# The HUP, simultaneous measurements, and eigenfunctions

1. Oct 14, 2012

### jmcelve

Hi everyone,

I know this topic has been discussed quite a bit -- and in particular it's been done in this thread and this thread. But there are still some things I want to talk about in order to (hopefully) clarify my own thoughts.

One of the threads discusses this Ballentine article in which Ballentine presents an argument against the statement that the HUP requires that simultaneous measurements of conjugate variables have a lower bound in uncertainty. I find Ballentine's argument convincing, but in one of the threads, Demystifier makes a convincing argument that Balletine's experiment requires identifying the probability of the momentum with the probability of the position measurement $|\langle y | \psi \rangle|^2$ when it is experimentally the case the the momentum is identified with the Fourier transform $|\langle p_y | \psi \rangle|^2$ of the coordinate space wavefunction. My first question then is: where does Balletine's experiment fail?

Additionally, I see a lot of people make the argument that the HUP does not say anything about simultaneous measurements of conjugate variables, but that it is rather a statement about a given prepared state (which describes an ensemble of particles). I understand the argument and find it convincing since it doesn't make sense to talk about deviations for a single particle. But then what about the obvious manifestation of the HUP in single particle systems where simultaneous eigenfunctions of position and momentum cannot exist? Indeed, doesn't this suggest that the HUP *does* imply that, for a given particle, its position and momentum cannot be simultaneously identified up to arbitrary accuracy -- that there *is* a lower bound?

Many thanks,
jmcelve

Last edited: Oct 14, 2012
2. Oct 15, 2012

### Jano L.

Hello jmcelve,

the important thing is to free oneself from the false idea that mathematical function $\psi(\mathbf r)$ is a complete description of the particle. Unless this is done, the particle can never be thought to be in one place nor to have, in any situation, to have momentum. This is because there are no corresponding normalized $\psi$ that would be able to describe the particle along Born's rule for $|\psi|^2$.

The natural thing to do is to expect that the particle has both position and momentum simultaneously, but this position and momentum are not properties of some wave function. Wave function can be naturally thought as describing probabilities, or frequencies in finite ensemble.

3. Oct 15, 2012

### Naty1

not again!! [LOL]

Good for you for reading those excellent and lengthy discussions. Consider taking notes for you own clarification.

Different people have different interpretations about identical mathematics...and some assign
preference to some math and others to other math. These have been going on for 90 years.
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My first question then is: where does Ballentine's experiment fail?

Fredrik: “Ballentine's argument in the article discussed in this thread seemed to prove that you could measure both [position, momentum] with accuracies Δx and Δp such that ΔxΔp is arbitrarily small. And I wasn't able to see what was wrong with it. But Demystifier was. I think that what he said here is a very good reason to not define QM in a way that makes what Ballentine described a "momentum measurement":

See post # 40 here:

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"But then what about the obvious manifestation of the HUP in single particle systems where simultaneous eigenfunctions of position and momentum cannot exist?"

I don't think that assumption is correct. QM formalism: an observable is represented by a self adjoint operator on a Hilbert space, and a state, represented by a state operator [also called a statistical operator or density matrix]. The only values which an observable may take on are its eigenvalues and the probabilities of each of the eigenvalues can be calculated.

Fredrick said it this way: "It is possible to measure position and momentum simultaneously…a single measurement of a particle. In fact, we often measure the momentum by measuring the position and interpreting the result as a momentum measurement. (Check out figure 3 in this pdf). What we can't do is to prepare an identical set of states…. such that we would be able to make an accurate prediction about what the result of a position measurement would be and an accurate prediction about what the result of a momentum measurement would be….for an ensemble of measurements...."
and " "It's not true that every measurement puts the system in an eigenstate of the measured observable".

Zapper says it this way: "...It is possible to measure position and momentum simultaneously…a single measurement of a particle. In fact, we often measure the momentum by measuring the position and interpreting the result as a momentum measurement. What we can't do is to prepare an identical set of states…. such that we would be able to make an accurate prediction about what the result of a position measurement would be and an accurate prediction about what the result of a momentum measurement would be….for an ensemble of measurements.

from the prior post:
that is, if one is thinking classically; such is not the case with quantum mechanics!! Particles may have well-defined positions at all times, or they may not ... the statistical interpretation does not require one condition or the other to be true."

"I think we're closing in on an answer to my original question: There is no known argument or experiment that can completely rule out the possibility that particles have well-defined positions at all times, but we can rule out the possibility that the only significance of the wavefunction is to describe the statistical distribution of particles with well-defined positions."

Here are my own summary notes from those earlier discussions: [Some will likely disagree with the descriptions I have chosen.]

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Last edited by a moderator: May 6, 2017
4. Oct 15, 2012

### Fredrik

Staff Emeritus
I think maybe you forgot what I said here.

