What Intuitive Insights Explain Heisenberg's Uncertainty Principle?

[matphysik]

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?

NO. QM is a mathematical theory that happens to correctly model (in reasonable approximation) nature at that scale.

 

[Fredrik]

NO. QM is a mathematical theory that happens to correctly model (in reasonable approximation) nature at that scale.

I guess you could say that, but it's also a fact that this theory doesn't attribute the property of "having a position" to all particles. Same thing with momentum. So the fact that this theory makes accurate predictions about results of experiments isn't an argument in favor of the claim that both can be known. The theory doesn't even say that a particle has a position and a momentum at all times, so it certainly isn't immediately obvious that it has both and that there's nothing that prevents us from knowing them.

 

[matphysik]

Yes, that`s aside of what i said. What you`re saying follows immediately from the interpretation of |ψ(·,t)|² as the probability distribution of position in x-space, with its momentum analogue in k-space. Where ψ∈L₂, is the so-called `wave function`.

 

[fuesiker]

Dickfore, this may be what you mean, but it's not what Landau Lifgarbagez mean. I explained what they mean in an earlier post.

Frederik, note that the paper you are referencing is merely a thought experiment and the way they measure momentum is classical, and they themselves admit that others may have doubts about that. Experimentally, I am not familiar with any work that claims people measure an observable only to find the state after measurement not being an eigenstate of that observable.

Now, instead of citing the thought experiment you keep citing, just ask yourself this. And your premise of differentiating between these observables is wrong, because in QM any observable is a Hermitian matrix, and hence its mathematical behavior is similar to other observables, although of course physically they are propagators of different things. But, again, mathematically they are all Hermitian matrices, so whether you talk about momentum, position or the number operator it does not matter.

Let's take the latter. Consider an optical lattice that you have that starts with initial state |1,0,1,0,1,0,1,0>. Now, quench it with the Bose-Hubbard Hamiltonian, and it evolves according to that to around that state |0.5,0.5,0.5,...>. Now this is not an eigenstate of the number operator. In the lab, you go measure the number of bosons on this lattice, will you get 0.5? No you won't, you will measure something like |1,1,0,0,1,0,0,1>, an eigenstate of the number operator. This is because if you DO NOT get an eigenstate, then your measuring device is telling you you have a superposition of at least two eigenstates of the number operator (that would be what YOU are saying), which can look something like c1*|1,1,0,0,1,0,0,1>+c2*|1,1,0,0,1,0,1,0>. This is ridiculous because it means YOU in the lab measure on the last site of your lattice 1 boson and 0 bosons at the same time. This is EXACTLY similar to another example I gave you earlier about position.

I believe the paper you reference is wrong, and I bet you you would can find people who refuted the thought experiment therein, perhaps using simple arguments like mine above.

So now please, instead of writing me back again telling me the same thing from that same paper, please tell me how you think my above example is wrong. If you can prove to me that I'm wrong, i.e. if you can tell me of an experiment that measured different value of an observable at the same time for the same particle, , please tell me and let's go make millions of dollars in quantum computing :D

 

[fuesiker]

I believe that when two observables don't commute, the reason is always that the corresponding state preparation devices (not measuring devices) are incompatible in the sense that they would (at least) interfere with each other.

This is madness. It has nothing to with the preparation (or measuring) devices why two observables do not commute. Let's take your two favorite observables, position and momentum. YOU CANNOT MEASURE THEM BOTH SIMULTANEOUSLY, IT'S IMPOSSIBLE DESPITE WHAT THAT PAPER SAYS.

Sakurai's Modern Quantum Mechanics, Page 54, "Momentum Operator in Position Basis", he starts with the fact that the momentum operator is the propagator of translation, and from that he derives that the wavefunction in momentum space is the Fourier transform of the wavefunction in position space. As you know, the Fourier transform makes a broad function from a sharp one, and vice versa, making position and momentum INHERENTLY noncommutative, NO MATTER WHAT DEVICES ARE THERE, for preparation of measurement.

Please do make sure your statements are well established before making them because this will confuse a lot of the beginners here. Heisenberg was criticized for his Heisenberg microscope thought experiment to explain the uncertainty principle, specifically because it led to the reader erroneously thinking that the uncertainty lies in the measurement. Here you claim its the preparation, and that is equally wrong. The uncertainty is inherent here due to the fact that momentum is the propagator of position. Intuitively, one can simply see it this way: p is changing x continuously, and so in phase-space (plotting p against x), if you look at the interval [x,x+dx], you can see a range of momenta [p,p+dp], which would allow for such translation dx. But now consider a translation of dx=0, i.e. look exactly at the point x. You see based on the translation operator U=1-i*P*dx (capital P to differentiate operator from eigenvalue), that p can then take any value it wants, in phase-space that will have no effect since we are "frozen" in space and there is no translation. Now when dx=Inf, let's say, your only possible momentum value is 0, otherwise your translation operator is divergent. These are thus the two extreme cases that INTUITIVELY (surely not rigorously) show when dx=0, dp=Inf and vice versa. The rigorous proof is the one I cite above from Sakurai.

 

[Fredrik]

please tell me how you think my above example is wrong.

It's not. It's just not relevant to anything I've been saying. You keep refuting a claim I haven't made. Why are you doing that? In my previous post (after the last time you refuted the wrong idea) I told you explicitly that the idea you're refuting is the wrong one, and here you go refuting the same idea once more.

This is madness. It has nothing to with the preparation (or measuring) devices why two observables do not commute.

I might be wrong about that. If I am, I'd like to know about it. You can prove me wrong by finding an example of two commuting observables such that the corresponding state preparation devices interfere with each other, or an example of two non-commuting observables such that the corresponding state preparation devices don't interfere with each other.

Let's take your two favorite observables, position and momentum. YOU CANNOT MEASURE THEM BOTH SIMULTANEOUSLY, IT'S IMPOSSIBLE DESPITE WHAT THAT PAPER SAYS.

You can't refute an argument by shouting.

You need to ask yourself what sort of physical interaction an experimental physicist would consider a "momentum measurement". To say that Ballentine's thought experiment is wrong is to say that experimental physicists would reject the idea that momentum can be measured by detecting the particle and inferring the momentum from the location of the detection event. I don't think they would reject that idea. Do you?

As you know, the Fourier transform makes a broad function from a sharp one, and vice versa, making position and momentum INHERENTLY noncommutative, NO MATTER WHAT DEVICES ARE THERE, for preparation of measurement.

It's obvious from their mathematical definitions that they don't commute. The addition "no matter what devices there are for preparation" doesn't really make sense. What I'm saying (and this is the thing I might be wrong about) is that the correspondence between self-adjoint operators and measuring devices is such that operators commute precisely when the corresponding state preparation devices interfere. Your argument doesn't address this point.

Please do make sure your statements are well established before making them because this will confuse a lot of the beginners here.

I urge you to do the same. It's clear that you haven't made an effort to understand what I'm saying. You keep arguing against claims I haven't made, and you still haven't pointed out any flaws in my argument. ("It's wrong" doesn't count as an argument, no matter how loud you shout it).

 

[DrChinese]

This is madness. It has nothing to with the preparation (or measuring) devices why two observables do not commute. Let's take your two favorite observables, position and momentum. YOU CANNOT MEASURE THEM BOTH SIMULTANEOUSLY, IT'S IMPOSSIBLE DESPITE WHAT THAT PAPER SAYS...

