Heisenberg imcertainty principle (get it)

In summary, the conversation discusses the difficulty of finding the exact position and velocity of an orbiting electron and the purpose of knowing this information. It also touches on the Heisenberg Uncertainty Principle and the misconception that it is about the uncertainty in a single measurement. The conversation also mentions a website that perpetuates this misconception and a more accurate explanation of the HUP is provided. Finally, the conversation discusses the use of theories and technology in understanding and observing electrons.
  • #36
ZapperZ said:
Then if you think we have characterized it wrong, you should not fly commercially, drive your car, seek medical treatment, believe in the value of h and e, etc.

Zz.

I was just kidding. Your doing a great job!
:smile:
 
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  • #37
Again, nothing from the experiment above prevents me from obtaining a definite value of position and momentum from a single measurement. The uncertainty in these values are not governed by the HUP, nor are they related. If you are still claiming that they are, then tell me how my ability to change the width of the slit affects the density of the number of pixel on my CCD plate at the detector.

For the most part I agree with everything you say Zz. Let me be extremely
clear with the simplest example.

As you say, you certainly can measure position and momentum at the
same time and get definite values. And the uncertainties in the accuracy
would be related to your measuring technology exactly as you say.

Suppose you have a single free particle in a definite momentum state
(planewave, no localized position of the wavefunction.) Then you can
measure the momentum all day long and get the same definite value
each time you measure it without ever disturbing that momentum. It
is a predictable and definite value.

But if sometime along the way you also attempt to localize the particle
(that is, make a position measurement) then that position measurement
will introduce an uncertianty into any subsequent momentum measurement.

The uncertainty is proven out by the second measurement. But the
outcome (of a decreased predictability in the mometum value) was sealed
the instant you made the first position measurement.

In fact, the HUP applies even to wave packets that are not being
measured at all, like a minimum-uncertainty gaussian becasue it is a
purely mathematical consequence of the fact that specific positions are
built up of planewaves of all possible momenta, and a specific momentum
is a planewave which covers all of space (no particular position).
 
  • #38
Antiphon said:
But if sometime along the way you also attempt to localize the particle
(that is, make a position measurement) then that position measurement
will introduce an uncertianty into any subsequent momentum measurement.

In what form is the uncertainty introduced in a SINGLE measurement of the momentum? I have a very thin slit. At some time, your free particle passed through it. I say "Ah ha! At time t1, there was a particle going through position x1, and my uncertainty in the position is +- width/2!"

But being smart, I also put a CCD screen behind the slit. I can record, to ARBITRARY ACCURACY limited by my CCD technology, where the particle hit the detector. The lateral position (in the direction perpendicular to the slit) of where the particle hit tell me the momentum in this direction. The uncertainty in this momentum depends only on the uncertainty in determining where the electron hits the CCD. In fact, the higher the resolution of the CCD, the higher the accuracy with which I can determine the position the particle hits the detector. I have made measurement in which the accuracy is has high as 2 pixels on a CCD! This has zero to do with the width of the slit or the HUP. And voila, I have made a DEFINITE single measurement of position x AND momentum p_x of your particle.

I have just done what you said cannot be done.

In NONE of these have I said anything about "predictability". All I care about is the question "can I make as an accurate of a position and momentum measurement of a single particle?"

The answer is :yes. This is because that question depends on the technology of detection.

But the question: "if I make the position measurement VERY, VERY accurate (very small slit), then can I predict with equal accuracy where there the particle will hit the CCD and thus, determine it's transverse momentum?" The answer is NO.

These two are DIFFERENT QUESTIONS under different circumstances. You cannot say I cannot make accurate single measurement of position and momentum. I can, and HAVE done so. What limits me from doing it to infinite accuracy is the technology of detection, not the HUP. The HUP kicks in in my ability to PREDICT the outcome of such a measurement! That's a different question! Just because I have a plane wave and well-defined momentum, it doesn't mean any single measurement of position is undefined as if you will instead get a SMEAR all over your detector because THAT particle is supposed to be delocalized. No such thing has ever been detected. Instead, what is smeared is the position of REPEATED measurement of the identical system! When faced with such smearing of many position measurement, one will conclude that predicting where the particle is going to be is hopeless!

I have tried to explain this in painful detail in my journal entry on this. Obviously, if you have read it, I haven't done it with the clarity that I thought I did. A definite single measurement doesn't imply a definite knowledge of the system. I can make very good measurement of position and momentum - this not the HUP. However it doesn't mean that my knowledge of that can be used to predict with arbitrary accuracy the NEXT time I make the same measurement. Now THIS is the HUP!

Zz.
 
  • #39
HUP should always be placed in it's historicaly context as the basis of the
Copenhagen Interpretation: http://www.aip.org/history/heisenberg/p09.htm


HUP is a derived law.

Basically from Planks Law E=hf, p=h/λ relating Energy/Momentum with
Space/Time via the Fourier Transform, and applied on the Gaussian "Bell"
curve for statistical uncertainty.

Original page of Heisenberg's derivation plus explanation:
http://www.aip.org/history/heisenberg/p08a1.htm

It was only later corrected by a factor of 2.. ([itex]\hbar \rightarrow \hbar/2[/itex])

Often metaphysical talk creeps in here, when taken outside the original
scope, like: "One can choose to measure the position exact but then one
can not meaure the momentum and visa versa: One can choose to
measure momentum exact but one can not measure position.

There is no "mind over matter control" choise. When both are measured
simultanously they may be exact to all the digits of the (digital) meter.
Only repeated measurements will reveal how far they were off from the
statistical averages. The ranges (and the center values) are always pre-
determined by the describing wave-function. There's no way to influence
that afterwards.

Even worse is: "The measurement itself disturbs the precission"
Improving thechnology will continue to make more exact and less disturbing
measurements.



Regards, Hans
 
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  • #40
ZapperZ said:
In what form is the uncertainty introduced in a SINGLE measurement of the momentum?

The uncertainty is introduced as follows; when the planewave was coming
in it had a definite momentum which means you knew what the momentum
would measure as before you measured it or if you measured it 6 times, you got the same
value 6 times. But after a position measurement the momentum is now
a planewave spectrum, NOT a definite value. You can of course MEASURE
only a definite value so it's the NEXT moementum measurment that FIXES
some random definite value. But between your slit and detector there was
NO DEFINITE VALUE to the momentum. It just didn't exist as a particular number.

In what range can we expect the modified momentum to be the next time
we measure it? That depends exactly and only on how tightly the position
measurment localized the particle.

I have a very thin slit. At some time, your free particle passed through it. I say "Ah ha! At time t1, there was a particle going through position x1, and my uncertainty in the position is +- width/2!"

Ok. You now have an uncertainty in your ability to PREDICT the value of
any subsequent momentum measurement. This uncertainty is of the order
[tex] \delta p = \hbar / (2 \times slitwidth) [/tex]. This uncertainty in the momentum didn't exist in my
planewave but it now exists because of the dimensions of your slit and
only after you register the particle having passed through it at time [tex] t [/tex].

But being smart, I also put a CCD screen behind the slit. I can record, to ARBITRARY ACCURACY limited by my CCD technology, where the particle hit the detector. The lateral position (in the direction perpendicular to the slit) of where the particle hit tell me the momentum in this direction.

This is now a second measurement, and yes you can record to arbitrary
accuracy where the particle hits.

The uncertainty in this momentum depends only on the uncertainty in determining where the electron hits the CCD. In fact, the higher the resolution of the CCD, the higher the accuracy with which I can determine the position the particle hits the detector. I have made measurement in which the accuracy is has high as 2 pixels on a CCD! This has zero to do with the width of the slit or the HUP. And voila, I have made a DEFINITE single measurement of position x AND momentum p_x of your particle.

