Exploring Heisenberg's Uncertainty Principle: Intuition & Explanations

[fuesiker]

Correct. However, there is also no reason to believe that one can prepare the same photon over and over again.

Yes there is, it's called technology. The sky does it, preparing photons with the same polarized state with the same frequency (blue) at a 90-degree angle with the sun. Surely we can do it, and much better.

How is that silly? It is not required that you can tell which photon at the wall is which photon before the slit to perform a position measurement. The concept that every measurement is a position measurement after all, is widely accepted and also central to several interpretations of qm. It is also one of the reason why Bohmian mechanics works and is equivalent to standard qm as here particle positions are the central well-defined quantities of interest.

I don't know much about Bohmian mechanics except that it is experimentally very difficult to verify in many aspects of its features. However, back to our story, you seem not to understand the whole point about particle-wave duality. If you are able to tell which photon before the slit hit which position after the slit, your interference pattern is gone. This is akin to trying to detect which slit the photon went through. To do so, you must place a detector at one slit, which automatically kills your interference pattern since one slit is blocked.

 

[SpectraCat]

I'm tired of arguing with you about semantics. It seems to me you keep using semantics to cover your mistakes in physics.
This is trivial. There's no reason to believe one cannot prepare the same photon over and over again. Moreover, you need this in order to find out in any measurement process what the state of your system on which you're measuring the observable is. Your measurement is a probability value, and over many measurements you can reconstruct your state.
As mentioned before, read my post again. I never said you have two detectors.

Yes you did, because you specified an MZ interferometer, which uses two detectors .. you can ignore one if you want, but then you are not really doing interferometry.

Moreover, we do not know what the result will be a priori. Moreover, all you need is one detector, and the photon will always collapse onto it.

Please explain how you can make an MZ interferometer work in such a fashion. Remember an MZ interferometer is what *you* specified in setting up your example.

All you know is that you have a photon that you can prepare over and over again (which is what is done in all experiments that do projective measurements to determine a state), and you then measure its momentum to machine precision without inferring anything from its position.

That is simply not true .. you do NOT measure the momentum "of the photon". You measure clicks. You then repeat the experiment with different settings of the interferometer and *infer* the momentum of the photons from the differences in the "click measurements" (which are position sensitive) at different settings. This of course requires the additional step of abstraction of assuming that all of the photons have the same wavelength and initial phase.

Also, to counter your argument about measuring position (or your implicit belief that measuring position in this experiment "has something to do" with measuring momentum), I explained in the post how you can make the detector arbitrarily low in resolution, yet this will have no effect on your momentum measurement. You can make the position measurement uncertainty arbitrarily large, but your momentum measurement will always have machine resolution. So please tell me again, how are you inferring momentum from position here?

The photon collapses to a point on the detector, whether or not you have high or low spatial resolution on the detector. If this were NOT true, then the results of interferometry experiments would change when position sensitive detectors were used ... this is not the case as far as I am aware. The fact is that the measurements *are* position-sensitive, whether or not you care to take advantage of that fact in your analysis.

Think about one monochromatic wave of light instead of one photon, and you are passing it through a double-slit configuration. From the interference pattern, you will get a measurement of the momentum of that beam, without inferring anything about position. Now you can argue the pattern on the wall (where we see the interference pattern) is a position measurement, but that would be silly, because you cannot tell which photon on that wall is which photon before the slits.

The analogy does not hold, because this entire discussion is about measuring the properties of individual particles, NOT ensembles of particles.

 

[fuesiker]

I'm sure it seems that way to you, but if it hadn't been for your constant misreads and misinterpretations of plain English, you wouldn't have come to the incorrect conclusion that I'm making elementary physics mistakes. This entire discussion (which by the way is more insulting than anything I've seen in my seven years here) is a result of those mistakes.

It's certainly possible that some of the things I've said in this thread are wrong, especially those statements that start with "it seems to me...", "I believe...", or something like that. But your attacks on me have mostly been focused on two things: a) I used the word "measurement" to refer to what measuring devices do, instead of a projection operator. b) I interpreted the words "has something to do with" the way everyone but you does.

Fredrik, I'm sorry if I have insulted you. I clearly stated before I have respect for you. I don't dislike you and I don't mean to offend you. Let's move on.

However, I'm not the only one who thinks you have made statements at odds with standard QM. And it's very improbable that more than one person on here have "reading comprehension problems" as you accuse me of, but rather it may be that you're not wording yourself 100% correctly all the time, which many of us here do anyway.

Agreed. If you want to find out what the state is before the measurement, you need to do something like this. I'm just saying that that's not what I'm interested in. You said earlier that interferometers can be used to measure momentum, but so far you have only described a way to determine what wavefunction is produced by a state preparation procedure that's already known to give us a momentum eigenstate. So I'm asking you now, do you know a way to use interferometry to measure the momentum of a single photon that's in an unknown state (not necessarily a momentum eigenstate) before the measurement?

You seem to be confusing this. You automatically assume, like many people mistakenly do, that a photon has a fixed definite frequency, and that is not true because it would violate the uncertainty relation [itex]\Delta t \Delta E \geq h[itex (to some positive nonzero real constant). If you frequency is fixed (i.e. your wavelength is too, and hence, your momentum, meaning your photon is in a momentum eigenstate), that means your photon is an infinitely long packet, which is nonsense. So there is a spread of frequencies, and therefore you're in a superposition of momentum eigenstates. However, for all practical purposes, you can consider it classically as a single-frequency photon to understand what goes on from a wave theory perspective.

