Where is the mistake -- violation of Heisenberg uncertainty

In summary: This is due to the uncertainty principle, which states that the more precisely one property of a particle is measured, the less precisely another property can be known.2. The Heisenberg uncertainty principle states that it is impossible to prepare a particle in a state where both its position and momentum are known to a degree of uncertainty less than the HUP allows. This means that it is not possible to measure both the position and momentum of a particle at the same time with perfect accuracy. By attempting to minimize the uncertainty in one observable (such as momentum), the uncertainty in the other observable (position) will increase. This is a fundamental principle of quantum mechanics and cannot be violated
  • #1
MichPod
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The following is kind of a semiclassical reasoning which goes along the style of the discussions between Bohr and Einstein in Solvay Conference.

Suppose we have a single slit from which a particle may be emitted and, on a very large distance from it, a thin cloud chamber CC, which we use as a kind of a screen to detect the particle.

cc.png


Also suppose that we know the moment at which the particle passed via the slit (we can put some light source to fix that moment or we can use another small cloud area put at the slit so we could see a small bubble in the area of the slit when the particle passes).

As the particle travels from the slit to the CC, there is no force applied to it so it goes on the straight line with a constant speed and finally creates some bubble at the CC (the Cloud Chamber). We can calculate the speed of the particle as the distance from the slit to the point where the particle was detected at CC divided by the time between these two detections. By making the distance between the slit and the cloud chamber CC very big, we can reduce to any necessary minimum the contribution of all the uncertainties involved into the measurement of the speed as the size of the slit at which the first detection of the particle done, the size of the first bubble in CC at which the second detection done, the time measurement errors. Because we can reduce the error of the speed measurement to any minimum, so we can reduce to any minimum the momentum measurement. And the error in the coordinate measurement is determined by the size of the bubble it the CC. So in the end the product of the coordinate error and momentum error at the point at which the particle was detected at CC may be made (as it looks) arbitrary small which violates the Heisenberg uncertainty principle.

Could you help me to see where is the error in this reasoning?
 

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  • #2
Newton's first law doesn't apply literally in QM. If particles moved at constant speed in a straight line, you would know the position and momentum at all times.
 
  • #3
MichPod said:
The following is kind of a semiclassical reasoning which goes along the style of the discussions between Bohr and Einstein in Solvay Conference.

Suppose we have a single slit from which a particle may be emitted and, on a very large distance from it, a cloud chamber CC.

View attachment 218549

Also suppose that we know the moment at which the particle passed via the slit (we can put some light source to fix that moment or we can use another small cloud area put at the slit so we could see a small bubble in the area of the slit when the particle passes).

As the particle travels from the slit to the CC, there is no force applied to it so it goes on the straight line with a constant speed. We can calculate the speed of the particle as the distance from the slit to the point where the particle was detected at CC divided by the time between two detections. By making the distance between the slit and the cloud chamber CC very big, we can reduce to any necessary minimum the contribution of all the uncertainties involved into the measurement of the speed as the size of the slit at which the first detection of the particle done, the size of the first bubble in CC at which the second detection done, the time measurement errors. Because we can reduce the error of the speed measurement to any minimum, so we can reduce to any minimum the momentum measurement. And the error in the coordinate measurement is determined by the size of the bubble it the CC. So in the end the product of the coordinate error and momentum error at the point at which the particle was detected at CC may be made (as it looks) arbitrary small which violates the Heisenberg uncertainty principle.

Could you help me to see where is the error in this reasoning?

You reasonably know your argument is wrong, and must know the devil is in those little details. Saying you can minimize the uncertainty in an observable, and then doing it, those are not the same thing.

1. Narrowing the time span on the back end to detect the time of arrival (so you can calculate the speed): this is a problem as you can see in this image of a bubble chamber: https://en.wikipedia.org/wiki/File:AlphaTrackRutherfordScattering3.jpg

Exactly how do you use this to determine where the particle is at time T? The tracks prevent that.

2. Many folks here feel it is a more effective re-statement of the HUP in the following light: it is impossible to prepare a particle (in practice, an ensemble of particles) in a state where momentum and position are known to a degree of uncertainty less than the HUP allows. If that were possible, you could obviously individually measure your choice of the observables to confirm. We would have already rejected the HUP if we could do this.
 
  • #4
PeroK said:
Newton's first law doesn't apply literally in QM. If particles moved at constant speed in a straight line, you would know the position and momentum at all times.

