Exploring Heisenberg's Uncertainty Principle: Intuition & Explanations

[kith]

No I'm not confusing them. All I stated were basic postulates of quantum mechanics, not necessarily one of the tenants of the Copenhagen interpretation.

The set of postulates depends on the interpretation. The ensemble interpretation for example modifies the state-postulate and throws away the collapse postulate. So does the Bohmian interpretation which also adds an additional postulate for the dynamics. What you have in mind as "basic postulates of QM" are probably the postulates from the CI like you can find them in Cohen-Tannoudji.

As far as the interference pattern is concerned: how do you detect an interference pattern without putting a detector at a fixed spot, thus performing a position measurement? That this is a measurement of individual photons can be seen, if you lower the intensity. The corresponding experiment has first been performed in 1909 by Taylor, if wikipedia is to be trusted. Note the pictures which illustrate my point.

 

[atyy]

What I have not thought much about, is of measuring momentum without measuring position. I've first read about all measurements beeing position measurements from Demystifier and it does sound kind of reasonable to me. If this was true, I would conclude that it is simply impossible to directly measure momentum at all and not conclude that the collapse postulate is wrong (at least not the part you are willing to throw away ;)).

Fuesiker gives the interferometer as an example, and Greiner's QM text talks about scattering of plane waves. Since in real life all equipment is approximately localized, a momentum measurement must always measure position. The usual heuristic is that position is so poorly measured (the track in a cloud chamber is much bigger than the de Broglie wavelength), that momentum is more precisely measured. Although generally simultaneous sharp measurements of non-commuting observables isn't possible, there does seem to be formalism treating simultaneous approximate measurements of non-commuting observables (eg. http://www.sciencedirect.com/science/article/pii/0003491692900862, whose text I can't access)

 

[fuesiker]

I agree with Kith regarding his claim that measurements cannot be points but finite intervals. There is a finite interaction time with the device.

Anyway, I disagree that interferometers must somehow measure position to measure the interference pattern. This is absolutely not true. Let me ask you this, you can create an interference of water waves in a bath tub. How are you detecting position to detect the wave momenta? It does not even make sense to talk of position in such a case.

And kith, thanks a lot because finally someone other than me saw that some of Fredrik's claims, despite his consistent and incessant objections and denial, ARE at odds with standard QM.

 

[Cthugha]

This claim indeed contradicts standard QM. I think that the standard version is true, especially since this postulate can be viewed as a definition of measurement.

I do not think that one should define a measurement this way. For example you run into serious problems explaining weak measurements using this definition. Unless of course you do not consider weak measurements measurements at all.

When you detect the position of the photon like on a special photon-counter imager (which does exist), all you see is particle behavior (pixels being lit up where the photons fall on them, indicating a "sharp localization"). Interference is a wave phenomena and is not related to position.

I can assure you that you can also get information about interference by using position (well, to be be precise: relative position) measurements using SPADs or other single photon counting devices. It is a question of your exact setup and the data analysis you perform. In any case you can get more than just bare particle behavior.

 

[SpectraCat]

I have changed my mind about this. The first detection is a state preparation, and not part of the momentum measurement. The second detection measures the momentum of the particle whose state was prepared by the first detection. So here's a new attempt to define the term "momentum measurement".

A detection of a particle is a momentum measurement if the particle was prepared with a sharply defined position $\vec x$ at a known time $t$. If the detection event is $(t',\vec x')$, then the vector $$m\gamma\frac{\vec x'-\vec x}{t'-t}$$ is called the result of the measurement.

Comments? Obviously, these are just my first attempts to write down something that resembles a definition, so don't take what I said as a claim that this is the definitive 100% perfect definition that everyone should use. Have I missed something obvious? Have you seen a definition in a book or a published article? Can you think of a meaningful definition that applies to particles that haven't been prepared in localized states? Can you think of a definition that doesn't require us to measure the position of the particle?

I don't think that constitutes a momentum measurement in the quantum sense, since it is unclear in principle how such a scheme would collapse the wavefunction of the particle into a momentum eigenstate. The reason I suggested describing such a process as *inferring* the momentum is because you are not actually getting a momentum measurement at a distinct point in time. You cannot say that "the momentum at point $(\vec{x},t)$ is $\vec{p}$" from such a "measurement", which is actually a pair of measurements, as is clear from your formula. As far as I can tell, all that formula allows you to *calculate* is the average momentum as the particle traveled from point $\vec{x}$ to point $\vec{x'}$.

 

[Fredrik]

Fredrik, I can't quote (technically) for some reason. Everytime I click quote on one of your post, it quotes one of your much earlier quotes. I'm new on here, I just need time to get more fluent with the dynamics.

If you click the quote button next to a post, you should get quote tags around the text in that post, not including text that was displayed in quote boxes. If you click the little > symbol above a quote box, it should take you to the post where the quoted comment was made. Hm, maybe the problem is that you have some of the "multi quote" buttons checked. That would make the text from the first multi quoted post appear at the top. If you want to quote several posts at once, click "multi quote" next to all of them except the last one. (The buttons will change color if you click them once. If you click them again, the color will change back. This unselects the post). Then you click "quote" next to the last one. Alternatively, click "multi quote" next to all of them, and then click "reply".

...YES, the uncertainties depend on the wavefunction but NO the uncertainty relationship I write above (HUP inequality) has nothing to do with the wavefunction.
...
The uncertainties have to do with the wavefunction you prepare, but the uncertainty relations don't.

OK, we're getting into semantics here. I obviously agree that the uncertainty relation for x and p is a theorem of the form "for all wavefunctions $\psi$, we have blah-blah($\psi$) ≥ blah-blah". I don't think I have said anything that gave you a reason to think that the inequalities only hold for specific values of $\psi$. If I did, it was certainly unintentional. You really need to find the exact quote if you're going to insist that I'm the one who screwed up here.

