that the measurement uncertainty in a single is still the same as in the classical case"

What does this mean?
It means that someone is good at physics and bad at English. :smile:

I think the author means, "..for a single photon.." The "dot" refers to where the wave function collapses and triggers the detector (photo film, photoelectric sensor array, or simply the spot on a screen where it reflects from and into your eye).

As long as the slit is much wider than the size of the photon the certainty of exactly where within that slit the photon passes through will be low; and as a result the momentum certainty (speed and energy are unchanging, so this really just means direction) will be high-- it will (with high certainty) be going in a straight line as it passes through. But as the slit is narrowed we become more certain about exactly where, within this now smaller range or possibilities, the photon is travelling. So certainty of momentum (ie, direction) must become less; and they begin to veer off as they pass through. (Well, that's not strictly true. What we are seeing are the wave properties beginning to manifest. It propagates as an expanding wave without actually "choosing" a new direction. If you go by the Copenhagen Interpretation, the photon "chooses where it is, where it went" only when it finally interacts with something, such as a detector. Sort of.) So the result of narrowing the slit further results is a wider spread in the angles from which the photons emerge after passing through the slit. The band of light on the screen, quite curiously, begins to widen after the slit is narrowed past some particular amount.
 
  • Like
Likes bubblewrap
  • #7
Dear Zapped Z, i have read you article but i completely don' t agree with it.
"I have shown that there’s nothing to prevent anyone from knowing both the position and momentum of a particle in a single measurement that is limited only by our technology. ": you are wrong.
As you know, the commutator between X and P is different from zero: the meaning is that you can' t construct a base of common eigenfunctions to X and P, so you can' t measure both x and p, with an arbitrary precision that depends only by our technology: the statement about the commutator is a fact of nature, and it is about a single measurement.
The Heisenberg principle, as you say in the article, is about more than one measurement, you are right, but you have forgotten that this "principle" is not a principle, but a mathematical consequence of the value of the commutator between two operators (you can demonstrate it.)
To sum up: the statement about a single measurement (commutator) implies a statement about more measurements (Heisenberg), but the last one is not a "principle" but only a consequence of the behaviour of a single measurement, described by the value of the commutator.
 
  • #8
Andre' Quanta said:
Dear Zapped Z, i have read you article but i completely don' t agree with it.
"I have shown that there’s nothing to prevent anyone from knowing both the position and momentum of a particle in a single measurement that is limited only by our technology. ": you are wrong.
As you know, the commutator between X and P is different from zero: the meaning is that you can' t construct a base of common eigenfunctions to X and P, so you can' t measure both x and p, with an arbitrary precision that depends only by our technology: the statement about the commutator is a fact of nature, and it is about a single measurement.
The Heisenberg principle, as you say in the article, is about more than one measurement, you are right, but you have forgotten that this "principle" is not a principle, but a mathematical consequence of the value of the commutator between two operators (you can demonstrate it.)
To sum up: the statement about a single measurement (commutator) implies a statement about more measurements (Heisenberg), but the last one is not a "principle" but only a consequence of the behaviour of a single measurement, described by the value of the commutator.

ZapperZ and I have discussed this before. The way most people will understand his sentence, it is certainly wrong. However, apparently he did not mean what the sentence is most commonly understood to mean, so his intention is right.
 
  • #9
I think HUP is best understood with the concept of phase space. You know, the space that represents the set that consists of all possible states of a physical system, and in which each point represents a state. HUP is basically talking about coordinates in classical phase space, in which a point is defined by location and speed. But quantum mechanics has a different state space, Hilbert Space. And the state of a physical system is actually quantum state. Thus, there is no such thing as uncertainly, the phase space defined in classical physics is simply wrong, period.
 
  • #10
Andre' Quanta said:
Dear Zapped Z

That's "ZapperZ" not "Zapped Z". He's the one doing the zapping, not the one getting zapped.
 