Yet we can actually see a linear track in a Cloud Chamber (not in the experiment I am describing where the CC is very thin, but in a general experiment with a big thick CC). What I am doing it's like I am actually building a very long cloud chamber, putting some vapor at the point of the start and some vapor at the point of the arrival of the particle. Whatever quantum laws are, you cannot deny that the tracks in a cloud chamber are linear when no additional force is applied to the particle.
DrChinese said:
1. Narrowing the time span on the back end to detect the time of arrival (so you can calculate the speed): this is a problem as you can see in this image of a bubble chamber: https://en.wikipedia.org/wiki/File:AlphaTrackRutherfordScattering3.jpg

Exactly how do you use this to determine where the particle is at time T? The tracks prevent that.

The time of the traveling from the slit to the chamber is much much bigger than the time it takes the particle to travel through the CC on the right. Say, the particle is launched from the Pluto, it takes it one year to arrive to the chamber and then a tiny fraction of the second to cross it. So whatever small error can we make measuring the time of the arrival, it is canceled when we calculate the speed, taking the overall big time interval the particle has traveled from the slit to the CC.

DrChinese said:
We would have already rejected the HUP if we could do this.

My intention is only to understand where is the flaw in the reasoning I presented here. This reasoning is maybe anyway irrelevant as we know that the classical mechanics is not applicable to the particles, but this semiclassical way of reasoning was used on some first stages of QM creation, used in arguments between Bohr and Einstein, so, at least this reasoning could potentially be presented to Bohr as a violation of HUP. I am interested to see where is the mistake, on the semiclassical level, if there is one.

We know that Einstein presented rather sophisticated experiments designs on Solvay to demonstrate HUP is wrong and all of his puzzles were resolved by Bohr. Now what I present in this topic is rather trivial comparing to Einstein's puzzles, yet I do not see where the flaw is.
 
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  • #5
MichPod said:
The time of the traveling from the slit to the chamber is much much bigger than the time it takes the particle to travel through the CC on the right. Say, the particle is launched from the Pluto, it takes it one year to arrive to the chamber and then a tiny fraction of the second to cross it. So whatever small error can we make measuring the time of the arrival, it is canceled when we calculate the speed, taking the overall big time interval the particle has traveled from the slit to the CC.

The distance from source to target is relevant only within certain parameters (perhaps a low velocity). What matters is how sure ("certain") we are that it arrived at a specific place at a specific time. You will NEVER know both when the target is a chamber per the diagram I supplied. A mere nanosecond off and a micrometer off (in rough terms of course) could be enough.

Oh and you need to know it started at a specific place at a specific time so you can properly determine the distance traversed and the time it took to traverse. That goes back to my 2nd point above: you cannot prepare a particle in such a state to start with.

v=(d2-d1)/(t2-t1) [where d is distance and t is time, and we ignore the case where change in mass/energy is a factor]

You can easily see that we need to know 4 terms above with great certainty to make your program work out. But we can only ever get 2. If you know its starting position with great accuracy, its velocity becomes very uncertain and trying to say it is there at a specific time becomes difficult. You can't say when it left its starting position. Was it a little earlier or later?
 
  • #6
DrChinese said:
v=(d2-d1)/(t2-t1) [where d is distance and t is time, and we ignore the case where change in mass/energy is a factor]

You can easily see that we need to know 4 terms above with great certainty to make your program work out. But we can only ever get 2. If you know its starting position with great accuracy, its velocity becomes very uncertain and trying to say it is there at a specific time becomes difficult. You can't say when it left its starting position. Was it a little earlier or later?

I put some vapor in the area of the slit or I put some light at the slit - that helps me to know d1 and t1. I can know d1,t1,d2,t2 with some uncertainties which are basically determined by the local conditions (bubble size, time of bubble forming etc) and then by putting the slit and a final CC at a very long distance one from another I get the velocity with the arbitrary small uncertainty. Say, I have uncertainty of the coordinate 1 mm and of the time 1 s (which are both huge), then I make a distance between the slit and the target 10^35 m? Yes, I know that is bigger than the visible universe size, but I am speaking theoretically, just to show how the errors of measurement could be compensated by selecting of the large distance.

Practically, if I did not make any mistake, for an electron moving with a speed ~1000 km/s we'll need 100000km distance to arrive close to the HUP violation in the context of my reasoning.
 
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  • #7
MichPod said:
I put some vapor in the area of the slit or I put some light at the slit - that helps me to know d1 and t1. I can know d1,t1,d2,t2 with some uncertainties which are basically determined by the local conditions (bubble size, time of bubble forming etc) and then by putting the slit and a final CC at a very long distance one from another I get the velocity with the arbitrary small uncertainty. Say, I have uncertainty of the coordinate 1 mm and of the time 1 s (which are both huge), then I make a distance between the slit and the target 10^35 m? Yes, I know that is bigger than the visible universe size, but I am speaking theoretically, just to show how the errors of measurement could be compensated by selecting of the large distance.