 

[SpectraCat]

Anyway, I disagree that interferometers must somehow measure position to measure the interference pattern. This is absolutely not true.

Ok, then please describe in detail how you would construct an interferometer for *single photons* (since that is what we are discussing here), that can give information about relative phases without measuring position. I have thought about it for a while now, and I am not sure that it can be done. (@Cthugha: please feel free to shoot me down on that statement ). A typical interferometer measures intensity on it's detectors, and so is insensitive to position, but of course that setup will not give information about single particle momenta, only ensemble averages.

Let me ask you this, you can create an interference of water waves in a bath tub. How are you detecting position to detect the wave momenta? It does not even make sense to talk of position in such a case.

I am not sure what you mean by that ... how can you begin to describe an interference pattern without reference to position? Don't you need to know the relative spacings of the constructive and destructive peaks? How can you talk about that without position?

 

[Fredrik]

I don't think that constitutes a momentum measurement in the quantum sense, since it is unclear in principle how such a scheme would collapse the wavefunction of the particle into a momentum eigenstate.

It's clear that it won't collapse the wavefunction into a momentum eigenstate. If the final detection doesn't destroy the particle, it will prepare it with a wavefunction with a sharp peak at the location of the detector. This of course implies that its Fourier transform isn't sharply peaked around the "result of the measurement".

Why would you want collapse to be a part of the definition of "measurement"? The definition of "momentum measurement" (possibly a very different one) must be considered part of the definition of "quantum mechanics", because if no such definition is part of the theory, the theory doesn't make any testable predictions about momentum. Since this is the reason why we need a definition, it seems very strange to require anything else from a "momentum measurement" than that it makes the theory agree with experiments.

...the average momentum...

I still haven't had time to read up on momentum measurements based on interferometry, but aren't we going to have the same problem there (and in every other kind of experiment that might possibly be called a momentum measurement)?

 

[fuesiker]

Fredrik, thanks a lot for your "forum advice". I think my problem was the I did click multi-quote once, and now every time I click to quote someone, it gives me a bunch of threads. I will fix it later ;). But for now, I want to say I did actually quote one of your sentences when you said, and I quote you now "I reject any claim that the uncertainty relations have nothing to do with preparations" (your post number 94). Had you said only uncertainties, I would agree. But you explicitly said uncertainty relations, which I understand as the HUP. I guess its semantics striking once again ;)

SpectraCat, of course you can do interferometry on single photons, even on one photon with itself. You can pass a photon (or even an electron, or with the proper cooling method, a whole car - this is way too difficult obviously, but can be done) and make that object (photon or car) interfere with itself. We're all waves, my friend.

The spatial grating between interference patterns has nothing to do with photon positions per se, but it does tell you the wavelength of your wave and from that you get momentum, which is hbar*k, and k=2*pi/wavelength.

 

[atyy]

Why would you want collapse to be a part of the definition of "measurement"? The definition of "momentum measurement" (possibly a very different one) must be considered part of the definition of "quantum mechanics", because if no such definition is part of the theory, the theory doesn't make any testable predictions about momentum. Since this is the reason why we need a definition, it seems very strange to require anything else from a "momentum measurement" than that it makes the theory agree with experiments.

The collapse is part of the measurement definition, so that we know how to evolve the wavefunction after the measurement. It is true that a projective measurement is not the most general measurement, but I think that is what is meant by a "sharp" measurement. Even the more general definition of measurement (which includes projective measurements) specifies the state after the measurement (eg. http://books.google.com/books?id=65...mmary_r&cad=0#v=onepage&q=von neumann&f=false , p85)

Also, my understanding is that Ballentine, Park and Margenau sort of "precise simultaneous momentum and position measurements" only work if the second time is taken to infinity and the initial state is restricted. If the second time is finite, the momentum distribution recovered is not the same as that predicted by the observables that commute with the observable conjugate to position. I believe their setup has state preparation, two position measurements and time of flight (so it's a bit different if you consider the first position measurement to be state preparation).

I don't have direct access to Park and Margenau's paper, but in #89 I posted an excerpt from a review by Bell which describes their stuff. Also, Busch and Lahti's 1984 paper (10.1103/PhysRevD.29.1634) describes the Park and Margenau argument (my very, very rough transcription) as "Take a state vector initially localized in some region of space and 0 in another region. Let it evolve freely. Then there is an observable F(Q)= mQ/t which approaches the observable P in the sense of the distribution of measured values being the same as t approaches infinity."

 

[Fredrik]

My claim is

"It's not true that every measurement puts the system in an eigenstate of the measured observable".​

This claim indeed contradicts standard QM. I think that the standard version is true, especially since this postulate can be viewed as a definition of measurement.

You're obviously right that if we take the collapse postulate as the definition of a measurement, my claim is false. However, this raises two interesting questions:

1. Why should we include collapse as part of the definition of "measurement"? (See my reply to SpectraCat above).

2. Don't experimental physicists who test the theory's predictions by measuring momenta already use a definition that contradicts the collapse postulate? (If a momentum measurement really is an inference of momentum from the coordinates of the event where the particle was detected, then the collapse postulate would make QM inconsistent).

 

[atyy]

2. Don't experimental physicists who test the theory's predictions by measuring momenta already use a definition that contradicts the collapse postulate? (If a momentum measurement really is an inference of momentum from the coordinates of the event where the particle was detected, then the collapse postulate would make QM inconsistent).

My understanding is that there isn't any true momentum measurement. Actually some other observable is measured, and marginalizing its distribution gives approximate position and momentum distributions. This other observable is chosen to make the approximate momentum distribution very close to the "ideal" momentum distribution. I'm not entirely sure that is correct, but I think these are the ideas behind eg. http://www.sciencedirect.com/science/article/pii/0003491692900862 and http://arxiv.org/abs/quant-ph/0609185. I think this would correspond to usual heuristic that the cloud chamber track is far wider than the de Broglie wavelength, so position is poorly measured, and momentum well measured.