  • Like
Likes dlgoff
  • #11
The video is crystal clear but your text…

Well, ok, the English is not perfect (please see http://web2.uvcs.uvic.ca/elc/studyzone/330/grammar/poss.htm for the difference between the contraction “it’s” = “it is” and the possessive adjective or pronoun “its”), but I think I can see through it. In the end it boils down to three things:

First, you imply, as you have done in the past, that HUP has nothing to do with the so-called measurement problem (the fact that at microscopic level the measurement display somehow interferes with the measured object). It is not that you deny that such things actually happen from time to time. For example, in the experiment where we observe a particle by making a photon rebound against it: people claim that “the act of position measurement will simply destroy the accurate information of that electron’s momentum” and “this is true…”, you concede. But you also claim that this phenomenon “isn’t really a manifestation of the HUP”.

Second, you seem to state that in other single experiments, however, neither the measurement problem nor the HUP are present. In this sense, you say: “there’s nothing to prevent anyone from knowing both the position and momentum of a particle in a single mesurement (sic) with arbitrary accuracy that is limited only by our technology”. In particular, what is such fortunate experiment? It is not so clear, but the answer seems to lie in this other sentence of yours: “If I shoot the particle one at a time, I still see a distinct, accurate “dot” on the screen to tell me that this is where the particle hits the detector.” Could you please confirm that in this single experiment you consider that both position and momentum can be determined with arbitrary accuracy?

Third, we identify another situation where the HUP does arise and (I gather) the measurement problem does not (right?). It is a situation where we want “to make a dynamical model that allows us to predict when and where things are going to occur in the future”. Which one? It seems to be the same single experiment of the shot described above, but with some differences. Now there is a series of shots, the experimenter is narrowing down the slit between each firing and she wants to predict what will happen after each narrowing operation. In the classical case (the particle is a marble ball), prediction is possible: more narrowing means that the experimenter gets more certainty about both position and momentum. Instead in the quantum case (the particle is an electron), the gist of HUP would be that momentum predictability worsens with the narrowing, because the electron beam diffracts. Apparently, however, this funny or “strange” part would only come about after a certain narrowing threshold is reached. A threshold that you identify as the one when “the width of the slit is comparable to the deBroglie wavelength” of the electron. Right? And what about before that? Before that, as certainty over position increases, does certainty about momentum decline as well or on the contrary does it also increase? Could you clarify your opinion in this respect?
 
  • #13
bahamagreen said:
Check this - Experimenters violate Heisenberg's original version of the famous maxim, but confirm a newer, clearer formulation.

So, in actual experimental practice, have you all been employing Earle Kennard's formulation for uncertainty all along?

"The one that physicists use in everyday research and call Heisenberg's uncertainty principle is in fact Kennard's formulation."

In fact it is possible to define things so that even the informal Heisenberg principle has a formal form like the Kennard-Robertson form. http://arxiv.org/abs/1306.1565
 
  • #14
@ZapperZ
This makes the explicit assumption that no external forces acts on the particle at and after it passes through the slit, so that it’s momentum remains constant from the slit to the screen (which is a reasonable assumption).Let’s say the particle drifts from the center, straight-through line and hits the screen at a distance Y.
Actually, this part seems too classical. You're assuming that you can assign a definite momentum to the particle for its whole time of flight from the slit to the screen. Also you're assuming that the motion of the particle is rectilinear. How can we talk about the particle's path? How can we say its a straight line?
The extreme version of this question, can be asked by considering the Feynman's thought experiment about infinite planes with infinite slits that leads to path integrals. How can you say that the particle's path is this one among that infinite number of possible ones?
 
  • #15
Shyan said:
@ZapperZ

Actually, this part seems too classical. You're assuming that you can assign a definite momentum to the particle for its whole time of flight from the slit to the screen. Also you're assuming that the motion of the particle is rectilinear. How can we talk about the particle's path? How can we say its a straight line?
The extreme version of this question, can be asked by considering the Feynman's thought experiment about infinite planes with infinite slits that leads to path integrals. How can you say that the particle's path is this one among that infinite number of possible ones?