Now let's go to what you have in mind (or at least what I think you do), and this is a multi-chromatic wave, like the usual illumination you have on the street. Of course you can't measure momentum then, because you are working with different non-interacting systems here (those being all those different single photons with different frequency spreads). This is a different story than the above. You want to measure only one quantum system, not many non-interacting ones. The thing is, the photon always has a spread in its position and in its momentum, and those collapse upon measurement. This is different from having five photons with different position and momenta.

You can't make the detector arbitrarily large, because then it would cover the region of space where the other beam emerges. You need to keep the resolution high enough to distinguish between those two positions. As long as the setup can distinguish between two positions, it's a position measurement.

You can make the arms infinitely large. There is no limit on the arm length. But that's not the point. And you seem to have again ignored my detailed explanation above, so I won't waste my time.

 

[Cthugha]

Yes there is, it's called technology. The sky does it, preparing photons with the same polarized state with the same frequency (blue) at a 90-degree angle with the sun. Surely we can do it, and much better.

What does this have to do with preparing the same photon? To be the same, every essential property of the photon would have to be exactly the same in every repeated preparation step. However, you will find that the properties will differ as governed by Heisenberg's uncertainty principle. There is no way to test whether it is that property of the particle that is not well defined and each prepared photon is therefore the same or whether it is in principle impossible to prepare these properties to a precision better than governed by the uncertainty principle, but the properties are in principle well defined. There is no experiment that could distinguish between these two scenarios.

I don't know much about Bohmian mechanics except that it is experimentally very difficult to verify in many aspects of its features.

It is hard to verify or falsify because its predictions are fully equivalent to standard qm. That also means that this kind of interpretation is perfectly valid (although I personally dislike it) and so is the Bohmian interpretation of the uncertainty principle in which it is indeed a consequence of being unable to know the initial conditions exactly (although I personally also dislike that interpretation). Nevertheless it is a tenable position.

However, back to our story, you seem not to understand the whole point about particle-wave duality. If you are able to tell which photon before the slit hit which position after the slit, your interference pattern is gone. This is akin to trying to detect which slit the photon went through. To do so, you must place a detector at one slit, which automatically kills your interference pattern since one slit is blocked.

Of course I understand that, but it has once again nothing to do with the question at hand. The question was, whether the measurement mentioned (or in principle any measurement) can be interpreted as "a" position measurement, not whether it is a measurement of the position of some photon before or at some slit of a double slit or similar. It is of course not a measurement of the photon position at the slit, but I did not claim that. Nevertheless it is a position measurement.

 

[Fredrik]

This is what I mean that you ignore my rebuttals that you can't refute. I will say it here again, Fredrik.
...

This is not something that I need to refute, because this argument is still as irrelevant as the first three times (at least) you posted it. The first few times, I was baffled as to what your mistake might be, but now that you have explicitly demonstrated that you don't know how to negate a statement (posts 130 and 137), I'm fairly certain that I understand what you're doing wrong. You actually think that the negation of

"All measurements project the state vector onto an eigenspace of the measured observable."​

is

"No measurement projects the state vector onto an eigenspace of the measured observable."​

It's not. The correct negation is

"There exists a measurement that doesn't project the state vector onto an eigenspace of the measured observable."​

No matter how many examples you give of experiments where the system ends up being projected onto an eigenspace, it can't possibly refute the last statement, which is what you need to refute to prove me wrong. I can certainly admit that you have refuted the middle statement at least four times. Do it a fifth time if you want to, but stop claiming that doing so proves me wrong.

You have had a few days to think. Didn't it at some point occur to you that there might be something wrong with your logic?

Now, in your above quote, you seem to imply that von Neumann's axiom is a "very bad reason" (exactly and 100% as you put it) to reject the notion that you can measure both momentum and position simultaneously to arbitrary precision (which, whether you acknowledge it or not, is contradictory to the HUP).

von Neumann's axiom obviously implies that you can't measure both at the same time. (It does so by disallowing us from calling what measuring devices do "measurements"). What I said is that the desire to preserve his axiom in its original form is a very bad reason to refuse to call what measuring devices do "measurements".

 

[DrChinese]

If you are able to tell which photon before the slit hit which position after the slit, your interference pattern is gone. This is akin to trying to detect which slit the photon went through. To do so, you must place a detector at one slit, which automatically kills your interference pattern since one slit is blocked.

Not sure this is a good analogy, you do not have to block the slit to get that information. You could place polarizers in front of each slit instead. As you vary their relative orientation, the interference pattern appears or disappears.

 

[Fredrik]

Fredrik, I'm sorry if I have insulted you. I clearly stated before I have respect for you.

Yes, but even that statement was insulting.

Moreover, despite what you may think, I have respect for you. In the end, you do seem to love physics, though yourself more.

 

[fuesiker]

Yes you did, because you specified an MZ interferometer, which uses two detectors .. you can ignore one if you want, but then you are not really doing interferometry.



Please explain how you can make an MZ interferometer work in such a fashion. Remember an MZ interferometer is what *you* specified in setting up your example.

It is not written in law that the MZ interferometer has to use two detectors. One is enough. Read below.

That is simply not true .. you do NOT measure the momentum "of the photon". You measure clicks. You then repeat the experiment with different settings of the interferometer and *infer* the momentum of the photons from the differences in the "click measurements" (which are position sensitive) at different settings. This of course requires the additional step of abstraction of assuming that all of the photons have the same wavelength and initial phase.

The photon collapses to a point on the detector, whether or not you have high or low spatial resolution on the detector. If this were NOT true, then the results of interferometry experiments would change when position sensitive detectors were used ... this is not the case as far as I am aware. The fact is that the measurements *are* position-sensitive, whether or not you care to take advantage of that fact in your analysis.

Like I mentioned, this is experimentally really, really well established that you can prepare photons in the same polarization state and with the same frequency over and over again. As I also mentioned before, the sky does it. It's that simple.