 

[fuesiker]

Fredrik, please do not ignore what I will say here which is repetition of what I have said before: our measuring devices are classical, and hence, when you measure position, let's say, your meter should give you a sharp position value (an eigenvalue) as your state has to collapse onto an eigenstate of the position operator. Otherwise, your state is a superposition of two or more distinct position eigenstates, which you cannot measure on a classical device because that would mean your particle has two distinct positions at once (and this is not a spread, it literally means two sharp peaks at the same time). Sure, an electron in a hydrogen atom does have simultaneously infinitely many positions, but when you measure its position, it will just be one of them, hence the collapse. I like the boson number example better, because first of all bosons are cool, second of all, an example where you count particles is better suited to intuitively explain collapse in that you cannot count 1.5 bosons. In this example, if your state does not collapse to an eigenstate of the boson number operator, then you will HAVE to count a non-integer number of bosons.

 

[grzz]

... I've been wondering if there is any reason to intuitively expect difficulties when trying to simultaneously know both quantities. What I mean is, is there anything about the nature of "position" and "momentum" that hints that we should not be able to know both simultaneously?...

Did anyone mention the example which David Bohm gives in his QM book?
It goes something like this. If one uses a somewhat large interval of time to take a photo of a car which is moving very fast, then the car appears blurred. That is the photo shows clearly that the car has momentum but it lost most information about its position. Using a much shorter interval of time, the photo is not blurred anymore, showing better information about the position of the car but one barely notices that the car is moving, ie almost all information re momentun is lost.

 

[fuesiker]

My understanding is that there isn't any true momentum measurement. Actually some other observable is measured, and marginalizing its distribution gives approximate position and momentum distributions. This other observable is chosen to make the approximate momentum distribution very close to the "ideal" momentum distribution. I'm not entirely sure that is correct, but I think these are the ideas behind eg. http://www.sciencedirect.com/science/article/pii/0003491692900862 and http://arxiv.org/abs/quant-ph/0609185. I think this would correspond to usual heuristic that the cloud chamber track is far wider than the de Broglie wavelength, so position is poorly measured, and momentum well measured.

But this is not just the case with momentum, its the case with every observable and kith mentioned this in an earlier post. For example, if you want to measure the frequency (or energy) of a photon, you will not get a sharp frequency, but a spread of frequencies. This is because the photon interacts with your CCD camera (or SPAD, whatever) for a finite period of time dt, and we come back to HUP where dt*dE >= h (times some constant), and this means there will always be uncertainty in anything you measure. Quantum squeezing is a cool idea where you can "squeeze" one quadrature at the expense of the other. This is useful for metrology purposes.

 

[atyy]

But this is not just the case with momentum, its the case with every observable and kith mentioned this in an earlier post. For example, if you want to measure the frequency (or energy) of a photon, you will not get a sharp frequency, but a spread of frequencies. This is because the photon interacts with your CCD camera (or SPAD, whatever) for a finite period of time dt, and we come back to HUP where dt*dE >= h (times some constant), and this means there will always be uncertainty in anything you measure. Quantum squeezing is a cool idea where you can "squeeze" one quadrature at the expense of the other. This is useful for metrology purposes.

I agree.

 

[Fredrik]

The collapse is part of the measurement definition, so that we know how to evolve the wavefunction after the measurement.

It doesn't make any sense to me to include it as a part of the definition of "measurement". (See my reply to SpectraCat above). If we use a definition of measurement that allows us to consider what Ballentine is describing a "joint measurement of y and p_y", the collapse postulate is logically inconsistent. But we still know the wavefunction after the measurement, so we don't need the collapse postulate for that. It would still be true that a non-destructive measurement collapses the state (approximately) to an eigenstate of one of the measured observables, so it's not like we're throwing all of QM out the window.

I haven't completely thought this through, but it seems to me that we would just have to change the collapse postulate from a "for all observables" statement to a "for all observables that commute with position" statement.

Regarding the other things you said, I haven't had time to think about them yet, but I'll try to take the time later.

 

[atyy]

It doesn't make any sense to me to include it as a part of the definition of "measurement". (See my reply to SpectraCat above). If we use a definition of measurement that allows us to consider what Ballentine is describing a "joint measurement of y and p_y", the collapse postulate is logically inconsistent. But we still know the wavefunction after the measurement, so we don't need the collapse postulate for that. It would still be true that a non-destructive measurement collapses the state (approximately) to an eigenstate of one of the measured observables, so it's not like we're throwing all of QM out the window.

I haven't completely thought this through, but it seems to me that we would just have to change the collapse postulate from a "for all observables" statement to a "for all observables that commute with position" statement.

Regarding the other things you said, I haven't had time to think about them yet, but I'll try to take the time later.

I wonder how that would generalize to the relativistic case, where position observables have weird properties, but momentum is more-or-less ok (not worse than the non-relativistic case)?

 

[vanhees71]

Hi vanhees71, I don't think I understand what you're asking me, but I will give it a shot. Born's rule is straightforward. It simply says that when you measure an observable on a system (be it a single-particle of many-body system), the only possible measurement result is an eigenvalue of your observable. That is all he says basically, and it does not stem from any statistical properties of the ensembles. Rather, the statistical properties of the ensemble are resultant from Born's rule. Preparing a system the exact same way and performing a measurement on it again and again will give you different eigenvalues (unless your system is prepared in an eigenstate of the observable to begin with) of the measured observable. In the end and after a sufficiently large number N of measurements, you can infer the state of your system from the statistics of the measurement. However, you will never be able to know the exact state of your system unless N is infinity, but you can come incrementally close.