Then you need to write a paper and tell all those people who do angle-resolved photoemission spectroscopy that their deduction of the particle energy and momentum in those electron analyzers are wrong. In case you don't know, they consider the trajectory of the photoelectron from the entrance slit of the analyzer all the way to the CCD screen, through all those electronic optics, to be the same classical trajectory.

Zz.
 
  • #16
ZapperZ said:
Then you need to write a paper and tell all those people who do angle-resolved photoemission spectroscopy that their deduction of the particle energy and momentum in those electron analyzers are wrong. In case you don't know, they consider the trajectory of the photoelectron from the entrance slit of the analyzer all the way to the CCD screen, through all those electronic optics, to be the same classical trajectory.

Zz.

I'm not sure why you interpreted my post that way. But I'm not another physicist who is criticizing your approach in one of your papers, I'm a physics student asking about your approach as a professor teaching something! I didn't say you are wrong. I asked for clarification.
 
  • #17
Shyan said:
I'm not sure why you interpreted my post that way. But I'm not another physicist who is criticizing your approach in one of you papers, I'm a physics student asking about your approach as a professor teaching something!

But reread your post. The problem here is that you seem to not be aware that what I described is exactly the approach that has been taken in deducing the momentum of electrons and many other particles, i.e. from the point of interaction to the point of detection, it has a classical trajectory!

You don't have to believe this. All you need to do is look up mass or energy spectrometer when they measure the energy of particles such as electron, protons, etc. Do you think they used all these "infinite" possible paths?

Zz.
 
  • #18
ZapperZ said:
But reread your post. The problem here is that you seem to not be aware that what I described is exactly the approach that has been taken in deducing the momentum of electrons and many other particles, i.e. from the point of interaction to the point of detection, it has a classical trajectory!

You don't have to believe this. All you need to do is look up mass or energy spectrometer when they measure the energy of particles such as electron, protons, etc. Do you think they used all these "infinite" possible paths?

Zz.

Yeah, I understand. That's how people measure things.
But this brings this question to my mind that how is it that all those quantum effects let you do that? How is it that you actually can assume the particle has a classical trajectory and still retain consistency with QM? Is it an approximation?
I mean...we're really supposed to work with wave-functions and probability amplitudes here. So is it that those calculations give us classical results with good approximation?

So to state my question clearly. I know that experimentally its OK to do that. But how can you justify theoretically that you actually can do that?
 
  • #19
Shyan said:
Yeah, I understand. That's how people measure things.
But this brings this question to my mind that how is it that all those quantum effects let you do that? How is it that you actually can assume the particle has a classical trajectory and still retain consistency with QM? Is it an approximation?
I mean...we're really supposed to work with wave-functions and probability amplitudes here. So is it that those calculations give us classical results with good approximation?

So to state my question clearly. I know that experimentally its OK to do that. But how can you justify theoretically that you actually can do that?

Because there is no longer a superposition of path or superposition of trajectory after it passes through the slit. The "uncertain" quantity now is the momentum. ONCE the particle is detected, THEN, and not before that, can you reconstruct its trajectory from where it came from. This tells you the transverse momentum of that particle when it hits the detector.

We do this in all the detectors, big and small. How do you think high energy physics detectors able to do its path reconstruction after it detects the particle? All those "lines" you see drawn from ATLAS and CMS, do you think those were all there in the detector, or do you think they were reconstructed AFTER the fact?

Zz.
 
  • Like
Likes ShayanJ
  • #20
ZapperZ said:
Because there is no longer a superposition of path or superposition of trajectory after it passes through the slit. The "uncertain" quantity now is the momentum. ONCE the particle is detected, THEN, and not before that, can you reconstruct its trajectory from where it came from. This tells you the transverse momentum of that particle when it hits the detector.

We do this in all the detectors, big and small. How do you think high energy physics detectors able to do its path reconstruction after it detects the particle? All those "lines" you see drawn from ATLAS and CMS, do you think those were all there in the detector, or do you think they were reconstructed AFTER the fact?

