What Intuitive Insights Explain Heisenberg's Uncertainty Principle?

[fuesiker]

I asked you to describe the design of an interferometer that could provide phase data about single photons so as to constitute a momentum measurement that does not also involve position measurement. You claimed in an earlier post that interferometry was capable of measuring momentum of individual photons without measuring their positions. I cannot see how that is possible, so I asked you to explain it.

OK I decided to answer this quickly, but next time ask in a more polite way ;). Moreover, I think this is something many were keen about here because Fredrik's propaganda machine had wrongly (yet again) made people on here think you can't measure momentum unless you measure position (I don't have the time or the motivation to dig this up in his many posts, but he did say something in that color). I may add that I heard drug addicts at the local train station here make more sense than that. I hope the below experimental description (which is way too simplified in terms of technical, yet not conceptual, matters) will put an end to this myth of having to measure position to measure momentum.

As I mentioned before, a photon is a wave and a particle. Now, you want to measure a wave phenomenon, so you better not know anything about the particle behavior of the photon. You can do this using either a double-slit experiment or the Mach-Zehnder interferometer. Let's take the latter, because it is easier, in my opinion. I think this is done in experimental quantum optics courses as a lab task too.

Now, your Mach-Zehnder interferometer has two arms of lengths l_1 and l_2. The whole essence of interference is the fact that you cannot tell which path the photon takes, whether the first arm or the second arm. You can think of it in a naive way as the photon takes both at the same time (wave nature of the photon). Now you don't know what the wavelength \lambda of your photon is, but we assume that you are sure only one photon is going through the interferometer at any given time (this is experimentally laborious, but it is done). When l_1=l_2, then the photon interferes contructively with itself, and you hear a click at a detector that you have placed at the output of the interferometer. Now don't fool yourself into thinking this is measuring the position of the photon by the detector. We are inferring nothing from the measurement on the detector other that a photon has hit it. We make no further use of this knowledge, nor do we need to. Now as you vary l_2 (in some non-random process, such as a linear ramp), while keeping l_1 fixed, you start going away from constructive interference. Then at a certain value of l_2, you don't hear any clicks on the detector anymore -> destructive interference (of course, you are sending similarly-prepared photons each time you make a measurement). This allows you to determine the momentum k=\frac{2\pi}{\lambda} of the photon you have up to machine precision, without knowing anything about its position. To see this more clearly, intentionally make your detector one huge pixel whereby anywhere the photon hits it, it gives you a click, yet you can have this pixel occupy a huge spatial extent such that the uncertainty in your position can be considered, for all practical purposes, infinite. I call such an extension "the Fredik-propaganda negator factor".

As you can see, this is pretty much the same procedure for interfering two photons, or two laser beams. That's why I thought you did not know that a single photon can interfere with itself, and I still think you could not have actually believed this when you did not think it could be done experimentally, because then obviously you when you think that is not possible you also have not seen it in nature (otherwise that would be your experiment), and hence, it does not make sense to say you do know a single photon interferes with itself when you believe it is impossible to see this effect.

But I really suggest Loudon's book. It's my favorite quantum optics book.

 

[Fra]

What a long thread.

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.

I personally think this is the best answer to the OT.

The rest of the discussion seems to be about howto get from single datapoints to knowledge, and howto distinguish information from single measurements from information from an acquired history of measurments (also called preparation). This tends to be very interpretation dependent as always.

I in particular like the perspective Fredrik added that you can INFER momentum from position (and vice versa). But at least in my picture, such an inference does not constitute a measurement, it constitutes an internal process in the observing system. What would however be extremelt interesting (but also off topic) is to discuss if there is such a thing as primary and secondary observables, which are linked by such inferences. This reminds me of The principle of relative locality http://arxiv.org/abs/1101.0931 where they argue that spacetime is the result of an inference made from local momentum measurements.

One can take such a thing more seriously, and question wether this "inference" is internal processes or not. After all, NO local observer, can possibly make measurements of remote points. EVERYTHING is ultimately a local inference (and extrapolation) fro locally acquired information.

/Fredrik

 

[Fredrik]

Instead, could one or more of the principals in the discussion maybe take a crack at stating points of mutual agreement??

The thing is, we agree about essentially everything except the meaning of the words "has something to do with" and whether it makes sense to define a "measurement" as what a measuring device does in an experiment that tests the accuracy of the theory's predictions. (The alternative is to define it in purely mathematical terms as a projection onto an eigenspace). Feusiker has a very strange interpretation of "has something to do with" and thinks anyone who uses the word measurement in the former sense is a complete idiot who doesn't know anything about physics and deserves to be insulted and ridiculed at every possible opportunity. (I think I've been extremely polite to him under the circumstances). He hasn't offered any arguments to support his position, but instead insists that he's right because he has a PhD and because (according to him) Cohen-Tannoudji and Sakurai would both agree with him and Ballentine is an idiot.

We obviously also disagree about whether he has successfully refuted the things I said. He has repeatedly made arguments that are essentially correct, but don't even come close to addressing the things I actually said.

 

[fuesiker]

The thing is, we agree about essentially everything except the meaning of the words "has something to do with" and whether it makes sense to define a "measurement" as what a measuring device does in an experiment that tests the accuracy of the theory's predictions. (The alternative is to define it in purely mathematical terms as a projection onto an eigenspace). Feusiker has a very strange interpretation of "has something to do with" and thinks anyone who uses the word measurement in the former sense is a complete idiot who doesn't know anything about physics and deserves to be insulted and ridiculed at every possible opportunity. (I think I've been extremely polite to him under the circumstances). He hasn't offered any arguments to support his position, but instead insists that he's right because he has a PhD and because (according to him) Cohen-Tannoudji and Sakurai would both agree with him and Ballentine is an idiot.

Fredrik, I was aggressive with you as much as you've been with me. And no, we don't agree on everything. Cohen-Tannoudji and Sakurai (if the latter was alive) would agree with me unless their books are wrong. Theirs are my quantum mechanics bibles. Anyway, read the post I made, it is 2 before your post that I am quoting here. Here's a way to measure the momentum of a single photon without knowing anything about its position, something you claimed in earlier posts (and do not deny this) cannot be done. And I hope you won't say this method is crap or not right. I am certain it is done experimentally in basic quantum optics lab work, and I do not feel I need to go searching for the proof. Moreover, if you truly understand quantum mechanics, my description will make perfect sense to you.

Moreover, despite what you may think, I have respect for you. In the end, you do seem to love physics, though yourself more.

 

[Fredrik]

Fredrik, I was aggressive with you as much as you've been with me.

This is easily the most absurd statement you have made so far.

Since you have made me spend almost the entire day defending myself, I haven't yet had time to get up to speed about momentum measurements with interferometers. I will as soon as I can.

 

[Naty1]

Is the following agreed upon:

Atyy:

The uncertainty relation is defined as the non-commutation of position and momentum operators. A state with definite momentum is an eigenstate of momentum, and state with definite position is an eigenstate of position. The commutation relation prevents an eigenstate of momentum from being an eigenstate of position, so there is no state with definite momentum and position, and so it cannot be prepared.

And Fra's post endorsing this:
Originally Posted by fuesiker

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.

These two,atyy and Fuesiker, don't conflict, right?

I thought I'd check Wikipedia for comparsion with all the discussion above; it was a bit discouraging (This is the first line) :

The framework of quantum mechanics requires a careful definition of measurement. The issue of measurement lies at the heart of the problem of the interpretation of quantum mechanics, for which there is currently no consensus...

http://en.wikipedia.org/wiki/Quantum_measurement

 

[fuesiker]

Is the following agreed upon:

Atyy:


And Fra's post endorsing this:
Originally Posted by fuesiker



These two,atyy and Fuesiker, don't conflict, right?

