The Uncertainty Principle Uncertainty

[UncertaintyAjay]

My friend and I had this argument about whether or not the uncertainty principle is applicable to stationary particles. I maintain that it is, because the principle is really about predictability ( isn't it?) But he maintains that it doesn't. So I would just like to clear things up . Does it or doesn't it? Thanks in advance.

 

[phinds]

My friend and I had this argument about whether or not the uncertainty principle is applicable to stationary particles. I maintain that it is, because the principle is really about predictability ( isn't it?) But he maintains that it doesn't. So I would just like to clear things up . Does it or doesn't it? Thanks in advance.

"stationary" is not a meaningful term in this context. Motion is relative. There is no such thing as absolute motion in which something can be absolutely stationary. You have to say stationary RELATIVE TO WHAT.

For example, you, yourself, right now as you read this, are moving at .9999999c and are massively time dilated. Does that have any impact on you?

 

[atyy]

The uncertainty principle does apply to stationary particles (particles with zero average velocity).

 

[UncertaintyAjay]

You are absoloutely right. I do have to specify relative to what. I meant stationary relative to the observer.That also answers my question , so thanks.

 

[mfb]

The uncertainty principle applies to all objects and for all observers.

 

[UncertaintyAjay]

Right, thanks a lot

 

[jtbell]

whether or not the uncertainty principle is applicable to stationary particles

You can never say for sure that a particle's momentum is exactly zero. All you can ever do is say that $p = 0 \pm \Delta p$ for some value of $\Delta p$.

 

[bhobba]

In QM you can't know both position and momentum simultaneously ie there is no observation that gives both at the same time. Its is a misconception to believe QM forbids measuring either with as great an accuracy as you like, but what you can't do is measure them both at the same time.

That being the case a particle at rest in QM is not possible because that would mean knowing it position and momentum simultaneously - unless of course you mean by at rest zero momentum but you don't know where it is - that is possible - but I don't think that's what's usually meant in saying an object is at rest.

Thanks
Bill

 

[DaveC426913]

In fact, HUP is so applicable to particles that it means there is no such thing as A "stationary" particle.

Attempts have been made to bring atoms to a (relative) stop by reducing their temperature to 0K. Presumably, an atom at 0K will cease all motion, and thus its position and momentum (zero) could be precisely known.

Atoms aren't havin' any of that.

As atoms are reduced to 0K, and their motion stops, the atoms lose their individual identity and become "smeared out" into what's called a Bose-Einstein Condensate, nicely cooperating with HUP.

 

[UncertaintyAjay]

Yeah thanks

 

[moriheru]

The uncertainty principle is not only true for delta p and delta x thus postion and momentum, but it is far more general. The reason why we have h(bar)/2 is because the comutators of the X and P ,namely[X,P], are equivalent to -ih(bar). The general uncertainty principle is ΔAΔB>=<[A,B]>*1/2.

 

[exponent137]

How to describe UP of a photon? It is at location 5 m from us and its momentum is 10^-20 kg m/s.

Because, the simplest explanation of UP is Fourier transformation of a position wave function of a rest particle. This gives a momentum wave function. Square of absolute value of both of them are gaussian distributions and this gives thicknesses of these distributions and then product of thicknesses of both distributions.

How to describe gaussian wave function of the photon.

 

[moriheru]

In QM you can't know both position and momentum simultaneously ie there is no observation that gives both at the same time. Its is a misconception to believe QM forbids measuring either with as great an accuracy as you like, but what you can't do is measure them both at the same time.

That being the case a particle at rest in QM is not possible because that would mean knowing it position and momentum simultaneously - unless of course you mean by at rest zero momentum but you don't know where it is - that is possible - but I don't think that's what's usually meant in saying an object is at rest.

Thanks
Bill

Yes,If the the observables do not comute. Else the uncertainty is zero in the product of the observables.

 

[Dale]

In QM you can't know both position and momentum simultaneously ie there is no observation that gives both at the same time. Its is a misconception to believe QM forbids measuring either with as great an accuracy as you like, but what you can't do is measure them both at the same time.