I don't think Fredrik is prone to madness. I think the point being made is getting sort of technical and the language gets in the way.

For example, you say that you can't measure 2 non-commuting properties simultaneously when in a way you can. If you measure them on a pair of entangled particles where the value for the other can be deduced, you effectively can measure them simultaneously (by inference). And yet the results won't violate the HUP.

I understand Fredrik's point to be that those particles cannot be prepared in a state where both momentum and position are known - that would violate the HUP. So I think that is probably a fair statement.

 

[San K]

For example, you say that you can't measure 2 non-commuting properties simultaneously when in a way you can. If you measure them on a pair of entangled particles where the value for the other can be deduced, you effectively can measure them simultaneously (by inference). And yet the results won't violate the HUP.

Trying to understand the above better.

Is it possible, in an entangled photon pair, to measure position of one twin and momentum of other twin simultaneously?

 

[vanhees71]

As usual, such "philosophical" debates lead to a lot of confusion since it depends on believes of the proponents of different interpretations rather than scientifically well defined statements.

E.g., I'm a proponent of the "Minimal Statistical Interpratation", and for me this means to just take the general mathematical structure of quantum mechanics about states and observables with the Born probality interpretation of the states (kets or more general statistical operators).

Given this mathematical framework, the uncertainty relations are a precise mathematical statement about the standard deviations of observables valid in any state of the system. It says

\Delta A \Delta B \geq \frac{1}{2} |\langle [A,B] \rangle|.

From the point of view of the minimal interpretation, quantum mechanics describes the statistical properties of measurements on an ensemble of equally and indpendently prepared physical systems. Then quantum theory predicts that the standard deviations of two observables, no matter how the systems are prepared (i.e., in which state they are set up) fulfill this uncertainty relation.

Thus, indeed the uncertainty relations are statements about any possible preparation of physical systems, i.e., one cannot prepare any ensemble of particles which at the same time have arbitrarily sharp positions and momenta. It doesn't say anything about a single particle within the ensemble. According to the minimal interpretation, the question whether or not a single particle can have sharp position and momentum simultaneously, cannot be answered within quantum theory. It doesn't even make sense to ask that question since quantum theory only makes statesments about the statistical properties of ensembles in the above given sense.

Of course, Heisenberg's microscope has been the first attempt to explain the uncertainty relation in more "physical terms". It's ironic that Bohr had another view on the interpretation of this gedanken experiment. It's telling that even the very inventers of the Copenhagen interpretation haven't agreed on this interpretation themselves. As far as I can see, only the minimal interpretation is free of such arbitrariness and thus the only scientifically proper way to think in terms of quantum theory as a decription of nature.

A pretty nice summary on this "interpretational issues" can be found here:

http://plato.stanford.edu/entries/qt-uncertainty/#MinInt

 

[Fredrik]

I don't think Fredrik is prone to madness.

Thanks.

For example, you say that you can't measure 2 non-commuting properties simultaneously when in a way you can. If you measure them on a pair of entangled particles where the value for the other can be deduced, you effectively can measure them simultaneously (by inference). And yet the results won't violate the HUP.

In Ballentine's single-slit experiment, the momentum of the particle is measured by detecting the particle and inferring the value of the momentum from the location of the detection event. But every detection is a position measurement, so we are in fact measuring both the position and the momentum of just one particle with a single detection.

I understand Fredrik's point to be that those particles cannot be prepared in a state where both momentum and position are known - that would violate the HUP.

That's certainly a statement that I agree with, but my main point is this thread is what I just mentioned: Momentum measurements are position measurements, so it doesn't make sense to say that we can't measure both at the same time. Another important point is that this doesn't contradict any uncertainty relations, precisely because if the particle survives the detection, it will be in a state of sharply defined position (which implies poorly defined momentum).

The uncertainty relations are statements about what sort of states can be prepared, or equivalently, statements about the statistical spread of measurement results around the mean in long series of measurements on identically prepared systems. The point of Ballentine's example is to show explicitly that the uncertainty relation for position and momentum doesn't apply to the result of a single detection of a particle.

 

[fuesiker]

Vanhees, the HUP is valid for a single particle. In a single particle, you cannot measure its position and momentum simultaneously. Again, please see my previous posts on how the wavefunction of a single particle (or a many-body quantum system) in momentum space is the Fourier transform of that particle's wavefunction in momentum space, effectively rendering it impossible to know both momentum and position at the same time. You can start with this argument and its corresponding equation, and derive the HUP.

Frederik, I do not mean to shout when I all-cap things, and I specifically states that I am all-capping to stress my views not to shout.

Moreover, not one statement I made on here is not consistent with the general postulate of quantum mechanics, including the wavefunction collapse onto an eigenstate of the measure observable: http://en.wikipedia.org/wiki/Measur...antities_.28.22observables.22.29_as_operators

Sorry you feel I didn't make an effort to understand your claims. I think I did, especially that I asked you for the paper you're referencing and actually read through it. To me, I feel we reached a point where there are no significant disagreement, and thus this discussion is over. Thanks for it!

 

[fuesiker]

Dr. Chinese, I do not agree with your argument on the entangled pair of photons. Let's say Alice on one side is measuring polarization of photon A and Bob on the other side is measuring the momentum of photon B, and photons A and B are entangled. Now it all depends on who measures first. Let's say Alice measure first, then the state of both photons has collapsed onto an eigenstate of the the polarization operator (i.e., the photon pair now has the polarization of the axis of the polarizer Alice used). From a practical point of you, you can forget about the entanglement from now on because this photon pair is no longer entangled the way it started with, due to entanglement between this pair and Alice's device. But for argument's sake, let's say this state, after measuring at Alice's end, is prepared such that photons A and B are again entangled such that both have the same polarization as the axis of Alice's polarizer (eigenstate of polarization observable). Then when Bob measures momentum, this state will collapse onto an eigenstate of momentum.

Hence in short, it always depends on who measures first. You can now argue that what if we make it happen such that both people measure their quantities at the exact same time. The answer is that you can't do that. In your measuring devices, there is always a finite nonzero positive time period of interaction between the state and the device, and as best as I know, no-one has been able to determine the time-dynamics of wavefunction collapse. But I do know that work on quantum decoherence deals with this in a very good way (Zurek).

 

[SpectraCat]

The problem here is that no one has ever made (as far as I know) a device that can measure momentum without simultaneously localizing the wavefunction, and thus inherently ALSO measuring position. I cannot think of a single momentum measurement that I can make experimentally that does not involve using timing (i.e. velocity) and position measurements.

So, I think both Frederik and fuesiker are partly right. Fuesiker is correct that you cannot simultaneously collapse the wavefunction in both momentum space and position space .. this is obvious from the Fourier formulation, and is the essential content of the HUP.

However, Frederik is correct that we can simultaneously measure position, and *infer* momentum information. The way this would usually be done would be to prepare a beam of particles with a well-characterized velocity distribution, pass them through a small pin-hole (which is a de-facto position measurement), and then measure their position along a direction orthogonal to their propagation direction. This position measurement collapses the wavefunction in position space, however since you also have an earlier position measurement from the pinhole, you can also *infer* the off-axis momentum of the particle, although you did not measure it directly in the experiment. Since this is a thought experiment, both position measurements (the pinhole and the final one) can be made to arbitrary accuracy, as can the velocity selection of the beam (this is straightforward to do using a series of beam choppers). Thus the off-axis momentum can also be calculated with arbitrary accuracy. That is what is typically meant by the claim that it is possible to "measure" both the position and momentum of a single particle with arbitrary accuracy.