Moreover, what is a point to you? You seem to not understand the idea of pixels. You can make your pixels REALLY huge such that on your screen one of those HUGE pixels light up when the photon hits. This is the same as what happens in your cell phone camera, that's the whole point of resolution. I said we can intentionally make those pixels huge in order to see there is no relationship between the uncertainty in position and the measurement result you are getting that would allow one to say we're inferring momentum from position.

Again, the detector is there to tell you that something hit. You always need to use two detectors when using classical beams for example because then there are no probabilities. With one photon, you use one detector, and this will ensure your photon collapse in its position on that detector, and no other detector, but the point you're failing to get is that we simply don't care. Because, like I mentioned, the uncertainty in position can be made arbitrarily large, and this will not change the value of momentum you're measuring. So tell me then, how are we inferring momentum from position. You seem to be getting or have gotten your PhD in chemical physics, so this shouldn't be so hard to get. Make your pixel really low in resolution, such that if it's 25 square meters of 25 square nanometers, all it would tell you is that the photon has hit it, no less, no more. The uncertainty in position here is equivalent to the area of the detector surface, which is just a one-pixel detector, let's say (and you can build this). Whichever area you have, you will get the same result. No clicks at a certain $\Delta l = l_2-l_1$, whichever detector you use. Hence, one has to be really mathematically impaired to still think we are inferring momentum from position here.

The analogy does not hold, because this entire discussion is about measuring the properties of individual particles, NOT ensembles of particles.

Usually when I discuss physics, I don't always talk about single photons. But in fact, this will hold for a single photon because if you try to detect the photon in the MZ inteferometer for example, before the arms cross, you will always get a reading that your photon hit that detector (again you have one detector but this is before the arms cross so it is not like our detector above which comes after the arms have crossed). Then no matter what $\Delta l$ is, you will always here a click on the detector that is before the arms crossing. Remove that, then you will hear or not hear click at the detector we originally had after the arms crossing depending on $\Delta l$.

You seem to not understand wave-particle duality so well. Again, I suggest Loudon's book.

 

[fuesiker]

Yes, but even that statement was insulting.

Please try to not comment on non-physics issues. Fredrik, come on, I was being totally nice.

As a last note, I have said all I said in this thread, and thank you all for a nice discussion that showed me why I love quantum mechanics so much. This is my last entry on here. All of you take care and have fun.

 

[Fredrik]

You automatically assume, like many people mistakenly do, that a photon has a fixed definite frequency,

No, I don't.

OK, based on what you're saying here and in several other posts, I have to ask: Do you think that in order to measure an observable, the system must be in an eigenstate of that observable before the measurement begins? (That is, before the interaction between the system and the measuring device becomes non-negligible).

 

[atyy]

von Neumann's axiom obviously implies that you can't measure both at the same time. (It does so by disallowing us from calling what measuring devices do "measurements"). What I said is that the desire to preserve his axiom in its original form is a very bad reason to refuse to call what measuring devices do "measurements".

So you agree that if we define measurement in the standard sense, then quantum mechanics forbids simultaneous sharp measurements of canonically conjugate observables? (For the purpose of this discussion, we can separate the projection postulate out, and define a sharp measurement of O to be one that when repeated on an ensemble of identically prepared states gives a distribution of measured values such that the expectation value is <state|O'|state>, for all observables O' that commute with O. In particular, a sharp measurement gives a sharp distribution if the state is an eigenstate of O).

 

[SpectraCat]

It is not written in law that the MZ interferometer has to use two detectors. One is enough. Read below.

I am aware that single-detector configurations can be used for MZ-interferometers for many applications, but it is so clear to me how such an interferometer can be used to analyze single photon experiments, where you need to have a result for every photon passing through the apparatus. You can of course describe *average* behavior of the photons by measuring click rates, but that is not really relevant to what we are discussing.

Please describe how a one-detector MZ interferometer can be used for single photon experiments in more detail. On which side of the second beam-splitter do you choose to place your detector? What mechanism do you use to know that a photon was emitted, yet failed to register a click on the single detector you do have? How do you extract the momentum information from your single-detector data? Most of the time, your detector will fail to click .. how will you decide in which time-intervals the failure to click was significant?

Like I mentioned, this is experimentally really, really well established that you can prepare photons in the same polarization state and with the same frequency over and over again. As I also mentioned before, the sky does it. It's that simple.

I have no idea what you mean by that last comment about the sky, but let's look at your other statement. I am pretty sure you don't mean what you said. You can prepare photons which are sampled from a particular frequency distribution, which can be made to have a narrow, but still finite, width. A similar statement holds for polarization. These photons cannot however be said to each have precisely the same frequency. I understand and accept your point however, that we can assume fair sampling from that narrow frequency distribution, and thus analyze the momentum of the photons. However, the point you are missing is that the information about the momentum comes from an ensemble of measurements where you measured the position of a photon (more on that below), using different instrumental settings. No information at all about the momentum of the photon is contained in any single measurement, and THAT is the sort of experiment we are discussing in this thread. One particle, one measurement. My statement still stands that there is no way to obtain information about momentum of a particle from such a measurement, without considering other information and then using it to *infer* the past momentum of the particle (I agree with jtbell's language on this point).

Moreover, what is a point to you? You seem to not understand the idea of pixels. You can make your pixels REALLY huge such that on your screen one of those HUGE pixels light up when the photon hits. This is the same as what happens in your cell phone camera, that's the whole point of resolution. I said we can intentionally make those pixels huge in order to see there is no relationship between the uncertainty in position and the measurement result you are getting that would allow one to say we're inferring momentum from position.

