What Intuitive Insights Explain Heisenberg's Uncertainty Principle?

[jeebs]

You are no doubt familiar with Heisenberg's uncertainty principle, putting a limit on the accuracy with which we can measure a particle's position and momentum, \Delta x \Delta p \geq \hbar/2
On my course I was shown the derivation, it popped out of a few lines of mathematics involving the Cauchy-Riemann inequality.

However, 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?

One explanation I heard was that if you, say, bounced a photon off an atom to measure its position, then the recoil would affect its momentum, thus giving rise to the uncertainty - this seems straightforward enough. However, I have also been told that this is apparently not a valid explanation, although I do not understand why.
Can anyone shed any light on this for me?

 

[CJames]

The intuitive explanation you present is called Heisenberg's microscope. There are two primary problems with it. The first is that it only results in an approximate expression of the equation that you cite.

The second is that it attacks its own premises. The thought experiment first assumes that the electron has a definite location and momentum, and then demonstrates why such a thing can't exist, which invalidates its own premises.

 

[nkadambi]

Here is a very general answer: From the axioms of QM and the math that is used to build observables and states of systems, it turns out that position and velocity (and also momentum, because momentum p = mv) are what are called "canonical conjugates", and they cannot be both be "sharply localized". That is, we cannot measure them both to an arbitrary level of precision. It is a mathematical fact that any function and its Fourier transform cannot both be made sharp.

This is a purely a mathematical fact and so has nothing to do with our ability to do experiments or our present-day technology. As long as QM is based on the present mathematical theory, it cannot be done using the mathematics we have.

 

[eaglelake]

There is nothing in Classical theory that prevents us from knowing both the momentum and the position of a particle with "certainty". i.e. we can repeat the classical experiment many times and always get the same result for momentum and the same result for position.

But in quantum mechanics, position and momentum are linear operators in a Hilbert space and, most importantly, the momentum operator and the position operator do not commute. This means that there is no wavefunction that is a common eigenfunction of both momentum and position. Further, we must know the wavefunction in order to calculate the uncertainties. For practice, make up a (simple) wavefunction and do the caculations for \Delta x and \Delta p and then take their product to convince yourself that the uncertainty principle is satisfied. I am trying to emphasize that this is quantum mechanics and not classical physics.

Bouncing a photon off an atom tells us nothing about any uncertainties. We must bounce many identically prepared photons off like atoms in order to get the statistical distributions of atomic position measurements and atomic momentum measurements. What we call "uncertainty" is a property of a statistical distribution. You cannot determine an uncertainty from a single measurement. I hope this helps.
Best wishes.

 

[Fredrik]

Good post nkadambi, but I must point out an inaccuracy in what you said. I didn't really understand this myself until recently. It is possible to measure position and momentum simultaneously. In fact, we often measure the momentum by measuring the position and interpreting the result as a momentum measurement. (Check out figure 3 in this pdf). What we can't do is to prepare a state such that we would be able to make an accurate prediction about what the result of a position measurement would be and an accurate prediction about what the result of a momentum measurement would be.

jeebs: Mathematically, the "uncertainty" is derived from the axioms of QM, and is only non-zero if the commutator of the two operators is non-zero. Physically, I think the problem is always that a device that prepares a state with a sharply defined value of one of the observables would interfere with a device that prepares a state with a sharply defined value of the other observable. So... non-zero uncertainty = non-commutativity = the state preparation devices would interfere with each other.

 

[nkadambi]

Thanks for the post, Fredrik. And thanks for the paper: I'll check it out.

I am new to this forum (I just registered couple hours ago!) I mainly have a pure math background, and only just starting into mathematical physics. Are you a grad student or professor? At my school there is hardly anyone who ventures into math physics. I am looking to make a few friends online with similar background.

 

[CJames]

Wow Fredrik, that is incredibly helpful. Just when I thought I was starting to understand something...

 

[dlgoff]

It is possible to measure position and momentum simultaneously.

I've always liked how ZapperZ explains this in is blog http://physicsandphysicists.blogspot.com/2006/11/misconception-of-heisenberg-uncertainty.html" .

 

[Fredrik]

I am new to this forum (I just registered couple hours ago!) I mainly have a pure math background, and only just starting into mathematical physics. Are you a grad student or professor? At my school there is hardly anyone who ventures into math physics. I am looking to make a few friends online with similar background.