Practically, if I did not make any mistake, for an electron moving with a speed ~1000 km/s we'll need 100000km distance to arrive close to the HUP violation in the context of my reasoning.

You are thinking somehow in terms of lowering the % of error by increasing the distance. This is not the HUP. The issue is the standard deviation, which can be small relative to the measured average value - or not.

The formal inequality relates the standard deviation of position σx and the standard deviation of momentum σp. This cannot be less than a very small number (depending on selection of units).
 
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  • #8
DrChinese said:
You are thinking somehow in terms of lowering the % of error by increasing the distance

That is correct. Yet by lowering the % of the error of the distance and time measurement I do lower the error of the velocity measurement and so the error of the momentum measurement. If I were measuring the length and was getting the low % of the error on long distances, you could argue that the absolute error does not change. 1mm of the error measuring 1000km is still 1mm. But I measure the velocity having long distances and long times and here low % of error of time and distance measurement does give me (arbitrarily) low absolute error of velocity and momentum.

Suppose you have distance and time with velocity 1 m/s
distance=1m+1mm=1001mm, time=1s-1ms=999ms velocity = 1001/999=1.002 m/s (2 mm/s error)
ditance=1km+1mm time = 1000s-1ms velocity = 1.000002 m/s ( 0.002 mm/s error)

So you see that if my distance and time measurement errors are constant, then by using bigger distance I reduce the error on the velocity.
 
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  • #9
MichPod said:
That is correct. Yet by lowering the % of the error of the distance and time measurement I do lower the error of the velocity measurement and so the error of the momentum measurement. ... So you see that if my distance and time measurement errors are constant, then by using bigger distance I reduce the error on the velocity.

Not hardly. You can have 1 m/s velocity +/- .001 m/s at SD=1, or 1,000,000 m/s velocity +/- .001 m/s at SD=1. Standard deviation has not changed, and neither has uncertainty. The uncertainty does not get smaller as you measure over longer and longer distances.

At the end of the day: no one has ever said you can't get a number for P and Q simultaneously. You can, and there are a lot simpler ways to do it than you have suggested. They just don't represent anything meaningful. For example: take a pair of entangled particles we'll call A and B. Measure P on A, X on B. There is nothing to stop you from doing that to unlimited uncertainty. So you now know both P and X for both A and B... violating the HUP! Or not, obviously. You think you know them, but subsequent measurements will not confirm this.
 
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  • #10
This is the uncertainty relation:

ef8e682fd2c381ebfee023d4dca01e4b6319c146


Where ħ is on the order of approximately 1.05457×10−34 joule seconds. You should be able to see that you are not going to have materially less variation (relative to this very small value) by measuring over progressively longer and longer distances.

Surely you have read that as you confine a particle to a progressively smaller region, that particle's momentum (and velocity) takes on a wide range of typical values. Those values can be wildly high, for example, and will not follow any particular pattern as you repeat the experiment.
 
  • #11
Here's the problem with your setup.

Suppose you place another bigger cloud chamber, twice as far from the slit as the first. If you were actually measuring the position and momentum of the particle to arbitrary accuracy, then once you saw the track in the first chamber you would be able to predict very precisely where to expect a track later in the second chamber. But what you will find is that it doesn't work: the particle will not hit the spot you predict. If you repeat the experiment many times, you will find that there is an inherent spread-out-ed-ness in where the particle will hit the second screen even after accounting for where it hit the first screen.

Basically you're doing an experiment and saying if the particle had a well-defined position and momentum, then given my results this is what they must be. The problem is that the particle doesn't actually have a sufficiently well-defined position and momentum, so it is incorrect to infer them the way you want to do from the experiment.
 
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  • #12
The error in your reasoning is that the Heisenberg uncertainty principle describes a different situation than what you are describing. In the HUP, one has two identically prepared ensembles. On the first ensemble, one makes position measurements. On the second ensemble, one makes momentum measurements. The HUP governs [(variance in position measurements on the first ensemble) X (variance in momentum measurements on the second ensemble)].

Also, take note of PeroK's reply in post #2, although it is true that in this case, at large distance, the momentum measurement you described in the OP is correct (classical and quantum calculations are essentially the same).

There are inequalities similar to the HUP. These are derived from the commutation relations. One has to be careful about what these inequalities say.
 
