Why is heisenberg uncertainty not a limit of technology?

[danphan323]

How do we know that the uncertainty principle is a property of an electron and not a limit of our measuring ability? I understand that photons striking an electron alter its momentum, but imagine an electron that is not being observed. Couldn't it have both a position and a momentum at a given point in time?

 

[Pythagorean]

Because it's not a problem of measurement, it's a problem of definition of time and space as we know it. It's a theoretical, mathematical problem that can be experimentally verified; it's not an experimental nuisance that we gave a name.

 

[jtbell]

There is a very general uncertainty principle that applies to all kinds of "wave packets", and can be derived from the mathematics of Fourier analysis:

$$\Delta x \Delta k \ge \frac{1}{2}$$

It applies to electromagnetic waves, sound waves, electrical signals in wires, etc. The HUP is simply the application of this principle to the wavelike behavior of particles. It is no more a reflection of technological limitations on measurement, than is the case with sound waves, electrical signal pulses, etc.

 

[Demystifier]

How do we know that the uncertainty principle is a property of an electron and not a limit of our measuring ability?

We don't.

I understand that photons striking an electron alter its momentum, but imagine an electron that is not being observed. Couldn't it have both a position and a momentum at a given point in time?

Yes, it could. In fact, the Bohmian interpretation of quantum mechanics proposes a very precise value of both position and momentum at a given time. See e.g.
http://xxx.lanl.gov/abs/quant-ph/0611032

 

[DrChinese]

Couldn't it have both a position and a momentum at a given point in time?

Welcome to PhysicsForums, danphan323!

Demystifier made some comments which are accurate in a certain sense. However, the best answer is NO, particles do not have simultaneously well-defined values for non-commuting properties.

Notice that I said "non-commuting". Commuting properties CAN have simultaneously well-defined values. So for example: spin and momentum can both be known, but not position and momentum.

In addition, a well known paper from 1935 referred to as EPR (Einstein is the E) tackled this issue from your perspective. A series of works over nearly 50 years answered the question in the negative. See EPR, Bell (1965), Aspect (1981) for more on this.

 

[Dali]

Couldn't it have both a position and a momentum at a given point in time?

Yes, it could. In fact, the Bohmian interpretation of quantum mechanics proposes a very precise value of both position and momentum at a given time.

True. But maybe one should point out here also that the price you pay for this objectivity in Bohm-theory is that it is explicitly non-local, i.e. particles have to move (much!) faster than light.

We also know this is true in general from Bell's theorem which shows that any model insisting on particles "having" propertied even when they are not measured (i.e. objective theories) will have to be non-local (i.e. allow super-luminal signaling), otherwise that model will contradict experimental results in the setups made with entangled particle pairs.

 

[Fredrik]

Demystifier made some comments which are accurate in a certain sense. However, the best answer is NO, particles do not have simultaneously well-defined values for non-commuting properties.

I was thinking that too. Demystifier's comments are correct, as far as I can tell, but only because to "have a position" can mean something different from to "be prepared in a state represented by a sharply peaked wavefunction".

I started writing a much longer explanation, but it's taking too long. So this short comment will have to do, at least for now.

 

[DrChinese]

I was thinking that too. Demystifier's comments are correct, as far as I can tell, but only because to "have a position" can mean something different from to "be prepared in a state represented by a sharply peaked wavefunction".

I always love Demystifier's answers. Always sharp. Of course with that Bohmian edge as well.

 

[San K]

Because it's not a problem of measurement, it's a problem of definition of time and space as we know it. It's a theoretical, mathematical problem that can be experimentally verified; it's not an experimental nuisance that we gave a name.

are time and space non-commuting?

can all non-commuting properties can be broken down (reducible to or derived from) time-space?

 

[zonde]

There is a very general uncertainty principle that applies to all kinds of "wave packets", and can be derived from the mathematics of Fourier analysis:

$$\Delta x \Delta k \ge \frac{1}{2}$$

It applies to electromagnetic waves, sound waves, electrical signals in wires, etc. The HUP is simply the application of this principle to the wavelike behavior of particles. It is no more a reflection of technological limitations on measurement, than is the case with sound waves, electrical signal pulses, etc.

You have made similar posts in number of thread about Heisenberg uncertainty. But is there discussion where this explanation has been discussed in more details?

