Exploring Heisenberg's Uncertainty Principle: Intuition & Explanations

[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.

 

[fuesiker]

atyy, you don't understand "how position and momentum can be simultaneously sharply measured, and the particle collapses to a position eigenstate" because that's total nonsense, so your brain is working right. For a quantum system, you simply can never measure position and momentum at the same time with arbitrary precision. You measure the expectation values of position and momentum to relatively arbitrary precision over many measurements, but this is going to the classical limit and we are no longer talking about "clean closed quantum systems".

 

[kith]

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.

I think what matters for Frederik's point of view is that you need to detect the position of the photons to get the interference pattern. A Mach-Zehnder interferometer is very similar to the doubleslit if you use a delay stage to shift the phase between your two waves. Then, the only difference is that you get the interference pattern at one detector spot at different times in the case of the interferometer, instead of at many detector spots at one time like in a double slit experiment.

In general, I think you are sometimes confusing the Copenhagen Interpretation with the interpretation-independent essence of QM. Many textbooks consider only the Copenhagen viewpoint and don't mention interpretational issues. For example like Cthugha said before, the assumption that you can attribute a state vector to a single particle, is not necessary.

 

[fuesiker]

No I'm not confusing them. All I stated were basic postulates of quantum mechanics, not necessarily one of the tenants of the Copenhagen interpretation. It is completely wrong that you have to detect the position of the photons to get the interference pattern. Are you serious? When you detect the position of the photon like on a special photon-counter imager (which does exist), all you see is particle behavior (pixels being lit up where the photons fall on them, indicating a "sharp localization"). Interference is a wave phenomena and is not related to position. A wave has no position. You seem to be confusing that. For example, just consider light, it has a phase (wave phenomena) and an amplitude (energy or ray phenomena, also a wave phenomena because wave optics is more general than ray theory or geometric optics). In optics, the former is the wave vector, and the latter is the Poynting vector. The wave vector addition of several waves or the same wave interfering with itself is what gives you the interference pattern. The Poynting vector is just energy flux and its direction of propagation is that of the "photon" velocity. Hence, the position of your photon is not what in general gives you the interference.

 

[atyy]

atyy, you don't understand "how position and momentum can be simultaneously sharply measured, and the particle collapses to a position eigenstate" because that's total nonsense, so your brain is working right. For a quantum system, you simply can never measure position and momentum at the same time with arbitrary precision. You measure the expectation values of position and momentum to relatively arbitrary precision over many measurements, but this is going to the classical limit and we are no longer talking about "clean closed quantum systems".

Well, I don't know if have a brain;) but I do find Ballentine's claim (which is based on Park and Margenau's work) different from what most textbooks seem to say. My understanding is the same as yours - non-commuting observables cannot be simultaneously sharply measured, and sharp measurement of an observable will collapse the particle to an eigenstate of the observable (if the particle is not destroyed, or other complex scenario).

My understanding is that a momentum measurement does not consist of two sharp position measurements with finite time separation, and that a momentum measurement can be performed on arbitrary states. In the case where momentum is approximately measured by approximately measuring position (simultaneous non-sharp measurement), neither observable is precisely measured.

I think Ballentine, and Park and Margenau are talking about a special case where assigning a "momentum" by two sharp position measurements with the second infinitely far in time approaches the correct distribution for a true momentum measurement. Then the question is whether this special case can be called a "sharp measurement" of both momentum and position. It seems not, since it requires special preparation of the state, whereas measurements should be doable on arbitrary states.

By sharp measurement, I'm thinking of projective measurements.

(But I am a biologist, so am just trying to find out if my understanding of the textbooks is wrong.)

 

[kith]

My claim is

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

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

What I think, is that there is simply no way to measure the observables corresponding to the operators X and P. Real measurements can't measure points but only intervals with finite resolutions, so the corresponding operators should reflect this. X and P are just idealizations.

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