In the latter case, where the HUP comes in is that if you repeat the experiment for a large ensemble of particles (all selected with the same narrow velocity distribution), then you will observe that the width of the off-axis momentum distribution will conform to the expected HUP relationship, based on the uncertainties of the two position measurements. It's worth noting that if you make careful timing measurements, you can also get the same "inferred momentum measurement" for the longitudinal component of the particle momentum.

 

[DrChinese]

Dr. Chinese, I do not agree with your argument on the entangled pair of photons. Let's say Alice on one side is measuring polarization of photon A and Bob on the other side is measuring the momentum of photon B, and photons A and B are entangled. Now it all depends on who measures first. Let's say Alice measure first, then the state of both photons has collapsed onto an eigenstate of the the polarization operator (i.e., the photon pair now has the polarization of the axis of the polarizer Alice used). From a practical point of you, you can forget about the entanglement from now on because this photon pair is no longer entangled the way it started with, due to entanglement between this pair and Alice's device. But for argument's sake, let's say this state, after measuring at Alice's end, is prepared such that photons A and B are again entangled such that both have the same polarization as the axis of Alice's polarizer (eigenstate of polarization observable). Then when Bob measures momentum, this state will collapse onto an eigenstate of momentum.

Hence in short, it always depends on who measures first. You can now argue that what if we make it happen such that both people measure their quantities at the exact same time. The answer is that you can't do that. In your measuring devices, there is always a finite nonzero positive time period of interaction between the state and the device, and as best as I know, no-one has been able to determine the time-dynamics of wavefunction collapse. But I do know that work on quantum decoherence deals with this in a very good way (Zurek).

Again, yes and no.

I said simultaneously by inference. And I agree you can't really do anything at different places simultaneously. But the flip side is that you can't really say there is any difference in outcomes as a result of the ordering of measurements. There is no specific evidence to support that conclusion, it could just as easily be the second measurement that controls instead of the first. I could just as easily say (and I did, also without specific evidence) that you simultaneously know non-commuting observables about Alice & Bob, even though they are placed into different eigenstates for any subsequent measurement.

 

[atyy]

Here's what I don't understand about Ballentine's set up. He does p_y=psinθ. Doesn't the sinθ imply the particle has a definite position at the slit, and is no longer in a momentum eigenstate? If so, what is the meaning of p then?

I don't think momentum measurements are made by measuring positions precisely. The proper way would be to use http://www.kitp.ucsb.edu/members/PM/joep/Web221A/LSZ.pdf" , but heuristically, I think the track in a cloud chamber is so spatially large, that it is a much more precise measurement of momentum than position.

 

[Fredrik]

Here's what I don't understand about Ballentine's set up. He does p_y=psinθ. Doesn't the sinθ imply the particle has a definite position at the slit, and is no longer in a momentum eigenstate? If so, what is the meaning of p then?

It's been a while since I looked at the details of his argument, but I think he's saying that the particle has a definite energy at the slit, and therefore a definite value of \vec p^2. The direction of \vec p is however not sharply defined at the slit.

 

[Fredrik]

Vanhees, the HUP is valid for a single particle. In a single particle, you cannot measure its position and momentum simultaneously. Again, please see my previous posts on how the wavefunction of a single particle (or a many-body quantum system) in momentum space is the Fourier transform of that particle's wavefunction in momentum space, effectively rendering it impossible to know both momentum and position at the same time. You can start with this argument and its corresponding equation, and derive the HUP.

The uncertainty relation for position and momentum "applies to a single particle" in the sense that it describes a property of the state of a particle*. It also "applies to an ensemble of identically prepared systems", in the sense that it correctly predicts the distribution of results in a long series of measurements. So it looks like we have one argument in favor of "applies to a single particle" and one in favor of "applies to an ensemble". But there's also an argument against "applies to a single particle", even though you refuse to acknowledge it: If the uncertainty relation is interpreted as a prediction about the accuracies of simultaneous measurements, it's wrong. Ballentine showed that the product of the accuracies can be made as small as we want them to be.

The reason you don't see this is that you don't understand the distinction between measurement and state preparation. You said to vanhees71 that the stuff about Fourier transforms proves that you can't measure both at the same time. That's not true. It only shows that no preparation procedure will give you a state with a sharply defined momentum and a sharply defined position. You can still measure them at the same time, because you measure the momentum by measuring the position, and this prepares a state with a sharply defined position and a poorly defined momentum.

*) This argument in favor of "applies to a single particle" is actually very weak, since we were just able to say that it applies to the state of a single particle, and it's not at all clear that a state represents the properties of a particle. The only entirely uncontroversial thing we can say about a state of a particle is that it represents the statistical properties of an ensemble of identically prepared particles. (QM is consistent with several ideas about what it might represent in addition to that, but there's no evidence that favors any of them over the others).

 

[BWV]

the operators not commuting would also be true of position and velocity operators in classical mechanics (x d/dx - d/dx x) = 1 so the non-commutation does not inherently constitute a proof for the uncertainty principle? or do you just not care about the uncertainty at classical scales? (but in that case the math would pre-date the uncertainty principle?)

 

[SpectraCat]

Uo

The reason you don't see this is that you don't understand the distinction between measurement and state preparation. You said to vanhees71 that the stuff about Fourier transforms proves that you can't measure both at the same time. That's not true.

You really need to carefully define what you mean by "measure" to clarify by what you mean by the above statement. You are *not* using the typical QM definition of *measure* for the momentum, i.e. act upon the quantum state to project it into an eigenstate of the operator being measured, when you make the above statement. Obviously you cannot simultaneous project the particle into a position eigenstate and a momentum eigenstate, and I don't think you are claiming that you can. What you mean by "measure" in your statement is, "perform and experiment to obtain the value of the momentum", which is not necessarily the same thing, as you have been saying all along, and as I explained in my last post in detail.

It only shows that no preparation procedure will give you a state with a sharply defined momentum and a sharply defined position. You can still measure them at the same time, because you measure the momentum by measuring the position, and this prepares a state with a sharply defined position and a poorly defined momentum.

I would really suggest the following linguistic change to the phenomenon you are describing. Instead of saying that you are "measuring" the momentum, say instead that you are "measuring the position, and inferring (or calculating) the momentum". This is important, because you need (at least) two position measurements to infer the momentum in the manner you suggest, whereas if you actually *measured* the momentum directly in the QM sense, you would only need one measurement.

 

[atyy]

But if the collapse postulate (http://arxiv.org/abs/0903.5082" "outcomes correspond to eigenstates of the measured observable, and only one of them is detected in any given run of the experiment") is wrong, then why would we expect the particle to be in a position eigenstate after the measurement of position?

 

[SpectraCat]

But if the collapse postulate (http://arxiv.org/abs/0903.5082" "outcomes correspond to eigenstates of the measured observable, and only one of them is detected in any given run of the experiment") is wrong, then why would we expect the particle to be in a position eigenstate after the measurement of position?