I understand just fine ... you seem to be missing the point that resolution is completely irrelevant to this example. If you use a detector that is just one big pixel, you get precisely the same results as if you use a high-resolution CCD for your detector, assuming that you do not overlap another spatial region where photons traveling along a different path can also be detected. Do you agree with that statement? Assuming that you do, do you not see how this shows that you are in fact measuring position when you register the click? If you were not, then using a high-resolution detector would change the results of your interferogram, and I am almost 100% certain that it does not.

Again, the detector is there to tell you that something hit. You always need to use two detectors when using classical beams for example because then there are no probabilities. With one photon, you use one detector, and this will ensure your photon collapse in its position on that detector, and no other detector, but the point you're failing to get is that we simply don't care. Because, like I mentioned, the uncertainty in position can be made arbitrarily large, and this will not change the value of momentum you're measuring.

I agree, but you interpretation of the significance of that information is exactly backwards, as I explained above.

So tell me then, how are we inferring momentum from position. You seem to be getting or have gotten your PhD in chemical physics, so this shouldn't be so hard to get.

You really need to quit it with the condescending side remarks about how your points should be "obvious" based on other people's level of education. For one thing, what if your "obvious" point is wrong, or irrelevant to the argument at hand (as in this case)? Then your choice to make personal remarks just makes you look all the more foolish, and irritates people who are just trying to have a scientific discourse.

Make your pixel really low in resolution, such that if it's 25 square meters of 25 square nanometers, all it would tell you is that the photon has hit it, no less, no more. The uncertainty in position here is equivalent to the area of the detector surface, which is just a one-pixel detector, let's say (and you can build this). Whichever area you have, you will get the same result. No clicks at a certain $\Delta l = l_2-l_1$, whichever detector you use. Hence, one has to be really mathematically impaired to still think we are inferring momentum from position here.

As you yourself pointed out (correctly), measurement precision is completely irrelevant to the HUP. The HUP defines the intrinsic limit on the relationship between the widths of the momentum and position distributions associated with a quantum state. Simply choosing to ignore positional resolution when it is available cannot change the results of an experiment .. unless of course you change the fundamental nature of the experiment by doing so. As long as only one photon-path intersects the detector, then its size and or resolution are completely irrelevant to our discussion.

Usually when I discuss physics, I don't always talk about single photons. But in fact, this will hold for a single photon because if you try to detect the photon in the MZ inteferometer for example, before the arms cross, you will always get a reading that your photon hit that detector (again you have one detector but this is before the arms cross so it is not like our detector above which comes after the arms have crossed). Then no matter what $\Delta l$ is, you will always here a click on the detector that is before the arms crossing. Remove that, then you will hear or not hear click at the detector we originally had after the arms crossing depending on $\Delta l$.

Are you still talking about some large detector that intersects both paths? Because otherwise you need two detectors .. one for each arm, to ensure that a click is always registered (and only one detector will click for any given photon). This is because of what I have been saying all along .. registering a click on a detector requires localization of the photon .. i.e. detection of its particle-like properties. This is commonly called "which-path" information .. if you put a detector in one of the arms, it will click half the time. If you put detectors in both arms, then one or the other will click .. the photon can never be directly observed taking both paths through the interferometer. You can only *infer* that it did from the interference pattern that is recorded after the second beam-splitter, when you do not look at which-path information.

If I got it wrong, and you *were* talking about having a large detector that intersects both arms, then your comments above seem trivial and I don't understand their significance at all.

You seem to not understand wave-particle duality so well. Again, I suggest Loudon's book.

I have just about had it with your statements about what you think other people do and don't understand. You have no idea who you are talking to when you post on these threads, and your incredibly arrogant asides serve no purpose than to antagonize the other participants of this thread. Please confine your comments to the discussion at hand.

 

[fuesiker]

No, I don't.

OK, based on what you're saying here and in several other posts, I have to ask: Do you think that in order to measure an observable, the system must be in an eigenstate of that observable before the measurement begins? (That is, before the interaction between the system and the measuring device becomes non-negligible).

Again, you deny the obvious. And of course not, I don't believe the system has to be in an eigenstate of the observable before measuring that observable begins. It is laughable you concluded this from any of my posts. I clearly said many times the state COLLAPSES onto an eigenstate upon measurement. Hence I obviously assumed the state to be in general a superposition of eigenstates of the observable.

 

[Fredrik]

So you agree that if we define measurement in the standard sense, then quantum mechanics forbids simultaneous sharp measurements of canonically conjugate observables?

I don't agree that it's standard to define "measurement" as "projection onto an eigenspace", or at least not that it's the only standard. But I certainly agree that the wavefunction and its Fourier transform can't both be sharply peaked.

(For the purpose of this discussion, we can separate the projection postulate out, and define a sharp measurement of O to be one that when repeated on an ensemble of identically prepared states gives a distribution of measured values such that the expectation value is <state|O'|state>, for all observables O' that commute with O. In particular, a sharp measurement gives a sharp distribution if the state is an eigenstate of O).

I'm not sure I understand this. You didn't say that O is position, so I wonder if you meant that O is arbitrary. Can it be momentum? In that case, doesn't the momentum "inference" in Ballentine's thought experiment satisfy this definition, at least approximately? It seems to me that it does, but I might have misunderstood you. If it does, then this definition means that you can make sharp measurements of position and momentum at the same time, in contradiction with what I believed that you were referring to as "standard".

 

[Fredrik]

Again, you deny the obvious.

I can't tell if you're just lying to try to discredit me or if you honestly believe it. I don't really care at this point.