As I pointed out in the earlier post, the uncertainty relation does not
address the two-measurement case. Einstein pointed this out himself
with an arrangement similar to the one you have here. A definite value
for the momentum is actualized by your sensor. The particle has an
indefinite momentum after going through your slit whereas it had a definite
mometum before the slit.

I have just done what you said cannot be done.

No, you have made two measurements. I said after a single precise position
measurement it no longer made sense to say that the particle has a definite
momtum. It will ACTUALIZE a definite mometum if and only if a momentum measurment is made, like your CCD device.

In NONE of these have I said anything about "predictability". All I care about is the question "can I make as an accurate of a position and momentum measurement of a single particle?"

The answer is :yes. This is because that question depends on the technology of detection.

- but not at the same time. For a single measurment, the answer is no.


But the question: "if I make the position measurement VERY, VERY accurate (very small slit), then can I predict with equal accuracy where there the particle will hit the CCD and thus, determine it's transverse momentum?" The answer is NO.

Correct, because of the HUP.

These two are DIFFERENT QUESTIONS under different circumstances. You cannot say I cannot make accurate single measurement of position and momentum. I can, and HAVE done so.

Let's be clear. If your slit is tiny then: You made a position measurement
at the slit and have accurate position information. But now you don't
know the momentum because THERE ISN'T a DEFINITE momentum anymore.
Then you made a moementum measurement and got an UNPREDICTIBLE but
definate result. No problem with this.

What limits me from doing it to infinite accuracy is the technology of detection, not the HUP. The HUP kicks in in my ability to PREDICT the outcome of such a measurement!

Yup, and the fact that there was no definte value for the momntum
between slit and CCD, hence the reason for the Heisenberg unpredictability.

That's a different question!

It's the only case in which HUP makes any sense here.

Just because I have a plane wave and well-defined momentum, it doesn't mean any single measurement of position is undefined as if you will instead get a SMEAR all over your detector because THAT particle is supposed to be delocalized. No such thing has ever been detected.

Of course not, because that's all wrong. You WILL get a definite position
outcome from a position measurment of a planewave. But it will cease
to be a planewave thereafter which is the whole point of the HUP.

A definite single measurement doesn't imply a definite knowledge of the system.

I can make very good measurement of position and momentum - this not the HUP.

...I've said, not at the same time becuase this is the HUP.
 
  • #41
This is getting mind-boggling by the minute.

Here's the starting point. A and B are 2 non commutating operators. The HUP that most people can recite is that if you measure observable A, let's say, then the greater the certainty you know about A, the LESS certain you know about B.

Now so far so good.

However, here's where things are bastardized. They then go on by saying that if one were to measure A very precisely, then B is undefined and one can't measure B with any kind of accuracy. I have seen this repeated many times, even on PF. This is what drove me to write the lengthy explanation on why this is utterly false. A single value of A and a single value of B can be measured with arbitrary accuracy that does not depend on each other's accuracy. This is an instrumentation accuracy.

Now you brought up the "at the same time" issue, which I find rather strange. The whole concept of commutation relation IS the order of measurement, that AB is not the same as BA. How and where does this imply a "simultaneous" measurement? The presence of the slit is similar to causing the wavefunction to "collapse" to a single position eigenstate. NO ONE, not even me, ever argued that the momentum part is still undermined. However, it CAN be measured and when we do, it will reveal a SINGLE value the same way we measured the "single" value for the position! When I do that, I have 2 values - position with its uncertainty, and momentum with its uncertainty.

The question is, in that single measurement, are the uncertainties associated with those two values THE uncertainties in the HUP? I say no. Now is THIS what you are disputing?

Zz.
 
  • #42
ZapperZ said:
The question is, in that single measurement, are the uncertainties associated with those two values THE uncertainties in the HUP? I say no. Now is THIS what you are disputing?

Zz.
After the brief discussion related to HUP on the other thread I was almost convinced that I had simply misinterpreted your statement about HUP being a derived consequence rather than a fundamental principle, so I followed up by reading the dialog over here to better understand your interpretation. Now I see that we really do have a fundamentally different understanding of the HUP.

As I (and I believe everyone else with whom I have ever discussed this topic) understand it, the answer to your last question above is YES. What I believe is being disputed by Antiphon, as I understand his posts, and certainly by myself is your notion that simultaneous measurements of the position and momentum of a single particle can be made to any degree of precision, limited only by detection technology. The reason you see so many instances of a different interpretation of HUP from yours is simply that a lot of people have a different understanding of what it means. So either there are a lot of us misguided folk out here, or your interpretation is incorrect.

I have to agree with Antiphon that your description of the position and momentum measurements at the slit and the screen do not constitute simultaneous measurements of position and momentum. Your argument appears to rest on the assumption that particles have a highly localized momentum that is determined when they pass through the slit, which becomes manifiest when they are are detected by the screen. Implicit in that assumption is that the particle follows a direct path from the slit to the screen, never deviating from that path by more than a slit width and forcing it to hit a localized spot on the screen.

I find this interpretation to be contrary to the postulates of QM. The wave function does not confine the particle to a specific path between the slit and the screen, or give it a nearly specific momentum. In fact, the measurements you have described do not directly measure the momentum of the particle at all. What you have described is two position measurements, one at the slit and a second one at the screen. You then deduce the momentum by assuming you know how the particle got from the first location to the second. The interpretation that Antiphon has spelled out is that the first position measurement that localizes the particle at the slit necessarily leaves that particle in a state of uncertain momentum, centered in the forward direction with momentum components equally distributed to either side. The wave function does not tell us how the particle got from the slit to the screen. It only gives the probability that when the second measurement is made, the particle will be found in any localized area of the screen.

In wave mechanics, the wave function characterizes the state of individual particles. It is not just a distribution function for an ensemble of particles. An eigenfunction of the momentum operator has infinite spatial extent. The wave function of a localized particle is a wave packet consisting of a distribution of momentum eigenfunctions. The more localized the wave packet, the wider the distribution of momenta, and the wider the distribution of momenta the faster the wave packet spreads with time. The manifestation of this in the slit experiment is that the smaller you make the slit and the farther away you place the screen, the wider the wave packet will be when it reaches the screen, resulting in a wider pattern of particle hits on the screen.
 
  • #43
Heisenberg's Uncertainty Principle (HUP)

Some claim that it should be labeled Heisenberg's Indeterminacy Principle (HIP) unfortunately for them alternative views are still valid (HIP is intepretation laden with Copenhagenism and related views/interpretations). Thus from the beginning we must make a clear distinction between the different interpretations of HUP, as of now there are two main interpretations:

1. The 'soft' version, which is compatible with all valid interpretations of QM, simply says that we cannot measure complementary parameters (such as position-momentum or energy-time) simultaneously with infinite precision. The minimal explanation of this is the 'interactionist' hypothesis, according to it this principial limitation is due to the existence of a finite quanta of action (different from 0), the perturbations which appear being unfathomable (irrespective of the technology used). No assumption about the intrinsic nature of the universe (deterministic or not) is made.

2. The 'strong' version, theory ladden (with the Copenhagen interpretation and related views), goes way further by saying that a quantum particle does not have simultaneously a precise momentum and position (this being also the explanation for our principial incapacity to know). In other words nature is inherently random.