Zz.

Its clear now. Thanks a lot.
 
  • #21
ZapperZ said:
Then you need to write a paper and tell all those people who do angle-resolved photoemission spectroscopy that their deduction of the particle energy and momentum in those electron analyzers are wrong. In case you don't know, they consider the trajectory of the photoelectron from the entrance slit of the analyzer all the way to the CCD screen, through all those electronic optics, to be the same classical trajectory.

Zz.

No. It means your justification of the calculation is wrong.
 
  • #22
Shyan said:
Yeah, I understand. That's how people measure things.
But this brings this question to my mind that how is it that all those quantum effects let you do that? How is it that you actually can assume the particle has a classical trajectory and still retain consistency with QM? Is it an approximation?
I mean...we're really supposed to work with wave-functions and probability amplitudes here. So is it that those calculations give us classical results with good approximation?

So to state my question clearly. I know that experimentally its OK to do that. But how can you justify theoretically that you actually can do that?

ZapperZ's explanation is wrong. It is an approximation or uses assumptions that he has failed to state. Of course one must use the quantum mechanical formulation.

There are certain special cases under which the classical method of calculation is very good, or even exact.

For example, for the case of a free Gaussian wave packet (it could be more general than this) the Wigner quasidistribution is not negative, and so can be considered a distribution. Furthermore, the quantum Liouville evolution for a free Gaussian is the same as the classical Liuoville evolution. I might be remembering some conditions wrongly here, look up http://dspace.mit.edu/bitstream/handle/1721.1/49800/50586846-MIT.pdf?sequence=2 (Eq 4.10 and following).

Another special case is (I believe) the use of a position measurement (as ZapperZ has done) to perform a momentum measurement of the wave function just after the slit - this is correct in the special case in which the screen on which the position measurement is performed is an infinite distance from the slit. It is not correct at any finite distance. At large screen distance, it is a good approximation. Essentially, the momentum is the Fourier transform of position. The large screen limit is the Fraunhofer limit in which the time evolution takes the Fourier transform for you.

One way to see that ZapperZ's explanation is wrong is to remember that if a particle in non-relativistic quantum mechanics has a trajectory, the trajectory is governed by de Broglie-Bohm like equations, and not Newtonian dynamics.
 
Last edited:
  • Like
Likes ShayanJ
  • #23
atyy said:
One way to see that ZapperZ's explanation is wrong is to remember that if a particle in non-relativistic quantum mechanics has a trajectory, the trajectory is governed by de Broglie-Bohm like equations, and not Newtonian dynamics.
Good point!
atyy said:
For example, for the case of a free Gaussian wave packet (it could be more general than this) the Wigner quasidistribution is not negative, and so can be considered a distribution. Furthermore, the quantum Liouville evolution for a free Gaussian is the same as the classical Liuoville evolution. I might be remembering some conditions wrongly here, look up http://dspace.mit.edu/bitstream/handle/1721.1/49800/50586846-MIT.pdf?sequence=2 (Eq 4.10 and following).
I'm trying to read that. But actually I don't know enough about the phase space formulation of QM(Which goes back to not knowing enough about the corresponding classical equations, But I plan to change that). Can you explain it in terms of other formulations?
It seems to me that if you calculate the trajectories in dBB, its easy to see in what limit(s) they give you the Newtonian trajectories. Then I think we can translate those limit(s) to other formulations.
 
  • #24
Shyan said:
Can you explain it in terms of other formulations?
It seems to me that if you calculate the trajectories in dBB, its easy to see in what limit(s) they give you the Newtonian trajectories. Then I think we can translate those limit(s) to other formulations.

Not off the top of my head, but these are all different formulations of the same thing, so it should be in principle possible.
 