And of course not, I don't believe the system has to be in an eigenstate of the observable before measuring that observable begins.
...
I clearly said many times the state COLLAPSES onto an eigenstate upon measurement. Hence I obviously assumed the state to be in general a superposition of eigenstates of the observable.

Good for you.

It is laughable you concluded this from any of my posts.

And I'm sure it's not laughable that you somehow concluded that I think that every photon has a well-defined frequency (which implies a well-defined momentum) just because I asked if you know a way to perform a momentum measurement on a photon which is not necessarily in a momentum eigenstate.

 

[atyy]

I don't agree that it's standard to define "measurement" as "projection onto an eigenspace", or at least not that it's the only standard. But I certainly agree that the wavefunction and its Fourier transform can't both be sharply peaked.

I'm not sure I understand this. You didn't say that O is position, so I wonder if you meant that O is arbitrary. Can it be momentum? In that case, doesn't the momentum "inference" in Ballentine's thought experiment satisfy this definition, at least approximately? It seems to me that it does, but I might have misunderstood you. If it does, then this definition means that you can make sharp measurements of position and momentum at the same time, in contradiction with what I believed that you were referring to as "standard".

Yes, including that O can be momentum. To take care of Ballentine's objection, let me try adding "for unknown arbitrary states", so how about:

Quantum mechanics forbids simultaneous sharp measurements of canonically conjugate observables upon unknown arbitrary states. A sharp measurement of O is one that when repeated on an ensemble of identically prepared states gives a distribution of measured values such that the expectation value is <state|O'|state>, for all observables O' that commute with O. In particular, a sharp measurement gives a sharp distribution if the state is an eigenstate of O.

 

[LostConjugate]

It is the measurements that do not commute.

A measurement of position is taken at a point in time and a measurement of momentum is taken over a period of time. Only in the special case where the particle's momentum is zero for the entire period of time will the two measurements commute.

 

[SpectraCat]

It is the measurements that do not commute.

A measurement of position is taken at a point in time and a measurement of momentum is taken over a period of time. Only in the special case where the particle's momentum is zero for the entire period of time will the two measurements commute.

This is an interesting point, but seems to employ the same vague definition for "measurement". Can you be a little more specific? What is the character of the momentum measurement that you are talking about? Does it involve a time-ordered sequence of position measurements, or something else? If it is the former, then what is the source of the non-commutative property of the "measurement"? Is it that individual position measurements taken at different times don't commute (I don't see why that would be so, but perhaps I am missing something), or is it that the time-ordered sequence that doesn't commute with a single position measurement?

 

[atyy]

I don't agree that it's standard to define "measurement" as "projection onto an eigenspace", or at least not that it's the only standard. But I certainly agree that the wavefunction and its Fourier transform can't both be sharply peaked. I'm not sure I understand this. You didn't say that O is position, so I wonder if you meant that O is arbitrary. Can it be momentum? In that case, doesn't the momentum "inference" in Ballentine's thought experiment satisfy this definition, at least approximately? It seems to me that it does, but I might have misunderstood you. If it does, then this definition means that you can make sharp measurements of position and momentum at the same time, in contradiction with what I believed that you were referring to as "standard".

BTW, is known whether Ballentine's method recovers the exact expectation values for all observables that commute with Py (ie. is Ballentine's method a sharp measurement of Py that is canonically conjugate to Y)?

 

[LostConjugate]

This is an interesting point, but seems to employ the same vague definition for "measurement". Can you be a little more specific? What is the character of the momentum measurement that you are talking about? Does it involve a time-ordered sequence of position measurements, or something else? If it is the former, then what is the source of the non-commutative property of the "measurement"? Is it that individual position measurements taken at different times don't commute (I don't see why that would be so, but perhaps I am missing something), or is it that the time-ordered sequence that doesn't commute with a single position measurement?

The latter. In order to have a value of momentum measured you must have multiple positions measured which requires measuring over a period of time.

Now if you try to say that your object has an exact position and an exact momentum what you have just said is that it has an exact position over the period of time that you measured the momentum. That is simply not true. The measurements do not commute.

 

[romsofia]

I would be interested in seeing that page number reference. This discussion (or whatever I should call it) has made me curious about the terminology in standard textbooks.

Sure thing.

On page 112 he mentions that "Only if the wavefunction were simultaneously an eigenstate of both observables would it be possible to make the second measurement without disturbing the state of the particle (the second collapse wouldn't change anything in that case). But this is only possible, in general, if the two observables are compatible."

 

[Fra]

This turned out a heated discussion. There is little doubt that probably everyone here understands what the HUP really means, ie. it refers to standard deviations and is thus probabilistic to it's nature.

The rest of the discussion seems to me to be interpretationally rooted. Not entirely unexpectedly, the ensemble interpretations indeed seems to struggle with the ontological meaning of "single datapoints".

Are single datapoints, more than just datapoints? Personally I stronlgy object to the soundness in thinking of single datapoints (single detector counts) in terms of PARTICLE hits. That whole thing has a realist flavour to it that I do not prefer. No matter what is sometimes practice, I am extremely biased myself towards my own thinking but I think of detector counts simply as "evidence counts". And a single evidence counts always comes with finite confidence.

Anyway, the way I see it I still think that the non-commutativity has nothing to do with how the body of evidence (information state) is acquired. It's a feature of the internal state of the information state. This is why I personally prefer to think of it in terms of a constraint on the information state, rather than something that has anything to do with relations between single data points.

Also the information state; although a function of interaction history, encodes nothing but the expectations of the future. Ballentines example does not seem to have much physical relevance as basis for decisions. And in my crazy view, a theory should not postdict a recorded past; it should - provide rational expectations of the future based on the past. This is the survival value of a theory.