You have come to the right place. There are plenty of people here with all sorts of backgrounds.

I've always liked how ZapperZ explains this in is blog http://physicsandphysicists.blogspot.com/2006/11/misconception-of-heisenberg-uncertainty.html" .

Yes, ZZ understood this a long time before I did. I would have understood it much sooner if I had read his posts in these threads more carefully. I was naive enough to think I already understood these things.

 

[Naty1]

(sorry this is so long but I have just been struggling through the same concepts.)

I hope the essence of Zapper's HUP explanation is here:

The HUP isn't about a single measurement and what can be obtained out of that single measurement. It is about how well we can predict subsequent measurements given the identical conditions.

and

What I am trying to get across is that the HUP isn't about the knowledge of the conjugate observables of a single particle in a single measurement. I have shown that there's nothing to prevent anyone from knowing both the position and momentum of a particle in a single mesurement with arbitrary accuracy that is limited only by our technology. However, physics involves the ability to make a dynamical model that allows us to predict when and where things are going to occur in the future. While classical mechanics does not prohibit us from making as accurate of a prediction as we want, QM does!

Somebody in the recent past posted this...my boldface.. (I did not record the poster, maybe even Zapper??..was a trusted source here.) I'm posting this to confirm that it is an equivalent description, that it matches Zappers blog...

...to measure a particle's momentum, we need to interact it with a detector, which localizes the particle. So we actually do a position measurement (to arbitrary precision). Then we calculate the momentum, which requires that we know something else about the position of the particle at an earlier time (perhaps we passed it through a narrow slit). Both of those position measurements, and the measurement of the time interval, can be done to arbitrary precision, so we can calculate the momentum to arbitrary precision. From this you can see that in principle, there is no limitation on how precisely we can measure the momentum and position of a single particle.

Where the HUP comes into play is that if you then repeat the same sequence of arbitrarily precise measurements on a large numbers of identically prepared particles (i.e. particles with the same wave function, or equivalently particles sampled from the same probability distribution), you will find that your momentum measurements are not all identical, but rather form a probability distribution of possible values for the momentum. The width of this measured momentum distribution for many particles is what is limited by the HUP. In other words, the HUP says that the product of the widths of your measured momentum probability distribution, and the position probability distribution associated with your initial wave function, can be no smaller than Planck's constant divided by 4 times pi

So what I think these mean is that you can get precise but not necessarily ACCURATE simultaneous measurements...that is, you cannot REPEAT the exact measurement results as is possible to arbitrary precision in classical measurements. What had me confused, and I hope I understand better, was that commutativity and non commutativity of operators applies to the distribution of results, not an individual measurement.

In Quantum Mechanics, Albert Messiah provides an interpretation for the inability to repeat the measurements :

Immediately after the operation of a measurement the system is in a dynamical state with the arbitrarily precise position measured; Such a state cannot be represented by a wave function (psi). The state psi in general corresponds to a probability distribution of finding some value x, not a precise value of x…..or of measuring momentum. The function psi does not give more than the statistics of positions…or momenta... During the process of observation the measured system can not be considered as separate from the observed phenomena. The intervention of the measuring instrument destroys all causal connection between the state of the system before and after the measurement; this explains why one cannot in general predict with certainty in what state the system will be found after the measurement.

 

[Fredrik]

This "inability to repeat measurements" is in my opinion better described as an inability to prepare a state with the desired properties (or as the non-existence of such a state in the mathematical part of QM). Since you measure the momentum by measuring the position, you can measure both with an accuracy that's only limited by the size of the detector.

 

[Naty1]

This "inability to repeat measurements" is in my opinion better described as an inability to prepare a state with the desired properties (or as the non-existence of such a state in the mathematical part of QM).

I'm surprised, if I understand what you posted: Zapper's blog which I quoted above seems to me a bit different:

Where the HUP comes into play is that if you then repeat the same sequence of arbitrarily precise measurements on a large numbers of identically prepared particles (i.e. particles with the same wave function, or equivalently particles sampled from the same probability distribution), you will find that your momentum measurements are not all identical, but rather form a probability distribution of possible values for the momentum.

But I haven't quite been able to figure out exactly what "prepare a state" means which Messiah in QUANTUM MECHANICS also mentions but doesn't explain. Where is Zapper?