I thought that it was more that you could not prepare a state where both were well defined, but you could measure them as accurately as you like.

 

[phinds]

I thought that it was more that you could not prepare a state where both were well defined, but you could measure them as accurately as you like.

Yes, that's my understanding as well, but there are strong proponents here on this forum for both points of view --- (1) you can measure both simultaneously to any degree of accuracy but the next measurement from the same setup won't give the same values, and (2) setup doesn't matter since you can't measure them simultaneously to arbitrary accuracy anyway. I tried once to get it pinned down but the thread I started didn't achieve that, just made it clear that there ARE two points of view here on this forum, so I've been unable to tell from this forum which it is.

 

[atyy]

I thought that it was more that you could not prepare a state where both were well defined, but you could measure them as accurately as you like.

Yes, that's my understanding as well, but there are strong proponents here on this forum for both points of view --- (1) you can measure both simultaneously to any degree of accuracy but the next measurement from the same setup won't give the same values, and (2) setup doesn't matter since you can't measure them simultaneously to arbitrary accuracy anyway. I tried once to get it pinned down but the thread I started didn't achieve that, just made it clear that there ARE two points of view here on this forum, so I've been unable to tell from this forum which it is.

ZapperZ and I had a discussion about this, and I believe there is only one point of view (I can't find the post now) - non-commuting observables cannot be simultaneously and accurately measured on an arbitrary unknown state. People often read ZapperZ's blog post as meaning that simultaneously accurate measurement of position and momentum is possible, but that is not what he means. What he means is that in a double slit experiment, if one puts the screen far away, and measures position accurately at the screen, then one can use that position result to accurately calculate (the "simple" method he uses gives the right answer, but one can also obtain it by a correct quantum mechanical calculation using the Fraunhofer limit), and thus to accurately measure the transverse momentum of the particle immediately after the slit. Thus a later accurate position measurement can be used to accurately measure an earlier, non-simultaneous, momentum. As the slit is narrowed, the initial position does become more certain, but the narrowing of the slit changes the wave function immediately after the slit, so it is not a more accurate measurement of "the same" wave function.

So to summarize:

(1) The textbook uncertainty relations do not refer to simultaneous measurement. They say that a state cannot have simultaneously well-defined position and momentum. This means that in accurate measurements of position on an ensemble of particles in a state and separate, non-simultaneous accurate measurements of momentum on a different ensemble of particles in the same state will yield position and momentum spreads that satisfy the textbook uncertainty relation.

(2) The textbook uncertainty relations are derived from non-commuting observables. It is not possible to simultaneously and accurately measure non-commuting observables on an arbitrary unknown state. There are exceptions if the state is not arbitrary and we know something about it. For example, if the state is known completely, we can simply write down simultaneous, accurate "measurement results". Also, if we know (for discrete variables) that the state is one of several eigenstates of an observable, then a measurement of that observable will leave the state undisturbed, so that the same state is still available for a non-commuting observable to be measured. Another famous special state is the EPR state that allows simultaneous measurement of position and momentum because of entanglement. (There's a "controversy" in the literature, but it's semantics about how one should define the "measurement error". Let's just say that for the various choices of "measurement error", I believe these papers are correct.)

http://arxiv.org/abs/1306.1565
Proof of Heisenberg's error-disturbance relation
Paul Busch, Pekka Lahti, Reinhard F. Werner

http://arxiv.org/abs/1304.2071
How well can one jointly measure two incompatible observables on a given quantum state?
Cyril Branciard

http://arxiv.org/abs/1212.2815
Correlations between detectors allow violation of the Heisenberg noise-disturbance principle for position and momentum measurements
Antonio Di Lorenzo

There are a couple of famous, earlier papers in the literature. Park and Margenau's 1968 paper is correct, because it shows the simultaneous measurement of conjugate position and momentum for a restricted, non-arbitary set of states. Ballentine's 1970 review is wrong, because position and momentum are not conjugate in his example.

