Exploring Heisenberg's Uncertainty Principle: Intuition & Explanations

[fuesiker]

In principle, once you measure a quantity on a certain wavefunction, this wavefunction will collapse onto an eigenstate of that observable quantity, say O. Then, it depends on the dynamics of your system. In general, there is time evolution going on propagated by some Hamiltonian H, and then it depends on whether or not H and O commute. If they do, then your wavefunction will remain an eigenstate of O acquiring merely a phase shift. If O and H do not commute, then this wavefunction will no longer remain an eigenstate of O.

 

[Fredrik]

In principle, once you measure a quantity on a certain wavefunction, this wavefunction will collapse onto an eigenstate of that observable quantity, say O.

I don't think that's true, for the reasons already mentioned above: Momentum is measured by measuring the position, and there are detectors that let the particle pass through it. After the measurement, if the relevant components of the detector are of size L in units such that $\hbar=1$, then the position will be spread out of a region smaller than L and that means that the momentum will be spread out over a region (in momentum space) roughly of size 1/L. So when L is small, the momentum will have a large spread immediately after a momentum measurement, and the state will therefore be very different from a momentum "eigenstate".

See figure 3 in the article I linked to in my first post in the thread for a description of how to measure momentum by measuring position

 

[fuesiker]

Hi Frederik, I can't find that article, sorry I am new to this forum. Perhaps you can email me the link or repost it?

I believe what you are talking about is what I would call a "statistical" measurement or perhaps a QND measurement. I heard of an experiment where they were able to measure the momentum and position of photons or electrons (forgot which) passing through some detectors in the form of a Young's double slit experiment, but they clearly stated that this only works for a statistical ensemble of photons, never for one photon. I will try to find that paper in my files. It was also reported on BBC.

One has to be very careful about what we mean by measurement. Are we measuring the observable on an ensemble, or a particle? For example, let's measure the polarization and intensity of a 45-degree polarized laser beam using a 90-degree axis polarizer. The outgoing beam will be 90-degree polarized but will be at half its input intensity. Does that mean that half of each photon passed through? No. At the quantum scale, things are much different, each photon can either fully go in as a 90-degree polarized photon (wavefunction collapses onto one of two basis eigenstates) or get absorbed (collapses onto second basis eigenstate, 0-degree polarization which effects absorption).

There are a lot of quantum non-demolition experiments going on in the group of Serge Haroche in Paris, where they "count" the number of photons in a cavity without destroying these photons by passing Rydberg atoms through the cavities. The matter of fact is, you can only do that over many many measurements, each time preparing the field inside the cavity in the exact same coherent state (a Glauber state, which is what laser is, but they used microwave frequencies). However, if you take a single measurement, as soon as a Rydberg atom goes through the first time, you automatically destroy the coherent field, though not the photons in it. This is due to entanglement. When you measure, there is entanglement between the state of your system and the so-called "pointer states" of your environment (your measuring device). Now, if you want to measure a certain observable precisely on one system, its state will collapse to an eigenstate of the observable. If you want a QND measurement, then this is like what you describe or the photon-counting experiment whereby you have to prepare your state several times and just perturb it instead of hard-measure the observable on it once.

Moreover, reading your statement again, you are not really measuring the exact momentum of the particle, you are only measuring what its momentum may be. Surely then, the resultant state is not an eigenstate of the momentum operator. Again, if you want to measure the momentum of a particle (NOT the range of momenta it can have), your particle will have to collapse in its state to that of a momentum eigenstate.

 

[Fredrik]

The link was included in post 5 and reposted in post 17. Here it is again: http://www.kevinaylward.co.uk/qm/ballentine_ensemble_interpretation_1970.pdf

Edit: The article describes an experiment where the particle is destroyed (absorbed) by the detector, but in principle, the position detectors in figure 3 (the little boxes on the right) could be of the kind that let's a particle pass through them.

 

[fuesiker]

Thanks. However, this kind of makes the point I was trying to make. If you look at Section 3.2 (right to the right of Fig. 3 in the version you linked above), the authors clearly state that "Clearly the statistical dispersion principle and the common statement of the uncertainty principle are not equivalent or even closely related. The latter refers to the error of simultaneous measurements of q and p on ONE system,... On the other hand, the former refers to statistical spreads in ensembles of measurements on similarly prepared systems".

So basically this exactly what I said above about preparing the coherent field over and over the same way in the QND experiments, and so on. Moreover, it is also exactly what I said about this not working when you want to measure the momentum or position of ONE system. Then, to measure the exact position of this particle, its state must collapse onto a position eigenstate. Same goes for momentum.