For me Born's rule is the probability interpretation of states, i.e., that the squared modulus of a wave function is the probability (or probability distribution) for the outcome of measurements of the observables, defining the Hilbert-space basis wrt. which you take the wave function. And I think at this point we agree fully about the consequence of this interpretation. So, what you have to do is to measure N times the observable on N identically prepared systems. If N is large enough you get the probabilities with a certain statistical accuracy, and this accuracy can be made as good as one wishes by taking large enough N.

Be careful observables are not correlated when you have many-body interactions, it is rather the particles in your many-body system are. This means if you perform a measurement on one of them, it will affect the other particles. In a sense, all particles are dancing to the same tune to a certain degree, and this depends on how strong the correlation is. If the particles are all fully entangled, then your many-body system can be thought of as just one big particle, because what you do to one particle will be equivalent to as if had you done it on another particle in your system.

Also true. That's why I said, one has to prepare always precisely the system one wants to measure (at least for our discussion). So it can be very different, if you prepare an N-particle state (be they interacting or not) and make measurements on each of them. Due to entanglement you can have correlations, which you don't want in the ensemble considered above.

I do fancy the Copenhagen interpretation, but I don't understand your EPR question. I do not see how you can address the EPR problem with just one particle to begin with. I think you are talking about superluminal transfer of information which Einstein wrongly implied from quantum mechanics because he believed nature had to be real and local at the same time, which experimentally has been shown with strong evidence to not be the case.

Sure, you can refer to the original EPR example or more modern realizations as the Aspect type experiments with entangled photons. In any case the problem with the Copenhagen interpretation in the sense that the wave function is a real property of the considered system and that thus the state collapse is a real physical process, there's trouble with Einstein causality, becaus that would mean when Alice measures the polarization of her photon by putting a polarizer in her beam instantaneously also the polarization of Bob's photon is determined. Now Alice's act is a local operation on her photon, and this cannot instantaneously alter any physical property of bob's photon.

Within the minimal interpration there's no problem. The photon pair is prepared in the entangled state and describes the probability to find a certain polarization of Alice's or Bob's photon. It also contains the description of the strong correlations between the outcomes of polarization measurements. That's all one can say about the photons when prepared in this state, and when Alice measures the polarization of her photon, I don't need any instantaneous collapse of the state to understand Bob's findings, which turn out to be totally correlated with Alice's photon-polarization state. This correlation has been there all the time with the preparation of the photon pair in its entangled state. That's it, and there's no trouble with Einstein causality.

I said preparation has nothing to do with the collapse in that collapse will always happen when you measure regardless of what state you prepare your system in. The preparation may only affect the resultant eigenstate onto which the state on which the observable is measured collpases to. Is this what you wanted to ask?

I don't understand this last statement. Why do I have to assume that the state collapses to the eigenstate of the measured value of my observable? This is an unnecessary complication. All I can say, given the quantum theoretical description of the system as being prepared in a certain state, are the probabilities to find any of the possible values of the measured observable. These probabilities I can measure with any desired accuracy by repeating the measurement on independently identically prepared systems as discussed at the beginning of this posting.

 

[Fredrik]

But for now, I want to say I did actually quote one of your sentences when you said, and I quote you now "I reject any claim that the uncertainty relations have nothing to do with preparations" (your post number 94). Had you said only uncertainties, I would agree. But you explicitly said uncertainty relations, which I understand as the HUP. I guess its semantics striking once again ;)

Maybe my choice of words in that particular sentence wasn't the greatest, but even a statement of the form "for all wavefunctions, blah-blah" is a statement that involves wavefunctions, so I don't think it's wrong to reject a suggestion that such a statement has nothing to do with wavefunctions.

Next time you have a problem with something I said, you should try to find the original statement or a clarification of it that I made later. Things will get weird if you quote a comment of mine where I'm referring to a statement of yours where you're referring a statement of mine where I'm referring to a statement of yours, and it's even worse if you don't quote me at all and just paraphrase what I said. For example, if I say that uncertainty relations are statements where wavefunctions are involved in some way, and you paraphrase by saying that I have said that uncertainty relations "depend on" wavefunctions, it's going to be very hard for me to reply.

Fredrik, please do not ignore what I will say here which is repetition of what I have said before: our measuring devices are classical, and hence, when you measure position, let's say, your meter should give you a sharp position value (an eigenvalue) as your state has to collapse onto an eigenstate of the position operator. Otherwise, your state is a superposition of two or more distinct position eigenstates, which you cannot measure on a classical device because that would mean your particle has two distinct positions at once (and this is not a spread, it literally means two sharp peaks at the same time). Sure, an electron in a hydrogen atom does have simultaneously infinitely many positions, but when you measure its position, it will just be one of them, hence the collapse. I like the boson number example better, because first of all bosons are cool, second of all, an example where you count particles is better suited to intuitively explain collapse in that you cannot count 1.5 bosons. In this example, if your state does not collapse to an eigenstate of the boson number operator, then you will HAVE to count a non-integer number of bosons.

I still don't know why you think this has anything to do with the things I've been saying.

 

[kith]

1. Why should we include collapse as part of the definition of "measurement"?

Because of the measurement problem. We observe collapse but when does it occur? The collapse postulate says, when a measurement is performed. This formal distinction between "measurement" interactions and common, unitary interactions seems very unnatural to me. Collapse-like behaviour can be explained in principle by decoherence on a short time scale (yet the implications are different in different interpretations). So I'm inclined to take the viewpoint that measurements are certain interactions with environments, which cause so fast decoherence, that it seems like an instantaneous collapse to the observer. This was the motivation for my remark. But I don't think, that a discussion about decoherence should enter this thread. It's already complex enough. ;-)

What is your definition of "measurement" in QM? What is the common ground, on which all measurements stand if not collapse?

2. Don't experimental physicists who test the theory's predictions by measuring momenta already use a definition that contradicts the collapse postulate?

With my definition, they perform finite resolution position measurements and calculate momentum from those measurements. What I have not thought much about yet, is if and how "actual" momentum measurements can be performed. Clearly, approximate momentum eigenstates can be prepared in experiments.