Right! Any two observables that do not commute cannot have common eigenstates.

 

[Fredrik]

Regarding atyy's statement: I disagree with the words "defined as", but the commutator is certainly involved in the derivation. The rest of it is OK.

Regarding Fra's Feusiker quote: I disagree with the words "has nothing to do with", but I agree with everything else in that quote. Edit: Uh wait, I didn't look at it closely enough. Give me a minute to think.

OK, I have thought about it. I have some issues with the statement that the two non-commuting observables can't be determined simultaneously. I interpret "determined" as "measured", and I consider a measurement to be what a measuring device does in an experiment that tests the accuracy of the theory's predictions. If experimentalists consider a detection of a particle, followed by a calculation of a momentum from the coordinates of the detection event, a valid way to measure momentum, then position and momentum can be measured simultaneously. In fact, we would be measuring momentum by measuring position.

Also, "time" isn't an observable.

 

[atyy]

Naty1, the interpretation of quantum mechanics has no consensus, but I think we aren't discussing that. We are discussing things within the simple-minded textbook interpretation of say Cohen-Tannoudji, Sakurai, or Chuang and Nielsen.

The statements at issue:

1. Non-commutativity of position and momentum means that sharp measurements (projective measurements) of position and momentum cannot be simultaneous.

2. A sharp momentum measurement is the same as two sharp position measurements plus timing.

3. After simultaneous sharp measurements of position and momentum, the state collapses to a position eigenstate.

In my current understanding (remember I am a biologist, and I claim no authority in physics - nor biology for that matter): 1 is true, if we also add that nothing about the initial position and momentum of the particle is known (ie. for arbitrary initial states). 2 is not true, again for arbitrary initial states, since if successive position measurements are used to approach an accurate momentum measurement, the position measurements must not be sharp. 3 is not true since it directly contradicts the projection postulate. I believe the Ballentine, Park and Margenau argument that 1 is not true only works for certain initial states and if in using 2 to measure momentum the second sharp position measurement is taken at infinite time, which is why I have added the qualification "arbitrary states" (see also the quote from Bell in post #89).

 

[fuesiker]

Regarding atyy's statement: I disagree with the words "defined as", but the commutator is certainly involved in the derivation. The rest of it is OK.

Regarding Fra's Feusiker quote: I disagree with the words "has nothing to do with", but I agree with everything else in that quote.

It indeed has nothing to do with observation in the sense that you do not need to observe for the HUP to hold. You keep dodging this. What effect does observation have to do with the HUP? Please explain it to me. What does observation have to do with the relation

\langle (\Delta x)^2\rangle \langle(\Delta p)^2\rangle \geq \frac{\hbar^2}{4}

Perhaps you mistakenly think it is because of the expectation values involved which "have to do something with wavefunctions" as you put it? Well, the thing is, these expectation values are there whether you observe or not. They can take on any value, whether one you can measure, or one you cannot (expectation values of the momentum operator with respect to a position wavefunction). In such a sense, the HUP has nothing to do with observation. Your observation, if you make it and you don't have to for the HUP to hold, will only alter \Delta x and \Delta p, but not the HUP, and I believe you understand this. But it is perfectly correct to say the HUP has nothing to do with observation, since, as I discuss above, the expectation values in the HUP need not be those resulting from an observation. I insist you kindly address this post before claiming again that it is wrong to say the HUP has nothing to do with observation. It indeed has nothing to do with it.

 

[Fredrik]

It indeed has nothing to do with observation in the sense that you do not need to observe for the HUP to hold.

Of course. It's a theorem. (That's the reason why I can't make myself call it a "principle", no matter how standard that term is). But it doesn't make much sense to say that it has nothing to do with observation when of the things it tells us is how the results of measurements (i.e. observations) will be statistically distributed around the expectation values.

 

[fuesiker]

Of course. It's a theorem. (That's the reason why I can't make myself call it a "principle", no matter how standard that term is). But it doesn't make much sense to say that it has nothing to do with observation when of the things it tells us is how the results of measurements (i.e. observations) will be statistically distributed around the expectation values.

Yes, but this is better stated as observation has something to do with the Heisenberg Uncertainty Principle rather than the Heisenberg Uncertainty Principle has something to do with observation. It is the nature of things, embedded in the HUP, that makes observation the way it is, and not the other way around. Moreover, like I mentioned, even if you're not "looking", the HUP is there. In this sense alone, the HUP has nothing to do with observation, because, like I said repeatedly (which you keep ignoring), the HUP is there whether you observe or not.

 

[jtbell]

This thread is temporarily closed pending moderation.

 

[jtbell]

This thread is now open for discussion again. My apologies for the delay. It took me a long time to read and digest the whole thread, because I had not been following it closely, and it sort of "blew up" rapidly.

I'd like to issue a general reminder that it's OK to disagree with other people, but PF's policy is that you need to at least "play nice" while doing so. Personal attacks and other ways of "personalizing" an argument only serve to degrade the tone of discussion and detract from facts and logical arguments. (Click the "Rules" link at the top of any page and note the section Guidelines on Langauge and Attitude.)

This is an interesting subject, and I'd hate to see the discussion dragged downhill again.

 

[jtbell]

Much of this thread has revolved around a thought experiment in which one sends a particle through a single slit, and then observes its transverse position as it hits a screen further on. This also allows you to infer the particle's transverse momentum by reconstructing its path from the slit. The question is, how is this consistent with the Heisenberg uncertainty principle?

Here's a variation of the experiment that might shed some light on it. Imagine the screen as a collection of small pixel-detectors. Remove one of the pixels to create a small hole in the screen. If the particle passes through the hole, we know its transverse position immediately after the screen, just as if it had hit a detector there. But its momentum immediately after the screen is uncertain, because it has once again undergone diffraction.

It seems to me that in this case the HUP applies to the "future" momentum of the particle and not to the "past" momentum of the particle which we inferred from the position measurement, and that inferring the "past" momentum doesn't qualify as a "measurement" in the sense of the HUP and the common statement that we cannot "measure" two non-commuting variables simultaneously.

 

[atyy]

Much of this thread has revolved around a thought experiment in which one sends a particle through a single slit, and then observes its transverse position as it hits a screen further on. This also allows you to infer the particle's transverse momentum by reconstructing its path from the slit. The question is, how is this consistent with the Heisenberg uncertainty principle?

Here's a variation of the experiment that might shed some light on it. Imagine the screen as a collection of small pixel-detectors. Remove one of the pixels to create a small hole in the screen. If the particle passes through the hole, we know its transverse position immediately after the screen, just as if it had hit a detector there. But its momentum immediately after the screen is uncertain, because it has once again undergone diffraction.

It seems to me that in this case the HUP applies to the "future" momentum of the particle and not to the "past" momentum of the particle which we inferred from the position measurement, and that inferring the "past" momentum doesn't qualify as a "measurement" in the sense of the HUP and the common statement that we cannot "measure" two non-commuting variables simultaneously.

But why should the "past momentum" be disqualified? Is it because the past momentum cannot be measured for arbitrary initial states, ie. for certain initial states, if one has some knowledge of the initial state, then one can perform "single meaurements" that yield a distribution of (p,q) values that when marginalized give the correct p and q distributions of non-simultaneous measurements? (I'm not sure that's what's really happening, but I think that is Park and Margenau's definition of simultaneous precise measurement, as well as Busch et al's.)

 

[Fredrik]

I think it's a little bit misleading to say that we're "inferring the past momentum", because the state before the measurement was a state with a sharply localized position, not momentum. The particle didn't have a well-defined momentum before this inference.

It also seems to me that this is precisely the type of "inference" that measuring devices do when they test the accuracy of the theory's predictions, so how can anyone not call it a measurement? These are the answers I can think of right now:

a) Because if we do, the common statement that we can't simultaneously measure position and momentum wouldn't be true.

b) Because if we do, von Neumann's axiom that all measurements project the state onto an eigenspace of the measured observable would contradict itself.

c) Because if we do, we would have to distinguish between the terms "measurement" and "state preparation".​

Those are, in my opinion, very bad reasons to not use the term "measurement" for what measuring devices do when they test the accuracy of the theory's predictions.

 

[atyy]

I think it's a little bit misleading to say that we're "inferring the past momentum", because the state before the measurement was a state with a sharply localized position, not momentum. The particle didn't have a well-defined momentum before this inference.

It also seems to me that this is precisely the type of "inference" that measuring devices do when they test the accuracy of the theory's predictions, so how can anyone not call it a measurement? These are the answers I can think of right now:

a) Because if we do, the common statement that we can't simultaneously measure position and momentum wouldn't be true.

b) Because if we do, von Neumann's axiom that all measurements project the state onto an eigenspace of the measured observable would contradict itself.

c) Because if we do, we would have to distinguish between the terms "measurement" and "state preparation".​

Those are, in my opinion, very bad reasons to not use the term "measurement" for what measuring devices do when they test the accuracy of the theory's predictions.

Could you clarify what your procedure is (is it to measure two positions sharply, and divide by the time between measurements), and how you use it to measure the momentum of a state with sharply defined momentum?

 

[Fredrik]

Could you clarify what your procedure is (is it to measure two positions sharply, and divide by the time between measurements), and how you use it to measure the momentum of a state with sharply defined momentum?

I'm certainly no expert in experimental methods, so I don't have a general definition that I'm satisfied with yet. For the moment, I don't have anything better than what I said in #91, and I don't find that definition satisfactory at all. (It might be OK as a definition of what it means to measure the momentum of a particle that has a sharply defined position, but I would like to find a procedure that we can use to test the theory's predictions about momentum regardless of what state the particle is in).

I don't see how two position measurements would constitute a momentum measurement. The first measurement prepares a new state that can be very different from the one we want to measure.

 

[Fredrik]

Yes, but this is better stated as observation has something to do with the Heisenberg Uncertainty Principle rather than the Heisenberg Uncertainty Principle has something to do with observation.

There's no difference between those statements. "has something to do with" is a reflexive relation.

Moreover, like I mentioned, even if you're not "looking", the HUP is there. In this sense alone, the HUP has nothing to do with observation, because, like I said repeatedly (which you keep ignoring), the HUP is there whether you observe or not.

You say that I "keep ignoring it", but you're quoting a post where I'm not only acknowledging it but also agreeing with it.

Now, your Mach-Zehnder interferometer has two arms of lengths l1 and l2. The whole essence of interference is the fact that you cannot tell which path the photon takes, whether the first arm or the second arm. You can think of it in a naive way as the photon takes both at the same time (wave nature of the photon). Now you don't know what the wavelength λ of your photon is, but we assume that you are sure only one photon is going through the interferometer at any given time (this is experimentally laborious, but it is done). When l1=l2, then the photon interferes contructively with itself, and you hear a click at a detector that you have placed at the output of the interferometer. Now don't fool yourself into thinking this is measuring the position of the photon by the detector. We are inferring nothing from the measurement on the detector other that a photon has hit it. We make no further use of this knowledge, nor do we need to. Now as you vary l2 (in some non-random process, such as a linear ramp), while keeping l1 fixed, you start going away from constructive interference. Then at a certain value of l2, you don't hear any clicks on the detector anymore -> destructive interference (of course, you are sending similarly-prepared photons each time you make a measurement). This allows you to determine the momentum k=2πλ of the photon you have up to machine precision, without knowing anything about its position. To see this more clearly, intentionally make your detector one huge pixel whereby anywhere the photon hits it, it gives you a click, yet you can have this pixel occupy a huge spatial extent such that the uncertainty in your position can be considered, for all practical purposes, infinite. I call such an extension "the Fredik-propaganda negator factor".

What you're describing here is a way to find out which momentum eigenstate your photons are in when you already know that they're all in the same momentum eigenstate. I would be more interested in a way to measure the momentum of any given photon (a photon that's in an arbitrary unknown state before the measurement).

Now don't fool yourself into thinking this is measuring the position of the photon by the detector. We are inferring nothing from the measurement on the detector other that a photon has hit it.

How is this not a position measurement? Is it just because we already know what the result will be? You have a detector at position 1 and a detector at position 2. The theory predicts that detector 1 will click 100% of the time that one photon is sent through. You test that prediction by sending photons through one at a time to see what percentage of times each detector clicks. What could you possibly use the term "position measurement" for if not the experiments you would use to test the theory's predictions about which detectors are going to click?

 

[Fredrik]

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 #.

I would be interested in seeing that page number reference. This discussion (or whatever I should call it) has made me curious about the terminology in standard textbooks.

 

[fuesiker]

There's no difference between those statements. "has something to do with" is a reflexive relation.

I'm tired of arguing with you about semantics. It seems to me you keep using semantics to cover your mistakes in physics.

What you're describing here is a way to find out which momentum eigenstate your photons are in when you already know that they're all in the same momentum eigenstate. I would be more interested in a way to measure the momentum of any given photon (a photon that's in an arbitrary unknown state before the measurement).

This is trivial. There's no reason to believe one cannot prepare the same photon over and over again. Moreover, you need this in order to find out in any measurement process what the state of your system on which you're measuring the observable is. Your measurement is a probability value, and over many measurements you can reconstruct your state.

How is this not a position measurement? Is it just because we already know what the result will be? You have a detector at position 1 and a detector at position 2. The theory predicts that detector 1 will click 100% of the time that one photon is sent through. You test that prediction by sending photons through one at a time to see what percentage of times each detector clicks. What could you possibly use the term "position measurement" for if not the experiments you would use to test the theory's predictions about which detectors are going to click?

As mentioned before, read my post again. I never said you have two detectors. Moreover, we do not know what the result will be a priori. Moreover, all you need is one detector, and the photon will always collapse onto it. All you know is that you have a photon that you can prepare over and over again (which is what is done in all experiments that do projective measurements to determine a state), and you then measure its momentum to machine precision without inferring anything from its position. Also, to counter your argument about measuring position (or your implicit belief that measuring position in this experiment "has something to do" with measuring momentum), I explained in the post how you can make the detector arbitrarily low in resolution, yet this will have no effect on your momentum measurement. You can make the position measurement uncertainty arbitrarily large, but your momentum measurement will always have machine resolution. So please tell me again, how are you inferring momentum from position here?

Think about one monochromatic wave of light instead of one photon, and you are passing it through a double-slit configuration. From the interference pattern, you will get a measurement of the momentum of that beam, without inferring anything about position. Now you can argue the pattern on the wall (where we see the interference pattern) is a position measurement, but that would be silly, because you cannot tell which photon on that wall is which photon before the slits.

 

[fuesiker]

I think it's a little bit misleading to say that we're "inferring the past momentum", because the state before the measurement was a state with a sharply localized position, not momentum. The particle didn't have a well-defined momentum before this inference.

It also seems to me that this is precisely the type of "inference" that measuring devices do when they test the accuracy of the theory's predictions, so how can anyone not call it a measurement? These are the answers I can think of right now:

a) Because if we do, the common statement that we can't simultaneously measure position and momentum wouldn't be true.

b) Because if we do, von Neumann's axiom that all measurements project the state onto an eigenspace of the measured observable would contradict itself.

c) Because if we do, we would have to distinguish between the terms "measurement" and "state preparation".​

Those are, in my opinion, very bad reasons to not use the term "measurement" for what measuring devices do when they test the accuracy of the theory's predictions.

b) Because if we do, von Neumann's axiom that all measurements project the state onto an eigenspace of the measured observable would contradict itself.

