# Heisenberg's uncertainty principle invalidated?

1. Feb 21, 2013

### quantumphilosopher

Heisenberg's uncertainty principle asserts a fundamental limit to the precision with which certain pairs of physical properties of a particle, such as position and momentum, can be known simultaneously. The more you know about one, the less you know about the other.

Why does the following thought experiment not violate Heisenberg's uncertainty principle?

Suppose you have a source that releases simultaneously two identical particles in opposite directions. The particles have the same momentum: being identical they have the same mass, and they were ejected from the source with the same speed. (You may think of the particles as being two protons resulting from the "explosion" of a dihydrogen molecule; because of the conservation of momentum, the two protons fly in opposite directions with the same speed).

Now, at t=1s after the particles have been released from the source you carry out two measurements:
1) You measure the position of particle A and find a value, xA. Because of the conservation of momentum, you can deduce the position of particle B at that time, which will be -xA
2) You measure the momentum of particle B and find a value, pB. Because of the conservation of momentum, you can deduce the momentum of particle A, which will be -pB

As a result, you have acquired precise knowledge at a given time of both the position and the momentum of particle A (and similarly for particle B). Therefore, Heisenberg's uncertainty principle is false.

What's wrong with this thought experiment? It resembles a lot EPR's thought experiment, but is not identical. I'm pretty sure that there must be a mistake somewhere, but I can't figure out where. Any thoughts?

Last edited: Feb 22, 2013
2. Feb 21, 2013

### phinds

That's as far as I got. I'm confident the answer is no.

If you had asked "Why does the following thought experiment not violate the HUP" I would have continued reading.

3. Feb 22, 2013

### quantumphilosopher

I agree it was badly phrased. I changed it.

Tell me why if you know the answer.

4. Feb 22, 2013

### my_wan

Your reasoning roughly follows some classic arguments in physics.

1) The more accurately you measure the position of particle A the less you know about when it held that position.
2) The more accurately you measure the momentum of particle B the less you know about where it was when it possessed that momentum.

This means you can't possibly have precise knowledge "at a given time" of both the position and the momentum of particle, since you don't know when one was measured or where the other was measured.

Also, position and momentum is subject to quantum fluctuations over time, such that its momentum at one moment may differ slightly in another moment. The more precise the time period the greater this short term fluctuation can be.

5. Feb 22, 2013

### Cthugha

The uncertainty principle is about states of particles or statistical ensembles if you prefer. That means it says something about standard deviations of many such measurements and therefore the uncertainties you get are the minimal statistical uncertainties you get for repeated measurements of absolutely identically prepared states.

In other words: You cannot even prepare an ensemble or even just two particles with smaller uncertainties than those given by the uncertainty principle. Therefore your proposal can be summarized as: let us assume we could prepare two particles in a manner which violates the uncertainty principle and I can show you that these two particles then violate the uncertainty principle. That is not too surprising. Wrong assumptions give wrong consequences.

6. Feb 22, 2013

### quantumphilosopher

I don't need to prepare it myself. I let Nature prepare it. In alpha decay the atomic nucleus emits simultaneously two protons (and two neutrons).

7. Feb 22, 2013

### quantumphilosopher

I'm measuring the position of one and the momentum of the other exactly at t=1s after the particles left the source. Are you suggesting measurement is a process that happens during a time interval? How much time can the measurement take?

8. Feb 22, 2013

### my_wan

Note that momentum is the product of the mass and velocity. It you want to know the velocity you need to know how long it took to get from here to there. At exactly t=1s it never went anywhere because no time has passed. So how can you know the momentum?

Yes, measuring momentum requires a time interval to be operationally meaningful.

Edit: Position becomes fuzzy if viewed over any finite time interval.

Last edited: Feb 22, 2013
9. Feb 22, 2013

### quantumphilosopher

Forget about measuring them 1 second after they left the source. I place a device that measures the particle's speed on the path of the particle, 1 meter away from the source. When the particle passes through the device I detect its speed. And I know its position (1 meter away from the source).

10. Feb 22, 2013

### Cthugha

That does not make the slightest difference. Also any ensemble of identically (by nature) prepared protons will show a standard deviation according to the uncertainty principle. Also you seem to have the impression that the uncertainty principle is about the outcome of a single experiment. It is not.

Nothing prevents you from doing so. Also, this is not against the uncertainty principle. You can measure position and momentum of that particle at different times with any precision you like. But you will not be able to predict what it will do the next time if you repeat exactly the same experiment. THIS is what the uncertainty principle is about.

11. Feb 22, 2013

### quantumphilosopher

If I can obtain a precise value for both the position and the momentum of a single particle at a given time, then Heisenberg's uncertainty principle is false. I don't need to repeat the experiment. That's NOT what Heisenberg's principle is about!

12. Feb 22, 2013

### phinds

That is NOT what Cthugha said. He said DIFFERENT times.

You are correct in that if you COULD get the exact measurements at the same time, then the HUP would be violated, but all of the responses in this thread have been trying to help you understand that that is exactly what you CANNOT do.

13. Feb 22, 2013

### Cthugha

Of course it is about repeating things. Also "at a given time" is not "at different times".