  • #25
atyy said:
For example, for the case of a free Gaussian wave packet (it could be more general than this) the Wigner quasidistribution is not negative, and so can be considered a distribution. Furthermore, the quantum Liouville evolution for a free Gaussian is the same as the classical Liuoville evolution. I might be remembering some conditions wrongly here, look up http://dspace.mit.edu/bitstream/handle/1721.1/49800/50586846-MIT.pdf?sequence=2 (Eq 4.10 and following).

Does it mean that in this case and similar cases, we can interpret QM the same way we interpret statistical mechanics?
 
  • #26
Shyan said:
Does it mean that in this case and similar cases, we can interpret QM the same way we interpret statistical mechanics?

In all cases in which Bohmian mechanics works, QM can be like statistical mechanics.

In the special cases where the Wigner function is positive, and the quantum Liouville equation is identical to the classical one, we can additionally say that the particle has simultaneously-defined position and momentum. (I think, not entirely sure, I think there are some tricky points about Wigner functions and Bell inequality violations)

Edit: some references about Wigner functions and Bell inequality violations

http://arxiv.org/abs/quant-ph/0405100v1
Bell's Inequality Violation (BIQV) with Non-Negative Wigner Function
M. Revzen, P. A. Mello, A. Mann, L. M. Johansen

http://arxiv.org/abs/1409.0437
Violations of entropic Bell inequalities with coarse-grained quadrature measurements for continuous-variable states
Zeng-Bing Chen, Yao Fu, Yu-Kang Zhao
 
Last edited:
  • #27
Sorry for reviving an old(ish) thread but I just had a thought while reading a new paper.

One interesting way of thinking about this in the context of qubit error correction. A direct consequence of the HUP is that any measurement of a bit-flip error will produce a random flip in phase, making it impossible to directly stabilize a single qubit. This is a direct consequence of the fact that it is impossible to simultaneously measure both the phase and amplitude of a qubit.
Hence, this is nice practical (as opposed to philosophical) example of when the HUP limits our ability to measure something single-shot (i.e. in a "non-statistical" way),

Btw, the paper I was reading was Martinis new qubit metrology paper/comment (not sure if he intends to publish this)

http://arxiv.org/abs/1510.01406
 

1. What is the Heisenberg Uncertainty Principle?

The Heisenberg Uncertainty Principle is a fundamental principle in quantum mechanics that states that the more precisely the position of a particle is known, the less precisely its momentum can be known, and vice versa. In other words, it is impossible to know both the exact position and velocity of a particle at the same time.

2. What is the misconception surrounding the Heisenberg Uncertainty Principle?

The most common misconception about the Heisenberg Uncertainty Principle is that it is a limitation of measurement tools or techniques. Many people believe that the principle means that we are unable to measure the position and momentum of a particle accurately. However, the principle is not a limitation of measurement but rather a fundamental property of quantum mechanics.

3. Can the Heisenberg Uncertainty Principle be violated?

No, the Heisenberg Uncertainty Principle is a fundamental principle of quantum mechanics and cannot be violated. It is a consequence of the wave-particle duality of matter and is supported by experimental evidence. Any attempt to violate the principle would result in a violation of the laws of quantum mechanics.

4. How does the Heisenberg Uncertainty Principle affect everyday life?

The Heisenberg Uncertainty Principle only applies to particles on a microscopic level, so it does not have any significant impact on everyday life. It is only relevant in the study of quantum mechanics and has no practical implications in our daily lives.

5. Is the Heisenberg Uncertainty Principle still relevant in modern physics?

Yes, the Heisenberg Uncertainty Principle is still a fundamental principle in modern physics and is used in various fields such as quantum computing, particle physics, and atomic and molecular physics. It is a crucial concept in understanding the behavior of particles on a microscopic level and is still actively studied and applied in current research.

Similar threads

Replies
13
Views
1K
  • Quantum Physics
Replies
2
Views
882
  • Quantum Physics
Replies
16
Views
947
Replies
1
Views
816
Replies
2
Views
2K
Replies
5
Views
737
Replies
14
Views
1K
Replies
52
Views
4K
Replies
10
Views
1K
Replies
15
Views
1K
Back
Top