I do not think that one should define a measurement this way. For example you run into serious problems explaining weak measurements using this definition. Unless of course you do not consider weak measurements measurements at all.

Yes, when I write measurement, I imply projective measurement. The motivation is in the answer to Frederik's first question.

 

[fuesiker]

Maybe my choice of words in that particular sentence wasn't the greatest, but even a statement of the form "for all wavefunctions, blah-blah" is a statement that involves wavefunctions, so I don't think it's wrong to reject a suggestion that such a statement has nothing to do with wavefunctions.

Next time you have a problem with something I said, you should try to find the original statement or a clarification of it that I made later. Things will get weird if you quote a comment of mine where I'm referring to a statement of yours where you're referring a statement of mine where I'm referring to a statement of yours, and it's even worse if you don't quote me at all and just paraphrase what I said. For example, if I say that uncertainty relations are statements where wavefunctions are involved in some way, and you paraphrase by saying that I have said that uncertainty relations "depend on" wavefunctions, it's going to be very hard for me to reply.


I still don't know why you think this has anything to do with the things I've been saying.

It does because you say when you measure something on a state the latter does not have to collapse to an eigenstate. If by measurement you mean inferred measurement, then you're not wrong, but that's not what you call a measurement, at least not in the labs I work for.

Fredrik, I am not cutting corners here. You did say what I put in quotations, and that statement is wrong. Please don't be stubborn. A sign of intelligence is acknowledging your mistakes. None of us knows everything here. You said the *uncertainty relations* depend on the wavefunction, and that is wrong. The uncertainties do, not the uncertainty relations.

 

[Fredrik]

Fredrik, I am not cutting corners here. You did say what I put in quotations, and that statement is wrong. Please don't be stubborn. A sign of intelligence is acknowledging your mistakes. None of us knows everything here. You said the *uncertainty relations* depend on the wavefunction, and that is wrong.

Show me where I said that they depend on the wavefunction. It's possible that I've said something different that you have paraphrased using the words "depend on", and since I can't tell what that even means, I may have just assumed that you were expressing yourself poorly and didn't just completely change the meaning of what I had said.

In this case, the text you surrounded by quotation marks is "I reject any claim that the uncertainty relations have nothing to do with preparations". I stand by that comment, because to me, a statement of the form "for all wavefunctions $\psi$, blah-blah($\psi$)" certainly has something to do with wavefunctions. A statement that doesn't have anything to do with wavefunctions would be something like "x=2". Anyway, that statement is not what started this complete waste of time.

How should I interpret the fact that you haven't admitted any of your mistakes? There have been lots of them.

 

[Fredrik]

Because of the measurement problem. We observe collapse but when does it occur? The collapse postulate says, when a measurement is performed. This formal distinction between "measurement" interactions and common, unitary interactions seems very unnatural to me. Collapse-like behaviour can be explained in principle by decoherence on a short time scale (yet the implications are different in different interpretations). So I'm inclined to take the viewpoint that measurements are certain interactions with environments, which cause so fast decoherence, that it seems like an instantaneous collapse to the observer. This was the motivation for my remark. But I don't think, that a discussion about decoherence should enter this thread. It's already complex enough. ;-)

What is your definition of "measurement" in QM? What is the common ground, on which all measurements stand if not collapse?

The common ground is that they are the experimental procedures that are used to test the predictions of the theory. It seems very odd to me to have a purely mathematical definition of the term. A full definition of "quantum mechanics" must include for each observable (that we want to be able to make predictions about), a specification of what sort of device we would consider a measuring device for that observable. These specifications are the "correspondence rules" that tell us how to interpret the mathematics as predictions about results of experiments. I would define "measurement" by saying that an interaction+calculation that gives us a number that we can think of as "the result", is a measurement if and only if that interaction+calculation is mentioned in the correspondence rules.

If it's standard procedure for experimentalists to test predictions about momentum by detecting a particle and inferring a value of the momentum from the coordinates of the detection event, then I see no reason why this shouldn't be called a "measurement" of momentum. (No reason has been offered in this thread...except maybe atyy's comments, which I haven't had time to think about yet). Since it gives the particle a sharply localized position, it can't also give it a sharply localized momentum.

With my definition, they perform finite resolution position measurements and calculate momentum from those measurements. What I have not thought much about yet, is if and how "actual" momentum measurements can be performed.

I doubt that there is such a thing. The only thing a measurement device does is to produce a "signal" that informs us that an interaction of a specific kind has taken place. (I'm using the term "signal" very loosely. I would consider any approximately classical record of the result a "signal" here). The signal is by definition approximately classical, so it has a well-defined position. Doesn't this mean that we have determined the position of the interaction that caused the "signal"? If the particle participated in that interaction, haven't we determined (i.e. measured) the position of the particle?

 

[fuesiker]

Fredrik, when you say "I reject any claim that the uncertainty relations have nothing to do with preparations" is equivalent to saying "The claim is uncertainty relations have nothing to do with prepaprations => Fredrik rejects the claim". Its contrapositive is "Fredrik does not reject (ergo accepts) the claim => the claim is that uncertainty relations have something to do with preparations". If you cannot agree with this, then you really are at a mental level such as religious fanatics and I'm just wasting my time arguing with you when you have axiomatic fallacies in your reasoning.

Now, preparation is preparation of wavefunction, hence, any human would understand from your words that you are saying the HUP depends on the wavefunction when you say "I reject any claim that the uncertainty relations have nothing to do with preparations", and that is wrong and at odds with standard QM, just like your other statement on measurement and collapse that Kith quotes above and says it is at odds with standard QM.