Personally, I don't think about things this way, and so avoid the whole issue . It seems clear to me that the delta-functions corresponding to position eigenstates have no relevance to actual experiments. In an actual experiment, there is always a width to any position measurement ... the best you can do is say that the wavefunction of the particle became localized within an region less than the width of a single pixel on your detector.

I guess my question would be whether or not it matters in some fundamental way whether the position measurement resolves to a "position eigenstate" (i.e. an infinitessimally small region), or whether it just results in localization of the particle within some small by finite region. To me, the latter seems much more likely (perhaps the paper you linked says something similar .. I have not had time to read it yet), particularly since localization to a "position eigenstate" would imply that the momentum became completely undefined, which also seems unphysical.

 

[atyy]

Personally, I don't think about things this way, and so avoid the whole issue . It seems clear to me that the delta-functions corresponding to position eigenstates have no relevance to actual experiments. In an actual experiment, there is always a width to any position measurement ... the best you can do is say that the wavefunction of the particle became localized within an region less than the width of a single pixel on your detector.

I guess my question would be whether or not it matters in some fundamental way whether the position measurement resolves to a "position eigenstate" (i.e. an infinitessimally small region), or whether it just results in localization of the particle within some small by finite region. To me, the latter seems much more likely (perhaps the paper you linked says something similar .. I have not had time to read it yet), particularly since localization to a "position eigenstate" would imply that the momentum became completely undefined, which also seems unphysical.

Hmmm, I wonder if there are any other observables other than position and momentum for which Ballentine's argument can be made.

This is important, because you need (at least) two position measurements to infer the momentum in the manner you suggest, whereas if you actually *measured* the momentum directly in the QM sense, you would only need one measurement.

It does seem like Ballentine's momentum measurement is very non-local in time.

 

[Fredrik]

You really need to carefully define what you mean by "measure" to clarify by what you mean by the above statement. You are *not* using the typical QM definition of *measure* for the momentum, i.e. act upon the quantum state to project it into an eigenstate of the operator being measured, when you make the above statement. Obviously you cannot simultaneous project the particle into a position eigenstate and a momentum eigenstate, and I don't think you are claiming that you can. What you mean by "measure" in your statement is, "perform and experiment to obtain the value of the momentum", which is not necessarily the same thing, as you have been saying all along, and as I explained in my last post in detail.

A measuring device is a physical system that interacts with the system of interest for a finite time and then settles into a final state that's macroscopically distinguishable from the other possible final states. Associated with the physical device there's also an assignment of a numerical value to each of the possible final states. I consider this assignment to be a part of the "measuring device".

The correspondence between measuring devices and the mathematical "things" that represent them in a theory is postulated, not calculated. So we must at some point define what we mean by a "momentum measurement". Since we can't even begin to test the theory's predictions about momentum until we have defined the term, this definition must be considered part of the full definition of "quantum mechanics". In a perfect world, this would mean that every introductory QM book defines the term. In our world, these issues are swept under the rug. I think that's why it's so hard to discuss these things.

It seems to me that all techniques used by experimentalists (who claim to be measuring momentum) involve the detection of a particle followed by inferring the momentum from the location (in space or in spacetime) of the detection event. So it seems that the definition of "momentum measurement" they're using is consistent with what Ballentine is doing.

I agree that the terminology is important, but I don't believe that anything I've been saying is non-standard. I don't think it's standard to define "measurement" as a projection onto an eigenspace, but I do think it's very common to claim (like Fuesiker does) that this always happens when a measurement is performed. If I'm right about how the term "momentum measurement" must be defined for the theory to agree with experiments, then it follows that this claim is false.

I would really suggest the following linguistic change to the phenomenon you are describing. Instead of saying that you are "measuring" the momentum, say instead that you are "measuring the position, and inferring (or calculating) the momentum".

I don't think that change is appropriate. I think what's appropriate is to define the term "momentum measurement". I'm not going to try to write down a perfect definition here, but I believe it should say that a measurement of all components of momentum involves two detection events (the first one must obviously be non-destructive) and a calculation of \gamma m\vec v from the spacetime coordinates of the detection events. The fact that a detector of the kind that let's the particle pass through it located at Ballentine's single slit wouldn't change the state significantly, leads me to believe that the detection+calculation that he considers a p_y measurement is consistent with this definition. Note that there's only one measurement in his thought experiment. The "squeezing" of the wavefunction as the particle passes through the slit is a state preparation, but not a measurement, since no approximately classical record of the result is produced.

...you need (at least) two position measurements to infer the momentum in the manner you suggest, whereas if you actually *measured* the momentum directly in the QM sense, you would only need one measurement.

How do you "actually measure the momentum directly"? You can of course apply a projection operator to a state vector, but I don't believe that there's a physical device that can "actually measure the momentum directly". So with your terminology, "measurement" would be a purely mathematical term, and experimentalists would have to find another word for what they're doing.

 

[atyy]

Of relevance, I think, so we can see what else is in the literature apart from Ballentine:
http://arxiv.org/abs/1008.4591

 

[fuesiker]

Fredrik, you're now being funny. The quantum state of a particle is a complete description of that particle. That's one of the main things that distinguishes a quantum state from a classical state. It is surely not the state of an ensemble of similarly prepared particles. You make me laugh hard saying Ballentine "showed" you can measure position and momentum to arbitrary precision. This is ridiculous. Did he experimentally prove his thought experiment? Show us some papers where he does. The EPR paradox is a cool thought experiment, but is it right? No, check out Bell's inequalities and how they are experimentally violated.

Moreover, accusing me of not understanding the difference between measurement and preparation makes you look very silly, especially since I'm a published PhD researcher in quantum many-body physics and quantum optics. I don't like throwing that around, but when there is someone like you on here who accuses people of not understanding stuff you have no clue about, and throwing around false statements such as the quantum state is that of an ensemble, I really have to make things clear.

By the way, if you prepare a bunch of atoms identically, for example, your quantum state for this system is much different from that of one of them or any other number of them. There's a concept called entanglement, buddy, check it out. That's the whole problem with why we cannot do quantum simulations on a classical computer so efficiently, because the computational cost scales with the number of quantum bits (or atoms) that you use, even when you "prepare" them identically.

You seem hung up on the preparation issue. I understand that quite well because that's what I have done repeatedly in many experiments, preparing BEC's and doing various measurements on them. So far in my EXPERIMENTS, never did a state not collapse to the eigenstate of the observable being measured.

Instead of launching low attacks against me by pathetically accusing me of not understanding things I understand quite well, try to find arguments to support your claim other than a thought experiment that was never experimentally proven. One idea would be to actually use the scientific method and argue analytically against my arguments, those being that the state of a system collapses onto an eigenstate of the measured observable, negating your totally false argument that the HUP has to do with the preparation of the system when in fact it DOES have to do with the FT relationship between position and momentum based on their definition, and also that the quantum state of a particle, based on quantum mechanics, is the most complete description of that particle, and has nothing to do with an ensemble. It's also not "very weak" of an argument, it's the only argument QM makes about the state of a particle (http://en.wikipedia.org/wiki/Quantum_state).

PS. Let is be clear that these are my stances that Fredrik (though he keeps changing his word) keeps saying are "very weak" or that they show I don't understand certain things. To me, if you, Fredrik, have a problem with any of the above, you really need to go back to the basics. I recommend books such as Cohen-Tannoudji or Sakurai, and then you can actually begin to learn quantum mechanics from well-established concepts and not peak-a-boo though experiments never proven experimentally.