Anyways I would like to understand to what extent this explanation works so let me ask some questions.
You can't make f(x) and \hat{f}(\xi) peak sharply at the same time so it seems very elegant explanation for uncertainty principle in QM. Now the question I have is what would be physical interpretation of functions \hat{f}(\xi). It takes as an argument frequency and produces amplitude and phase for particular frequency.
It seems like dimension of frequency can't span real space or time so it should be something more complex and indirect, right?

 

[Pythagorean]

are time and space non-commuting?

can all non-commuting properties can be broken down (reducible to or derived from) time-space?

no (pretty sure time and space straight forward operators) and I don't know about all non-commuting properties. I know a lot of properties in classical physics can be reduced to time and space (but only if you forgive that mass is a ratio of distances).

 

[Dead Boss]

Time is not an operator in QM.

 

[Pythagorean]

Time is not an operator in QM.

hrm, so then it's not an observable?

 

[Pythagorean]

so then how do they commute time and energy for the uncertainty principle involving them? I thought commutation was for observable operators.

 

[Dead Boss]

No, time is not an observable and the time-energy uncertainty is a different principle than the position-momentum relation (it doesn't help that they look the same).

Unfortunately I don't really know how it's derived or why it should be true (but it kinda feels right in the light of special relativity).

 

[Fredrik]

so then how do they commute time and energy for the uncertainty principle involving them? I thought commutation was for observable operators.

It's explained here.

 

[Pythagorean]

Apparently, that's exactly where it came from ("it kinda feels right")

The energy-time uncertainty relation is not, however, an obvious consequence of the general Robertson–Schrödinger relation. Since energy bears the same relation to time as momentum does to space in special relativity, it was clear to many early founders, Niels Bohr among them, that the following relation should hold:[8][9]

http://en.wikipedia.org/wiki/Uncertainty_principle#Energy.E2.80.93time_uncertainty_principle

addendum:

thanks Fredrik

 

[jtbell]

You can't make f(x) and \hat{f}(\xi) peak sharply at the same time so it seems very elegant explanation for uncertainty principle in QM. Now the question I have is what would be physical interpretation of functions \hat{f}(\xi).It takes as an argument frequency [...]

I think most practical applications of Fourier analysis (e.g. signal processing) use time and frequency as the conjugate variables. However, the same mathematics applies when you use position and wavenumber $k = 2 \pi / \lambda$ as the conjugate variables:

$$\psi(x) = \frac{1}{\sqrt{2\pi}} \int^{+\infty}_{-\infty} {A(k)e^{ikx} dk}$$
$$A(k) = \frac{1}{\sqrt{2\pi}} \int^{+\infty}_{-\infty} {\psi(x)e^{-ikx} dx}$$

Wavenumber and momentum are of course related by $p = \hbar k$.

 

[GW Leibniz]

How do we know that the uncertainty principle is a property of an electron and not a limit of our measuring ability?

We don't.

I thought this is something we do know. My understanding is that the suggestion that the uncertainty principle is merely a limit on our ability to measure (a particle that does in fact have a fixed position and velocity) is known as the "hidden variable interpretation." And whenever I see mention of the hidden variable interpretation, it is always coupled with a statement along the lines of "convincing empirical evidence has now falsified the hidden variable interpretation." I have no real idea what evidence they are referring to, but if this is still an active controversy, it certainly isn't represented as such in anything I've read...

Am I mistaken? I only have a lay understanding of QM...

 

[DrewD]

As far as I know, there aren't experiments that can falsify hidden variables. What the experiments have done is put extreme requirements on any hidden variable theory. I have no professors that buy into a hidden variable interpretation, but there are a number of very smart people out there that are still making good arguments. It isn't mainstream, but it is far from crackpottery.

According to the standard interpretation (the one that every undergrad textbook that I know of and most grad textsbooks buy into), the HUP is definitely a fundamental principle. That being said, 150 years ago galilean relativity was a fundamental principle.

So... you are pretty much right, but these are very complicated things and it is hard to prove that an interpretation of statistical data is wrong.

 

[tom.stoer]

The first problem is to ascribe with certainty a value a of a property A to a quantum state ψ w/o having done a measurement i.e. w/o knowing that the state ψ is an eigenstate of A. Of course there are some obvious but contradictory ways to do that, but as said some (non-local) hidden variable theories may be clever enough to manage this.

The second problem is to do that simultaneously for two non-communiting observables like X and P. As can be seen this is not possible on the basis of the wave function alone, especially due to mathematimal theorems applying to the Fourier transformation of wave functions. Therefore in order to achieve that one must abandon the idea that the wave function as we know it is the only entity to which one can ascribe a kind of reality (= to ascribe with certainty a value a of a property A to a quantum state ψ). So the notion of "quantum state" must be adapted as well.

 

[Demystifier]

True. But maybe one should point out here also that the price you pay for this objectivity in Bohm-theory is that it is explicitly non-local, i.e. particles have to move (much!) faster than light.

Since people like my sharp answers, I will try to be sharp again.
It is explicitly non-local, but not because particles move faster than light. It is nonlocal because entangled particles DON'T need messenger particles at all to know something about other particles far away.

By the way, for those who like particle trajectories of Bohmian interpretation but also locality of Copenhagen interpretation, I have constructed a sort of a hybrid interpretation:
http://xxx.lanl.gov/abs/1112.2034

 

[Demystifier]

Of course with that Bohmian edge as well.

Not always. In fact, I have even proposed an anti-Bohmian interpretation, according to which even in classical mechanics particle trajectories don't exist and wave function collapses when a measurement is performed:
http://xxx.lanl.gov/abs/quant-ph/0505143 [Found.Phys.Lett. 19 (2006) 553-566]
http://xxx.lanl.gov/abs/0707.2319 [AIPConf.Proc.962:162-167,2007]

Also, when the weak measurement of particle trajectories, coinciding with Bohmian trajectories, was discussed on this forum, nobody insisted more than me that these measurements do NOT prove that Bohmian interpretation is correct.

 

[zonde]

I think most practical applications of Fourier analysis (e.g. signal processing) use time and frequency as the conjugate variables. However, the same mathematics applies when you use position and wavenumber $k = 2 \pi / \lambda$ as the conjugate variables:

$$\psi(x) = \frac{1}{\sqrt{2\pi}} \int^{+\infty}_{-\infty} {A(k)e^{ikx} dk}$$
$$A(k) = \frac{1}{\sqrt{2\pi}} \int^{+\infty}_{-\infty} {\psi(x)e^{-ikx} dx}$$

Wavenumber and momentum are of course related by $p = \hbar k$.

Well, yes then the question is how do we measure momentum of particle. I have seen this question asked but haven't seen answer that would satisfy me.

Then maybe I can ask this. You said that uncertainty principle the same way applies to sound waves and electrical signals. Can you describe example with sound waves or electrical signals where there would be source and two types of measurements? And with particular attention to measurement that measures signal in frequency (or wavenumber) dimension?

 

[zonde]

Well, maybe is not reasonable to ask for classical example when it is unclear what things it should clear up.

So let me try to explain my thoughts on this.
In older thread you posted this:

Further, we assume that a superposition of waves with different wavelengths (momenta) represents a particle that can have any of the momenta, with a probability determined by the square of the amplitude of each wave in the superposition.

So to determine momenta we should use resonator that would filter particles based on frequency.
To model random response we can model resonator as having some idle oscillations. Then based on phase of these idle oscillations resonator can give different responses for the same amplitude of driving oscillation.

Does it seems fine so far?

 

[tom.stoer]

The problem which does not exist with classical waves is that you don't interpret them as particles, so you don't have something like x and p.

 

[jtbell]

Then maybe I can ask this. You said that uncertainty principle the same way applies to sound waves and electrical signals. Can you describe example with sound waves or electrical signals where there would be source and two types of measurements? And with particular attention to measurement that measures signal in frequency (or wavenumber) dimension?

In the days before cheap digital computers that can quickly do Fourier analysis on digitized signals, electronic engineers used spectrum analyzers based on analog electronics. I've never used one of those, so you'd have to ask an EE (probably an older one ) for more details about how they work. I suspect that they use, in effect, a variably-tunable resonating circuit like the resonators that you mention in your next post.

I think I vaguely remember reading something about devices for analyzing sound or mechanical waves in an analog fashion, but I haven't found anything specific yet.