But probably the discriminator between all these views will be when we see which of these abstractions that are most fit to solve the remaining open question in physics regarding unification etc. Somehow, some of this things that are here almost purely intepretational and seemingly meaningless, I think will be more pronounced when situations such as comosological models yields abstractions such as "ensembles" at least IMHO unviable, and we do need a different way to understand "single instances", without embedding them into ensembles that can't be realized.

/Fredrik

 

[Fredrik]

On page 112 he mentions that "Only if the wavefunction were simultaneously an eigenstate of both observables would it be possible to make the second measurement without disturbing the state of the particle (the second collapse wouldn't change anything in that case). But this is only possible, in general, if the two observables are compatible."

What he's saying here is that you can't for example first measure position and then measure momentum without disturbing the system. That's not the sort of thing that most of this discussion has been about, but thank you for posting this text so that I could see that for myself. We're talking about detecting a particle that was prepared in a state with sharply defined position, and then interpreting this as a simultaneous measurement of position and momentum.

 

[Fredrik]

Yes, including that O can be momentum. To take care of Ballentine's objection, let me try adding "for unknown arbitrary states", so how about:

Quantum mechanics forbids simultaneous sharp measurements of canonically conjugate observables upon unknown arbitrary states. A sharp measurement of O is one that when repeated on an ensemble of identically prepared states gives a distribution of measured values such that the expectation value is <state|O'|state>, for all observables O' that commute with O. In particular, a sharp measurement gives a sharp distribution if the state is an eigenstate of O.

I still don't understand. It doesn't look like you have changed the definition. You just preceded it by a statement that, as far as I can tell, contradicts the definition.

BTW, is known whether Ballentine's method recovers the exact expectation values for all observables that commute with Py (ie. is Ballentine's method a sharp measurement of Py that is canonically conjugate to Y)?

I don't think any kind of "measurement" satisfies a requirement that strong. Wouldn't it have to be infinitely accurate? Anyway, I suppose you would also be interested in the answer to a related question: If we turn Ballentine's thought experiment into an actual experiment, and perform this momentum "inference" over and over on identically prepared systems, how accurately will the distribution of results agree with the values of $|u(p)|^2$ where $u$ is the Fourier transform of the wavefunction $\psi$.

Unfortunately I don't know the answer. Let's face it, this is a part of QM that we all suck at, because it's not covered in books. As far as I know, none of the standard books define what sort of device to call a "momentum measuring device". So I can only say that I expect that the distribution of "inferred" values would agree very well with the predicted distribution, but not exactly. The discrepancy doesn't really have anything to do with QM. The problem is with the definition of a "momentum measuring device". I don't think it's possible to even describe a hypothetical measuring device that would get results that are distributed exactly as predicted, at least not if the description doesn't involve some limit that isn't possible to actually take.

Think e.g. about length measurements in SR. I would define a length measurement by describing a radar device with a clock that measures the roundtrip time and multiplies it with c/2. But unless this device is doing inertial motion, it will fail to measure the length of the spacelike curve it's designed to measure. The measurement is however exact in the limit when the size of the device goes to zero.

 

[Ben Niehoff]

I am jumping in here after having not read most of the thread, so forgive me if this has already been mentioned or is slightly tangential to the current discussion.

To construct a quantum system (in the mathematical sense), we take a classical system and follow a certain prescription on phase space that promotes the coordinates on phase space (i.e., the position and momentum) to operators which do not necessarily commute, turning the phase space geometry into a noncommutative one. The commutators of the quantum operators are set equal to their classical Poisson brackets (modulo some factors of $i \hbar$). The Hamiltonion operator generates time translations in the same way it did in classical mechanics, but with the Poisson bracket replaced by the commutator. All of these operators are taken to act linearly on some abstract "state space" with inner product; this inner product gives a notion of "conservation of total probability" under the action of unitary operators.

One way to view the HUP is as the mathematical consequence of non-commuting observables (i.e. operators with real eigenvalues). Since an observation consists of picking out some eigenvalue of the operator in question, thereby changing the state to the corresponding eigenstate. Two operators that cannot be simultaneously diagonalized do not have the same set of eigenvectors (this is a basic fact of linear algebra), and operators can be simultaneously diagonalized precisely when they commute. Hence if $[A,B] \neq 0$, a state which is sharply peaked near an eigenvalue of A (i.e., close to an eigenstate of A), cannot also be sharply peaked near an eigenvalue of B, since A and B do not share eigenstates. One can mathematically derive the relationship between the variances of A, B, and [A,B] directly from linear algebra.

Another way to explain the HUP, however, is from merely looking at the Schrodinger equation. In classical mechanics, the equations of motion typically have two time derivatives; the Schrodinger equation, however, has only one. As you well know from basic differential equations, you may specify as many initial conditions as you have derivatives. In classical mechanics, we have two time derivatives, and hence we can specify two initial conditions: position and velocity. But in quantum mechanics, we have only one time derivative. Hence position and velocity cannot be independently specified. They are not independent quantities, and so it should not be too surprising that they cannot be measured to arbitrary accuracy at the same time. In the Schrodinger picture, "position" and "momentum" are measurements we make by taking certain weighted averages over a wavefunction; one can show in this case that position and momentum are complementary variables, in the Fourier transform sense, in which case the mathematical statement of the HUP comes from signal analysis.

In either case, it is the state itself which cannot contain sharply-peaked information about position and momentum simultaneously. So the HUP is not a a statement about our measurement ability; it is a statement about what information actually exists.

It is worth noting that one can construct states of minimal uncertainty; i.e. wave functions for which the inequality in the HUP becomes an equality. These are wave packets that resemble our classical notion of "particles". They have a position and a momentum that are specified "pretty well", but not perfectly. These wavefunctions look like gaussians in both position and momentum space.