 

[Fredrik]

It's not different. The "desired" properties are precisely those properties that would ensure that the results of all the momentum measurements (on different members of an ensemble of identically prepared systems) are essentially the same, and that the results of all the position measurements (on different members of the same ensemble) are essentially the same. ZZ's statement explains what my statement means.

I just don't like the phrase "repeated measurements", because it sounds like it might be referring to something you do repeatedly to the same particle (without re-preparing it between measurements) rather than to the members of an ensemble of identically prepared particles.

To prepare a state is just to bring a particle on which we intend to do a measurement to the measuring device. Different ways of doing that may give us different average results. Two ways of doing it (two preparation procedures) are considered equivalent if no series of measurements can distinguish between them (i.e. if they give us the same wavefunction, or more generally, the same state operator/density matrix). These equivalence classes are often called "states".

 

[Naty1]

Fredrik: thanks for the assistance...I have a bit more thinking to do, but I "get" the last two of your three paragraphs...

"It's not different". well, THAT's a relief! maybe wording semantics got in the way...

Your explanation of "to prepare a state" clarifies what that means...I sure do not like that terminology, but maybe that's just me...

 

[Naty1]

eaglelake:

I just read your early post...

Bouncing a photon off an atom tells us nothing about any uncertainties. We must bounce many identically prepared photons off like atoms in order to get the statistical distributions of atomic position measurements and atomic momentum measurements. What we call "uncertainty" is a property of a statistical distribution. You cannot determine an uncertainty from a single measurement. I hope this helps.

BRAVO!...concise, well stated...

 

[Roo2]

I've always liked how ZapperZ explains this in is blog http://physicsandphysicists.blogspot.com/2006/11/misconception-of-heisenberg-uncertainty.html" .

Is there a peer-reviewed article which presents this argument (that the limit of accuracy for the measurement of a single electron is technology rather than HUP)? It makes sense to me intuitively and mirrors what I was taught, but I'd like to see it addressed formally rather than via blog.

 

[CJames]

Is there something wrong with the link Fredrik posted earlier?

http://www.kevinaylward.co.uk/qm/ballentine_ensemble_interpretation_1970.pdf

Page 365

I don't mean this to be sarcastic if it comes across that way.

 

[abaio]

Great informoation. Thanks. It would have been awsome to be there when the Quantum giants were discussing and racing to find new discoveries in the new mysterious quantum world. Bohr vs. Einstein was a great duel...kind of that like Edison vs. Tesla.

 

[Roo2]

Is there something wrong with the link Fredrik posted earlier?

http://www.kevinaylward.co.uk/qm/ballentine_ensemble_interpretation_1970.pdf

Page 365

I don't mean this to be sarcastic if it comes across that way.

Thanks! I was skimming this thread at work and didn't notice the link. That's exactly what I was looking for. I was actually amused to find this thread; I had a discussion with my room mate a couple weeks ago about HUP. Both of us are college undergrads, and our experience with quantum mechanics is limited to one semester of physical chemistry apiece (at different universities). I mentioned that I had been taught that HUP applied to sets of measurements of p and r and didn't apply to the position and momentum of a single particle. He replied that since the particle is itself a wave, and a wave has no defined absolute position, then HUP applies to a single particle. I don't have the physics background to respond to that, so I let it be. That paper is clearing up the picture for me. Forgive me for asking again for references, but I don't have the physics education to know off the bat... have there been any revisions to the interpretation of the statistical model in regard to a single particle measurement in the 40 years since this paper was published?

 

[fuesiker]

The most intuitive way, imho, to view the uncertainty principle is to recognize that the wavefunction of some system in momentum space is the Fourier transform of the wavefunction of that same system in position space. You can arrive at this by honoring the fact that the momentum operator is the propagator of translation, and that is how you can relate both wavefunctions to each other to arrive at the equation showing they are related by the Fourier transform.

Now that we are convinced of this, we can look at the properties of the Fourier transform. If a function K is the Fourier transform of a function F, then the sharper F is, the broader is K.