 

[anorlunda]

My friend and I had this argument about whether or not the uncertainty principle is applicable to stationary particles.

The uncertainty principle is not only true for delta p and delta x thus postion and momentum, but it is far more general.

Moriheru directly answered the original question.

Just to be more specific, I would point out the example that energy-time as a form of uncertainty that has nothing to do with particle motion.

 

[UncertaintyAjay]

Right, its just that position- momentum uncertainty is a specific example.

 

[UncertaintyAjay]

Yes, that's my understanding as well, but there are strong proponents here on this forum for both points of view --- (1) you can measure both simultaneously to any degree of accuracy but the next measurement from the same setup won't give the same values, and (2) setup doesn't matter since you can't measure them simultaneously to arbitrary accuracy anyway. I tried once to get it pinned down but the thread I started didn't achieve that, just made it clear that there ARE two points of view here on this forum, so I've been unable to tell from this forum which it is.

Yeah, so, which is it, then? Just wondering, of the two viewpoints which one is correct?

 

[phinds]

Yeah, so, which is it, then? Just wondering, of the two viewpoints which one is correct?

If you are asking me, the I have to assume you did not understand the last sentence of the post you just quoted. If you're asking the rest of the world. read the posts in this thread. We continue to have both views, although atvy has put forth citations for one side of the argument.

 

[atyy]

Yeah, so, which is it, then? Just wondering, of the two viewpoints which one is correct?

See post #16.

 

[bhobba]

I thought that it was more that you could not prepare a state where both were well defined

That's true.

But given any state you can measure position as accurately as you like, you can measure momentum as accurately as you like, but there is no observation that will simultaneously allow you to measure both as accurately as you like.

Both views of course are related in that a filtering type measurement of either position or momentum is in fact a state preparation procedure.

Thanks
Bill

 

[Demystifier]

The uncertainty principle does apply to stationary particles (particles with zero average velocity).

In QM, stationary state us usually defined as a Hamiltonian eigenstate. It may have a non-zero average velocity.

 

[atyy]

In QM, stationary state us usually defined as a Hamiltonian eigenstate. It may have a non-zero average velocity.

Yes, I was just using a definition that I thought might be closer to what the OP had in mind. But of course, the definition doesn't really matter, since the uncertainty principle applies to all states.

 

[Godparicle]

If uncertainty in position is zero, uncertainty in momentum is infinity (and vice versa).

According to Measure theory, zero times infinity is zero, i.e you can't have uncertainty of either position or momentum to be equal to zero.

According to non-measure theories, zero times infinity is undefined, i.e you are not allowed to draw any conclusion.


Personally, I will wait for experimental result.

 

[bhobba]

If uncertainty in position is zero, uncertainty in momentum is infinity (and vice versa).

You are misinterpreting the uncertainty principle.

Its a statistical statement - not a statement about individual measurements. You can measure position or momentum to any degree of accuracy you like - and that can be for all practical purposes exact. Indeed the operators of the QM formalism assume they are. That obviously can't occur in practice which is why the associated wave-functions are not square integrable - or in the case of the Dirac Delta function - not even a function - but are useful theoretical fictions introduced for mathematical convenience.

Here is what it says. If you have a large number of similarly prepared systems and exactly measure the position in half of them, and exactly measure momentum in the other half, then the standard deviation of the statistical spread of the results will be as per the uncertainty principle.

Its a crucial point many, myself included, often are sloppy about, but what I said above is the correct statement. You can find a discussion on page 225 of Ballentine - QM - A Modern Development - where he discusses how these misconceptions may have arose - which he he thinks dates back to some early work by Heisenberg before the foundations of QM were better understood.

Thanks
Bill

 

[phinds]

...
I believe there is only one point of view (I can't find the post now) - non-commuting observables cannot be simultaneously and accurately measured on an arbitrary unknown state. People often read ZapperZ's blog post as meaning that simultaneously accurate measurement of position and momentum is possible, but that is not what he means ...

... You can measure position or momentum to any degree of accuracy you like - and that can be for all practical purposes exact. Indeed the operators of the QM formalism assume they are ...

Here we go again.

This is the kind of thing I was referring to in post #15.

 

[bhobba]

This is the kind of thing I was referring to in post #15.

I didn't say anything at the time, and its not what you are saying in your two options, but there seems to be some confusion about exact measurements. The formalism says you can measure them to any degree you like, and indeed exact measurements exist in it even though they are fictions in practice. That's the precise reason the Rigged Hilbert Space formalism is introduced so that such fictions exist in the mathematics. To be exact for a state with exact position you have a Dirac Delta function which is a load of bunkum as a function in the usual sense, and for a state with exact momentum you have the exponential phasor which extends to infinity in both directions and is not square integrable. But they are useful fictions mathematically as analysing many problems in applied math, not just QM, readily attests to - which is the precise reason the area, called distribution theory, is an extremely important part of the mathematical armoury of all applied mathematicians.

I have mentioned it a number of times in the past, but its so important I will mention it again. IMHO every applied mathematician should have a book in their library on it - my favourite is:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

Fourier transforms done that way is much easier than the usual approach and much more general - and shows exactly what's going on in QM in transforming from the position to momentum representation.

And if you don't allow, in principle, exact positions or momentum you can't interpret either representation.

Yes, that's my understanding as well, but there are strong proponents here on this forum for both points of view --- (1) you can measure both simultaneously to any degree of accuracy but the next measurement from the same setup won't give the same values, and (2) setup doesn't matter since you can't measure them simultaneously to arbitrary accuracy anyway. I tried once to get it pinned down but the thread I started didn't achieve that, just made it clear that there ARE two points of view here on this forum, so I've been unable to tell from this forum which it is.

What I said above is to do with exact measurements in the theory - they exist in principle, and the formalism accommodates it. You can't measure both simultaneously because no observable exists that does that so your first option is out.

Thanks
Bill

 

[atyy]

Here we go again.

This is the kind of thing I was referring to in post #15.

As far as I can tell, bhobba and I don't disagree. The quote you mentioned from bhobba does not say the measurements are simultaneous and accurate. My post #16 and bhobba's posts #22 and #26 are in agreement.

Edit: I might disagree with a tiny point in bhobba's post #26, where the reason he gives for perfectly accurate measurements of position in which the particles survive the measurement not being possible is that the eigenfunction of the position operator is not square integrable. It turns out that that there is a generalization of the projection postulate, and a Hamiltonian devised by Ozawa that will carry out a perfectly accurate position measurement, and leave the particle in a state that is square integrable. Ozawa's Hamiltonian is given in http://arxiv.org/abs/0706.3526 (section 2.3.2).

 

[bhobba]

As far as I can tell, bhobba and I don't disagree. The quote you mentioned from bhobba does not say the measurements are simultaneous and accurate. My post #16 and bhobba's post #22 are in agreement.

We don't.

I think there are some misconceptions floating about.

As I pointed out in my previous post you can't exactly measure both position and momentum simultaneously because there no observable that allows that. I am scratching my head where Phinds got that idea from. Even exact measuring either in QM leads to mathematical difficulties which I discussed in the same post - but they are accommodated in the Rigged Hilbert Space formalism - which is admittedly rather deep and advanced mathematically.

Thanks
Bill

 

[phinds]

Guys, I don't really know squat about this. I had believed up until several months ago that simultaneous exact measurements were not possible and that that was the basis for the HUP and then several threads on this forum convinced me that that was incorrect and the HUP really is about not getting the same results for identical setups and that therefore while you could in theory make simultaneous measurements, the results follow a statistical distribution constrained by the HUP.

I'm glad to hear that you two are actually in agreement. I must have misinterpreted the bolded "can" and "cannot" in my previous post. You DO, I hope, see how it would seem to an innocent bystander that "can" and "cannot" does not sound like agreement.

 

[bhobba]

Guys, I don't really know squat about this

Its nothing to worry about.