 

[Fredrik]

The quantum state of a particle is a complete description of that particle.

I will return to this later, when I have more time.

the quantum state of a particle, based on quantum mechanics, is the most complete description of that particle, and has nothing to do with an ensemble.

You might want to read Ballentine's textbook, or at least the relevant chapter of it. (From memory, I think it's chapter 9).

That's one of the main things that distinguishes a quantum state from a classical state. It is surely not the state of an ensemble of similarly prepared particles. You make me laugh hard saying Ballentine "showed" you can measure position and momentum to arbitrary precision. This is ridiculous. Did he experimentally prove his thought experiment? Show us some papers where he does.

There's nothing controversial about this thought experiment. The only thing that can even be criticized is that he might be relying on an incorrect definition of the term "momentum measurement" (which he doesn't include in the article). If that's your argument, then please show us your definition and how it implies that what Ballentine is describing isn't a momentum measurement.

The EPR paradox is a cool thought experiment, but is it right? No, check out Bell's inequalities and how they are experimentally violated.

I'm very familiar with this, and it has absolutely nothing to do with what we've been talking about.

especially since I'm a published PhD researcher in quantum many-body physics and quantum optics.

Argument by authority is a logical fallacy. This is not the first fallacy in your posts, since you have repeatedly refuted absurd claims that did not follow from the ones I did make.

By the way, if you prepare a bunch of atoms identically, for example, your quantum state for this system is much different from that of one of them or any other number of them.

I know the difference between |\psi\rangle and |\psi\rangle\otimes\cdots\otimes|\psi\rangle if that's what you're getting at.

There's a concept called entanglement, buddy, check it out. That's the whole problem with why we cannot do quantum simulations on a classical computer so efficiently, because the computational cost scales with the number of quantum bits (or atoms) that you use, even when you "prepare" them identically.

If they're entangled, they're not identically prepared. State preparation by definition erases all information about what happened to the system before it. The silver atoms that emerge in an upward direction after passing through a Stern-Gerlach magnet are prepared in identical spin states.

So far in my EXPERIMENTS, never did a state not collapse to the eigenstate of the observable being measured.

Have you ever measured the momentum of a particle? How did you do it? Were you able to do it without detecting the particle (which would put it in a state of sharply defined position, if the detection is non-destructive)?

Instead of launching low attacks against me by pathetically accusing me of not understanding things I understand quite well, try to find arguments to support your claim other than a thought experiment that was never experimentally proven. One idea would be to actually use the scientific method and argue analytically against my arguments,

Excuse me? You're the one who turned to personal attacks. You're the one who made a long series of posts claiming that I'm wrong without ever even trying to point out a flaw in the thought experiment (not counting arguments where you refuted absurd claims I had never made).

...my arguments, those being that the state of a system collapses onto an eigenstate of the measured observable

Seriously? Do you still believe that those arguments had any relevance to anything I've said?

...your totally false argument that the HUP has to do with the preparation of the system when in fact it DOES have to do with the FT relationship between position and momentum based on their definition,

There's at least one more logical fallacy in this quote. You're suggesting that I've said that it doesn't have anything to do with Fourier transforms, when I have in fact repeatedly said that it does. This is called a straw man. You're also suggesting that if the statement "it has to do with the preparation of the system" is true, then the statement "it has to do with Fourier transforms" can't also be true. I'm not sure what to call that fallacy, but what you're suggesting is certainly wrong.

(though he keeps changing his word)

What? The only thing I've "changed" is that my later posts have been emphasizing the definition of "momentum measurement" as a key issue.

 

[fuesiker]

Yeah man, you go have fun with Ballentine. I'm out of this discussion since I made all my points clear as day, providing mathematical and conceptual arguments that are well-established by scientists far more accomplished than your beloved Ballentine. Do me a favor and go argue with Cohen-Tannoudji and Sakurai before making extremist remarks that are at complete odds with standard QM theory. I will ignore all your further posts, unless I find them offensive or provocative.

 

[vanhees71]

It's a bit strange for me that somebody as fuesiker, who is working in quantum optics, doesn't like the minimal interpretation. So what's your interpretation of quantum theory?

I'm myself a theoretical high-energy nuclear physicist (heavy-ion collisions, QGP, and all that), and I don't know any single experiment, which cannot be very clearly be described within the minimal statistical interpretation, which takes nothing than the fundamental postulates on states and observables within the mathematical formalism of quantum theory. The main point of the interpretation, i.e., the map between mathematical formalisms and physically observable phenomena is the probabilistic interpretation of (pure or mixed) states (Born's rule).

If you accept this minimal skeleton of QT as a physical theory, and I think all physicists applying quantum theory to real phenomena do so, the remaining question concerning "interpretation" is, how to measure probabilities. I know, there's a whole industry of philosophy around about this, but for an experimental physicist, it's simply statistics, i.e., you do the measurement, say the momentum distribution of a certain sort of particles in a proton-proton collision experiment, on many as well as possible identically prepared setups of the experiment. In our example of proton-proton collisions that means to shoot a lot of protons with as well as possibly defined momenta on top of each other and then measure as well as you can the identified particles's momentum and their numbers in each collision and make a statistics about the outcome of this measurement. Then you obtain, with a certain statistical and systematic uncertainty, a probability distribution (or, as is more common in collision experiments cross sections) about the measured process of interest. Then the statistical probability distribution of this ensemble can be compared with the probabilities predicted by quantum theory, and as you know, up to today quantum theory has been very successful in discribing all kinds of measurements.

That also holds true in other branches of physics as condensed matter physics and also, of course, quantum optics. Also in quantum optics you do experiments with a lot of identically prepared photon states, e.g., with entangled photon pairs to make sure that you have only a single electron or to do checks of Bell's inequality, etc. That you know better than me.

Now my question is, why do you fight so vehemently against this minimal interpretation or, in other words, how do you test quantum theory in real-world experiments if not with statistics on an ensemble of identically and independently (i.e., uncorrelatedly) prepared systems.

 

[Cthugha]

The quantum state of a particle is a complete description of that particle. That's one of the main things that distinguishes a quantum state from a classical state. It is surely not the state of an ensemble of similarly prepared particles.

The latter is called the ensemble interpretation. It is the standard minimalist interpretation of qm. It is rather strange that a German (well at least your name sounds like you took a humorous variant of the german word Physiker as your name) PhD student never came across it. Have a look at the German Wikipedia entry on The Heisenberg uncertainty principle. It is even mentioned there.

Do me a favor and go argue with Cohen-Tannoudji and Sakurai before making extremist remarks that are at complete odds with standard QM theory.

That statement was nowhere near extremist or at odds with qm.

 

[fuesiker]

Cthugha, how on Earth did you conclude that I am not familiar with the ensemble interpretation? Or do you just want to sound smart on here? I am quite familiar with ensembles, both pure and mixed, and their representation, and the time evolution of their von Neumann entropies, etc...

Again, all I am doing is the crazy notion of believing in the uncertainty principle (it's funny I have to repeat this over and over), and the fact that the state of a system (whether that system is one particle or N particles, whatever) collapses onto the eigenstate of the measured observable.

However, go ahead and fight Fredrik's battles who describes the state of a system of N identical particles as |\psi>|\psi>|\psi>... rather than |\psi_1>|\psi_2>|\psi_3>...