Well in spite of the actual preference in the scientific field I find the first interpretation the most rational one (though of course the latter is also rational) based on all evidence we have and the fact that there is subdetermination at quantum level. Indeed the true nature of our universe (deterministic or not) is still an open problem, all the experiments done so far are also totally compatible with fully deterministic interpretations of QM, involving non local hidden variables. There is no experimental way, as of now at least, which to make a compelling difference between existing interpretations, the Copenhagen Interpretation (and further improvements) is not at all really superior empirically to the others. Simply some variants of copenhagenism at least, have a higher degree of coherence with GR, that is with the main body of actually accepted theories, but this is not at all a necessary sign of truth.

Some touched the problem wether HUP (at least the 'softer' version) is really a prediction of the standard formalism of QM. Well it can be argued that HUP can be deduced from the standard formalism of QM but we must be very careful here for we should interpret those results statistically (consistent with Born's interpretation of wavefunction) or Heisenberg claims that the principle holds for every singular case...this does not follow (strictly deductively) from quantum mechanics and moreover there are problems with the 'frequentist' interpretations of probability.

To expand a bit the last idea is the 'softer' version of HUP (Heisenberg's Uncertainty Principle) really an universally valid prediction of the standard formalism? As I've already undelined the usual 'frequentist' interpretation of probabilities is incompatible with the claim of Heisenberg that we have the right to extend the probabilistic meaning of HUP, as derived from standard QM, to small samples or single events. The predictions made by the standard formalism of QM referrs at statistically relevant samples, identically prepared (indistinguishable practically), thus the uncertainty relations as derived from the standard formalism of QM are valid only statistically.

To extend them at singular particles would imply to assume that the wavefunction defines the status of singular particles in the sample too, in contradiction with Born's statistical interpretation assumed initially. It is compatible with a bayesian interpretation of probabilities (vastly involved in scientific practice) indeed but in the absence of any clear interpretation of probabilities neither is there a clear answer regarding its real meaning (Bayesianism has its own problems, important ones). Unfortunately some thought experiments/'Gedanken' experiments and great coherence with other (still) accepted parts of science are not enough to back Heisenberg's strong claim that HUP holds in the case of all imaginable experiments/ thought experiments.

So that, at limit, we cannot even say that HUP (the 'softer', general definition) is a prediction of the standard formalism of QM valid for all cases (including single particles). Anyway even accepting that the 'softer' version of HUP is an universal prediction of the standard mathematical formalism (this is a fully descriptive/formal deduction not a causal deduction the only one which can give real understanding by answering the 'why HUP' question!) the 'hard' fact remain: the 'indeterminacy' is totally related to the copenhagenist, rather positivist, approach. I think it would be safer to always mention after saying that Nature is inherently indeterministic the disclaimer 'at least according to the Copenhagen Interpretation and further 'improvements', preferred currently by a majority of scientists'.
 
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  • #44
OlderDan said:
As I (and I believe everyone else with whom I have ever discussed this topic) understand it, the answer to your last question above is YES. What I believe is being disputed by Antiphon, as I understand his posts, and certainly by myself is your notion that simultaneous measurements of the position and momentum of a single particle can be made to any degree of precision, limited only by detection technology. The reason you see so many instances of a different interpretation of HUP from yours is simply that a lot of people have a different understanding of what it means. So either there are a lot of us misguided folk out here, or your interpretation is incorrect.

I have to agree with Antiphon that your description of the position and momentum measurements at the slit and the screen do not constitute simultaneous measurements of position and momentum. Your argument appears to rest on the assumption that particles have a highly localized momentum that is determined when they pass through the slit, which becomes manifiest when they are are detected by the screen. Implicit in that assumption is that the particle follows a direct path from the slit to the screen, never deviating from that path by more than a slit width and forcing it to hit a localized spot on the screen.

And this is where I am baffled. Maybe I blanked out when I wrote "simultaneous measurement" earlier in this thread (did I ever?). However, if you read either my journal entry on the Misconception of the HUP, or my last entry regarding what really is meant by non-commutation, I don't think I ever implied making an "instantaneous" measurement of BOTH non-commuting observables!

The WHOLE issue that prompted my original essay on this in my journal was in direct response to the repeated claims that once one has made a measurement of one observable, one can no longer THEN make any accurate measurement of the other! This means that in the single-slit case, AFTER I have determined the position with very "good" certainty (particle passing through a slit), what that statement is saying is that I can no longer determine with any reasonable accuracy the momentum of the particle. This is FALSE, and I can demonstrate EXPERIMENTALLY, not just in principle, that this is false. All I need to know is where the particle hits the detector AFTER the slit. I have then determined its momentum. The accuracy to which I measure that momentum depends ENTIRELY on the accuracy of my detector. This accuracy has nothing to do with the width of the slit.

So up to this point, which part is in dispute? I have made ZERO assumption about "superposition of many momentum", etc... etc. ALL I care about is determining the position FIRST, and then determining the momentum NEXT. This is purely an act of measurement, something experimentalists like me like to do.

Now if you ASK me to tell you what the momentum of the NEXT particle that passes through the slit will be, then my ability to accurately give you the answer depends VERY MUCH on the width of the slit, i.e. the degree of certainty of the position measurement. The smaller the slit (the smaller Delta(x)), the less certain is my ability to predict where the particle will hit the detector, and thus, the larger the Delta(p_x) will be.

So again, up to this point, which part is in dispute?

Note that I HAVE done similar measurments that I have described. The electrons in the conduction bands of metals are described almost by sum of plane waves. In a photoemission experiment, the in-plane momentum of these electrons are conserved upon being photoemissted from the surface plane. This means that BEFORE a measurement, it has the same sum of various plane waves. If I use a hemispherical electron analyzer such as the Scienta SES, I can make as accurate determination of the momentum as I want, limited ONLY by the pixel size on my CCD screen! It is accepted that what is detected IS the in-planed momentum of the electron while it is in the material - there is a clear one-to-one correspondence! How do we know this? The band structure we obtained agrees with theoretical band structure of "standard metals"![1]

The accuracy of determining where one particle is, and its momentum is detector dependent. This is not the HUP. This is INSTRUMENTATION!

Zz.

[1] T. Valla et al., PRL v.83, p.2085 (1999).
 
  • #45
ZapperZ said:
And this is where I am baffled. Maybe I blanked out when I wrote "simultaneous measurement" earlier in this thread (did I ever?). However, if you read either my journal entry on the Misconception of the HUP, or my last entry regarding what really is meant by non-commutation, I don't think I ever implied making an "instantaneous" measurement of BOTH non-commuting observables!

The accuracy of determining where one particle is, and its momentum is detector dependent. This is not the HUP. This is INSTRUMENTATION!

Zz.

I'm truly sorry if I keep misinterpreting you, but in fact you have made statements that sound like you are making a claim for simultaneously meausring both quantities. One of them is the last paragraph in the quote above, and another is the quote that follows

ZapperZ said:
But being smart, I also put a CCD screen behind the slit. I can record, to ARBITRARY ACCURACY limited by my CCD technology, where the particle hit the detector. The lateral position (in the direction perpendicular to the slit) of where the particle hit tell me the momentum in this direction. The uncertainty in this momentum depends only on the uncertainty in determining where the electron hits the CCD. In fact, the higher the resolution of the CCD, the higher the accuracy with which I can determine the position the particle hits the detector. I have made measurement in which the accuracy is has high as 2 pixels on a CCD! This has zero to do with the width of the slit or the HUP. And voila, I have made a DEFINITE single measurement of position x AND momentum p_x of your particle.

I can't see how you can make this statement, but I believe it is representative of the statements you have made that make some of us think you are talking about "simultaneous measurement" of both observables. You have not made a single measurement that determines the position AND the momentum of the particle. You have made two measurements of the postion of the particle at two different times, and from those deduced what the momentum of the particle must have been between those two measurements to get the particle from the first position to the second position.