That pdf is Ballentine's article. My claim that you can measure both at the same time was based on the validity of Ballentine's argument. But Demystifier's argument convinced me that what Ballentine considered a momentum measurement shouldn't have a distribution of results that are consistent with the predictions of QM.

Honestly, I'm still unsure about some of this. ZZ posted an article somewhere about a technique called ARPES that's used to measure momentum components. I still haven't tried to really understand the article, so I don't know what to make of it, but my first impression (based on things I no longer remember) was that this technique is similar enough to what Ballentine described, that Demystifier's argument might apply to it. But this is a technique that experimental quantum physicists use, and some of them should know what a momentum measurement is better than I do. So I have a feeling that there's something about this that I still don't understand.

5. Oct 15, 2012

### jmcelve

It's nice to see that there's still some uncertainty (!!) regarding these questions. I don't feel so alone in pondering them now.

I'll have to consider what you've said more thoroughly, Naty1, but thank you for the response.

The same thanks go to you Jano, though I think Naty1 is correct in his response to your statement:

I suppose the next place to start is to read Ballentine's discussion of measurements for a quantum system.

Last edited: Oct 16, 2012
6. Oct 16, 2012

### strangerep

Probably better to read his more modern textbook instead of his much older paper on the statistical interpretation, imho.

7. Oct 16, 2012

### Naty1

I finally found Zappers blog.....on HUP:

Misconception of the Heisenberg Uncertainty Principle.

http://physicsandphysicists.blogspot.com/2006/11/misconception-of-heisenberg-uncertainty.html

Fredrik: "I think maybe you forgot what I said here....

yeah, well that was post #287!!! [LOL]
and my notes are so long I never got that far this time around.....[I amended my notes this time!]

Fredrik: " Honestly, I'm still unsure about some of this."

But I did already post: "...Different people have different interpretations ...going on for over 90 years" so at least I got that part right.

Last edited: Oct 16, 2012
8. Oct 16, 2012

### Naty1

jmcleve:

Just note I was quoting from earlier discussions; I'm lucky when I THINK I understand what
the experts are posting.

Not only are QM conceptual interpretations elusive and subtle, but QM also suffers from people drawing different conclusions about a single statement. These forms are rife with
QM statements being dissected...as,say, three different people drawing three different interpretations from a single statement.

I even bought a huge Albert Messiah 'Quantum Mechanics' combined two volumes hoping to find carefully edited explantions that would be clear and unambiguous.....no such luck!
People here objected to some wording there,too!! Glad it was used and cheap!!

I still like the 'Shut up and calculate.' explanation.

9. Oct 16, 2012

### Naty1

While I was skimming my notes to update based on Fredrik's comment above, I came across this statement from the previous discussion:

I'm not sure I interpret this as did the original poster.....nor the same as when I recorded it. I read this to say, paraphrasing, 'Even if you could prepare an ensemble of identically prepared, precisely the same, particle states [which you can't] the nature of Fourier synthesis prevents one from make simultaneous sharp measurements.

Anybody else see it this way??

10. Oct 16, 2012

### Jano L.

Of course, but you deviated from the original question of jmcelve. The question was

I have tried to explain that the uncertainty relation based on $\psi$ has no bearing on the individual particle. I can try to explain better, if my argument was not clear.

I did not say that the statistical interpretation REQUIRES that the electron HAS position and momentum simultaneously; there is no point on insisting what electron HAS or IS.

However, it is natural to think r,p can be used to describe the electron, for many reasons. The statistical interpretation in the sense of Bohm or Ballentine showed that this is not in contradiction to experiments or the rest of the probabilistic theory based on the wave function.

The expression "statistical interpretation" easily gets unclear. In 30's Heisenberg would say that the Copenhagen interpretation is the statistical interpretation and deny applicability of simultaneous position and momentum to one particle. But, after works of Einstein, Bohm and Ballentine we have every reason to believe that the results of the theory do not rest on the denial of existence of r,p for particles. Nowadays, I think it is more appropriate to understand the expression "statistical interpretation" as referring to a theory in which $\psi(\mathbf r)$ gives probability in coordinate space but says nothing about the r,p of the individual particle.

11. Oct 16, 2012

### Naty1

nope....the only part I tohught MIGHT need clarification I commented upon. In fact my
last post matches yours, I think, nicely.

12. Oct 16, 2012

### Jano L.

I almost agree with your last post, apart from the sentence

What can be measured depends on theory one adopts. If one denies r,p in theory, of course in that theory one cannot measure them. This is the case when we adopt the idea "the particle is a wave and since the wave has neither position nor momentum the particle does not have them either".

But if we assume that the wave function is just an auxiliary function (lives in configuration space!) used to describe particles, and allow r,p for particles, they may be measured and no nature of Fourier transformation can prevent this.