And if you think my last statement that you quote as extremist, then there's no point arguing with someone like you. Suggesting books to read that someone is arguing against their main well-established principles is nothing extreme, and it's dumb you even claim it's at odds with qm. what the hell does my statements have to say that is at odds with qm?

 

[atyy]

Reading this review http://arxiv.org/abs/quant-ph/0609185, they describe a model of simultaneous measurement by Arthurs and Kelly and say "Arthurs and Kelly showed that this constitutes a simultaneous measurement of position and momentum in the sense that the distributions of the outputs reproduce the quantum expectation values of the object’s position and momentum."

Is it known whether Ballentine's measurements produce the right expectation values for position and momentum given the wavefunction?

Now that I try to think of it this way, which wavefunction's expectation position and momentum are being measured by Ballentine? The one before the slit? The one immediately after the slit? Or the wavefunction at infinity?

 

[Cthugha]

Cthugha, how on Earth did you conclude that I am not familiar with the ensemble interpretation?

You explicitly stated that the Ensemble interpretation of qm is wrong by pointing out "It [the quantum state of a particle] is surely not the state of an ensemble of similarly prepared particles."

This statement is plain wrong and therefore the assumption that you do not know much about the ensemble interpretation is natural.

Or do you just want to sound smart on here? I am quite familiar with ensembles, both pure and mixed, and their representation, and the time evolution of their von Neumann entropies, etc...

Unfortunately all of that does not have much to do with the ensemble interpretation vs. Kopenhagen-like interpretation issue at hand. I could also comment on the sounding smart remark, but ad hominem arguments are never sensible basics of a discussion.

Again, all I am doing is the crazy notion of believing in the uncertainty principle (it's funny I have to repeat this over and over), and the fact that the state of a system (whether that system is one particle or N particles, whatever) collapses onto the eigenstate of the measured observable.

However, go ahead and fight Fredrik's battles who describes the state of a system of N identical particles as |\psi>|\psi>|\psi>... rather than |\psi_1>|\psi_2>|\psi_3>...

I am not interested in fighting any battles and in particular I am not interested in that discussion you mention here. I just pointed out that one of the things you wrote above is obviously wrong.

And if you think my last statement that you quote as extremist, then there's no point arguing with someone like you. Suggesting books to read that someone is arguing against their main well-established principles is nothing extreme, and it's dumb you even claim it's at odds with qm. what the hell does my statements have to say that is at odds with qm?

Ehm, I do not know at all what you mean. You accused someone else of making extremist remarks and I just commented that the remark in question was not extremist. It seems English is not your native language, so maybe just rereading my post might already help.

 

[fuesiker]

You cannot say that the state of a particle can be used to describe the state of an ensemble of similarly prepared particles or that of another similarly prepared particle. Your statement in this regard is plain wrong because there is a concept of time evolution. These particles start out with the same initial state (in principle and only if you neglect many-body effects, which is not the general case), but they time-evolve differently depending on what Hamiltonian propagates this time evolution. Now you can argue that they all time-evolve with the same Hamiltonian, well still you cannot say the state of the single particle is the same as that of the ensemble of particles due to many-body interactions in the ensemble, which is the general case. So in essence, my statement is not "plain wrong" but rather very correct if you want to consider many-body interactions, which you should, because you know, there's a whole darn field dealing with this stuff. Danke! ;)

 

[Cthugha]

You cannot say that the state of a particle can be used to describe the state of an ensemble of similarly prepared particles or that of another similarly prepared particle.

Ok, this makes it necessary to get into semantics. It is clear that you cannot take the quantum state of some particle and claim others behave similarly. That is true. The ensemble interpretation, however, goes the other way round and does not really have something like a quantum state for a single particle, just a quantum state of an ensemble, maybe of repeated realizations of the same state or whatever. So in this case there is just the state of the ensemble.
Of course it is easy to address this problem theoretically and propose a single particle quantum state. However, in experiments you can never measure it anyway without measuring an ensemble of repeated experiments or performing measurements on particles prepared nominally in the same state. So, in a nutshell you said: "The quantum state of a particle is a complete description of that particle.". This is the central point that can be questioned and whether it is more like "The quantum state of an ensemble is a complete description of what happens if you perform experiments on some ensemble of nominally identical particles." I am not aware of any experiment that can really distinguish between these situations.

Your statement in this regard is plain wrong because there is a concept of time evolution. These particles start out with the same initial state (in principle and only if you neglect many-body effects, which is not the general case), but they time-evolve differently depending on what Hamiltonian propagates this time evolution. Now you can argue that they all time-evolve with the same Hamiltonian, well still you cannot say the state of the single particle is the same as that of the ensemble of particles due to many-body interactions in the ensemble, which is the general case.

Well, as I said above. Sure, this is a valid interpretation and within this interpretation your statement as formulated in your last post is correct , but it is not necessarily the only possible interpretation and I still consider the "It is surely not the state of an ensemble of similarly prepared particles." as wrong. If you consider the state of the ensemble as the most fundamental level accessible, the state of an ensemble of similarly prepared particles is the best you can get and there is no separate single particle quantum state.

So in essence, my statement is not "plain wrong" but rather very correct if you want to consider many-body interactions, which you should, because you know, there's a whole darn field dealing with this stuff.

The point about many-body physics is clear, but I still assume it is not the central point here.

Danke! ;)

Öhm, Bitte.

 

[fuesiker]

Sure you can prepare a single particle. This is heavily done in the groups of Hommelhof and Bloch at the Max Planck Institute of Quantum Optics. The former plays with single electrons, the latter with single bosons (Rubidium-87 atoms) on lattices. My understanding of an ensemble is that it would make most sense to speak of an ensemble, or a collection of particles, when there is many-body interaction. Otherwise, the problem reduces to a single-particle one, which is "simple" quantum mechanics principles since entanglement plays no role.

Your explanation of the ensemble interpretation makes me think that this is a quasi-quantum mechanical treatment of things, because you say you do not have the resolution available to probe single particles, and therefore you kind of average over the ensemble.

On the other hand, if by ensemble you mean we prepare single particles in the same state and repeat the experiment over and over, then this is different. Of course, to determine the state of a particle you must repeat the measurement N many times, and then for sufficiently large N, you can determine what the state of the particle is (or was) in the eigenbasis of your measured observable.

 

[Cthugha]

Sure you can prepare a single particle. This is heavily done in the groups of Hommelhof and Bloch at the Max Planck Institute of Quantum Optics.

Hmm, why did I know you were associated with I. Bloch. However, I never objected to being able to prepare single particles. However, any experiment I am aware of is nevertheless ensemble based (see next comments).

Your explanation of the ensemble interpretation makes me think that this is a quasi-quantum mechanical treatment of things, because you say you do not have the resolution available to probe single particles, and therefore you kind of average over the ensemble.

Well, with the meaning of ensemble as given below, it is rather a minimalist treatment. While it seems natural to assume that the properties you measure in repeated experiments on some single particle is inherent to the particle and its state, it is not a necessary assumption or one that can be tested experimentally. Therefore the ensemble interpretation takes one step back and leaves that question more or less open.

On the other hand, if by ensemble you mean we prepare single particles in the same state and repeat the experiment over and over, then this is different. Of course, to determine the state of a particle you must repeat the measurement N many times, and then for sufficiently large N, you can determine what the state of the particle is (or was) in the eigenbasis of your measured observable.