If you had chosen the word OR instead of the word AND when making your statements about measuring both position and momentum, then I would have no problem agreeing with what you are saying.

ZapperZ said:
The WHOLE issue that prompted my original essay on this in my journal was in direct response to the repeated claims that once one has made a measurement of one observable, one can no longer THEN make any accurate measurement of the other! This means that in the single-slit case, AFTER I have determined the position with very "good" certainty (particle passing through a slit), what that statement is saying is that I can no longer determine with any reasonable accuracy the momentum of the particle. This is FALSE, and I can demonstrate EXPERIMENTALLY, not just in principle, that this is false. All I need to know is where the particle hits the detector AFTER the slit. I have then determined its momentum. The accuracy to which I measure that momentum depends ENTIRELY on the accuracy of my detector. This accuracy has nothing to do with the width of the slit.
I have no disagreement with the first part of this. I agree that the claims you are attributing to others are FALSE. Measuring the position of the particle at some point by passing it through a slit demands that its wave function be a superposition of a wide range of momentum eigenfunctions, anyone of which could become the momentum observed in a subsequent momentum measurement. The point that needs to be clear is that you cannot predict which of those momentum values is going to be measured before making the measurement. I agree with you that passing the particle through the slit does not preclude making a later momentum measurement with arbitrary precision.

The second part of your last paragraph however gives me pause. If measuring the position of the particle by passing it through a slit forces it into a state where its momentum is distributed over a wide range of values, how can detecting its arrival on a screen within one or two pixels be interpreted as a measurment of its momentum with high precision? The measurement that locates the particle on the screen should have exactly the same effect with regard to its momentum as the position measurement at the slit. The smaller you make the grid of your CCD to precisely locate the position of the particle, the less you know about its momentum at the time of that position measurement.

I question whether you are making momentum measurements at all in the quantum sense. A momentum measurement would require that you somehow determine the wave vector of the particle, and a relatively precise measurement of the wave vector can only be obtained if the position of the particle is relatively unknown. That would require something like a phased array that gives up precision of position in order to achieve precision in determining direction of arrival. I'm not convinced that a classical view which says that to get from one point to another in a certain amount of time the particle must have had a certain momentum during transit yields a valid measurement of the particle's momentum at the screen, or at the slit, or anywhere in between. I question your claim that you have measured the particle's momentum by determining where it hits the screen.
 
  • #46
OlderDan said:
I can't see how you can make this statement, but I believe it is representative of the statements you have made that make some of us think you are talking about "simultaneous measurement" of both observables. You have not made a single measurement that determines the position AND the momentum of the particle. You have made two measurements of the postion of the particle at two different times, and from those deduced what the momentum of the particle must have been between those two measurements to get the particle from the first position to the second position.

If you had chosen the word OR instead of the word AND when making your statements about measuring both position and momentum, then I would have no problem agreeing with what you are saying.

Ignoring that fact that I did not explictly mention the word "simultaneous" in that paragraph, what if I said instead "Voila! I have made a definite measurement of postion, and I have also made a definite measurement of momentum"?

There is never any "time period" in applying AB and then BA. All that implies is that one applies one AFTER the other, in sequence. I assumed that this is known. Obviously, I should have stressed that.

The second part of your last paragraph however gives me pause. If measuring the position of the particle by passing it through a slit forces it into a state where its momentum is distributed over a wide range of values, how can detecting its arrival on a screen within one or two pixels be interpreted as a measurment of its momentum with high precision? The measurement that locates the particle on the screen should have exactly the same effect with regard to its momentum as the position measurement at the slit. The smaller you make the grid of your CCD to precisely locate the position of the particle, the less you know about its momentum at the time of that position measurement.

I question whether you are making momentum measurements at all in the quantum sense. A momentum measurement would require that you somehow determine the wave vector of the particle, and a relatively precise measurement of the wave vector can only be obtained if the position of the particle is relatively unknown. That would require something like a phased array that gives up precision of position in order to achieve precision in determining direction of arrival. I'm not convinced that a classical view which says that to get from one point to another in a certain amount of time the particle must have had a certain momentum during transit yields a valid measurement of the particle's momentum at the screen, or at the slit, or anywhere in between. I question your claim that you have measured the particle's momentum by determining where it hits the screen.

At the risk of repeating what I have written in my journal, here we shall go again.

1. Slit is at postion y1 with a width of Delta(y), slit oriented along x-direction. Plane wave particles moving along z-direction impinging normal to the slit.

2. Particle passes through the slit. When this occurs, all I can say is that at that instant, the particle is at y1 location where the slit is, and my uncertainty in its position corresponds to the slit width Delta(y).

3. Particle leaves the slit, hits a detector at location L after the slit. The detector produces an image at the a y-location, call that y2.

4. If the slit is small enough, y2 can differ, and differ greatly from y1. So it has "drifted" off center along the y-direction by an amount y2-y1, i.e. it has a y-component of momentum, something it didn't have before.

5. You know the z-momentum before, and since you didn't do any position constaints along the z-direction (and x-direction), the p_z momentum remains unchanged (there can be relativistic corrections here if necesary which is way too long to describe). You then know how long it takes for the particle to travel from the detector.

6. It is then a matter of geometry and straightforward mechanics to find the y-component of velocity and momentum. I now have p_y.

The accuracy of measuring y1 and y2 are instumentation accuracy. They are not dictated by the HUP. As I know more about y1, the pixel size on my detector do not automatically becomes larger, nor is the spot size on the detector when the particle hits it blows up.

Nowhere in here did I claim to measure ALL the momentum components, only the one affected by the presence of the slit (position measurer). In the Scienta analyzer, the orientation of the slit determines in which crystallographic direction are measuring the momentum. We NEVER measure the full momentum since (i) the out-of-plane momentum is not conserved and (ii) the slit is only along one particular direction.

So again, what is in dispute here?

Zz.
 
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  • #47
ZapperZ said:
6. It is then a matter of geometry and straightforward mechanics to find the y-component of velocity and momentum. I now have p_y.

The accuracy of measuring y1 and y2 are instumentation accuracy. They are not dictated by the HUP. As I know more about y1, the pixel size on my detector do not automatically becomes larger, nor is the spot size on the detector when the particle hits it blows up.

Nowhere in here did I claim to measure ALL the momentum components, only the one affected by the presence of the slit (position measurer). In the Scienta analyzer, the orientation of the slit determines in which crystallographic direction are measuring the momentum. We NEVER measure the full momentum since (i) the out-of-plane momentum is not conserved and (ii) the slit is only along one particular direction.

So again, what is in dispute here?

Zz.
The dispute is that a classical mechanical computation of a change in position divided by a change in time is being used to deduce the momentum of a quantum particle everywhere along a trajectory between two position measurements. The wave function of the particle that is squeezed through the first slit is collapsed to a localized wave packet in the y direction, but not into a y-momentum eigenstate, or even a narrow spectrum of y-momentum eitenstates. The smaller you make the slit, the less you can know about the y-momentum of the particle. The second measurement of the postion of the particle collapses the wave function a second time, but it does not determine the momentum of the particle. If it did, then if the screen had a single second slit the trajectory of the particle would continue along the extended line between the slits. That is not what QM says will happen. If there were a second slit you would have the same uncertainty in the momentum of the particle passing through the second slit that you had when it left the first slit. If there were multiple slits in the screen, the particle would not be forced to pass through anyone of them and you would have a multi-slit interference distribution on another screen farther along the path. I submit that you have not measured the momentum of a single particle by capturing it on the screen. By repeating the experiment with many particles you have found the spread of the wave packet at the screen, from which you can infer the momentum distribution of the wave function of the particles that made it through the first slit. That distribution is consistent with the degree of momentum uncertainty suggested by the HUP.
 
  • #48
OlderDan has articulated everything I was thinking but has done a much better job
of expressing it, and using the correct terminology as well. ZapperZ, I think you do
have a handle on the HUP. But there must be a subtle miscommunication between
us all about it which still leaves me feeling uneasy as I cannot spot it.

Nevertheless I think the discussion in this thread of the HUP speaks for itself for
those who are able to follow it, so I will not add to it any further. I encourage
the rest of you to go on though and I will follow the thread from a distance.
 
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  • #49
OlderDan said:
The dispute is that a classical mechanical computation of a change in position divided by a change in time is being used to deduce the momentum of a quantum particle everywhere along a trajectory between two position measurements. The wave function of the particle that is squeezed through the first slit is collapsed to a localized wave packet in the y direction, but not into a y-momentum eigenstate, or even a narrow spectrum of y-momentum eitenstates. The smaller you make the slit, the less you can know about the y-momentum of the particle. The second measurement of the postion of the particle collapses the wave function a second time, but it does not determine the momentum of the particle. If it did, then if the screen had a single second slit the trajectory of the particle would continue along the extended line between the slits. That is not what QM says will happen. If there were a second slit you would have the same uncertainty in the momentum of the particle passing through the second slit that you had when it left the first slit. If there were multiple slits in the screen, the particle would not be forced to pass through anyone of them and you would have a multi-slit interference distribution on another screen farther along the path. I submit that you have not measured the momentum of a single particle by capturing it on the screen. By repeating the experiment with many particles you have found the spread of the wave packet at the screen, from which you can infer the momentum distribution of the wave function of the particles that made it through the first slit. That distribution is consistent with the degree of momentum uncertainty suggested by the HUP.

But you somehow are ignoring the fact that I DID measure a location on the detector and should be able to deduce THAT particular momentum of THAT particle for THAT instant. I didn't say that this is the ONLY possible momentum for the NEXT particle. In fact, if I repeat this a gazillion times, I will get a spread in the measured value of momentum. This spread would correspond exactly to the HUP as dictated by the slit width!

If you have a problem with this, then you should also have a problem with Bell-type experiments. According to your position, the polarization that I would measure is simply ONE of all the possible superposition of polarizations and therefore, not the "true" polarization.

Again, I would appeal to the ALREADY established method that is used in angled-resolved photoemission spectroscopy.[1] What I had described is no different than what is done is this technique. If we are getting the "wrong" momentum, then the published results are nonsense and faulty and have nothing to do with any kind of "momentum". If you think this is so, then it is imperative that you alert the various publications on this.

Zz.

[1] Valla et al., Science v.285, p.2110 (1999).
 
  • #50
ZapperZ said:
But you somehow are ignoring the fact that I DID measure a location on the detector and should be able to deduce THAT particular momentum of THAT particle for THAT instant. I didn't say that this is the ONLY possible momentum for the NEXT particle. In fact, if I repeat this a gazillion times, I will get a spread in the measured value of momentum. This spread would correspond exactly to the HUP as dictated by the slit width!
I am not ignoring the fact that you measured the location of the particle on the detector screen. What I said, very explicitly, is that the measurements you made are both position measurements at two different moments in time and that you then did a classical calculation to deduce the momentum of the particle from those two measurements. You did not measure the momentum of the particle in either location. You assumed that because you knew where the particle was in two different locations that the particle must have taken a direct route to get from one place to the other. In post #24 in this thread you stated

ZapperZ said:
1. How well I know the location of a photon depends on how wide a slit I make, no? The smaller the slit, the more I know about where that photon was when it passed by it.

2. When it passed by the slit, what is its transverse momentum perpendicular to that slit? Unless its momentum changed between the slit and the detector, then I can make the assumption that this momentum remained the same between the moment it passed through the slit and the moment it hits the detector. All I need to do is figure out WHERE on the detector it hits. Then, using simple geometry, I know it's momentum in that direction. How well I determine that momentum depends on the resolution of my detector, i.e. how well can I determine where it hits the detector. The larger the number of pixels on my CCD camera, for example, the finer I can determine this location.
I can't quite pin down what you are saying about "simultaneous" measurements of position and momentum of a siongle particle. When I bring up the word, you seem to not want it attributed to yourself as in your remark in #46, but this is what you said in #6 (the bold highlights are yours, not mine)

ZapperZ said:
The HUP is NOT about the uncertainty in INSTRUMENTATION or measurement. One can easily verify this by looking at HOW we measure certain quantities. It is silly for Heisenberg to know about technological advances in the future and how much more accurate we can measure things. This is NOT what the HUP is describing. The HUP is NOT describing how well we know about the quantities in a single measurement. I can make as precise of a measurement of the position and momentum of an electron as arbitrary as I want simultaneously, limited to the technology I have on hand. I can make improvements in my accuracy of one without affecting the accuracy of measurement of the other.
The issue I have raised is whether or not the assumption of "constant" momentum from the moment the particle passes through the slit until the time it reaches the screen that you claim you can make is valid. It is not valid from the point of view a single-particle interpretation of the wave function or state variable of the particle, and that is the argument I have been presenting. If you believe that you can measure the momentum of the particle in this manner, you must reject the single-particle interpretation and replace it with an ensemble average interpretation, which seems to be the theme of your posts on this issue.

Maybe this puts you in good company. I make no pretense of being current or anywhere near the forefront of the development of thinking on these issues. Far greater minds than mine have debated the interpretation for decades, and apparently the debate continues. Maybe the tide is flowing your way. I did make some effort to review what others have said before responding here again and came across this site

http://www.phys.tue.nl/ktn/Wim/muynck.htm#quantum

I am too far removed from the front to judge where the ideas presented here fit into the mainstream of thinking these days. I have not read all of it, and if I did I would not fully grasp all of the implications, but at least it exposed me to a point of view that seems closer to yours than to what this author refers to as the "standard formalism" that is the basis for my arguments. But even here, while the problem of simultaneous measurement of "incompatible observables" (like position and momentum) under the standard formalism is viewed as being somewhat out of touch with real world "nonideal" measurements, I don't see a claim being made for unlimited precision of simultaneous measurements of two such observables.

I have a problem dismissing the single-particle interpretation in favor of a pure ensemble interpretation. It goes back to what got me involved in this discussion in the first place on the other thread. I don's see how you can relegate the HUP to a statement about the inability to predict the measurement of the momentum of the NEXT particle, as opposed to future measurements of the momentum of the particle you whose position you just measured, and still use it to make arguments abut why electrons cannot be confined to nuclei or postulate the existence of "virtual particles" as quantum fluctuations involved in the forces of interaction between material particles as other great thinkers have done.

I've done as much as I am going to do to pursue this discussion. I will leave it to guys like you who are actually doing science instead of just talking about it to sort it all out in the end.
 
  • #51
I truly do not understand what the problem here is.

I have ALREADY explained what I meant by "simultaneous", so I don't understand why you still bringing it up. It ISN'T what you accused me of doing.

Secondly, and this is even more puzzling because the practice is very rampant. How do you think we "detect" the properties of electrons, or other particles in the first place? More often than not, we detect them by observing where they are! We do this in SEM, STM, etc. I measure the energy of an electron by how much it bends in a magnetic field, and then I look at WHERE it lands on a detector! This tells me how much it has bent! Yet, from the way you are tell me here is that this is NOT what its energy is as a free, plane-wave particle, that between the moment it enters the magnetic field till it is detected, its momentum and energy are still in a superposition of values and so what is being detected is some "detection values".

I have repeated this many times, that I make no assumption of what happened between the slit and the detector. All I'm saying is that THAT electron that hit the detector has THAT momentum when it hits the detector. If by looking at the image on the detector and deducing the momentum is WRONG, then we have been wrong in MANY, MANY other techniques and detection schemes, especially in high energy physics because they make even MORE strong assumptions about the trajectory of the particle from the collision point to the detectors.

Zz.
 
  • #52
For those with a little bit of linear algebra under their belts :

Let A, B be a pair of non-commuting operators. What this means is that A and B do not posses the same eigenfunctions (eigenkets). We first define the operator [itex]\Delta A \equiv A - \langle A \rangle I [/itex]. Where the expectation value of A with respect to some state [itex]|a \rangle [/itex], is defined as [itex]\langle A \rangle = \langle a | A | a \rangle [/itex]. This number tells you what A will be measured as, on average, over several repeated measurements performed on the system, when prepared identically.

Now, we define an important quantity - the variance or mean square deviation, which is [itex]\langle ( \Delta A ) ^2 \rangle [/itex]. This quantity is no different from the variance in any statistical collection of data. Plugging in from above :

[tex]\langle ( \Delta A ) ^2 \rangle = \langle (A - \langle A \rangle I)^2 \rangle = \langle A^2 \rangle - \langle A \rangle ^2 ~~~-~(1)[/tex]

Let [itex]|x \rangle [/itex] be any arbitrary (but normalized) state ket. Let :

[tex]|a \rangle = \Delta A ~ |x \rangle [/tex]
[tex]|b \rangle = \Delta B ~ |x \rangle [/tex]

First we apply the Schwarz inequality (which is essentially a result that is two steps removed from saying that the length of a vector is a positive, real number) : [itex] \langle a |a \rangle \langle b |b \rangle \geq | \langle a |b \rangle |^2 [/itex] to the above kets (keeping in mind that [itex]\Delta A~, ~\Delta B [/itex] are Hermitian), giving :

[tex]\langle (\Delta A)^2 \rangle \langle (\Delta B)^2 \rangle \geq |\langle \Delta A \Delta B \rangle | ^2 ~~~-~(2)[/tex]

Next we write

[tex]\Delta A \Delta B = \frac{1}{2}(\Delta A \Delta B - \Delta B \Delta A) + \frac{1}{2}(\Delta A \Delta B + \Delta B \Delta A) = \frac{1}{2}[\Delta A, \Delta B] + \frac{1}{2}\{ \Delta A, \Delta B \} ~~~-~(3) [/tex]

Now, the commutator

[tex][\Delta A,~ \Delta B] = [A - \langle A \rangle I,~B - \langle B \rangle I] = [A,B] ~~~-~(4)[/tex]

And notice that [itex] [A,B] [/itex] is anti-Hermitian, giving it a purely imaginary expectation value. On the other hand, the anti-commutator [itex]\{ \Delta A,~ \Delta B \} [/itex] is clearly Hermitian, and so, has a real expectation. Thus :

[tex]\langle \Delta A \Delta B \rangle = \frac{1}{2}\langle [A,B] \rangle + \frac{1}{2} \langle \{ \Delta A,~ \Delta B \} \rangle ~~~-~(5)[/tex]

Since the terms on the RHS are merely the real and imaginary parts of the expectation on the LHS, we have

[tex]| \langle \Delta A \Delta B \rangle |^2 = \frac{1}{4}| \langle [ A,B] \rangle |^2 + \frac{1}{4} | \langle \{ \Delta A,~ \Delta B} \rangle |^2 \geq \frac{1}{4}| \langle [A,B] \rangle |^2~~~-~(6)[/tex]

Using the result of (6) in (2) gives :

[tex]\langle (\Delta A)^2 \rangle \langle (\Delta B)^2 \rangle \geq \frac{1}{4} |\langle [A,B] \rangle | ^2 ~~~-~(7)[/tex]

The above equation (7), is the general statement of the Heisenberg Uncertainty Principle. So far, it is nothing more than the statement of a particular property of certain specifically constructed hermitian matrices - namely, those of the form[itex]\Delta H [/itex], constructed as above.

Notice that if the operators A, B comute (ie: [A,B] = 0), then the product of the variances vanish, and there is no uncertanity in measuring their observables simultaneously. It is only in the case of non-commuting (or incompatible) operators, that you see the more popular form of the Uncertainty Principle - the one where the product of the variances does not vanish. Specifically, in the case where [itex]A = \hat{x_i}~,~~B = \hat{p_i} [/itex], we have

[tex] [\hat{x_i},\hat{p_i}] = i \hbar ~~~-~(8)[/tex]

This result follows from essentially two observations :

(i) The infinitesimal translation operator, [itex]\tau (d \mathbf{x}) [/itex], defined by [itex]\tau (d \mathbf{x}) |\mathbf{x} \rangle \equiv |\mathbf{x} + d \mathbf{x} \rangle [/itex] can be written as

[tex] \tau (d \mathbf{x}) = I - i\mathbf{K} \cdot d \mathbf{x} [/tex]

(ii) K is an operator with dimension length -1, and hence, can be written as [itex] \mathbf{K} = \mathbf{p} / [action] [/itex]. The choice of this universal constant with dimensions of action (energy*time) comes from the de broglie observation [itex]k = p/ \hbar [/itex]. So, writing [itex] \tau (d \mathbf{x}) = I - i\mathbf{p} \cdot d \mathbf{x} /\hbar [/itex] leads to the expected commutation relation , [itex][\hat{x_i}, \hat{p_i} ] = i \hbar [/itex]. Plugging this into (7) gives the correct version of the popular form of the HUP :

[tex]\langle (\Delta x_i)^2 \rangle \langle (\Delta p_i)^2 \rangle \geq \frac{1}{4} \hbar ^2 ~~~-~(HUP)[/tex]

Illegally taking square roots above and forgetting that we are talking about variances and expectations, is what leads to popular misconceptions about intrinsic uncertainties in single measurements.
 
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  • #53
Another common misconception is that the HUP for position-momentum is merely a restatement of the x,p commutator.

No, [itex]~ \Delta x \Delta p \neq [x,p] [/itex].
 
  • #54
Gokul43201 said:
...
We first define the operator [itex]\Delta A \equiv A - \langle A \rangle [/itex].
...

Just a note to avoid (possible) unnecessary confusion: [itex]\Delta A \equiv A - \langle A \rangle I [/itex], where I is identity operator :smile: (you have been perfectly clear in your derivation everywhere else). I know this is not usually explicitly written, but it cannot hurt to say it. o:)
 
  • #55
Thanks, Igor. I'll make the changes to ensure that operators do not look like scalars anywhere.
 
  • #56
Gokul43201 said:
Illegally taking square roots above and forgetting that we are talking about variances and expectations, is what leads to popular misconceptions about intrinsic uncertainties in single measurements.
Would you care to elaborate on that? I don't think there is any real problem about what is going to happen if repeated measurements are made under the same conditions. Repeated trials are going to result in a pattern of hits on the screen. The issue is focused on a single particle passing through a single slit being detected one time on a screen. You know the particle made it through the slit, and you know, to a precision determined by the detector, where the particle hit the screen. What do you know about the particle? What have you measured?
 
  • #57
Gokul43201 said:
Illegally taking square roots above and forgetting that we are talking about variances and expectations, is what leads to popular misconceptions about intrinsic uncertainties in single measurements.

Wener Heisenberg said:
The more precisely the position is determined, the less precisely the momentum
is known in this instant, and vice versa.


Also, it would seem the two of you have a disagreement.
 
  • #58
OlderDan :

The Uncertainty Principle says nothing about what happens in a single measurement. What does one mean by the error or imprecision or uncertainty of a single data point ? It is meaningless. Even for a somewhat crude (although possibly illustrative) thought experiment one considers two data points at the extrema of the detection window and relates the spread between these data points to the spread between calculated values of the conjugate variable.

See the Gamma Ray Microscope thought experiment that Heisenberg came up with. When Heisenberg first proposed the thought experiment, he got it wrong, and had to be corrected by Bohr.

Antiphon : I shall not respond to claims based upon absolutely referenceless quotes. You'll have to do better than that.
 
  • #59
The Gamma Ray Microscope

http://www.aip.org/history/heisenberg/p08b.htm
 
  • #60
Well according to this thread all the problems I have done in high school are stupid.Like this in this thread:
https://www.physicsforums.com/showthread.php?t=83213 :
poolwin2001 said:
According to HUP [itex]\Delta x.\Delta p \geqq \frac{h}{4 \pi} [/itex]. Pluging in values where [itex]\Delta x[/itex] will be the size of the nucleus, we get
[itex]\Delta v[/itex] greater than c ! We should not have velocities above c while here even the uncertainity in velocity is greater than c which indicates that our [itex]\Delta x[/itex] is incorrect ==> e- can't be confined to the nucleus.
As the neutrons and protons differ in mass by about 10e3 the [itex]\Delta v[/itex] for n/p doesn't come above c ! So they may exist inside the nucleus.

If this is the case,Why isn't HUP denoted as
[itex]\sigma_x.\sigma_p \geqq some k [/itex] or something ?
where [itex]\sigma[/itex] stands for standard deviation.
 
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  • #61
Gokul43201 said:
OlderDan :

The Uncertainty Principle says nothing about what happens in a single measurement. What does one mean by the error or imprecision or uncertainty of a single data point? It is meaningless. Even for a somewhat crude (although possibly illustrative) thought experiment one considers two data points at the extrema of the detection window and relates the spread between these data points to the spread between calculated values of the conjugate variable.

See the Gamma Ray Microscope thought experiment that Heisenberg came up with. When Heisenberg first proposed the thought experiment, he got it wrong, and had to be corrected by Bohr.

Antiphon : I shall not respond to claims based upon absolutely referenceless quotes. You'll have to do better than that.
This is not an answer to my question. The only reference I made to measurement in my previous post was in asking the question about what is being measured given one particle directed toward one slit, and one detection of the particle striking the screen on the other side of the slit.

Nevertheless, the "uncertainty" of a single measurement is not a meaningless concept. It means the same thing here that it means when you get out your ruler to find out how long to cut a board. You measure it once, and write down a number or mark the board, and you know that the measurement lacks precision because the instrument you used and your ability to read it are imperfect. You may not know exactly how to quantify the uncertainty, but if it's a good ruler and your eyes are better than mine you can be pretty confident that the measurement is accurate to far better than an inch, and probably not much better than 1/32. If you are good at using the saw to actually cut the board to the length you measured, you can make a living as a carpenter. If you are a precision machinist, you use better tools and get more precise results. You don't have to take a statistical average of many measurements to have a pretty good idea of the potential error of your one measurement.

In the problem at hand, the particle passed through a slit that is presumed to have a known width to arbitrary precision. Given our belief that in order to arrive at the screen the particle must have passed through the slit, we conclude that the particle was localized when it passed through the slit. We don't claim specific knowledge of where the particle passed through the slit, just that it made its way through the gap somewhere. We must accept that we have not located the position of the particle with greater precision than one slit width, but we can be confident that the slit width is a measure of the potential uncertainty in the position of the particle when passing through the slit.

When the particle hits the screen it leaves a mark, or excites one or a few cells of an array of detectors. The size of the cell apertures and/or collection of excited cells in the array gives us the same sense of uncertainty in the location of the strike that the width of the slit gives us about where the particle came through the slit. We don't have to run the same particle through the apparatus again and again to take a statistical average of hits to have a sense of the precision of the location of one hit. We all know that in the scanario at issue repeating the measurement would only defeat the purpose.

I see no value in bringing yet another thought experiment into the discussion, especially given the last paragraph of the link you posted. The only connection between the microscope and the slit apparatus is the collection apertures, and that's covered in the previous paragraphs.

A couple of us have attempted to talk about the wave or state function of the particle as it pertains to what can be know about one particle, and the response has been as if we are talking in some foreign language. I will readily accept that I am out of touch with current thinking on these issues, but I have read a lot of stuff in the last couple of days about particle tracks and quantum measurement theory and I have yet to find a single reference that dismisses the fundamental problem of limited knowledge of the "observables" of a single particle, and I have found repeated references to the HUP in connection with those limitations.

Nowhere have I said that it is impossible for any single measurement of one observable to yield a precise value. Nowhere have I implied that a particle passing through a single slit is going to be smeared across the detection screen when it gets there. What I have done is to make reference to the fact that a wave function that localizes a particle can be represented as a spectrum of momentum eigenfunctions, but most certainly not as a single momentum eigenfunction. Represent that in whatever equivalent way you want with non-commuting operators, state functions that are not simultaneously eigenstates of position and momentum- I don't care. The representation is not the issue.

The issue is, can I do something that will tell me with unlimited precision both the position of one particle and its momentum at the same moment in time? If the answer is yes, then I want to know where to find the supporting evidence, because I cannot find it. If the answer is no then I want the broadly accepted reference that dismisses that fundamental property of individual microscopic particles as unrelated to the HUP, because everything I have found so far keeps saying that it is.
 
  • #62
OlderDan said:
The issue is, can I do something that will tell me with unlimited precision both the position of one particle and its momentum at the same moment in time? If the answer is yes, then I want to know where to find the supporting evidence, because I cannot find it. If the answer is no then I want the broadly accepted reference that dismisses that fundamental property of individual microscopic particles as unrelated to the HUP, because everything I have found so far keeps saying that it is.

But OlderDan, is this really the issue that we started with? And is this really what *I* initially started with?

I mean, look at what is going on. A particle that initially only had a momentum in the z-direction moving towards the slit, after passing the slit, now has gained a y-component of the momentum, something it did NOT have before it went through the slit. The act of using the slit to determine its position has CHANGED the system in the sense that it introduced an added momentum component. This should not be a surprise to anyone.

I THEN measure this momentum. And as far as I can tell, it is THIS ability that you are disputing. It would be silly for me to insist that this is the SAME momentum of the single particle when it enters the slit, because I'm measuring the component of momentum that it didn't have before! And I've given you at least a couple of citations and a slew of experimental techniques in which the momentum, energy, and other characteristics of an electron are deduced using the knowledge of where on a detector that electron hits. In fact, the Valla et al. Science paper was one of the top 20 most cited paper in 2001 primarily due to the introduction of the energy and momentum distribution curve in a photoemission measurement.

BTW, your explanantion of the uncertainty in your "ruler" is identical to what I have said about the uncertainty in instrumentation. This isn't the HUP. The tick marks and ability to read your ruler doesn't change just because you make something else smaller or larger. This uncertainty is not what Gokul has derived.

Zz.
 
  • #63
poolwin2001 said:
Well according to this thread all the problems I have done in high school are stupid.Like this in this thread:
https://www.physicsforums.com/showthread.php?t=83213 :


If this is the case,Why isn't HUP denoted as
[itex]\sigma_x.\sigma_p \geqq some k [/itex] or something ?
where [itex]\sigma[/itex] stands for standard deviation.

If that is what you have "deduced" from this thread, then you have severely misread it.

Zz.
 
  • #64
ZapperZ said:
Secondly, and this is even more puzzling because the practice is very rampant. How do you think we "detect" the properties of electrons, or other particles in the first place? More often than not, we detect them by observing where they are! We do this in SEM, STM, etc. I measure the energy of an electron by how much it bends in a magnetic field, and then I look at WHERE it lands on a detector! This tells me how much it has bent! Yet, from the way you are tell me here is that this is NOT what its energy is as a free, plane-wave particle, that between the moment it enters the magnetic field till it is detected, its momentum and energy are still in a superposition of values and so what is being detected is some "detection values".

I have repeated this many times, that I make no assumption of what happened between the slit and the detector. All I'm saying is that THAT electron that hit the detector has THAT momentum when it hits the detector. If by looking at the image on the detector and deducing the momentum is WRONG, then we have been wrong in MANY, MANY other techniques and detection schemes, especially in high energy physics because they make even MORE strong assumptions about the trajectory of the particle from the collision point to the detectors.

I think the detection techniques are actually measuring classical quantities. When you measure how much particle has bent in a magnetic field, you use R = mv / eB (at least in a simple cyclotron). This is classical equation. I don't think HUP says anything involving classical momentum (p = mv). Classical momentum appears in QM only as expectation value of QM momentum. Proper way to determine QM momentum would be from interference pattern (like I think OlderDan said before), by measuring the distances between max or minumum intensity points. I'm not 100% sure, so correct me if I'm wrong.

But why can we then see particle track in detector, I am not sure, but maybe it's because a particles in this experiments have so much bigger momentum in certain direction ? (so you can actually approximate that other components of momentum are zero)
 
  • #65
Igor_S said:
I think the detection techniques are actually measuring classical quantities. When you measure how much particle has bent in a magnetic field, you use R = mv / eB (at least in a simple cyclotron). This is classical equation. I don't think HUP says anything involving classical momentum (p = mv). Classical momentum appears in QM only as expectation value of QM momentum. Proper way to determine QM momentum would be from interference pattern (like I think OlderDan said before), by measuring the distances between max or minumum intensity points. I'm not 100% sure, so correct me if I'm wrong.

But why can we then see particle track in detector, I am not sure, but maybe it's because a particles in this experiments have so much bigger momentum in certain direction ? (so you can actually approximate that other components of momentum are zero)

But ALL "measurements" are "classical". We are trying to determine classical properties such as "position", "momentum", "energy", etc. When we make a measurement, we force the system to interact with large degree of freedom that causes certain degree of decoherence. What I indicated is no different.

Electrons in solids such as metals have a superpostion of momentum/energy, etc. The principle of photoemission says that we CAN measure accurately the in-plane momentum and energy of the emitted photoelectrons, and that what we detect on our detector are those two values. And not only that, these two values represent the in-plane energy and momentum of that electron while it was in that material!

http://arxiv.org/abs/cond-mat/0209476
http://arxiv.org/abs/cond-mat/0208504

Zz.
 
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  • #66
I think the confusion comes from the interpretation of zapper and older Dan words (and sometimes in forgetting we live in a 3D/4D world).
A measurement by the slit, defines a position measurement (x,y) in the “transversal” direction of the moving particle (direction z). From this measurement result we deduce the momentum of the particle thanks to the preparation (e.g. energy conservation, all particles have the same energy: see note below).
We have 2 commuting observables: the transversal position (x,y) and the momentum pz of the wave packet => the Heisenberg principle keeps working (what zapper and older dan says are ok, provided we understand their mapping into the physical reality and QM formalism).
If we have a detector behind the slit, it will also do a z position measurement (on the preparation given by the slit plate and the particle state). And we can say, yes “before” the measurement result of this detector, the particle has a momentum pz defined by the position (x,y) of the slit (but not the position z) (hence a position (x,y) and momentum pz) and yes “after” the detector measurement (the click), the particle is located at (x,y,z) with an undefined momentum.

I hope, that my words have not added more confusion :biggrin: .

Note: we can infer the momentum pz from the transversal position measurement because the preparation of the particle separates spatially (transversal direction) with ~100% confidence –the wave packets of different momentums pz. This is why we put a plate with a pin hole before the double slit plate of the double slit experiment for example (we can say that the two plates measure or prepare the momentum of the particle).

Hoping my contribution may help in ansering the different questions,

Seratend.

EDIT: in the note above, different momentums pz, mean different z axes (sorry for the confusion). I hope one has assumed the correct meaning : )
 
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  • #67
seratend said:
Note: we can infer the momentum pz from the transversal position measurement because the preparation of the particle separates spatially (transversal direction) with ~100% confidence

This is the point of the confussion. It's because this is an average momentum.

From the orthodox Copenhagen interpretation one might argue that this is not
the same as the instantaneous momentum. In the line of: "a particle may have
a ΔE for a certain Δt", one might argue that it may have a fluctuating Δp which
averages out over a longer traject.


Regards, Hans
 
  • #68
There is a common idea that it is impossible to determine the energy E and
momentum p of a particle from an infinitesimal small region of the wave
function because of the Fourier Transform relation.



However, If we write for a wave-packet at rest:

[tex]\ Q_{\{x\}}\ \ e^{ -iE_0t/\hbar}[/tex]

Were Q is the shape of the wave function (e.g. the Gaussian) the we can
determine the value of E at each point in space time simply by differentiating
in time.


I we then handle this as a moving wave-packet via the Lorentz transform:

[tex]\ Q'_{\{x,t\}}\ \ e^{ipx/\hbar}\ \ e^{ - iEt/\hbar }, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{with: } Q'_{\{x,t\}} = Q \{ \gamma (x-vt) \}[/tex]

Then we can infer E and p locally exact at each point in space and
time by differentiating in x and t.


Regards, Hans
 
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  • #69
Hans de Vries said:
There is a common idea that it is impossible to determine the energy E and
momentum p of a particle from an infinitesimal small region of the wave
function because of the Fourier Transform relation.



However, If we write for a wave-packet at rest:

[tex]\ Q_{\{x\}}\ \ e^{ -iE_0t/\hbar}[/tex]

Were Q is the shape of the wave function (e.g. the Gaussian) the we can
determine the value of E at each point in space time simply by differentiating
in time.


I we then handle this as a moving wave-packet via the Lorentz transform:

[tex]\ Q'_{\{x,t\}}\ \ e^{ipx/\hbar}\ \ e^{ - iEt/\hbar }, \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \mbox{with: } Q'_{\{x,t\}} = Q \{ \gamma (x-vt) \}[/tex]

Then we can infer E and p locally exact at each point in space and
time by differentiating in x and t.


Regards, Hans

On the other hand, E (actually H, the Hamiltonian) and p commute for a free particle. So it should be of no mystery that there are instances where they both can be known with the same uncertainty "simultaneously".

Zz.
 
  • #70
Hans de Vries said:
This is the point of the confussion. It's because this is an average momentum.

From the orthodox Copenhagen interpretation one might argue that this is not
the same as the instantaneous momentum. In the line of: "a particle may have
a ΔE for a certain Δt", one might argue that it may have a fluctuating Δp which
averages out over a longer traject.

Regards, Hans

Yes, however I have to emphasize we are playing with words (a matter of interpretation): delta pz. delta z ~ hbar => having 2 plates with 2 slits with an infinite distance between plates (=> delta z --> +oO) allows one to have a delta pz as small as one wants (provided some technical feasibilities).
In the formal limit delta pz=0 => the mean value equals the value of the momentum, we cannot distinguish them.
(what zapper said: we can measure the momentum pz with the precision we want depending only on the technology feasibility).

Seratend.
 

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