 

[Fredrik]

That's a good post Ben. Some of it had not been mentioned in this thread.

Most of this thread has been about what a "measurement" is and specifically what a "momentum measurement" is. The article by Ballentine that was linked to a few times early in this thread describes a single-slit experiment, where the particle has a wall of detectors in front of it after going through the slit. One of the detectors will signal detection. This is obviously a position measurement, but Ballentine argues that it's also a momentum measurement. We can certaintly calculate a value of momentum that we can call "the result". If we accept this as a valid way to measure momentum, i.e. if we in fact measure momentum by measuring the position, then von Neumann's axiom that all measurements project the state vector onto an eigenspace of the measured observable contradicts itself (since there's no state with a sharply defined position and a sharply defined momentum).

So we either have to modify that axiom, or refuse to call this a "measurement" of momentum.

 

[dextercioby]

[...]
Another way to explain the HUP, however, is from merely looking at the Schrodinger equation. In classical mechanics, the equations of motion typically have two time derivatives; the Schrodinger equation, however, has only one. As you well know from basic differential equations, you may specify as many initial conditions as you have derivatives. In classical mechanics, we have two time derivatives, and hence we can specify two initial conditions: position and velocity. But in quantum mechanics, we have only one time derivative. Hence position and velocity cannot be independently specified. They are not independent quantities, and so it should not be too surprising that they cannot be measured to arbitrary accuracy at the same time. [...]

I don't find your argument valid. The state in the SE encodes the information about the system on equal footing. A valid physical state is the one on which all possible observables of the system can be measured (momentum, energy, position, spin, electric charge, parity, etc.), which mathematically translates into the state vector being in the Garding domain of the maximal symmetry algebra (the $\Phi$ space in a rigged Hilbert space $\Phi$,$\mathcal{H}$, $\Phi '$).
A valid physical state is part of the space of all solutions to the Schroedinger equation which is nothing but a merger between the principle of temporal conservation of observable statistics and the need to have the time translations as a subgroup of the maximal symmetry group.

 

[dextercioby]

[...] The article by Ballentine that was linked to a few times early in this thread describes a single-slit experiment, where the particle has a wall of detectors in front of it after going through the slit. One of the detectors will signal detection. This is obviously a position measurement, but Ballentine argues that it's also a momentum measurement. We can certaintly calculate a value of momentum that we can call "the result". If we accept this as a valid way to measure momentum, i.e. if we in fact measure momentum by measuring the position, then von Neumann's axiom that all measurements project the state vector onto an eigenspace of the measured observable contradicts itself (since there's no state with a sharply defined position and a sharply defined momentum).

So we either have to modify that axiom, or refuse to call this a "measurement" of momentum.

Based on personal preferences, I'd say we should leave aside von Neumann's axiom, because it forces us to add the following words to SE <In the absence of measurement, all possible physical states are solutions to the following 1st order differential equation:...>
which automatically puts a severe and artificial restriction to the time-evolution postulate itself.

Any valid argument against Ballentine's ?

 

[Ben Niehoff]

I don't find your argument valid. The state in the SE encodes the information about the system on equal footing. A valid physical state is the one on which all possible observables of the system can be measured (momentum, energy, position, spin, electric charge, parity, etc.), which mathematically translates into the state vector being in the Garding domain of the maximal symmetry algebra (the $\Phi$ space in a rigged Hilbert space $\Phi$,$\mathcal{H}$, $\Phi '$).
A valid physical state is part of the space of all solutions to the Schroedinger equation which is nothing but a merger between the principle of temporal conservation of observable statistics and the need to have the time translations as a subgroup of the maximal symmetry group.

I'm not sure that what I said contradicts this, but maybe I'm not understanding you correctly. "Cannot be measured" was a poor choice of words; of course position and momentum can be measured simultaneously. What I mean is that given that they are not independent, it stands to reason that they can't have precise values at the same time (because if we could imagine a state with definite position X and definite momentum P, then it seems we could specify X and P independently, which is not allowed).

This was just meant to be a heuristic argument for understanding "why" X and P happen to be noncommuting observables. It fails in the relativistic case, because the Klein-Gordon equation has two time derivatives.

 

[dextercioby]

I was argueing about the not-allowed part. Why can't they be specified independently ?

Because I think they are essentially independent, and independent measurements of them can be made with arbitrary precision, regardless of (a however properly chosen) state. It's only that the statistics of measurements (mean squared deviations) are related by an inequality which could very well have had 0 in the right hand side.

 

[Ben Niehoff]

My heuristic argument was that X and P can't be specified independently because the Schrodinger equation has only one time derivative, and hence the entire time evolution is determined by one initial condition.

As for the rest, you are making statements about measurements, while I was making statements about the quantum state. The quantum state cannot be both an X eigenstate and a P eigenstate at the same time, because there is no such state. And the more concentrated the state vector is around a particular eigenstate in the X basis, the more spread out it must necessarily be among eigenstates of the P basis.

One can argue whether the quantum state has any real existence independent of measurements, but that's not something I want to get into. Consider my statements to be merely about the mathematical formalism if it makes you uncomfortable to think of them as statements about reality.

 

[Fredrik]

Based on personal preferences, I'd say we should leave aside von Neumann's axiom, because it forces us to add the following words to SE <In the absence of measurement, all possible physical states are solutions to the following 1st order differential equation:...>
which automatically puts a severe and artificial restriction to the time-evolution postulate itself.

That's a good point. I didn't even think about that during this discussion, but we should be able to drop it completely and derive it (a version of it that's not inconsistent) as a theorem using decoherence theory.

Uh, now that I think about it even more, I think we can also derive a version of it from the following two correspondence rules: a) the average value in a series of measurements goes to the expectation value, as the number of measurements go to infinity, b) if f is a polynomial function, and A is an observable that represents measuring device M, then the measuring device that outputs the value f(a) when M outputs a, corresponds to the operator f(A). (I may remember this all wrong, but I think Isham did this in his QM book).

The way I see it, a set of statements isn't a theory unless it falsifiable. So the purely mathematical part of QM isn't a theory. It has to be supplemented by a specification of what sort of devices that should be used to test the accuracy of the theory's predictions. When we make these specifications, we have the option to only define position measurements, or to define measurements of every member of some set of interesting observables (like a list of generators of the theory's symmetry group).

If we choose the former option, then von Neumann's axiom isn't logically inconsistent, but it would still be preferable (at least in my opinion) to derive it as a theorem. If we instead choose the latter option, then von Neumann's axiom is logically inconsistent in its standard form, but we probably only need to change it to say that all measurements of observables that commute with position project the system onto an eigenspace of the measured observable. I would still prefer to derive it.

It might actually be better (and by that I mean that it would make this stuff simpler) to go with the former option. This is what I'm thinking: Since we would be measuring every variable by measuring position anyway, we wouldn't really be able to make a wider range of predictions. An experiment that tries to falsify a prediction about momentum for example, has no chance of doing that as long as the position measurements that are performed in the process are consistent with the predictions about position. So if someone wants to try to falsify the theory, it doesn't seem like he has any reason to look at anything other than the results of position measurements.

A slightly different option is to drop the concept of measuring device from the terminology altogether. Particles are detected, no properties are measured (not even position). They are all inferred from the experimental setup and the coordinates of the detection events.

 

[Fredrik]

I was argueing about the not-allowed part. Why can't they be specified independently ?

I assumed that he meant that to specify an initial position is to specify a sharply peaked wavefunction, and that to specify an initial momentum is to specify a wavefunction with a sharply peaked Fourier transform. Since no wavefunction has both of these properties, you would have to specify two wavefunctions in order to specify both an initial position and an initial momentum. But that gives us two initial conditions, and we only need one. So this argument is just a different aspect of the usual stuff about Fourier transforms.

 

[atyy]

Anyway, I suppose you would also be interested in the answer to a related question: If we turn Ballentine's thought experiment into an actual experiment, and perform this momentum "inference" over and over on identically prepared systems, how accurately will the distribution of results agree with the values of $|u(p)|^2$ where $u$ is the Fourier transform of the wavefunction $\psi$.

I suspect it isn't. Raymer, "Uncertainty principle for joint measurement of noncommuting variables", American Journal of Physics 62:986 (1994) does give Ballentine's method of measuring position at large L as a way of measuring momentum. However, he seems to use it as the momentum conjugate to position at small L, not large L as Ballentine would need to claim simultaneous measurement of conjugate variables.

But even if that's true, I'm not sure this would get me off the hook due to the comments of Bell I quoted in post #89. Perhaps the qualifier "if the initial state is arbitrary and unknown" is still needed.

 

[Fredrik]

I suspect it isn't. Raymer, "Uncertainty principle for joint measurement of noncommuting variables", American Journal of Physics 62:986 (1994) does give Ballentine's method of measuring position at large L as a way of measuring momentum. However, he seems to use it as the momentum conjugate to position at small L, not large L as Ballentine would need to claim simultaneous measurement of conjugate variables.

But even if that's true, I'm not sure this would get me off the hook due to the comments of Bell I quoted in post #89. Perhaps the qualifier "if the initial state is arbitrary and unknown" is still needed.

Bell's first comment is that it's "largely a question of semantics". That's consistent with what I've been saying, but I guess it's consistent with a lot of things. Then he starts talking about the (non-)existence of a joint probability distribution that, among other things, is linear in the wavefunction. That's something I don't see why we would need. At least i don't see why we would need it for Ballentine's thought experiment. We might need a joint probability distribution $(p,q)\mapsto\rho_\psi(p,q)$ for each $\psi$ with a sharply defined position, but we don't need (or want) the map $\psi\mapsto\rho_\psi$ to be linear (which would mean that we're forced to consider states that aren't localized). Hm, now that I think about it, since our p is a function of q, I'd say that all we need is a distribution $q\mapsto\rho_\psi(q)$, and we have that already: $\rho_\psi(q)=|\psi(q)|^2$.

Ballentine doesn't need a large L to claim simultaneous measurement. It's a simultaneous measurement of $y$ and $p_y$ regardless of L, and even regardless of the margins of error $\delta y$ and $\delta p_y$. The reason he mentioned a large L is that he didn't just want to show that you can measure both at the same time. (The margins of error don't even enter into that). He wanted to show that you can do it in a way that makes the product $(\delta y)(\delta p_y)$ smaller than what a naive application of the uncertainty relations suggests it can be. Choosing L large is just the easiest way to make that product small. A large L gives us a small $\delta p_y$, and it's much easier to believe that we can make L large enough than that we can make $\delta y$ small enough.

I haven't looked at the Raymer article yet, but I also feel that if any kind of limit is supposed to be a part of the definition of this momentum measurement, it's L→0, not L→∞. The reason is that what we're measuring is more like an average momentum than the "momentum right now". The position measurement is performed on a particle with a wavefunction that has had some time to spread out. To claim that we have really performed a simultaneous measurement, we should measure the momentum when the particle is in the same state as when we measure the position, but the momentum measurement involves two different times, and the wavefunction is spreading out over time. So it seems that we are closer to a "true" simultaneous measurement when L is small.

Hm, this could possibly be developed into an argument that Ballentine is wrong about how $\delta p_y$ depends on L.