You say also:

"a statement of the form "for all wavefunctions ψ, blah-blah(ψ)" certainly has something to do with wavefunctions", which is not true and shows how stubborn you are and insisting on arguing and not admitting your mistakes. You pathetically (yet again) accuse me of something with no evidence. I have "lots" of mistakes on here. Name one! I dare you. Back to yet another wrong statement of yours, the one I quote above. Consider the relation $\langle\psi|\hat{N}|\psi\rangle \geq 0$, where $\hat{N}$ is the number operator and $|\psi\rangle$ is a single-particle state (for simplicity's sake). Tell me, how does that have to depend on $|\psi\rangle$. This is always true regardless of your wavefunction unless you believe we can count a negative number of particle. I see nothing on this forum other than your fake behavior as a waste of time. You want to sound smart, go learn your basics. This statement should not make you angry, but rather you should take it as good advice.

I hope your reply will be better than your generic one of (and I paraphrase): "you're changing my words, you're wasting my time, Ballentine says so and so". Just for once try to negate one of my arguments using physics, and not citing one paper that is just a thought experiment. I can come up with a thought experiment that in Neverland corn flakes taste like fish. That does not mean it's true unless I can prove it experimentally.

 

[fuesiker]

I doubt that there is such a thing. The only thing a measurement device does is to produce a "signal" that informs us that an interaction of a specific kind has taken place. (I'm using the term "signal" very loosely. I would consider any approximately classical record of the result a "signal" here). The signal is by definition approximately classical, so it has a well-defined position. Doesn't this mean that we have determined the position of the interaction that caused the "signal"? If the particle participated in that interaction, haven't we determined (i.e. measured) the position of the particle?

Yet another demonstration of how uninformed you are. A well-defined position? Really? What's a well-defined position of a photon. So basically this means one can know the spatial dimension of the photon (please don't deny that this is a direct implication of your statement that we can find a well-defined position of some particle). A well-defined position is a geometric point of zero dimension. Let's consider SPAD or a general imager. They are made of pixels on which you have photoreceptor circuitry that changes EM energy into an electric signal.
Automatically, your position uncertainty is the spatial dimension of the pixel. And you can't make the pixel dimension arbitrarily small, so please don't argue on that. The signal is "approximately" classical? It's 100% classical because your device is a classical one, and AGAIN, your device never gives out superposition states.

And no, when there is a signal, that doesn't necessarily mean you know the position of a particle. Once again, read about the Mach-Zehnder interferometer and see how they measure momentum without any idea about position.

 

[atyy]

Fredrik, when you say "I reject any claim that the uncertainty relations have nothing to do with preparations" is equivalent to saying "The claim is uncertainty relations have nothing to do with prepaprations => Fredrik rejects the claim". Its contrapositive is "Fredrik does not reject (ergo accepts) the claim => the claim is that uncertainty relations have something to do with preparations". If you cannot agree with this, then you really are at a mental level such as religious fanatics and I'm just wasting my time arguing with you when you have axiomatic fallacies in your reasoning.

Now, preparation is preparation of wavefunction, hence, any human would understand from your words that you are saying the HUP depends on the wavefunction when you say "I reject any claim that the uncertainty relations have nothing to do with preparations", and that is wrong and at odds with standard QM, just like your other statement on measurement and collapse that Kith quotes above and says it is at odds with standard QM.

I believe all that is meant by "The uncertainty relations have to do with preparations" is that they prevent you from preparing a state with definite position and momentum. I don't think it is meant that the uncertainty relations change depending on what state you prepare.

I can come up with a thought experiment that in Neverland corn flakes taste like fish. That does not mean it's true unless I can prove it experimentally.

Delicious

 

[fuesiker]

I believe all that is meant by "The uncertainty relations have to do with preparations" is that they prevent you from preparing a state with definite position and momentum. I don't think it is meant that the uncertainty relations change depending on what state you prepare.
Delicious

That deduction is not at all clear to me. Moreover, this is trivial to say because the HUP does not just hold at the time of preparation, it is always true. This makes the mere mention of preparation very trivial. Let's be philosophical here, what is preparation? It's an experimental term. The only "true preparation" is the Big Bang I guess, and from then on any attempted preparation is just a "experimentalist-tailored" time-evolution process of your state rather than one effected on it by nature.

 

[atyy]

That deduction is not at all clear to me. Moreover, this is trivial to say because the HUP does not just hold at the time of preparation, it is always true. This makes the mere mention of preparation very trivial. Let's be philosophical here, what is preparation? It's an experimental term. The only "true preparation" is the Big Bang I guess, and from then on any attempted preparation is just a "experimentalist-tailored" time-evolution process of your state rather than one effected on it by nature.

Yes, it could be trivial to say, but correct anyway. The deduction I am thinking of is something like this. The uncertainty relation is defined as the non-commutation of position and momentum operators. A state with definite momentum is an eigenstate of momentum, and state with definite position is an eigenstate of position. The commutation relation prevents an eigenstate of momentum from being an eigenstate of position, so there is no state with definite momentum and position, and so it cannot be prepared.

 

[fuesiker]

Yes, it could be trivial to say, but correct anyway. The deduction I am thinking of is something like this. The uncertainty relation is defined as the non-commutation of position and momentum operators. A state with definite momentum is an eigenstate of momentum, and state with definite position is an eigenstate of position. The commutation relation prevents an eigenstate of momentum from being an eigenstate of position, so there is no state with definite momentum and position, and so it cannot be prepared.

Agreed.

 

[Fredrik]

Fredrik, when you say "I reject any claim that the uncertainty relations have nothing to do with preparations" is equivalent to saying "The claim is uncertainty relations have nothing to do with prepaprations => Fredrik rejects the claim". Its contrapositive is "Fredrik does not reject (ergo accepts) the claim => the claim is that uncertainty relations have something to do with preparations". If you cannot agree with this, then you really are at a mental level such as religious fanatics and I'm just wasting my time arguing with you when you have axiomatic fallacies in your reasoning.

LOL, no the contrapositive of "If C says that URHNTDWP*, then I reject C" is "If I don't reject C, then C doesn't say URHNTDWP". So if I accept the claim, you can conclude that it says something other than URHNTDWP. It might for example say that 1+1=2.