The uncertainty principle has nothing to do with observation. It does not arise due to observation. It is the nature of things that there are non-commuting observables (time and energy, momentum and position) that cannot be determined simultaneously to arbitrary precision. We find this unintuitive because in our universe, Planck's constant is way too small for quantum effects to have become intuitive to our senses, just like special relativisitic effects are also counter-intuitive to our senses since c is so big in our world. Imagine if we lived in a world where c is 30 mph ;)

 

[Dickfore]

@OP:

The canonical commutation relation:
<br /> [\hat{x}_{i}, \hat{p}_{k}] = i \, \hbar \, \delta_{i k}<br />

EDIT:
How to derive these relations? You need to identify momentum as the generator of translations in space (according to the correspondence principle and the fact that the physical quantity conserved due to homogeneity of space is linear momentum) and it acts on states that are eigenstates of position according to:
<br /> \hat{T}(\mathbf{a}) \, \left|\mathbf{x}\right\rangle = \left|\mathbf{x} + \mathbf{a}\right\rangle, \; \hat{T}(\mathbf{a}) = \exp{\left(-\frac{i}{\hbar} \, \mathbf{a} \cdot \hat{\mathbf{p}}\right)} <br />

 

[ZealScience]

I heard that these states of momentum and position are orthogonal in Hilbert space. Probably mathematics thinks that it is intuitive, but not us. Because in experiments that can be seen with naked eyes have no violation to classical mechanics which is described by definite phases...

 

[Dickfore]

<br /> \begin{array}{l}<br /> \hat{x}_{i} \, \hat{T}(\mathbf{a}) \, \left|\mathbf{x}\right\rangle = (x_{i} + a_{i}) \, \left|\mathbf{x} + \mathbf{a}\right\rangle \\<br /> <br /> \hat{T}(\mathbf{a}) \, \hat{x}_{i} \, \left|\mathbf{x}\right\rangle = x_{i} \, \left|\mathbf{x} + \mathbf{a}\right\rangle<br /> \end{array} \Rightarrow \left[\hat{x}_{i}, \hat{T}(\mathbf{a})\right] = a_{i} \, \hat{T}(\mathbf{a})<br />

Expanding to linear power in the translation vector \mathbf{a}, we get:
<br /> -\frac{i}{\hbar} \, a_{k} \, \left[\hat{x}_{i}, \hat{p}_{k}\right] = a_{i} = \delta_{i k} \, a_{k}<br />
Since this has to hold for arbitrary components a_{k}, we must have the commutation relations to hold.

After you have a commutation relation between two operators, you can apply the same math that you mentioned in the OP (Cauchy-Schwartz ineqality) to derive an Uncertainty relation.

 

[Dickfore]

I heard that these states of momentum and position are orthogonal in Hilbert space.

They are not. For example:
<br /> \langle x | p \rangle = \frac{1}{(2\pi \hbar)^{1/2}} \, \exp{\left(\frac{i}{\hbar} \, p \, x\right)} \neq 0<br />

 

[ZealScience]

They are not. For example:
<br /> \langle x | p \rangle = \frac{1}{(2\pi \hbar)^{1/2}} \, \exp{\left(\frac{i}{\hbar} \, p \, x\right)} \neq 0<br />

Can you specify the eigenvectors of position and momentum? I just started learning it and I am not familiar with it.

 

[Dickfore]

Can you specify the eigenvectors of position and momentum? I just started learning it and I am not familiar with it.

FFFFFFFFFFFFF

I pressed the page back button on my keyboard and my whole response is gone. I hate this glitch on PF.

 

[ZealScience]

FFFFFFFFFFFFF

I pressed the page back button on my keyboard and my whole response is gone. I hate this glitch on PF.

Accidents can happen... it is not deterministic~~~

 

[Fredrik]

FFFFFFFFFFFFF

I pressed the page back button on my keyboard and my whole response is gone. I hate this glitch on PF.

I have found that you can usually get it back if you press the page forward button and answer yes to the question if you want to send the information again.

 

[Dickfore]

I have found that you can usually get it back if you press the page forward button and answer yes to the question if you want to send the information again.

For me the text gets erased.

 

[atyy]

Good post nkadambi, but I must point out an inaccuracy in what you said. I didn't really understand this myself until recently. It is possible to measure position and momentum simultaneously. In fact, we often measure the momentum by measuring the position and interpreting the result as a momentum measurement. (Check out figure 3 in this pdf).

If I remember correctly, after measuring position, the particle collapses to an eigenstate of position, then spreads out due to time evolution. After measuring momentum, it collapses to an eigenstate of momentum, then remains in the same state since it is an eigenstate of the free particle Hamiltonian. What state does it collapse to after position and momentum are simultaneously measured?