Its just getting the exact meaning of the principle pinned down.

Even textbooks like Griffiths excellent text doesn't explain it with care. Most of the time a simple intuitive idea they are a bit fuzzy is all you need.

Its only in exact discussions like this care is needed.

Thanks
Bill

 

[atyy]

Guys, I don't really know squat about this. I had believed up until several months ago that simultaneous exact measurements were not possible and that that was the basis for the HUP and then several threads on this forum convinced me that that was incorrect and the HUP really is about not getting the same results for identical setups and that therefore while you could in theory make simultaneous measurements, the results follow a statistical distribution constrained by the HUP.

The confusing thing perhaps is that it is true that (A) simultaneous exact measurements of non-commuting observables like position and momentum are not possible on an arbitrary unknown state. It is also true that statement (A) is not the what the textbook HUP is about: the textbook HUP is about (B) non-simultaneous exact measurements of position and non-simultaneous exact measurements of momentum on identically prepared particles, for which the exact measurement of position (or momentum) on each particle in an "ensemble" of identically prepared particles will yield a different observed position (or momentum).

Although (A) and (B) may appear contradictory, they are not. In the above, only (B) is the textbook uncertainty principle. They both stem from the same underlying principle, which is the commutation relation (C). The commutation relation (C) implies (A) and (B), but (B) alone does not imply (A). To add to the confusion, when speaking colloquially, one can call the commutation relation (C) the "uncertainty principle". In the literature, (A) is also called the "uncertainty principle". It also probably doesn't help that textbooks will in early chapters about the history of quantum mechanics refer to (A) as the "uncertainty principle", but then in the mathematical chapters (A) is usually not derived, but (B) is derived and called "uncertainty principle". Anyway, the upshot should be that the different statements (A), (B) and (C) are all correct, and are all called the "uncertainty principle" depending on taste or convention, and the only thing that is hard to get straight for the innocent bystander is which implies what. :)

Also, one often finds (A) stated for convenience in short form without the qualification "arbitrary unknown state". The qualification is needed, because there are cases in which it is possible to measure conjugate observables simultaneously and exactly if we have information about the state.

I'm glad to hear that you two are actually in agreement. I must have misinterpreted the bolded "can" and "cannot" in my previous post. You DO, I hope, see how it would seem to an innocent bystander that "can" and "cannot" does not sound like agreement.

Some of the confusion is probably generated by wrong arguments about this in the literature. An error which is famous is in Ballentine's 1970 review (the review is famous, that the review has fundamental errors is less famous). Ballentine claimed to show a case in which position and momentum can be simultaneously and exactly measured, but his example was wrong because the position and momentum in his example were not canonically conjugate.

 

[prithu roy]

My friend and I had this argument about whether or not the uncertainty principle is applicable to stationary particles. I maintain that it is, because the principle is really about predictability ( isn't it?) But he maintains that it doesn't. So I would just like to clear things up . Does it or doesn't it? Thanks in advance.

Motion itself is not fundamental,it is just relative.So is the predictability and uncertainty are also a relative term,my example will elaborate it further.
as we know momentum and position of electrons or any subatomic particles are not known accurately using our technique(keeping in mind we are in relative motion when observing them),but what happen when electron observe an electron(relatively in rest) what we get out then is 100% predictability,using same circumstance in real life i.e when we are at rest (relative) to observer then we too are predictable and in motion we are not.So above thought experiment infers that uncertainty and predictability are also relative(depends on observer) not absolute.

 

[phinds]

Atyy, thanks for that clarification.

Motion itself is not fundamental,it is just relative.So is the predictability and uncertainty are also a relative term,my example will elaborate it further.
as we know momentum and position of electrons or any subatomic particles are not known accurately using our technique(keeping in mind we are in relative motion when observing them),but what happen when electron observe an electron(relatively in rest) what we get out then is 100% predictability,using same circumstance in real life i.e when we are at rest (relative) to observer then we too are predictable and in motion we are not.So above thought experiment infers that uncertainty and predictability are also relative(depends on observer) not absolute.