Well, yes of course. Repeating an experiment on particles prepared nominally identical is what I mean by an ensemble. Sorry if that was unclear.

 

[vanhees71]

On the other hand, if by ensemble you mean we prepare single particles in the same state and repeat the experiment over and over, then this is different. Of course, to determine the state of a particle you must repeat the measurement N many times, and then for sufficiently large N, you can determine what the state of the particle is (or was) in the eigenbasis of your measured observable.

Yes, that's what I mean when I say, I make repeated measurements on an ensemble of single particles. If I investigate the statistics of single-particle observables, for each measurement, for each measurement I have to prepare this single particle in the state, I wantt to investigate, and I must make sure that the particle in one measurement is not somehow correlated with the particle in another measurement. That's what's meant by independent preparations of one particle in a certain state.

If you prepare a many-body system, they might interact with each other such that the single-particle observables are correlated, but then this is not a measurement to determine the properties of a single-particle state, but you still haven't answered clearly my question, what's your interpretation of Born's probability interpretation, if not related to the statistical properties of ensembles.

In one of your postings it seems as if you are a follower of the Copenhagen interpretation. But then, how do you resolve the good old EPR problem: If the wave function is a physical property of a single quantum system (e.g., a single particle or photon) and if then "the collapse of the state" when measuring an observable that is not determined by the state the system is prepared in, is thus a physical process, how can this be compatible with Einstein causality?

 

[fuesiker]

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.

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.

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.

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?

 

[atyy]

Reading this review http://arxiv.org/abs/quant-ph/0609185, they describe a model of simultaneous measurement by Arthurs and Kelly and say "Arthurs and Kelly showed that this constitutes a simultaneous measurement of position and momentum in the sense that the distributions of the outputs reproduce the quantum expectation values of the object’s position and momentum."

Is it known whether Ballentine's measurements produce the right expectation values for position and momentum given the wavefunction?

Now that I try to think of it this way, which wavefunction's expectation position and momentum are being measured by Ballentine? The one before the slit? The one immediately after the slit? Or the wavefunction at infinity?

There is an discussion of Ballentine's paper in section 7.3 of the above-linked paper by Busch, Heinonen, and Lahti. Their opinion seems to be that "this scheme is appropriately understood as a sequential measurement of first sharp position and then sharp momentum, and does therefore not constitute even an approximate joint measurement of position and momentum."

In a 1972 book review "Quantum Mechanical Ideas" in Science, Bell's opinion was that Park and Margenau demonstrated that position and momentum can be measured for special states, but that Wigner had shown it was impossible for arbitrary states (but he added "it seems to me").

Here's a longer quote from Bell's review:

"The most iconoclastic of the authors are Park and Margenau. In an analysis of the measurement of "incompatible" observables, they conclude that it is the quantum mechanical axioms of von Neumann which are incompatible. However, this turns out to be largely a question of the semantics of the words "measurement" and "observation". It illustrates the danger of using as technical terms words in common use with rather wide meanings. Narrowing down the issue, Margenau and Park end by posing a fairly well-defined question: can one find, assuming conventional quantum mechanics, a procedure which (for an arbitrary state of the system) will yield a joint probability distribution ρ(p, q) such that integration over q or p yields the conventional probability distributions for the other of the pair of incompatible observables? Margenau and Park illustrate procedures which work in this way for special states of a system, but leave the question open for arbitrary states.

As it happens just this question (it seems to me) is answered by Wigner in his elegant contribution "Quantum mechanical distribution functions revisited." This includes a proof (previously unpublished) that there is no such positive distribution bilinear in the wave function and its complex conjugate. Assuming that all possible statistical predictions about a system are contained in the density matrix, the question of Park and Margenau is then answered in the negative."

 

[Fredrik]

section 7.3 of the above-linked paper by Busch, Heinonen, and Lahti.

These guys seem to have a very different idea than I about what a measurement is. To me, measurements are what measuring devices do in experiments that test the accuracy of the theory's predictions. These guys also require that the result "can be used to predict the outcomes of future measurements". The inference of momentum in Ballentine's thought experiment doesn't satisfy this definition of "measurement", since the obtained value is useless for future predictions. The additional requirement is obviously something that should be part of the definition of "state preparation", but why should it be part of the definition of "measurement"? The authors don't seem to address this issue.

I don't think section 7.3 sheds any light on the single-slit experiment. It tells us that if we use their terminology, it's not a joint measurement and the momentum inference isn't even a measurement. But the real question is, why should we use their definitions? Is there e.g. a reason to think that the inferred value of the momentum is inconsistent with the predictions of QM? That would be a good reason to do things their way.

I don't think any article that doesn't include a definition of the term "momentum measurement" can shed any light on this issue.

 

[Fredrik]

I think what's appropriate is to define the term "momentum measurement". I'm not going to try to write down a perfect definition here, but I believe it should say that a measurement of all components of momentum involves two detection events (the first one must obviously be non-destructive) and a calculation of \gamma m\vec v from the spacetime coordinates of the detection events.

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?

 

[Fredrik]

I'm out of this discussion since I made all my points clear as day, providing mathematical and conceptual arguments that are well-established by scientists far more accomplished than your beloved Ballentine.

Nonsense. All you did was to knock down straw men. You still haven't presented any kind of argument against what I actually said.

Do me a favor and go argue with Cohen-Tannoudji and Sakurai before making extremist remarks that are at complete odds with standard QM theory.

You seem to be the only one here who believes that what I've been saying is at odds with standard QM theory.

 

[fuesiker]

Fredrik, I will ignore your ad hominem accusations and discuss physics.

You claimed earlier that there is "no way" to directly measure momentum (but I guess now you will deny you ever claimed or implied that?). One simple (though perhaps not the best) way to measure the momentum of a photon is through interferometry. Use a Mach-Zehnder interferometer and use two photons prepared in the same way and you can measure the phase shift these photons have when one passes through the sample-arm and one through the reference (bare) arm. That way you can measure momentum very accurately and you are not measuring the position of the photon.

Moreover, when you say that the Heisenberg uncertainty principle is based on preparation of your state, you indeed are at odds with standard QM theory. Preparation can only alter the HUP via squeezing, where you can "squeeze" the amplitude of light at the expense of its phase or vice versa, but the Heisenberg limit will never be violates, based on standard QM theory. Moreover, all the references you used are Ballentine's thought experiment, and you take that for an actual experiment. In physics, it's experiment that makes the last call, not theory. I love theory more than experiment, but the truth is that a lab experiment is the final judge, not a thought experiment. Don't be upset, I am just stating facts.

 

[Fredrik]

Fredrik, I will ignore your ad hominem accusations and discuss physics.

You're the one who's been attacking me. And you should probably look up what "ad hominem" means.

Use a Mach-Zehnder interferometer and use two photons prepared in the same way and you can measure the phase shift these photons have when one passes through the sample-arm and one through the reference (bare) arm. That way you can measure momentum very accurately and you are not measuring the position of the photon.

I will take a look at that tomorrow. If you can indeed measure the momentum without detecting the particle, that's very interesting, and I will certainly make an effort to understand it. However, this will not refute Ballentine's argument. (I explained what it would take to do that in an earlier post, and I'm doing it again at the end of this post).

Moreover, when you say that the Heisenberg uncertainty principle is based on preparation of your state, you indeed are at odds with standard QM theory.