  • #13
Strilanc said:
what you will find is that it doesn't work: the particle will not hit the spot you predict.

Can that be supported by any semiclassical reasoning? If not, I still am not sure what is the mistake. Well, maybe the mistake is I did not actually measure the momentum, just calculated it, but this looks a bit subtle and not clear for me.

I hope that what you said may be supported by a conventional qm calculations, it's only that I am interested here in how to look at these things semiclassically. As I said that were the actual terms of Solvay discussion...

Actually your post helped me much. I did not think of what will happen if I put another chamber... Thank you!
 
  • #14
MichPod said:
Can that be supported by any semiclassical reasoning? If not, I still am not sure what is the mistake. Well, maybe the mistake is I did not actually measure the momentum, just calculated it, but this looks a bit subtle and not clear for me.

There's no problem in doing the following:

1) Make a very accurate measurement of a particle's position.

2) Make a second very accurate measurement of a particle's position, some time later.

3) Calculate the (average) momentum of the particle in this time.

The HUP says essentially nothing about that.

More simply, you can:

1) Make a very accurate measurement of a particle's momentum.

2) A very short time afterwards, make a very accurate measurement of a particle's position.

The HUP says essentially nothing about that. Athough (in Copenhagen) measurement 2) will destroy your accurate knowledge of momentum. The HUP is a statistical law. To violate the HUP you would need a large sample of measurements and show the violation a certain confidence level.

If you follow this up with a third measurement:

3) Make another accurate measurement of the particle's momentum.

Now, for an ensemble of particles 3) should give a range of momenta, whose statistical variance is related to the accuracy of your measurement in 2). Note that it's entirely possible in one experiment to get the same value for momentum in 3) as in 1). This is, however, just one experiment and does not violate the statistical law. Ironically, of course, you are bound to get this result sooner or later, by the nature of probabilities.

Note that you will find many sources that claim that each pair of measurements in 1 and 3 must differ by an amount consistent with the HUP on every experiment. But, this is wrong!
 
  • #15
If you are not familiar with how a standard deviation is calculated, try this: https://en.wikipedia.org/wiki/Standard_deviation

00eb0cde84f0a838a2de6db9f382866427aeb3bf


The value is not affected by the mean (average) measured value. It is instead affected by the average difference from the mean, per the formula above.
 
  • #16
DrChinese said:
If you are not familiar with how a standard deviation is calculated, try this: https://en.wikipedia.org/wiki/Standard_deviation

00eb0cde84f0a838a2de6db9f382866427aeb3bf


The value is not affected by the mean (average) measured value. It is instead affected by the average difference from the mean, per the formula above.

The distance is 1 km. The deviation is 1 mm.
The time is 1000 s. The deviation is 1 ms.
What is the value and the deviation of the speed?
 
  • #17
MichPod said:
The distance is 1 km. The deviation is 1 mm.
The time is 1000 s. The deviation is 1 ms.
What is the value and the deviation of the speed?

The standard deviation applies to a sample of (many) measurements. Take, for example, the standard deviation in the throw of a die. The mean per throw is 3.5. Every throw is an exact number - there is no "error" or "deviation" in anyone throw. You always get exactly one of 1, 2, 3, 4, 5 or 6.

But, if you throw a die a large number of times, the mean of your sample will be approx 3.5 and the standard deviation can be calculated. This is statistics. And the HUP is a statistical law.

Just to repeat: no individual measurement can support or violate the HUP. Only the same measurements on a large sample of identical experiments can confirm or violate the HUP, with the key statistic being the standard deviation of position and momentum measurements.
 
  • #18
MichPod said:
The distance is 1 km. The deviation is 1 mm.
The time is 1000 s. The deviation is 1 ms.
What is the value and the deviation of the speed?

You must have the full data set to calculate SD. This is not a physics issue, it is one of statistics. It doesn't make sense to ignore the HUP's definition, and instead discuss something you made up.

After you put together a hypothetical realistic sample of observations, try calculating its SD. You will quickly see that your assumptions will not begin to violate the HUP.
 
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  • #19
DrChinese said:
If you are not familiar with how a standard deviation is calculated, try this: https://en.wikipedia.org/wiki/Standard_deviation

00eb0cde84f0a838a2de6db9f382866427aeb3bf


The value is not affected by the mean (average) measured value. It is instead affected by the average difference from the mean, per the formula above.

Just to say I have absolutely no idea why that has ##N-1## and not ##N## in that formula!
 