 

[Fredrik]

I'm not familiar with QND experiments, but the stuff you're quoting from the article is perfectly consistent with what I said. If the boxes on the right in the figure are detectors of the kind that records a detection event while letting the particle pass through it, the result can't possibly be a state with a sharply peaked momentum. The wavefunction after the measurement will be close to zero outside of that particular detector, and have a sharp peak at the location of the detector. That means that the Fourier transform of that wavefunction won't have a sharp peak. So the wavefunction after the measurement is nothing at all like a momentum eigenstate.

 

[fuesiker]

OK, like I mentioned before, you're not measuring momentum here, you're measuring a SPREAD of momentum. This does not violate the uncertainty principle. You said earlier that I was wrong stating that by measuring an observable on a state, the state will collapse to the eigenstate of that observable. That's where I do not agree with you. Measuring an observable, which does NOT mean measuring its spread, necessarily leads to the state on which this observable is being measured to collapse onto an eigenstate of that observable. If you measure a spread in momentum, which is not measuring the momentum of the state, then of course your state won't collapse onto an eigenstate of the momentum operator.

 

[Fredrik]

OK, like I mentioned before, you're not measuring momentum here, you're measuring a SPREAD of momentum.

Your comments closer to the end of your post suggest that what you mean by "measuring a spread of momentum" is to determine the Fourier transform of the wavefunction, which is of course equivalent to determining the wavefunction itself. To do this with any kind of accuracy, we would have to perform a large number of measurements on a large number of particles all prepared in the same way. That's not what I'm talking about at all, and it's not what Ballentine's thought experiment is about. A single particle is sent through a single slit, and then one member of the wall of detectors signals detection. The momentum of that particle is inferred from the location of that detector. So we have measured the momentum of a single particle by measuring its position.

This does not violate the uncertainty principle.

The simultaneous measurement of position and momentum that Ballentine describes doesn't violate any uncertainty relations, but it implies that your claim is false. We measured the momentum by measuring the position (once), so if your claim is true, the state after the measurement would be an eigenstate of both position and momentum, something that doesn't exist.

Measuring an observable, which does NOT mean measuring its spread, necessarily leads to the state on which this observable is being measured to collapse onto an eigenstate of that observable.

Again, this is easily seen to be false when we imagine that the detectors on the right are of the kind that let's the particle pass through. We measured the momentum by measuring the position. The wavefunction after detection will be sharply peaked after the detection, and that means that it's nothing like a momentum eigenstate. If the wavefunction after the detection would be a momentum eigenstate, then we could place a second wall of detectors behind the first one (to the right of it in the picture), and they would all be equally likely to signal detection after (say) the third one from the top in the first wall has signaled detection. Don't you think that after the third one from the top in the first wall has signaled detection, then the one directly to the right of it is more likely to detect the particle than the others?

 

[atyy]

I kinda think of [x,p]=iħ as analogous to [s_x,s_y]=iħs_z. But is it different, or could we say something like we simultaneously measure s_x and s_y, and we measure s_y by measuring s_x (except that that last thing sounds like a measurement of s_z)?

 

[fuesiker]

OK, maybe I need to read the paper in whole. But you had said from a sharp position-space wavefunction, they got a spread of momentum. In that case, my statement is not false at all, because he is basically measuring the position and the state is collapsing onto a position eigenstate...

My main point all throughout is that if you want to measure an observable on a state, immediately after that measurement, your state collapses onto an eigenstate of that observable. That is not false, contrary to what you say, and to me, it is a very simple and basic notion of quantum mechanics. Actually, if you are right (and you're not) that after measuring an observable the state does not collapse onto an eigenstate of the observable, then you removed one of the main hurdles standing in the way of quantum computing, namely quantum decoherence.

Another way to look at this mathematically: Your state can be expanded in the space basis which forms a complete orthogonal basis. Hence, your wavefunction in position space is nothing but a superposition of these vectors with some complex coefficients. You can choose any other quantity, not just position, that forms through its eigenstates a complete basis in the Hilbert space. This is the essence of the measurement process in that these coefficients, when you take their magnitudes and square them, they give you the probability that your state collapses onto one of the eigenstates of your observable. Now let's assume by way of contradiction that I am wrong, and ergo, that my statement that THE STATE OF A SYSTEM COLLAPSES ONTO AN EIGENSTATE OF THE OBSERVABLE WHEN THE OBSERVABLE IS MEASURED ON THAT SYSTEM (I all-cap this to stress that this has ALWAYS been my sentiment) is wrong, then you are basically saying that the resultant wavefunction is a superposition of at least two distinct position eigenstates, which means that your system exists simultaneously in 2 different positions at the same time, which is nonsense.

This is equivalent to saying that Schroedinger's cat is dead and alive at the same time. Before the opening the box (measurement), she may very well be, but after the measurement, she can only be one or the other in our universe which suffers from quantum decoherence.