If you're going to criticize something I've said, you need to quote it. I don't know what you're referring to here (and I probably spent 15 minutes trying to figure that out). I reject any claim that the uncertainty relations have nothing to do with preparations, since the uncertainties depend on the wavefunction and wavefunctions correspond (bijectively) to equivalence classes of preparations.

Preparation can only alter the HUP via squeezing, where you can "squeeze" the amplitude of light at the expense of its phase or vice versa, but the Heisenberg limit will never be violates, based on standard QM theory.

So? (I assume that you meant "alter the uncertainties", not "alter the HUP"). I haven't argued against what you're saying here.

In physics, it's experiment that makes the last call, not theory. I love theory more than experiment, but the truth is that a lab experiment is the final judge, not a thought experiment.

Right. Are you aware of any experiments that show that a definition of "momentum measurement" similar to what I described in post 91 would make the theory disagree with experiments?

You described one technique for momentum measurements that doesn't involve detection* but aren't there other techniques that experimentalists consider valid ways to measure momentum that do involve detection?

*) I haven't verified this but I'm operating under the assumption that you're right until I've checked it out.

 

[atyy]

But the real question is, why should we use their definitions? Is there e.g. a reason to think that the inferred value of the momentum is inconsistent with the predictions of QM? That would be a good reason to do things their way.

Yes, the critique about operational consequences seems not so strong. But what about Busch et al's definition that two observables are jointly measurable if they are the marginal observables of a joint observable (section 3.1)? This seems similar to the criterion that Bell discusses (see the quote in post #89), and says is due to Park and Margenau, whom Ballentine cites in his paper. Whether or not they are the same, Busch et al say that by their criterion there are no joint measurements of non-commuting observables, and Bell says that Wigner has shown the same for Park and Margenau's criterion.

but aren't there other techniques that experimentalists consider valid ways to measure momentum that do involve detection?

The Busch et al paper goes on to talk about approximate joint measurements of position and momentum, even though exact joint measurements are not possible. I would guess this pertains to measuring momentum by measuring position, like in a cloud chamber track. The usual heuristic is that the width of the track is so wide that the momentum measurement is far more precise than the position measurement, which seems consonant with Busch et al's discussion of uncertainty relations for approximate joint measurements.

Also, if a sharp momentum measurement causes the state to collapse to a position eigenstate, then the collapse postulate is wrong. In which case, why would we know that a sharp position measurement causes a collapse to a position eigenstate?

 

[romsofia]

Just going to put this out there, in Griffiths book "Introduction to QM" I remember him mentioning that theoretically you could measure position AND momentum; however, I don't have the book on hand but will comment tomorrow on the page #.

 

[atyy]

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 think even in Park and Margenau's case, they require t to approach infinity, otherwise the distributions are not recovered (similar to Ballentine's requirement of large L?).

 

[fuesiker]

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.

I still don't understand why you think the HUP depends on the wavefunction. I mean just look at the equation as in Sakurai: <dx^2><dp^2> >=h^2/16pi^2, where dx = x-<x> is an operator and dp = p-<p> is also an operator (in standard notation they take dx and dp as their expectation values). So you can see that there is a wavefunction involved in that there are expectation values of dx and dp (stemming from those of x and p) in the equation, BUT then no matter what your wavefunction is with respect to which you take the expectation values, you will never violate the above equation for position and momentum in one direction, which is the HUP. So you say "I reject any claim that the uncertainty relations have nothing to do with preparations, since the uncertainties depend on the wavefunction", and you're half right in that YES, the uncertainties depend on the wavefunction but NO the uncertainty relationship I write above (HUP inequality) has nothing to do with the wavefunction. Let me be more clear, the value of <dx^2><dp^2> of course depends on your choice of wavefunction prepared, but the uncertainty relation <dx^2><dp^2> >=h^2/16pi^2 surely holds NO MATTER what your prepared wavefunction is.

Now don't tell me I am arguing against something you didn't say. You specifically said "I reject any claim that the uncertainty relations have nothing to do with preparations" and you're wrong. The uncertainties have to do with the wavefunction you prepare, but the uncertainty relations don't.

Yeah, do look up the Mach-Zehnder interferometer. An interferometer is just what its name depicts: a device that interferes two waves. Think about it in a very simple way. Both photons traverse the same spatial length in each arm, albeit one crosses a distance x inside a sheet of glass, let's say, with refractive index n. This will cause a difference in the optical path length of the photons, where the reference photon has optical length L and the sample one has optical length L+nx. The difference, which you find from interferometry (the phase delay is deduced from the interference of the two photons) is phi = nx = c*k*x/omega, where omega is the frequency, c speed of light, k wavevector magnitude.

 

[fuesiker]

atyy, I like to think about collapse as this:

Of course, one cannot deny that all our measurement devices are classical. Hence, your outcome cannot be a quantum superposition. Now let's take a discrete operator, such as boson number on a three-site lattice, each of which is prepared initially with one boson. Now, there is tunneling as time goes on (this is the Bose-Hubbard model), and your state goes from |1,1,1> to say |0.5,1.2,1.3> depending on the Hamiltonian parameters that drive it. Now in the boson number space, this state can be expressed as a superposition of infinitely many states |n,m,o>, where n, m, and o are nonnegative integers, because these form a complete basis in the corresponding Hilbert space. Now assume that the state you are measuring the boson number on does not collapse to an eigenstate of the boson number operator. This is basically saying that your resultant state has to be a superposition of at least two distinct eigenstates, which amounts to you measuring a noninteger number of bosons on at least one site. The same applies to the position operator. When you measure it on a state, if that state does not collapse onto an eigenstate of position, then your result basically says that your object is in two different positions at the same time. So long as out world is classical (actually, it is quantum like everything in this world is quantum, however, decoherence removes quantum effects due to "averaging") the state we measure an observable on will collapse to an eigenstate of that observable.

 

[atyy]

atyy, I like to think about collapse as this:

Of course, one cannot deny that all our measurement devices are classical. Hence, your outcome cannot be a quantum superposition. Now let's take a discrete operator, such as boson number on a three-site lattice, each of which is prepared initially with one boson. Now, there is tunneling as time goes on (this is the Bose-Hubbard model), and your state goes from |1,1,1> to say |0.5,1.2,1.3> depending on the Hamiltonian parameters that drive it. Now in the boson number space, this state can be expressed as a superposition of infinitely many states |n,m,o>, where n, m, and o are nonnegative integers, because these form a complete basis in the corresponding Hilbert space. Now assume that the state you are measuring the boson number on does not collapse to an eigenstate of the boson number operator. This is basically saying that your resultant state has to be a superposition of at least two distinct eigenstates, which amounts to you measuring a noninteger number of bosons on at least one site. The same applies to the position operator. When you measure it on a state, if that state does not collapse onto an eigenstate of position, then your result basically says that your object is in two different positions at the same time. So long as out world is classical (actually, it is quantum like everything in this world is quantum, however, decoherence removes quantum effects due to "averaging") the state we measure an observable on will collapse to an eigenstate of that observable.

Yes that's my understanding too - measuring position sharply collapses to a position eigenstate (in the case where the particle is not destroyed), and measuring momentum sharply collapses to a momentum eigenstate. That's why I don't understand how position and momentum can be simultaneously sharply measured, and the particle collapses to a position eigenstate.