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  • #20
PeroK said:
Just to say I have absolutely no idea why that has ##N-1## and not ##N## in that formula!

I don't entirely follow the significance of the (N-1) either. Guess I need to bone up on my statistics. :smile:

On the same page, they also reference the use of the more familiar N as the denominator instead of N-1. Obviously, for any reasonable sample size, the distinction is quite small.
9a937016f00f1978197aa562c5f2d58619f90806
 
  • #21
PeroK said:
The HUP is a statistical law. To violate the HUP you would need a large sample of measurements and show the violation a certain confidence level.

PeroK said:
Just to repeat: no individual measurement can support or violate the HUP. Only the same measurements on a large sample of identical experiments can confirm or violate the HUP,

Yet in the argument between Bohr And Einstein they did discuss individual measurements.

Let me stress that again, I am trying to go along the same line of arguments which were between Bohr and Einstein in their very well known discussion in Solvay.
And they did discuss the error of measurement, not the statistical distribution of the particle coordinate and momentum (which is the current understanding of HUP).
 
  • #22
MichPod said:
Yet in the argument between Bohr And Einstein they did discuss individual measurements.

In individual measurements the informal "HUP" can be violated. The "HUP" is used differently by different people. People argue about terminology, not physics. In terms of physics, it is the inequalities derived from the commutation relations, which properly interpreted, cannot be violated.

BTW, in your OP, the use of a position measurement to measure momentum is not an accurate measurement of the canonically conjugate momentum. The momentum you measure is the an accurate measurement of the momentum at an earlier point in time. Thus the quantities you discuss in the OP are simply not canonically conjugate, and not governed by the HUP (even informal versions).

However, there are ways to correct the OP to violate the informal HUP, eg. see ZapperZ's variant of the setup in your OP: https://www.physicsforums.com/insights/misconception-of-the-heisenberg-uncertainty-principle/ (IIRC, I had problems with the language in ZapperZ's post, but after a long discussion it was a terminology difference)
 
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  • #23
MichPod said:
Yet in the argument between Bohr And Einstein they did discuss individual measurements.

Let me stress that again, I am trying to go along the same line of arguments which were between Bohr and Einstein in their very well known discussion in Solvay.
And they did discuss the error of measurement, not the statistical distribution of the particle coordinate and momentum (which is the current understanding of HUP).

You asked where the mistake is. That has been answered multiple times. You are batting at something which is not the HUP.

As we showed you: Increasing the traversed distance, in and of itself, does not affect the uncertainty (measured as SD, per the HUP) at all. We have shown you that bubble chambers do NOT record the ending position of a particle at a discrete point in time. And we explained that you cannot prepare an ensemble of test cases in a suitable manner to arrive at a suitable starting point either - each test sample will be different!

To have this conversation with Bohr and Einstein, you'll need to ignore almost everything we're learned since 1935 (or 1927 depending on your perspective). Good luck, but you are really just ignoring what you are being told in favor of stating your own opinions. DrChinese out... :smile:
 
  • #24
MichPod said:
I am trying to go along the same line of arguments which were between Bohr and Einstein in their very well known discussion in Solvay.

This is really a question of history, not physics. The physics has been sufficiently discussed.

Thread closed.
 

1. What is the Heisenberg uncertainty principle?

The Heisenberg uncertainty principle is a fundamental principle in quantum mechanics that states that it is impossible to know the exact position and momentum of a particle at the same time.

2. How does the Heisenberg uncertainty principle relate to the concept of measurement in quantum mechanics?

The Heisenberg uncertainty principle states that the act of measuring one observable property of a particle, such as its position, will inevitably affect the accuracy of measuring another observable property, such as its momentum. This is because the very act of measurement changes the state of the particle.

3. What is a violation of the Heisenberg uncertainty principle?

A violation of the Heisenberg uncertainty principle occurs when an attempt is made to measure both the position and momentum of a particle simultaneously with absolute precision. This is not possible due to the inherent uncertainty in quantum systems.

4. Can the Heisenberg uncertainty principle be proven experimentally?

Yes, the Heisenberg uncertainty principle has been proven experimentally through various experiments, such as the double-slit experiment and the Stern-Gerlach experiment, which demonstrate the inherent uncertainty of particles at the quantum level.

5. How does the Heisenberg uncertainty principle impact our understanding of the physical world?

The Heisenberg uncertainty principle has significant implications for our understanding of the physical world, as it challenges the classical notion of determinism and the ability to precisely predict the behavior of particles. It also highlights the inherent randomness and uncertainty present in the quantum world, which has led to the development of new theories and technologies.

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