 

[fuesiker]

I kinda think of [x,p]=iħ as analogous to [s_x,s_y]=iħs_z. But is it different, or could we say something like we simultaneously measure s_x and s_y, and we measure s_y by measuring s_x (except that that last thing sounds like a measurement of s_z)?

You can't measure Sx and Sy simultaneously. The equation that you wrote actually states that you can't do so. You can measure two observables A and B at the same time iff [A,B]=0.

 

[atyy]

Well, the example of momentum measurement from position is interesting. Could one say it isn't simultaneous since it is non-local in time - one has to remember when the initial position eigenstate was prepared (to get T)? (Ooops, that's in http://physicsandphysicists.blogspot.com/2006/11/misconception-of-heisenberg-uncertainty.html" 's example, not Ballentine's).

In http://www.kevinaylward.co.uk/qm/ballentine_ensemble_interpretation_1970.pdf" 's, he says the measured p_y is p(sinθ), where p is the initial momentum. But how can p be known, if he says that p changes when the particle passes through the slit? Couldn't one argue that the presence of the slit causes the particle not to be in a momentum eigenstate - consequently, initial p is not sharply defined? The use of sinθ to define p_y seems to require that the particle had a definite position at the slit. (His example also seems non-local in time, since one needs to know initial p to measure py)

Note: sorry, lots of edits.

 

[Fredrik]

But you had said from a sharp position-space wavefunction, they got a spread of momentum. In that case, my statement is not false at all, because he is basically measuring the position and the state is collapsing onto a position eigenstate...

What I said is that in Ballentine's thought experiment, the detector absorbs the particle, and therefore no new state is prepared. Then I said that if we imagine that the detectors are of the kind that let's a particle pass through it, the state after the measurement will have a sharply peaked (position-space) wavefunction. Yes, the detection of the particle is a position measurement, but it's also a momentum measurement. So if your claim is true, the state after the measurement is a simultaneous eigenstate of position and momentum, and as you know, such states don't exist.

My main point all throughout is that if you want to measure an observable on a state, immediately after that measurement, your state collapses onto an eigenstate of that observable. That is not false, contrary to what you say, and to me, it is a very simple and basic notion of quantum mechanics.

Again, Ballentine describes how to measure (the z components of) both position and momentum of the same particle, by detecting the particle only once. So if your claim is true, the state after the measurement is an eigenstate of both observables. This is a very simple argument, and you haven't pointed out any flaws in it. You have just repeated the claim that I'm wrong.

Now let's assume by way of contradiction that I am wrong, and ergo, that my statement that THE STATE OF A SYSTEM COLLAPSES ONTO AN EIGENSTATE OF THE OBSERVABLE WHEN THE OBSERVABLE IS MEASURED ON THAT SYSTEM (I all-cap this to stress that this has ALWAYS been my sentiment) is wrong, then you are basically saying that the resultant wavefunction is a superposition of at least two distinct position eigenstates, which means that your system exists simultaneously in 2 different positions at the same time, which is nonsense.

My claim is

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

You're arguing against the claim

"Position measurements never put the system in a position eigenstate"​

which isn't implied by mine. Your argument is correct, but you're arguing against a statement that I wouldn't support. Position measurements do put particles in position eigenstates. (To be more accurate: Approximate position measurements that don't destroy the particle leave it in a state represented by a wavefunction with a sharp peak at the location of the detector).

I believe that something similar to what I just said about position holds for all operators that commute with position, but not for any operators that don't commute with position. I haven't tried to prove that rigorously, so let's focus on momentum for now.

 

[Dickfore]

My claim is

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

This is what I was also saying by quoting Landau Lifgarbagez above.

 

[Fredrik]

The real reason why you can't measure $S_x$ and $S_y$ (of a silver atom) at the same time is that if you put two Stern-Gerlach magnets at the same location, their fields would add up to a field that correlates eigenstates of $(\vec e_x+\vec e_y)\cdot\vec S$ with eigenstates of $(\vec e_x+\vec e_y)\cdot\vec p$. The consequence is that when you measure the momentum of the particle, by measuring its position, you can interpret the result as a measurement of $(\vec e_x+\vec e_y)\cdot\vec S$.

You can measure position and momentum at the same time because you're measuring them with a single device, not with two devices that would interfere with each other or two devices that can't even exist at the same location.

However, a device that prepares a state with a sharply defined position interferes with a device that prepares a state with a sharply defined momentum. The former is something like a narrow slit, and the latter is something that ensures that the particle has an almost equal probability of detection at every location in a large region. (If someone can think of a specific device that puts an already existing particle in a state with a sharply defined momentum, let me know).

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.

 

[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?