I'm not 100% sure I understand what you are saying but to the extent that I understand it, it is wrong. For one thing, you say that uncertainty is frame dependent. That is not correct at the quantum level for the HUP. The HUP is not frame dependent and this thread is about the HUP, not about classical physics measurements, which may be what you are talking about.

 

[prithu roy]

i think you do understand what i said,i.e uncertainty principle is frame dependent.and i really mean ,for any measurement we do need a reference frame and since there is no universal reference frame,so i generalize that any measurement whatsoever it be is frame dependent.

 

[bhobba]

i think you do understand what i said,i.e uncertainty principle is frame dependent.and i really mean ,for any measurement we do need a reference frame and since there is no universal reference frame,so i generalize that any measurement whatsoever it be is frame dependent.

The uncertainty principle is the same regardless of frame. In fact the POR implies it must be the case - the laws of physics are the same in any inertial frame.

There is certainly no universal frame, at least as far as we can tell today, but regardless of what frame is used to make the measurements the uncertainty principle applies.

Thanks
Bill

 

[prithu roy]

The uncertainty principle is the same regardless of frame. In fact the POR implies it must be the case - the laws of physics are the same in any inertial frame.

There is certainly no universal frame, at least as far as we can tell today, but regardless of what frame is used to make the measurements the uncertainty principle applies.

Thanks
Bill

what u suggest is correct ,but there is small mistake in understanding what i said is principle of uncertainty remain invariant of frame what changes is outcomes of measurement , so in short outcomes of measurement are subjected to frame of reference not the principle itself (satisfying the P.O.R-law of physics does not change in any inertial frame. )

 

[phinds]

what u suggest is correct ,but there is small mistake in understanding what i said is principle of uncertainty remain invariant of frame what changes is outcomes of measurement , so in short outcomes of measurement are subjected to frame of reference not the principle itself (satisfying the P.O.R-law of physics does not change in any inertial frame. )

Ah, that's much more clear. It sounded as though you were saying that the uncertainty was frame dependent in a way that would allow for NO uncertainty, which would of course violate the HUP.

 

[exponent137]

How to describe UP of a photon? It is at location 5 m from us and its momentum is 10^-20 kg m/s.

Because, the simplest explanation of UP is Fourier transformation of a position wave function of a rest particle. This gives a momentum wave function. Square of absolute value of both of them are gaussian distributions and this gives thicknesses of these distributions and then product of thicknesses of both distributions.

How to describe gaussian wave function of the photon?

Because you are talking about reference frames, this is similar to my question. How How to describe gaussian wave function of the photon? And how to derive uncertainty for photon? It does not belong to the common derivation of uncertainty.

 

[atyy]

what u suggest is correct ,but there is small mistake in understanding what i said is principle of uncertainty remain invariant of frame what changes is outcomes of measurement , so in short outcomes of measurement are subjected to frame of reference not the principle itself (satisfying the P.O.R-law of physics does not change in any inertial frame. )

Usually we say that the measurement outcomes are events in the sense of classical special relativity, so they are frame invariant in the classical sense.

But I guess you are thinking about things like http://www.infres.enst.fr/~markham/QuPa/2010Sept26/Borzu-HIPTalk.pdf?

 

[bhobba]

what u suggest is correct ,but there is small mistake in understanding what i said is principle of uncertainty remain invariant of frame what changes is outcomes of measurement , so in short outcomes of measurement are subjected to frame of reference not the principle itself (satisfying the P.O.R-law of physics does not change in any inertial frame. )

But that is the same for any law. The law is invariant and particulars change.

Its a trivial and well known observation.

Thanks
Bill

 

[bhobba]

Because you are talking about reference frames, this is similar to my question. How How to describe gaussian wave function of the photon? And how to derive uncertainty for photon? It does not belong to the common derivation of uncertainty.

The photon does not have a wavefunction in the usual sense.