*) URHNTDWP = "uncertainty relations have nothing to do with preparations"

I have "lots" of mistakes on here. Name one! I dare you.

You failed to find the correct negation of a statement in the quote earlier in this post. I don't have time to try to locate all of your mistakes. You have wasted far too much of my time already. But these are a few that you made earlier:

Post #20: "The uncertainty principle has nothing to do with observation".
(Yes, you just expressed yourself really poorly, but it's still a mistake. If it has nothing to do with observation, i.e. measurement, then it wouldn't be a statement about how results of measurements will be statistically distributed around the mean).

Post #45: The third paragraph correctly refutes a statement that wasn't implied by any statement I had actually made, and you were still implying that it proves me wrong.

Post #54: Here you do exactly the same thing again.

Post #118: Here you do exactly the same thing again. It's fascinating that you keep making the same irrelevant argument over and over. If you had made it once, I would have assumed that you just don't want to admit to me that you made a mistake, but when you do it (at least) three times, I can only assume that you're so stubborn that you can't even admit it to yourself.

Post #54: You also made the mistake to think that my statement that observables fail to commute if and only if the corresponding state preparation devices would interfere with each other can be refuted by an obviously true statement about Fourier transforms. I might be wrong about what I said there (I started the statement with "I believe" just to make that clear), but if I am, it's not for the reason you said. If you're going to describe what someone else says as "madness", you should at least have a relevant counterargument.​

Back to yet another wrong statement of yours, the one I quote above. Consider the relation $\langle\psi|\hat{N}|\psi\rangle \geq 0$, where $\hat{N}$ is the number operator and $|\psi\rangle$ is a single-particle state (for simplicity's sake). Tell me, how does that have to depend on $|\psi\rangle$.

I didn't use the phrase "depend on". Those are the words you used to try to change the meaning of what I said. What I said is roughly that an inequality like that "has something to do with wavefunctions". It is after all a statement about a property shared by all wavefunctions, not all rational numbers, or all unicorns. Anyway, if I had in fact made statements like that, the problem would be at the level of language. It wouldn't have anything to do with physics.

I hope your reply will be better than your generic one of (and I paraphrase): "you're changing my words, you're wasting my time,

As long as you're doing exactly that, what else can I say?

Just for once try to negate one of my arguments using physics, and not citing one paper that is just a thought experiment.

You need to post a relevant argument first.

Yet another demonstration of how uninformed you are. A well-defined position? Really? What's a well-defined position of a photon.

Seriously, you need to learn how to read.

And no, when there is a signal, that doesn't necessarily mean you know the position of a particle. Once again, read about the Mach-Zehnder interferometer and see how they measure momentum without any idea about position.

I will, but I've had less than an hour today to do anything other than to answer posts in this thread.

 

[fuesiker]

LOL, no the contrapositive of "If C says that URHNTDWP*, then I reject C" is "If I don't reject C, then C doesn't say URHNTDWP". So if I accept the claim, you can conclude that it says something other than URHNTDWP. It might for example say that 1+1=2.

*) URHNTDWP = "uncertainty relations have nothing to do with preparations"

I don't see how my contrapositive is different from yours. I said if you don't reject a statement, that means the uncertainty relations have something to do with preparations, which is equivalent to your claim that one can conclude that it says something other than URHNTDWP, because any claim that you do not reject must not include URHNTDWP, hence every statement you make, for you not to reject must implicitly include not(URHNTDWP). But let's say I'm wrong about the contrapositive. Any person from your comments would understand that you believe the uncertainty relations have to do with preparations.

You failed to find the correct negation of a statement in the quote earlier in this post.

I said I will give you that, but at least I don't make a fool of myself by making foolish statements about physics that violate standard QM theory.

I don't have time to try to locate all of your mistakes. You have wasted far too much of my time already. But these are a few that you made earlier:

Post #20: "The uncertainty principle has nothing to do with observation".
(Yes, you just expressed yourself really poorly, but it's still a mistake. If it has nothing to do with observation, i.e. measurement, then it wouldn't be a statement about how results of measurements will be statistically distributed around the mean).

Post #45: The third paragraph correctly refutes a statement that wasn't implied by any statement I had actually made, and you were still implying that it proves me wrong.

Post #54: Here you do exactly the same thing again.

Post #118: Here you do exactly the same thing again. It's fascinating that you keep making the same irrelevant argument over and over. If you had made it once, I would have assumed that you just don't want to admit to me that you made a mistake, but when you do it (at least) three times, I can only assume that you're so stubborn that you can't even admit it to yourself.

Post #54: You also made the mistake to think that my statement that observables fail to commute if and only if the corresponding state preparation devices would interfere with each other can be refuted by an obviously true statement about Fourier transforms. I might be wrong about what I said there (I started the statement with "I believe" just to make that clear), but if I am, it's not for the reason you said. If you're going to describe what someone else says as "madness", you should at least have a relevant counterargument.​

None of these are physics mistakes, unlike your many laughable physics mistakes.. And you are lying implying I am twisting your words. You did imply that the HUP depends on preparation, which is identical, mathematically, to say it has something to do with preparation. Come on, have some intellectual honesty to take responsibility for your statements.

Regarding my statement "The uncertainty principle has nothing to do with observation", that is correct. One of the first things they teach you in QM is that the HUP is there regardless of measurement. You do not want to understand this because you don't seem to understand the Fourier transform argument I made, even though here you say you may agree with this now. This contradicts your belief that I made a mistake about the statement "The uncertainty principle has nothing to do with observation".

