The Uncertainty Principle Uncertainty

[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

 

[atyy]

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.

Actually, many key points I learned from bhobba - especially about Gleason's and noncontextuality!

 

[hmmmok]

of course. carbon statistically only has probability of its stationary atoms being exactly the same and over time there is also a statistically probable average that some will randomly change. Thus, the uncertainty principle applies to pretty much everything.

 

[carllooper]

The way I've always understood the uncertainty principle (which isn't necessarily correct of course) is in terms of textbook introductions, where the information (such as position or momentum) is dependant on the extent to which you can physically extract such information in a physical setup.

I've found this is typically explained in relation to macro-scopic observations, where any measurements on a macro-scopic object need not adversely affect what the object in question is otherwise doing (had you not made the measurement). But in terms of subatomic objects, they are so fragile that any measurement you made on them would adversely affect the object, ie. compared to the case where you did not make a measurement on it. At least that is the idea. So there would be a fundamental uncertainty on what kind of information you can physically extract from the object, ie. in terms of that which you might like to be representative of the object considered as otherwise unmeasured.

Now to say an object is stationary, it seems to me, presupposes that you could obtain the information necessary to make such a statement in the first place. But if I understand the uncertainty principle correctly, you can't make such an assumption in the first place, ie. you can't represent some object as stationary in the first place. The uncertainty principle would rule it out.

If I understand the principle.

Carl

 

[hmmmok]

The way I've always understood the uncertainty principle (which isn't necessarily correct of course) is in terms of textbook introductions, where the information (such as position or momentum) is dependant on the extent to which you can physically extract such information in a physical setup.

I've found this is typically explained in relation to macro-scopic observations, where any measurements on a macro-scopic object need not adversely affect what the object in question is otherwise doing (had you not made the measurement). But in terms of subatomic objects, they are so fragile that any measurement you made on them would adversely affect the object, ie. compared to the case where you did not make a measurement on it. At least that is the idea. So there would be a fundamental uncertainty on what kind of information you can physically extract from the object, ie. in terms of that which you might like to be representative of the object considered as otherwise unmeasured.

Now to say an object is stationary, it seems to me, presupposes that you could obtain the information necessary to make such a statement in the first place. But if I understand the uncertainty principle correctly, you can't make such an assumption in the first place, ie. you can't represent some object as stationary in the first place. The uncertainty principle would rule it out.

If I understand the principle.

Carl

what if the subatomic particles masses are balanced out (thus zero) by the force of them rotating around the sun.

 

[athosanian]

In my understanding the uncertainty principle is not about the predictability of the measurement setups, it si the intrinsic property of the quantum particles.it means that all of the properties of the particles do not change with the time continuously.

 

[bhobba]

In my understanding the uncertainty principle is not about the predictability of the measurement setups, it si the intrinsic property of the quantum particles.it means that all of the properties of the particles do not change with the time continuously.

Continuity really has nothing to do with it.

It got to do with the commutation properties of observables.

Thanks
Bill

 

[Qistina28]

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.

from my point of view, there is still apply uncertainty principle even the particles in stationary states. since we not live in isolated system therefore how can you be sure that the object or particles are in absolute stationary..

do correct me if I'm wrong ^_^

 

[exponent137]

Object is stationary in one inertial system. Photons are not stationary in any inertial system.

 

[carllooper]

there is still apply uncertainty principle even the particles in stationary states.

The original question is an ill-posed question. It's premise is faulty.

The Uncertainty Principle requires that we do not (or can not) represent a particle as in a stationary state in the first place.

To propose a particle at rest would require that we simultaneously represent both the position and momentum of the proposed particle. But the Uncertainty Principle rules out such a proposition.

Mathematically there would be no problem proposing a stationary particle since we can easily conceive such a particle, but mathematics doesn't have the same limitations as QM. The Uncertainty Principle requires the employment of Plank's constant as a limit.

C

 

[bhobba]

The Uncertainty Principle requires that we do not represent a particle as in a stationary state in the first place.

That's untrue.

Its a statement about non-commuting observables and applies to any state - stationary or otherwise.

Thanks
Bill