A little search on this forum will give the detail of this deep and mathematically advanced issue eg:
https://www.physicsforums.com/threads/why-photon-wave-function-does-not-exist.659614/

Thanks
Bill

 

[LitleBang]

How much influence does the observer have on the outcome of any measurement?

 

[atyy]

How much influence does the observer have on the outcome of any measurement?

The influence an observer has on the distribution of classical measurement outcomes that he obtains is summarized mathematically by the "observable" that is chosen. Thus for any initial given state, if the observer chooses to measure the position observable, he will get a distribution of classical positions that is determined entirely by the state and the position observable.

If the quantum system survives the measurement (ie. there is a classical measurement outcome and a quantum outcome), the distribution of quantum outcomes is summarized mathematically by the "instrument" that is chosen. A given instrument defines an observable. Thus for any initial given state, if the observer chooses to use a particular instrument that defines a position observable, he will get a distribution of classical positions and a distribution of quantum states that is determined entirely by the initial given state and the chosen instrument.

In general, an instrument used in a measurement will output a different quantum state than the initial given state, and it is usually in this sense that the observer is said to cause a disturbance.

 

[exponent137]

The photon does not have a wavefunction in the usual sense.

I little search on this forum will give the detail of this deep and mathematically advanced issue eg:
https://www.physicsforums.com/threads/why-photon-wave-function-does-not-exist.659614/

Thanks
Bill

Yes I read this thread. (Started with my question :) ) Thus, I can give a different question: How to define uncertainy principle of a photon, it its wave function does not exist. But UP exists always, also at photons?

 

[atyy]

Yes I read this thread. (Started with my question :) ) Thus, I can give a different question: How to define uncertainy principle of a photon, it its wave function does not exist. But UP exists always, also at photons?

I don't know the answer to your question directly. But the basic reason that a photon wave function doesn't exist (except as an approximation in some single photon cases) is that one has to do field quantization for the electromagnetic field. So the commutation relations and uncertainty relations are not between position and momentum, but between various field observables. Some examples are given in http://web.stanford.edu/~rsasaki/AP387/chap3 (Eq 3.50, 3.106, 3.107).

 

[bhobba]

Yes I read this thread. (Started with my question :) ) Thus, I can give a different question: How to define uncertainy principle of a photon, it its wave function does not exist. But UP exists always, also at photons?

The uncertainty principle is actually more general than between position and momentum applying to any two non-commuting observables. That of course still applies to photons. But since position is not an observable the usual form doesn't exist.

Thanks
Bill

 

[exponent137]

The uncertainty principle is actually more general than between position and momentum applying to any two non-commuting observables. That of course still applies to photons. But since position is not an observable the usual form doesn't exist.

Thanks
Bill

You and Atyy give persuasive explanations.

It is understandable to me, that photon location is not defined. But, how its momentum is defined? It seems to me, that it is defined?

It is known that background of QFT is spacetime of Minkowski. It is defined in everyone point. Why then location of photon is not defined? Maybe, its location is defined only at begin and end locations of photons? (vertex as it is said in Feynman's graphs).

 

[bhobba]

You and Atyy give persuasive explanations.

Thanks mate - but I generally find Atyy hones onto the key points better than me. I can meander a bit before the key thing jumps out.

It is understandable to me, that photon location is not defined. But, how its momentum is defined? It seems to me, that it is defined?

Momentum is a different matter - being pretty much always definable via Noethers theorem:
http://eduardo.physics.illinois.edu/phys582/582-chapter3.pdf

It is known that background of QFT is spacetime of Minkowski. It is defined in everyone point. Why then location of photon is not defined? Maybe, its location is defined only at begin and end locations of photons? (vertex as it is said in Feynman's graphs).

Its a difficult issue:
https://www.physicsforums.com/threads/is-position-not-an-observable-of-a-photon .

Intuitively I view it because it travels at the speed of light there is no frame where its at rest so it can't be stopped to find its position, but basically that view is a crock.

Unfortunately the correct answer is rather advanced:
http://arnold-neumaier.at/physfaq/topics/position.html

Don't you hate it when things are like that

Thanks
Bill