This is what I mean that you ignore my rebuttals that you can't refute. I will say it here again, Fredrik.

The number operator \hat{N} is that when you have a two-site lattice where you trap a number of atoms, where this site is described by the state |\psi\rangle = |n_l,n_r\rangle, then \langle\psi|\hat{N}|\psi\rangle=(n_l,n_r). Experimentally, you can measure the number observable on this site. Let's say the initial state is \psi(t=0)\rangle=|1,0\rangle and let's assume the lattice is bosonic. The eigenstates of \hat{N} form a complete basis of the Hilbert space of this state. We time-evolve this state using a Bose-Hubbard model, with nonzero tunneling and interaction factors. The density at each cite will oscillate and the amplitude of the oscillations with time will get smaller, with an offset at 0.5. If you want, I can refer you to a paper that experimentally deals with the above.

However, when you measure in the lab the number of atoms at any time t on the state |\psi(t), your device, which is classical, will give you non-negative integer numbers for n_l and n_r. Now experimentally, no-one has gotten non-integer values for the above measurement, and a lot has been done on this. This shows how somehow during the measurement process, your state collapses to an eigenstate of the observable being measure. If it doesn't, then your state is a superposition of at least two distinct eigenstates of your system and would look something like |\psi_m(t)\rangle=\frac{1}{\sqrt{2}}|1,0\rangle+\frac{1}{\sqrt{2}}|0,1\rangle, (subscript m to indicate state after measurement), which would literally mean that you are simultaneously counting one atom on the left site, zero atoms on the right site and zero atoms on the left site, one atom on the right site, which is ridiculous.

Now, in your above quote, you seem to imply that von Neumann's axiom is a "very bad reason" (exactly and 100% as you put it) to reject the notion that you can measure both momentum and position simultaneously to arbitrary precision (which, whether you acknowledge it or not, is contradictory to the HUP). To me, the von Neumann axiom is very intuitive considering the above explanation that I made. Moreover, this can be readily extended to any other observable.

 

[Cthugha]

This is trivial. There's no reason to believe one cannot prepare the same photon over and over again.

Correct. However, there is also no reason to believe that one can prepare the same photon over and over again.

Now you can argue the pattern on the wall (where we see the interference pattern) is a position measurement, but that would be silly, because you cannot tell which photon on that wall is which photon before the slits.

How is that silly? It is not required that you can tell which photon at the wall is which photon before the slit to perform a position measurement. The concept that every measurement is a position measurement after all, is widely accepted and also central to several interpretations of qm. It is also one of the reason why Bohmian mechanics works and is equivalent to standard qm as here particle positions are the central well-defined quantities of interest.

 

[Fredrik]

I'm tired of arguing with you about semantics. It seems to me you keep using semantics to cover your mistakes in physics.

I'm sure it seems that way to you, but if it hadn't been for your constant misreads and misinterpretations of plain English, you wouldn't have come to the incorrect conclusion that I'm making elementary physics mistakes. This entire discussion (which by the way is more insulting than anything I've seen in my seven years here) is a result of those mistakes.

It's certainly possible that some of the things I've said in this thread are wrong, especially those statements that start with "it seems to me...", "I believe...", or something like that. But your attacks on me have mostly been focused on two things: a) I used the word "measurement" to refer to what measuring devices do, instead of a projection operator. b) I interpreted the words "has something to do with" the way everyone but you does.

There's no reason to believe one cannot prepare the same photon over and over again.

I know.

Moreover, you need this in order to find out in any measurement process what the state of your system on which you're measuring the observable is. Your measurement is a probability value, and over many measurements you can reconstruct your state.

Agreed. If you want to find out what the state is before the measurement, you need to do something like this. I'm just saying that that's not what I'm interested in. You said earlier that interferometers can be used to measure momentum, but so far you have only described a way to determine what wavefunction is produced by a state preparation procedure that's already known to give us a momentum eigenstate. So I'm asking you now, do you know a way to use interferometry to measure the momentum of a single photon that's in an unknown state (not necessarily a momentum eigenstate) before the measurement?

I explained in the post how you can make the detector arbitrarily low in resolution, yet this will have no effect on your momentum measurement. You can make the position measurement uncertainty arbitrarily large, but your momentum measurement will always have machine resolution. So please tell me again, how are you inferring momentum from position here?

You can't make the detector arbitrarily large, because then it would cover the region of space where the other beam emerges. You need to keep the resolution high enough to distinguish between those two positions. As long as the setup can distinguish between two positions, it's a position measurement.

 

[fuesiker]

Correct. However, there is also no reason to believe that one can prepare the same photon over and over again.

Yes there is, it's called technology. The sky does it, preparing photons with the same polarized state with the same frequency (blue) at a 90-degree angle with the sun. Surely we can do it, and much better.

How is that silly? It is not required that you can tell which photon at the wall is which photon before the slit to perform a position measurement. The concept that every measurement is a position measurement after all, is widely accepted and also central to several interpretations of qm. It is also one of the reason why Bohmian mechanics works and is equivalent to standard qm as here particle positions are the central well-defined quantities of interest.

I don't know much about Bohmian mechanics except that it is experimentally very difficult to verify in many aspects of its features. However, back to our story, you seem not to understand the whole point about particle-wave duality. If you are able to tell which photon before the slit hit which position after the slit, your interference pattern is gone. This is akin to trying to detect which slit the photon went through. To do so, you must place a detector at one slit, which automatically kills your interference pattern since one slit is blocked.

 

[SpectraCat]

I'm tired of arguing with you about semantics. It seems to me you keep using semantics to cover your mistakes in physics.
This is trivial. There's no reason to believe one cannot prepare the same photon over and over again. Moreover, you need this in order to find out in any measurement process what the state of your system on which you're measuring the observable is. Your measurement is a probability value, and over many measurements you can reconstruct your state.
As mentioned before, read my post again. I never said you have two detectors.

Yes you did, because you specified an MZ interferometer, which uses two detectors .. you can ignore one if you want, but then you are not really doing interferometry.

Moreover, we do not know what the result will be a priori. Moreover, all you need is one detector, and the photon will always collapse onto it.

Please explain how you can make an MZ interferometer work in such a fashion. Remember an MZ interferometer is what *you* specified in setting up your example.

All you know is that you have a photon that you can prepare over and over again (which is what is done in all experiments that do projective measurements to determine a state), and you then measure its momentum to machine precision without inferring anything from its position.

That is simply not true .. you do NOT measure the momentum "of the photon". You measure clicks. You then repeat the experiment with different settings of the interferometer and *infer* the momentum of the photons from the differences in the "click measurements" (which are position sensitive) at different settings. This of course requires the additional step of abstraction of assuming that all of the photons have the same wavelength and initial phase.

Also, to counter your argument about measuring position (or your implicit belief that measuring position in this experiment "has something to do" with measuring momentum), I explained in the post how you can make the detector arbitrarily low in resolution, yet this will have no effect on your momentum measurement. You can make the position measurement uncertainty arbitrarily large, but your momentum measurement will always have machine resolution. So please tell me again, how are you inferring momentum from position here?

The photon collapses to a point on the detector, whether or not you have high or low spatial resolution on the detector. If this were NOT true, then the results of interferometry experiments would change when position sensitive detectors were used ... this is not the case as far as I am aware. The fact is that the measurements *are* position-sensitive, whether or not you care to take advantage of that fact in your analysis.

Think about one monochromatic wave of light instead of one photon, and you are passing it through a double-slit configuration. From the interference pattern, you will get a measurement of the momentum of that beam, without inferring anything about position. Now you can argue the pattern on the wall (where we see the interference pattern) is a position measurement, but that would be silly, because you cannot tell which photon on that wall is which photon before the slits.

The analogy does not hold, because this entire discussion is about measuring the properties of individual particles, NOT ensembles of particles.

 

[fuesiker]

I'm sure it seems that way to you, but if it hadn't been for your constant misreads and misinterpretations of plain English, you wouldn't have come to the incorrect conclusion that I'm making elementary physics mistakes. This entire discussion (which by the way is more insulting than anything I've seen in my seven years here) is a result of those mistakes.

It's certainly possible that some of the things I've said in this thread are wrong, especially those statements that start with "it seems to me...", "I believe...", or something like that. But your attacks on me have mostly been focused on two things: a) I used the word "measurement" to refer to what measuring devices do, instead of a projection operator. b) I interpreted the words "has something to do with" the way everyone but you does.

Fredrik, I'm sorry if I have insulted you. I clearly stated before I have respect for you. I don't dislike you and I don't mean to offend you. Let's move on.

However, I'm not the only one who thinks you have made statements at odds with standard QM. And it's very improbable that more than one person on here have "reading comprehension problems" as you accuse me of, but rather it may be that you're not wording yourself 100% correctly all the time, which many of us here do anyway.

Agreed. If you want to find out what the state is before the measurement, you need to do something like this. I'm just saying that that's not what I'm interested in. You said earlier that interferometers can be used to measure momentum, but so far you have only described a way to determine what wavefunction is produced by a state preparation procedure that's already known to give us a momentum eigenstate. So I'm asking you now, do you know a way to use interferometry to measure the momentum of a single photon that's in an unknown state (not necessarily a momentum eigenstate) before the measurement?

You seem to be confusing this. You automatically assume, like many people mistakenly do, that a photon has a fixed definite frequency, and that is not true because it would violate the uncertainty relation \Delta t \Delta E \geq h[itex (to some positive nonzero real constant). If you frequency is fixed (i.e. your wavelength is too, and hence, your momentum, meaning your photon is in a momentum eigenstate), that means your photon is an infinitely long packet, which is nonsense. So there is a spread of frequencies, and therefore you&#039;re in a superposition of momentum eigenstates. However, for all practical purposes, you can consider it classically as a single-frequency photon to understand what goes on from a wave theory perspective.<br /> <br /> Now let&#039;s go to what you have in mind (or at least what I think you do), and this is a multi-chromatic wave, like the usual illumination you have on the street. Of course you can&#039;t measure momentum then, because you are working with different non-interacting systems here (those being all those different single photons with different frequency spreads). This is a different story than the above. You want to measure only one quantum system, not many non-interacting ones. The thing is, the photon always has a spread in its position and in its momentum, and those collapse upon measurement. This is different from having five photons with different position and momenta.<br /> <br /> <blockquote data-attributes="" data-quote="Fredrik" data-source="post: 3432443" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Fredrik said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> You can&#039;t make the detector arbitrarily large, because then it would cover the region of space where the <i>other</i> beam emerges. You need to keep the resolution high enough to distinguish between those two positions. As long as the setup can distinguish between two positions, it&#039;s a position measurement. </div> </div> </blockquote><br /> You can make the arms infinitely large. There is no limit on the arm length. But that&#039;s not the point. And you seem to have again ignored my detailed explanation above, so I won&#039;t waste my time.

 

[Cthugha]

Yes there is, it's called technology. The sky does it, preparing photons with the same polarized state with the same frequency (blue) at a 90-degree angle with the sun. Surely we can do it, and much better.

What does this have to do with preparing the same photon? To be the same, every essential property of the photon would have to be exactly the same in every repeated preparation step. However, you will find that the properties will differ as governed by Heisenberg's uncertainty principle. There is no way to test whether it is that property of the particle that is not well defined and each prepared photon is therefore the same or whether it is in principle impossible to prepare these properties to a precision better than governed by the uncertainty principle, but the properties are in principle well defined. There is no experiment that could distinguish between these two scenarios.

I don't know much about Bohmian mechanics except that it is experimentally very difficult to verify in many aspects of its features.

It is hard to verify or falsify because its predictions are fully equivalent to standard qm. That also means that this kind of interpretation is perfectly valid (although I personally dislike it) and so is the Bohmian interpretation of the uncertainty principle in which it is indeed a consequence of being unable to know the initial conditions exactly (although I personally also dislike that interpretation). Nevertheless it is a tenable position.

However, back to our story, you seem not to understand the whole point about particle-wave duality. If you are able to tell which photon before the slit hit which position after the slit, your interference pattern is gone. This is akin to trying to detect which slit the photon went through. To do so, you must place a detector at one slit, which automatically kills your interference pattern since one slit is blocked.

Of course I understand that, but it has once again nothing to do with the question at hand. The question was, whether the measurement mentioned (or in principle any measurement) can be interpreted as "a" position measurement, not whether it is a measurement of the position of some photon before or at some slit of a double slit or similar. It is of course not a measurement of the photon position at the slit, but I did not claim that. Nevertheless it is a position measurement.

 

[Fredrik]

This is what I mean that you ignore my rebuttals that you can't refute. I will say it here again, Fredrik.
...

This is not something that I need to refute, because this argument is still as irrelevant as the first three times (at least) you posted it. The first few times, I was baffled as to what your mistake might be, but now that you have explicitly demonstrated that you don't know how to negate a statement (posts 130 and 137), I'm fairly certain that I understand what you're doing wrong. You actually think that the negation of

"All measurements project the state vector onto an eigenspace of the measured observable."​

is

"No measurement projects the state vector onto an eigenspace of the measured observable."​

It's not. The correct negation is

"There exists a measurement that doesn't project the state vector onto an eigenspace of the measured observable."​

No matter how many examples you give of experiments where the system ends up being projected onto an eigenspace, it can't possibly refute the last statement, which is what you need to refute to prove me wrong. I can certainly admit that you have refuted the middle statement at least four times. Do it a fifth time if you want to, but stop claiming that doing so proves me wrong.

You have had a few days to think. Didn't it at some point occur to you that there might be something wrong with your logic?

Now, in your above quote, you seem to imply that von Neumann's axiom is a "very bad reason" (exactly and 100% as you put it) to reject the notion that you can measure both momentum and position simultaneously to arbitrary precision (which, whether you acknowledge it or not, is contradictory to the HUP).

von Neumann's axiom obviously implies that you can't measure both at the same time. (It does so by disallowing us from calling what measuring devices do "measurements"). What I said is that the desire to preserve his axiom in its original form is a very bad reason to refuse to call what measuring devices do "measurements".

 

[DrChinese]

If you are able to tell which photon before the slit hit which position after the slit, your interference pattern is gone. This is akin to trying to detect which slit the photon went through. To do so, you must place a detector at one slit, which automatically kills your interference pattern since one slit is blocked.

Not sure this is a good analogy, you do not have to block the slit to get that information. You could place polarizers in front of each slit instead. As you vary their relative orientation, the interference pattern appears or disappears.

 

[Fredrik]

Fredrik, I'm sorry if I have insulted you. I clearly stated before I have respect for you.

Yes, but even that statement was insulting.

Moreover, despite what you may think, I have respect for you. In the end, you do seem to love physics, though yourself more.

 

[fuesiker]

Yes you did, because you specified an MZ interferometer, which uses two detectors .. you can ignore one if you want, but then you are not really doing interferometry.



Please explain how you can make an MZ interferometer work in such a fashion. Remember an MZ interferometer is what *you* specified in setting up your example.

It is not written in law that the MZ interferometer has to use two detectors. One is enough. Read below.

That is simply not true .. you do NOT measure the momentum "of the photon". You measure clicks. You then repeat the experiment with different settings of the interferometer and *infer* the momentum of the photons from the differences in the "click measurements" (which are position sensitive) at different settings. This of course requires the additional step of abstraction of assuming that all of the photons have the same wavelength and initial phase.

The photon collapses to a point on the detector, whether or not you have high or low spatial resolution on the detector. If this were NOT true, then the results of interferometry experiments would change when position sensitive detectors were used ... this is not the case as far as I am aware. The fact is that the measurements *are* position-sensitive, whether or not you care to take advantage of that fact in your analysis.

Like I mentioned, this is experimentally really, really well established that you can prepare photons in the same polarization state and with the same frequency over and over again. As I also mentioned before, the sky does it. It's that simple.

Moreover, what is a point to you? You seem to not understand the idea of pixels. You can make your pixels REALLY huge such that on your screen one of those HUGE pixels light up when the photon hits. This is the same as what happens in your cell phone camera, that's the whole point of resolution. I said we can intentionally make those pixels huge in order to see there is no relationship between the uncertainty in position and the measurement result you are getting that would allow one to say we're inferring momentum from position.

Again, the detector is there to tell you that something hit. You always need to use two detectors when using classical beams for example because then there are no probabilities. With one photon, you use one detector, and this will ensure your photon collapse in its position on that detector, and no other detector, but the point you're failing to get is that we simply don't care. Because, like I mentioned, the uncertainty in position can be made arbitrarily large, and this will not change the value of momentum you're measuring. So tell me then, how are we inferring momentum from position. You seem to be getting or have gotten your PhD in chemical physics, so this shouldn't be so hard to get. Make your pixel really low in resolution, such that if it's 25 square meters of 25 square nanometers, all it would tell you is that the photon has hit it, no less, no more. The uncertainty in position here is equivalent to the area of the detector surface, which is just a one-pixel detector, let's say (and you can build this). Whichever area you have, you will get the same result. No clicks at a certain \Delta l = l_2-l_1, whichever detector you use. Hence, one has to be really mathematically impaired to still think we are inferring momentum from position here.

The analogy does not hold, because this entire discussion is about measuring the properties of individual particles, NOT ensembles of particles.

Usually when I discuss physics, I don't always talk about single photons. But in fact, this will hold for a single photon because if you try to detect the photon in the MZ inteferometer for example, before the arms cross, you will always get a reading that your photon hit that detector (again you have one detector but this is before the arms cross so it is not like our detector above which comes after the arms have crossed). Then no matter what \Delta l is, you will always here a click on the detector that is before the arms crossing. Remove that, then you will hear or not hear click at the detector we originally had after the arms crossing depending on \Delta l.

You seem to not understand wave-particle duality so well. Again, I suggest Loudon's book.

 

[fuesiker]

Yes, but even that statement was insulting.

Please try to not comment on non-physics issues. Fredrik, come on, I was being totally nice.

As a last note, I have said all I said in this thread, and thank you all for a nice discussion that showed me why I love quantum mechanics so much. This is my last entry on here. All of you take care and have fun.

 

[Fredrik]

You automatically assume, like many people mistakenly do, that a photon has a fixed definite frequency,

No, I don't.

OK, based on what you're saying here and in several other posts, I have to ask: Do you think that in order to measure an observable, the system must be in an eigenstate of that observable before the measurement begins? (That is, before the interaction between the system and the measuring device becomes non-negligible).

 

[atyy]

von Neumann's axiom obviously implies that you can't measure both at the same time. (It does so by disallowing us from calling what measuring devices do "measurements"). What I said is that the desire to preserve his axiom in its original form is a very bad reason to refuse to call what measuring devices do "measurements".

So you agree that if we define measurement in the standard sense, then quantum mechanics forbids simultaneous sharp measurements of canonically conjugate observables? (For the purpose of this discussion, we can separate the projection postulate out, and define a sharp measurement of O to be one that when repeated on an ensemble of identically prepared states gives a distribution of measured values such that the expectation value is <state|O'|state>, for all observables O' that commute with O. In particular, a sharp measurement gives a sharp distribution if the state is an eigenstate of O).

 

[SpectraCat]

It is not written in law that the MZ interferometer has to use two detectors. One is enough. Read below.

I am aware that single-detector configurations can be used for MZ-interferometers for many applications, but it is so clear to me how such an interferometer can be used to analyze single photon experiments, where you need to have a result for every photon passing through the apparatus. You can of course describe *average* behavior of the photons by measuring click rates, but that is not really relevant to what we are discussing.

Please describe how a one-detector MZ interferometer can be used for single photon experiments in more detail. On which side of the second beam-splitter do you choose to place your detector? What mechanism do you use to know that a photon was emitted, yet failed to register a click on the single detector you do have? How do you extract the momentum information from your single-detector data? Most of the time, your detector will fail to click .. how will you decide in which time-intervals the failure to click was significant?

Like I mentioned, this is experimentally really, really well established that you can prepare photons in the same polarization state and with the same frequency over and over again. As I also mentioned before, the sky does it. It's that simple.

I have no idea what you mean by that last comment about the sky, but let's look at your other statement. I am pretty sure you don't mean what you said. You can prepare photons which are sampled from a particular frequency distribution, which can be made to have a narrow, but still finite, width. A similar statement holds for polarization. These photons cannot however be said to each have precisely the same frequency. I understand and accept your point however, that we can assume fair sampling from that narrow frequency distribution, and thus analyze the momentum of the photons. However, the point you are missing is that the information about the momentum comes from an ensemble of measurements where you measured the position of a photon (more on that below), using different instrumental settings. No information at all about the momentum of the photon is contained in any single measurement, and THAT is the sort of experiment we are discussing in this thread. One particle, one measurement. My statement still stands that there is no way to obtain information about momentum of a particle from such a measurement, without considering other information and then using it to *infer* the past momentum of the particle (I agree with jtbell's language on this point).

Moreover, what is a point to you? You seem to not understand the idea of pixels. You can make your pixels REALLY huge such that on your screen one of those HUGE pixels light up when the photon hits. This is the same as what happens in your cell phone camera, that's the whole point of resolution. I said we can intentionally make those pixels huge in order to see there is no relationship between the uncertainty in position and the measurement result you are getting that would allow one to say we're inferring momentum from position.

I understand just fine ... you seem to be missing the point that resolution is completely irrelevant to this example. If you use a detector that is just one big pixel, you get precisely the same results as if you use a high-resolution CCD for your detector, assuming that you do not overlap another spatial region where photons traveling along a different path can also be detected. Do you agree with that statement? Assuming that you do, do you not see how this shows that you are in fact measuring position when you register the click? If you were not, then using a high-resolution detector would change the results of your interferogram, and I am almost 100% certain that it does not.

Again, the detector is there to tell you that something hit. You always need to use two detectors when using classical beams for example because then there are no probabilities. With one photon, you use one detector, and this will ensure your photon collapse in its position on that detector, and no other detector, but the point you're failing to get is that we simply don't care. Because, like I mentioned, the uncertainty in position can be made arbitrarily large, and this will not change the value of momentum you're measuring.

I agree, but you interpretation of the significance of that information is exactly backwards, as I explained above.

So tell me then, how are we inferring momentum from position. You seem to be getting or have gotten your PhD in chemical physics, so this shouldn't be so hard to get.

You really need to quit it with the condescending side remarks about how your points should be "obvious" based on other people's level of education. For one thing, what if your "obvious" point is wrong, or irrelevant to the argument at hand (as in this case)? Then your choice to make personal remarks just makes you look all the more foolish, and irritates people who are just trying to have a scientific discourse.

Make your pixel really low in resolution, such that if it's 25 square meters of 25 square nanometers, all it would tell you is that the photon has hit it, no less, no more. The uncertainty in position here is equivalent to the area of the detector surface, which is just a one-pixel detector, let's say (and you can build this). Whichever area you have, you will get the same result. No clicks at a certain \Delta l = l_2-l_1, whichever detector you use. Hence, one has to be really mathematically impaired to still think we are inferring momentum from position here.

As you yourself pointed out (correctly), measurement precision is completely irrelevant to the HUP. The HUP defines the intrinsic limit on the relationship between the widths of the momentum and position distributions associated with a quantum state. Simply choosing to ignore positional resolution when it is available cannot change the results of an experiment .. unless of course you change the fundamental nature of the experiment by doing so. As long as only one photon-path intersects the detector, then its size and or resolution are completely irrelevant to our discussion.

Usually when I discuss physics, I don't always talk about single photons. But in fact, this will hold for a single photon because if you try to detect the photon in the MZ inteferometer for example, before the arms cross, you will always get a reading that your photon hit that detector (again you have one detector but this is before the arms cross so it is not like our detector above which comes after the arms have crossed). Then no matter what \Delta l is, you will always here a click on the detector that is before the arms crossing. Remove that, then you will hear or not hear click at the detector we originally had after the arms crossing depending on \Delta l.

Are you still talking about some large detector that intersects both paths? Because otherwise you need two detectors .. one for each arm, to ensure that a click is always registered (and only one detector will click for any given photon). This is because of what I have been saying all along .. registering a click on a detector requires localization of the photon .. i.e. detection of its particle-like properties. This is commonly called "which-path" information .. if you put a detector in one of the arms, it will click half the time. If you put detectors in both arms, then one or the other will click .. the photon can never be directly observed taking both paths through the interferometer. You can only *infer* that it did from the interference pattern that is recorded after the second beam-splitter, when you do not look at which-path information.

If I got it wrong, and you *were* talking about having a large detector that intersects both arms, then your comments above seem trivial and I don't understand their significance at all.

You seem to not understand wave-particle duality so well. Again, I suggest Loudon's book.

I have just about had it with your statements about what you think other people do and don't understand. You have no idea who you are talking to when you post on these threads, and your incredibly arrogant asides serve no purpose than to antagonize the other participants of this thread. Please confine your comments to the discussion at hand.

 

[fuesiker]

No, I don't.

OK, based on what you're saying here and in several other posts, I have to ask: Do you think that in order to measure an observable, the system must be in an eigenstate of that observable before the measurement begins? (That is, before the interaction between the system and the measuring device becomes non-negligible).

Again, you deny the obvious. And of course not, I don't believe the system has to be in an eigenstate of the observable before measuring that observable begins. It is laughable you concluded this from any of my posts. I clearly said many times the state COLLAPSES onto an eigenstate upon measurement. Hence I obviously assumed the state to be in general a superposition of eigenstates of the observable.

 

[Fredrik]

So you agree that if we define measurement in the standard sense, then quantum mechanics forbids simultaneous sharp measurements of canonically conjugate observables?

I don't agree that it's standard to define "measurement" as "projection onto an eigenspace", or at least not that it's the only standard. But I certainly agree that the wavefunction and its Fourier transform can't both be sharply peaked.

(For the purpose of this discussion, we can separate the projection postulate out, and define a sharp measurement of O to be one that when repeated on an ensemble of identically prepared states gives a distribution of measured values such that the expectation value is <state|O'|state>, for all observables O' that commute with O. In particular, a sharp measurement gives a sharp distribution if the state is an eigenstate of O).

I'm not sure I understand this. You didn't say that O is position, so I wonder if you meant that O is arbitrary. Can it be momentum? In that case, doesn't the momentum "inference" in Ballentine's thought experiment satisfy this definition, at least approximately? It seems to me that it does, but I might have misunderstood you. If it does, then this definition means that you can make sharp measurements of position and momentum at the same time, in contradiction with what I believed that you were referring to as "standard".

 

[Fredrik]

Again, you deny the obvious.

I can't tell if you're just lying to try to discredit me or if you honestly believe it. I don't really care at this point.

And of course not, I don't believe the system has to be in an eigenstate of the observable before measuring that observable begins.
...
I clearly said many times the state COLLAPSES onto an eigenstate upon measurement. Hence I obviously assumed the state to be in general a superposition of eigenstates of the observable.

Good for you.

It is laughable you concluded this from any of my posts.

And I'm sure it's not laughable that you somehow concluded that I think that every photon has a well-defined frequency (which implies a well-defined momentum) just because I asked if you know a way to perform a momentum measurement on a photon which is not necessarily in a momentum eigenstate.

 

[atyy]

I don't agree that it's standard to define "measurement" as "projection onto an eigenspace", or at least not that it's the only standard. But I certainly agree that the wavefunction and its Fourier transform can't both be sharply peaked.

I'm not sure I understand this. You didn't say that O is position, so I wonder if you meant that O is arbitrary. Can it be momentum? In that case, doesn't the momentum "inference" in Ballentine's thought experiment satisfy this definition, at least approximately? It seems to me that it does, but I might have misunderstood you. If it does, then this definition means that you can make sharp measurements of position and momentum at the same time, in contradiction with what I believed that you were referring to as "standard".

Yes, including that O can be momentum. To take care of Ballentine's objection, let me try adding "for unknown arbitrary states", so how about:

Quantum mechanics forbids simultaneous sharp measurements of canonically conjugate observables upon unknown arbitrary states. A sharp measurement of O is one that when repeated on an ensemble of identically prepared states gives a distribution of measured values such that the expectation value is <state|O'|state>, for all observables O' that commute with O. In particular, a sharp measurement gives a sharp distribution if the state is an eigenstate of O.

 

[LostConjugate]

It is the measurements that do not commute.

A measurement of position is taken at a point in time and a measurement of momentum is taken over a period of time. Only in the special case where the particle's momentum is zero for the entire period of time will the two measurements commute.

 

[SpectraCat]

It is the measurements that do not commute.

A measurement of position is taken at a point in time and a measurement of momentum is taken over a period of time. Only in the special case where the particle's momentum is zero for the entire period of time will the two measurements commute.

This is an interesting point, but seems to employ the same vague definition for "measurement". Can you be a little more specific? What is the character of the momentum measurement that you are talking about? Does it involve a time-ordered sequence of position measurements, or something else? If it is the former, then what is the source of the non-commutative property of the "measurement"? Is it that individual position measurements taken at different times don't commute (I don't see why that would be so, but perhaps I am missing something), or is it that the time-ordered sequence that doesn't commute with a single position measurement?

 

[atyy]

I don't agree that it's standard to define "measurement" as "projection onto an eigenspace", or at least not that it's the only standard. But I certainly agree that the wavefunction and its Fourier transform can't both be sharply peaked. I'm not sure I understand this. You didn't say that O is position, so I wonder if you meant that O is arbitrary. Can it be momentum? In that case, doesn't the momentum "inference" in Ballentine's thought experiment satisfy this definition, at least approximately? It seems to me that it does, but I might have misunderstood you. If it does, then this definition means that you can make sharp measurements of position and momentum at the same time, in contradiction with what I believed that you were referring to as "standard".

BTW, is known whether Ballentine's method recovers the exact expectation values for all observables that commute with Py (ie. is Ballentine's method a sharp measurement of Py that is canonically conjugate to Y)?

 

[LostConjugate]

This is an interesting point, but seems to employ the same vague definition for "measurement". Can you be a little more specific? What is the character of the momentum measurement that you are talking about? Does it involve a time-ordered sequence of position measurements, or something else? If it is the former, then what is the source of the non-commutative property of the "measurement"? Is it that individual position measurements taken at different times don't commute (I don't see why that would be so, but perhaps I am missing something), or is it that the time-ordered sequence that doesn't commute with a single position measurement?

The latter. In order to have a value of momentum measured you must have multiple positions measured which requires measuring over a period of time.

Now if you try to say that your object has an exact position and an exact momentum what you have just said is that it has an exact position over the period of time that you measured the momentum. That is simply not true. The measurements do not commute.

 

[romsofia]

I would be interested in seeing that page number reference. This discussion (or whatever I should call it) has made me curious about the terminology in standard textbooks.

Sure thing.

On page 112 he mentions that "Only if the wavefunction were simultaneously an eigenstate of both observables would it be possible to make the second measurement without disturbing the state of the particle (the second collapse wouldn't change anything in that case). But this is only possible, in general, if the two observables are compatible."

 

[Fra]

This turned out a heated discussion. There is little doubt that probably everyone here understands what the HUP really means, ie. it refers to standard deviations and is thus probabilistic to it's nature.

The rest of the discussion seems to me to be interpretationally rooted. Not entirely unexpectedly, the ensemble interpretations indeed seems to struggle with the ontological meaning of "single datapoints".

Are single datapoints, more than just datapoints? Personally I stronlgy object to the soundness in thinking of single datapoints (single detector counts) in terms of PARTICLE hits. That whole thing has a realist flavour to it that I do not prefer. No matter what is sometimes practice, I am extremely biased myself towards my own thinking but I think of detector counts simply as "evidence counts". And a single evidence counts always comes with finite confidence.

Anyway, the way I see it I still think that the non-commutativity has nothing to do with how the body of evidence (information state) is acquired. It's a feature of the internal state of the information state. This is why I personally prefer to think of it in terms of a constraint on the information state, rather than something that has anything to do with relations between single data points.

Also the information state; although a function of interaction history, encodes nothing but the expectations of the future. Ballentines example does not seem to have much physical relevance as basis for decisions. And in my crazy view, a theory should not postdict a recorded past; it should - provide rational expectations of the future based on the past. This is the survival value of a theory.

But probably the discriminator between all these views will be when we see which of these abstractions that are most fit to solve the remaining open question in physics regarding unification etc. Somehow, some of this things that are here almost purely intepretational and seemingly meaningless, I think will be more pronounced when situations such as comosological models yields abstractions such as "ensembles" at least IMHO unviable, and we do need a different way to understand "single instances", without embedding them into ensembles that can't be realized.

/Fredrik

 

[Fredrik]

On page 112 he mentions that "Only if the wavefunction were simultaneously an eigenstate of both observables would it be possible to make the second measurement without disturbing the state of the particle (the second collapse wouldn't change anything in that case). But this is only possible, in general, if the two observables are compatible."

What he's saying here is that you can't for example first measure position and then measure momentum without disturbing the system. That's not the sort of thing that most of this discussion has been about, but thank you for posting this text so that I could see that for myself. We're talking about detecting a particle that was prepared in a state with sharply defined position, and then interpreting this as a simultaneous measurement of position and momentum.

 

[Fredrik]

Yes, including that O can be momentum. To take care of Ballentine's objection, let me try adding "for unknown arbitrary states", so how about:

Quantum mechanics forbids simultaneous sharp measurements of canonically conjugate observables upon unknown arbitrary states. A sharp measurement of O is one that when repeated on an ensemble of identically prepared states gives a distribution of measured values such that the expectation value is <state|O'|state>, for all observables O' that commute with O. In particular, a sharp measurement gives a sharp distribution if the state is an eigenstate of O.

I still don't understand. It doesn't look like you have changed the definition. You just preceded it by a statement that, as far as I can tell, contradicts the definition.

BTW, is known whether Ballentine's method recovers the exact expectation values for all observables that commute with Py (ie. is Ballentine's method a sharp measurement of Py that is canonically conjugate to Y)?

I don't think any kind of "measurement" satisfies a requirement that strong. Wouldn't it have to be infinitely accurate? Anyway, I suppose you would also be interested in the answer to a related question: If we turn Ballentine's thought experiment into an actual experiment, and perform this momentum "inference" over and over on identically prepared systems, how accurately will the distribution of results agree with the values of |u(p)|^2 where u is the Fourier transform of the wavefunction \psi.

Unfortunately I don't know the answer. Let's face it, this is a part of QM that we all suck at, because it's not covered in books. As far as I know, none of the standard books define what sort of device to call a "momentum measuring device". So I can only say that I expect that the distribution of "inferred" values would agree very well with the predicted distribution, but not exactly. The discrepancy doesn't really have anything to do with QM. The problem is with the definition of a "momentum measuring device". I don't think it's possible to even describe a hypothetical measuring device that would get results that are distributed exactly as predicted, at least not if the description doesn't involve some limit that isn't possible to actually take.

Think e.g. about length measurements in SR. I would define a length measurement by describing a radar device with a clock that measures the roundtrip time and multiplies it with c/2. But unless this device is doing inertial motion, it will fail to measure the length of the spacelike curve it's designed to measure. The measurement is however exact in the limit when the size of the device goes to zero.

 

[Ben Niehoff]

I am jumping in here after having not read most of the thread, so forgive me if this has already been mentioned or is slightly tangential to the current discussion.

To construct a quantum system (in the mathematical sense), we take a classical system and follow a certain prescription on phase space that promotes the coordinates on phase space (i.e., the position and momentum) to operators which do not necessarily commute, turning the phase space geometry into a noncommutative one. The commutators of the quantum operators are set equal to their classical Poisson brackets (modulo some factors of i \hbar). The Hamiltonion operator generates time translations in the same way it did in classical mechanics, but with the Poisson bracket replaced by the commutator. All of these operators are taken to act linearly on some abstract "state space" with inner product; this inner product gives a notion of "conservation of total probability" under the action of unitary operators.

One way to view the HUP is as the mathematical consequence of non-commuting observables (i.e. operators with real eigenvalues). Since an observation consists of picking out some eigenvalue of the operator in question, thereby changing the state to the corresponding eigenstate. Two operators that cannot be simultaneously diagonalized do not have the same set of eigenvectors (this is a basic fact of linear algebra), and operators can be simultaneously diagonalized precisely when they commute. Hence if [A,B] \neq 0, a state which is sharply peaked near an eigenvalue of A (i.e., close to an eigenstate of A), cannot also be sharply peaked near an eigenvalue of B, since A and B do not share eigenstates. One can mathematically derive the relationship between the variances of A, B, and [A,B] directly from linear algebra.

Another way to explain the HUP, however, is from merely looking at the Schrodinger equation. In classical mechanics, the equations of motion typically have two time derivatives; the Schrodinger equation, however, has only one. As you well know from basic differential equations, you may specify as many initial conditions as you have derivatives. In classical mechanics, we have two time derivatives, and hence we can specify two initial conditions: position and velocity. But in quantum mechanics, we have only one time derivative. Hence position and velocity cannot be independently specified. They are not independent quantities, and so it should not be too surprising that they cannot be measured to arbitrary accuracy at the same time. In the Schrodinger picture, "position" and "momentum" are measurements we make by taking certain weighted averages over a wavefunction; one can show in this case that position and momentum are complementary variables, in the Fourier transform sense, in which case the mathematical statement of the HUP comes from signal analysis.

In either case, it is the state itself which cannot contain sharply-peaked information about position and momentum simultaneously. So the HUP is not a a statement about our measurement ability; it is a statement about what information actually exists.

It is worth noting that one can construct states of minimal uncertainty; i.e. wave functions for which the inequality in the HUP becomes an equality. These are wave packets that resemble our classical notion of "particles". They have a position and a momentum that are specified "pretty well", but not perfectly. These wavefunctions look like gaussians in both position and momentum space.