I didn't use the phrase "depend on". Those are the words you used to try to change the meaning of what I said. What I said is roughly that an inequality like that "has something to do with wavefunctions". It is after all a statement about a property shared by all wavefunctions, not all rational numbers, or all unicorns. Anyway, if I had in fact made statements like that, the problem would be at the level of language. It wouldn't have anything to do with physics.As long as you're doing exactly that, what else can I say?You need to post a relevant argument first.Seriously, you need to learn how to read. I will, but I've had less than an hour today to do anything other than to answer posts in this thread.

I am fluent in 3 languages and read a lot and really well, thank you. Maybe you can try to read yourself some more physics instead of making a joke out of yourself here. It's no crime if you don't understand QM unless you start acting like you do.

So far, and anyone with any intellectual honesty on here would agree, I made no physics mistakes, and you are the *only* one on this thread who's been making outrageous statements that violate QM theory, and what's worse is you make them sounding as if you're an expert. In my humble opinion, you need to ask yourself if you're a poser? Search deep in your soul and work on yourself. I appreciate your passion for physics, but you should have more respect for physics and learn its tenets really well before you make pseudo-informed statements about them.

Anyway, let's get back to the physics: From your posts, I and at least one other person (Kith) have recognized that you believe in at least the following:

1) The measurement on a state of an observable does not necessarily collapse that state onto an eigenstate of that observable. That's plain wrong.

2) The HUP has something to do with observation. This is also plain wrong. They teach you in quantum mechanics that the HUP is not an observational artifact, it's there no matter how sophisticated or accurate your measurement device so that even "God" (they really teach that in classes I took and tutored in from different teachers) cannot go beyond the HUP. It has NOTHING to do with observation. Again, look at the Fourier transform relationship between $\langle x|\psi\rangle$ and $\langle p|\psi\rangle$. How on Earth does that show any observation relationship?

3) The HUP has something to do with preparation or wavefunction. That is also wrong and you again avoided my challenge to actually address the relation I gave you related to the number operator except by claiming that you never said it "depends on". But mathematically depending on something is having something to do with it. Again, be honest with yourself. If I say x has something to do with the value of a function f, then definitely x is a variable upon which f depends either explicitly or at least implicitly. You not seeing this shows a sad mathematical reasoning within you.

Moreover, I am writing a paper to submit for publication this week, but I am still taking my time to read other stuff and this thread. So cut the crap regarding language and corresponding excuses. Face the fact that you made at least 3 statements completely at odds with QM theory, and what's worse you insist on them.

But the best question is: why am I wasting my time on someone like you?

 

[Fredrik]

I said I will give you that, but at least I don't make a fool of myself by making foolish statements about physics that violate standard QM theory.

No, you do it by aggressively and incorrectly accusing others of doing that.

unlike your many laughable physics mistakes..

You still haven't actually refuted any of my "laughable physics mistakes". All of your arguments have addressed something other than what I actually said.

And you are lying implying I am twisting your words.

I have been completely honest about everything. Your posts on the other hand are filled with outrageous lies and accusations.

Regarding my statement "The uncertainty principle has nothing to do with observation", that is correct. One of the first things they teach you in QM is that the HUP is there regardless of measurement. You do not want to understand this because you don't seem to understand the Fourier transform argument I made, even though here you say you may agree with this now. This contradicts your belief that I made a mistake about the statement "The uncertainty principle has nothing to do with observation".

This is just ridiculous. I haven't given you any reason to doubt that I understand the stuff about Fourier transforms. I also explained why your statement is wrong, but as always you don't address that in your reply.

1) The measurement on a state of an observable does not necessarily collapse that state onto an eigenstate of that observable. That's plain wrong.

No, it's a matter of definition (of "measurement"). So far, neither you nor anyone else have presented a valid argument for the definition that makes the above "plain wrong" should be preferred over the alternative.

2) The HUP has something to do with observation. This is also plain wrong. They teach you in quantum mechanics that the HUP is not an observational artifact, it's there no matter how sophisticated or accurate your measurement device so that even "God" (they really teach that in classes I took and tutored in from different teachers) cannot go beyond the HUP. It has NOTHING to do with observation. Again, look at the Fourier transform relationship between $\langle x|\psi\rangle$ and $\langle p|\psi\rangle$. How on Earth does that show any observation relationship?

3) The HUP has something to do with preparation or wavefunction. That is also wrong and you again avoided my challenge to actually address the relation I gave you related to the number operator except by claiming that you never said it "depends on". But mathematically depending on something is having something to do with it. Again, be honest with yourself. If I say x has something to do with the value of a function f, then definitely x is a variable upon which f depends either explicitly or at least implicitly. You not seeing this shows a sad mathematical reasoning within you.

This isn't about mathematical reasoning or physics. You just don't understand the meaning of the words "has something to do with".

 

[fuesiker]

When I accuse you of stating something that violates QM theory, I tell you what it is and I show you why it violates QM theory. You say I have made outrageous lies, please point them out (only the one regarding physics, and not our petty bickering).

If you're going to reply by saying you don't have time, then you just reinforce my belief that you're a joke who likes discussing physics to be smart and have some self-worth. I bet you're not a physicist, not published, and barely have respect for physics strong enough to actually read established papers that discuss QM. you've barely read standard QM books such as Sakurai or Cohen-Tannoudji, because if you have read them you surely wouldn't be making foolish statements.

And throwing the accusations I make about you right back on me shows what a little baby you can be when I call you on your mistakes.

 

[Fredrik]

When I accuse you of stating something that violates QM theory, I tell you what it is and I show you why it violates QM theory. You say I have made outrageous lies, please point them out (only the one regarding physics, and not our petty bickering).

What you said right now is an outrageous lie. You're lying about having refuted things I've said. If you think I have lied about anything, it's because you have reading comprehension issues and lack the honesty to admit your mistakes even to yourself. Your counterarguments are usually not wrong, but are always irrelevant.

When I throw things back at you, it's because you're falsely accusing me of exactly the things you're actually doing yourself. It's really bizarre. I haven't seen anyone behave this way since the last time I had an argument with someone who believes that there are people who can talk to ghosts.