"The relationship commonly referred to as the Heisenberg uncertainty principle (HUP)—in fact
proved later by Weyl [4], Kennard [3], and Robertson [2]—refers not to the precision and disturbance of a measurement,but to the uncertainties intrinsic in the quantum state. The latter can be quantified by the standard deviation
$\Delta \hat{A}=\sqrt{\langle \hat{A}^2 \rangle - \langle \hat{A}\rangle^2}$
, which is independent of any specific measurement."

This quote is taken from Phys. Rev. Lett. 109, 100404 (2012) (Violation of Heisenberg’s Measurement-Disturbance Relationship by Weak Measurements). It has been long known that an interpretation of the uncertainty principle in terms of a "disturbance-by-measurement" as you propose it is not the correct interpretation. However, this interpretation lasted for a while as a "common myth" because it was Heisenberg's own first interpretation.

Or do you have any peer-reviewed publication supporting your claim?

Also please note that measuring the position at one time, measuring the momentum at some other time and backtracking what the momentum might have been at the first time is not the same as measuring both simultaneously - even for the tiniest time differences between the measurements. There are obviously no simultaneous measurements for complementary quantities.

Last edited: Feb 22, 2013
14. Feb 22, 2013

### my_wan

As Cthugha has pointed out the uncertainty is intrinsic in the quantum
state, and not in the precision and disturbance of a measurement. This fact played a significant role in deriving the exact uncertainty principle.
Schrodinger equation from an exact uncertainty principle

15. Feb 22, 2013

### quantumphilosopher

I did not propose this. Why do you think I did?
Again, this is not what I described. I was explicit about the fact that the measurements are simultaneous. The fact that you are simply asserting the contrary is simply begging the question.

Anyway, here's an even simpler setup aiming to obtain definite values for both the position and the momentum of a given particle at a given time.

You have a particle-emitting source, and you turn it on briefly (enough to release a particle). At x meters from the source you have a screen. You measure the time when the particle arrives at the screen and you obtain the particle's speed. The position of the particle is the spot on the screen where it was detected.

What I'm questioning here is not the validity of Heisenberg's principle (which I regard as a law of nature). Rather, I want to know what's wrong with the experiment I propose.

16. Feb 22, 2013

### my_wan

If you turned the emitter on for a brief time, how do you know when in that brief time the particle was emitted? If you don't know precisely when it was emitted how can you tell how long it took to arrive at the screen?

Also, if this particle is undergoing random momentum fluctuations, defined by the uncertainty principle, on the way to the screen you didn't actually measure its momentum when it hit the screen, or its momentum at any particular time during its approach to the screen. What you measured was the average of all the different momentums it had at various times between the emission and detection.

17. Feb 22, 2013

### Demystifier

This description looks somewhat ambiguous to me, because it is not clear how exactly the wave function looks like for that situation. The solution of your puzzle critically depends on the resolution of this ambiguity, so a proper resolution of the puzzle is not possible before you write down the wave function. In particular, is the wave function localized in the position space? Or is it localized in the momentum space? Or perhaps you think that it is both (which is impossible)?

In fact, when you write down the wave function, it is very likely that you will be able to see the resolution of the puzzle by yourself.

18. Feb 22, 2013

### quantumphilosopher

I wondered the same thing. This is where the skill of the experimentalist comes in. Imagine a container full of particles ready to get out. And then you open the lid just enough for a particle to get out. I still worry that this may be more than just a practical problem.

You're right. But it matters how large the fluctuation is. It could be small enough to invalidate the HUP.

19. Feb 22, 2013

### Cthugha

Because you did. ;)

You said:"I don't need to repeat the experiment." and are referring to a uncertainty principle not based on the standard deviations, but meaningful for a single experiment. This is the "disturbance-by-measurement" version, even if it might not seem obvious.

My comment about simultaneous measurements was quite simple. There are no really simultaneous measurements for complementary quantities. A lot of apparent paradoxes in qm come from assuming that such things are possible, but one simply cannot measure non-commuting quantities simultaneously as they do not share any eigenstates. This was the main reason for that comment. These two measurements are never simultaneous and you will always backtrack from the time of one measurement to the time of the other.

So let me get this straight. You first perform a perfect position measurement at time zero as that is necessary to get the exact travel time and then you perform a second measurement at some other position to get position/momentum then, right? This is fine. It is basically equivalent to the single-slit experiment with a very narrow slit. By assuming that the momentum is constant in between you can backtrack what the (average) momentum has been after the first position measurement and before the second measurement. And if you repeat the experiment, the first position measurement will leave you with a rather undefined momentum and you will measure a different momentum every time you repeat the measurement. That is ok. Exactly this situation is described in detail in Ballentine's review article on the "statistical interpretation of quantum mechanics" from the 1970's (Rev. Mod. Phys. 42, 358–381 (1970)) or http://www.kevinaylward.co.uk/qm/ballentine_ensemble_interpretation_1970.pdf starting basically from the same phrasing of the UP you use.

This manuscript also explains in detail why the uncertainty relation is a question of ensembles and not of the results you get in a single run of "simultaneous" measurements (and why it might not be a good idea to phrase the UP the way you used it in your first post).

20. Feb 22, 2013

### vanhees71

Here again the link to my posting about this problem. It reviews an experiment, showing explicitly that the noise-disturbance relation by (weak) measurements is not identical with the the Heisenberg-Robertson uncertainty principle: