What Intuitive Insights Explain Heisenberg's Uncertainty Principle?

[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 ;)).

 

[kith]

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

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

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

 

[atyy]

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

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

 

[fuesiker]

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

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

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

 

[Cthugha]

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

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

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

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

 

[SpectraCat]

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

A detection of a particle is a momentum measurement if the particle was prepared with a sharply defined position \vec x at a known time t. If the detection event is (t&#039;,\vec x&#039;), then the vector $$m\gamma\frac{\vec x'-\vec x}{t'-t}$$ is called the result of the measurement.

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

I don't think that constitutes a momentum measurement in the quantum sense, since it is unclear in principle how such a scheme would collapse the wavefunction of the particle into a momentum eigenstate. The reason I suggested describing such a process as *inferring* the momentum is because you are not actually getting a momentum measurement at a distinct point in time. You cannot say that "the momentum at point (\vec{x},t) is \vec{p}" from such a "measurement", which is actually a pair of measurements, as is clear from your formula. As far as I can tell, all that formula allows you to *calculate* is the average momentum as the particle traveled from point \vec{x} to point \vec{x&#039;}.

 

[Fredrik]

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

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

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

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

 

[SpectraCat]

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

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

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

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

 

[Fredrik]

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

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

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

...the average momentum...

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

 

[fuesiker]

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

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

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

 

[atyy]

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

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

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

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

 

[Fredrik]

My claim is

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

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

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

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

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

 

[atyy]

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

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

 

[fuesiker]

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

 

[grzz]

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

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

 

[fuesiker]

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

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