 

[Fredrik]

If I remember correctly, after measuring position, the particle collapses to an eigenstate of position, then spreads out due to time evolution. After measuring momentum, it collapses to an eigenstate of momentum, then remains in the same state since it is an eigenstate of the free particle Hamiltonian. What state does it collapse to after position and momentum are simultaneously measured?

In this example, the particle is absorbed by the detector, so no new state is prepared. If it had been the kind of detector that let's the particle pass through it, the new state would have been one with a sharply defined position (the wavefunction will be close to zero far from the detector).

After measuring momentum, it collapses to an eigenstate of momentum

This is what I was taught as well, but I wonder if it really makes sense. Even if we disregard all the difficulties associated with the fact that an eigenfunction of -i\,\nabla isn't square-integrable, and interpret the statement as "after a momentum measurement, the wavefunction will have a sharply defined momentum", it still looks false to me. I don't know much about how things are measured, but it seems to me that momentum is always measured by measuring the position instead. It's possible that in most cases, the position measurement is inexact enough to give us something like exp(-x²) as the new wavefunction, so that both the wavefunction and its Fourier transform have a single peak, with widths of the same order of magnitude in units such that \hbar=1.

 

[atyy]

This is what I was taught as well, but I wonder if it really makes sense. Even if we disregard all the difficulties associated with the fact that an eigenfunction of -i\,\nabla isn't square-integrable, and interpret the statement as "after a momentum measurement, the wavefunction will have a sharply defined momentum", it still looks false to me. I don't know much about how things are measured, but it seems to me that momentum is always measured by measuring the position instead. It's possible that in most cases, the position measurement is inexact enough to give us something like exp(-x²) as the new wavefunction, so that both the wavefunction and its Fourier transform have a single peak, with widths of the same order of magnitude in units such that \hbar=1.

I haven't really thought this through myself, and am just going to ask questions as they come to mind (but OP please stop me if this is hijacking!). How about in the relativistic case, say like Compton scattering? Isn't momentum there measured without measuring position? OTOH, there is no "sensible" position operator in relativistic QFT, so maybe that's different?

 

[Fredrik]

Can you specify the eigenvectors of position and momentum? I just started learning it and I am not familiar with it.

Neither of those operators have eigenvectors in the semi-inner product space of square-integrable functions from \mathbb R into \mathbb C. If we view this space as a subspace of the vector space of all functions from \mathbb R into \mathbb C, then the functions u_p defined by u_p(x)=e^{ipx} for all x are eigenfunctions (with eigenvalue p) of the momentum operator. Note that they are not square-integrable. The position operator doesn't have any eigenfunctions in this space either, but there's a trick. The "functions" v_x such that v_x(y)=\delta(y-x) for all y, where \delta is the Dirac delta, can be thought of as "eigenfunctions" (with "eigenvalue" x) of the position operator, even though they aren't really functions.

 

[Fredrik]

How about in the relativistic case, say like Compton scattering? Isn't momentum there measured without measuring position?

I don't think so. The way I see it, a measuring device is just a device that produces a signal (that can be approximately described as classical) that informs us that an interaction has taken place. The location of the device (or the location of the relevant component of it) can always be interpreted as the result of a position measurement. So it's not possible to measure anything without measuring the position of the particle(s) that participated in the interaction that produced the signal.

Hm, I suppose it's possible that those particles aren't the same ones as the one we're trying to measure the momentum of.

 

[Dickfore]

Regarding measurement and falling back into the eigenstate of the observable corresponding to the observed eigenvalue, I have doubts

If you read section 7 in "Quantum Mechanics vol. 3" (the wave function and measurements) by Landau, Lifgarbagez, they discuss in the 12th paragraph the possibility that the wave function in which the electron drops after the measurement is not necessarily an eigenfunction of the observable being measured.

 

[fuesiker]

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

 

[Fredrik]

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

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

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

 

[fuesiker]

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

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

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

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

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

 

[Fredrik]

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

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

 

[fuesiker]

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

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

 

[Fredrik]

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

 

[fuesiker]

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

 

[Fredrik]

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

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

This does not violate the uncertainty principle.

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

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

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

 

[atyy]

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

 

[fuesiker]

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

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

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

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

 

[fuesiker]

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

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

 

[atyy]

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

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

Note: sorry, lots of edits.

 

[Fredrik]

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

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

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

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

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

My claim is

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

You're arguing against the claim

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

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

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

 

[Dickfore]

My claim is

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

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

 

[Fredrik]

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

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

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

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