For sound, here's a qualitative illustration of the frequency/time uncertainty relation: percussion (musical) instruments. I think it's safe to say that in general, the shorter the sound pulse the instrument produces, the less well-defined its pitch is. To produce a very short burst of sound, you have to combine waves with a wide range of frequencies (pitches).

 

[zonde]

In the days before cheap digital computers that can quickly do Fourier analysis on digitized signals, electronic engineers used spectrum analyzers based on analog electronics. I've never used one of those, so you'd have to ask an EE (probably an older one ) for more details about how they work. I suspect that they use, in effect, a variably-tunable resonating circuit like the resonators that you mention in your next post.

I think I vaguely remember reading something about devices for analyzing sound or mechanical waves in an analog fashion, but I haven't found anything specific yet.

For sound, here's a qualitative illustration of the frequency/time uncertainty relation: percussion (musical) instruments. I think it's safe to say that in general, the shorter the sound pulse the instrument produces, the less well-defined its pitch is. To produce a very short burst of sound, you have to combine waves with a wide range of frequencies (pitches).

Thank you for answers.

 

[zonde]

The problem which does not exist with classical waves is that you don't interpret them as particles, so you don't have something like x and p.

You can have amplitude modulation of classical wave. And if we say that in QM p is closely related to wavelength then certainly classical wave has that.

 

[tom.stoer]

No, for classical waves you don't have x as a position of a particle neither do you have p as the momentum of a particle. You don't have anything like a particle for sound waves, water waves, classical electromagnetic waves etc. So formally you can derive an "uncertainty relation" using Fourier analysis but it's not to be interpreted as something affecting particles.

The difference is that you interpret a QM wave function as something representing a particle; this you never do with classical waves.

 

[zonde]

You don't have anything like a particle for sound waves, water waves, classical electromagnetic waves etc.

I think that this is the key point (instead of arguments about x and p). Or in other words we don't have straight forward model for quantization of classical wave packet.

 

[TrickyDicky]

No, for classical waves you don't have x as a position of a particle neither do you have p as the momentum of a particle. You don't have anything like a particle for sound waves, water waves, classical electromagnetic waves etc. So formally you can derive an "uncertainty relation" using Fourier analysis but it's not to be interpreted as something affecting particles.

The difference is that you interpret a QM wave function as something representing a particle; this you never do with classical waves.

But aren't photons kind of interpreted that way (interpreting the classical EM wave as something particle-like), or how about phonons for sound, or solitons for water waves...

 

[tom.stoer]

Photons are particles (particle-like excitations) of the quantized electromagnetic field in QED; they are not particles of classical electromagnetic waves of Maxwell's theory; Maxwell's theory explicitly rules out particle-like behavior of the electromagnetic field!

 

[TrickyDicky]

Photons are particles (particle-like excitations) of the quantized electromagnetic field in QED; they are not particles of classical electromagnetic waves of Maxwell's theory; Maxwell's theory explicitly rules out particle-like behavior of the electromagnetic field!

You completely missed my point. I was not pointing to the evident ifferences between QED and Maxwell's EM, besides note we have the notion of photons since much earlier than QFT was imagined, or even QM-the Einstein photon- and you could even say from Newton and his "corpuscles".
I was giving examples of wave phenomena like light, sound and water waves (the three of them can still be regarded as "classical waves" in their macroscopic behaviour, can't they?) and when you say something like "The problem which does not exist with classical waves is that you don't interpret them as particles, so you don't have something like x and p" , I don't think is so clear cut that under certain very specific circumstances like the ones I mentioned you don't have something like x and p.

 

[Whovian]

How do we know that the uncertainty principle is a property of an electron and not a limit of our measuring ability? I understand that photons striking an electron alter its momentum, but imagine an electron that is not being observed. Couldn't it have both a position and a momentum at a given point in time?

I know this has been answered, but I feel like answering anyway.

That's what Einstein thought, if I'm remembering my history right (which I probably am not). Then we just get repeated experiments, proving his hypothesis wrong. There's even a famous misquote "God does not play dice with the Universe." (Again, a misquote.)

Even Einstein's susceptible to mistakes. (In fact, science is based around mistakes!)

 

[TrickyDicky]

Photons are particles (particle-like excitations) of the quantized electromagnetic field in QED; they are not particles of classical electromagnetic waves of Maxwell's theory; Maxwell's theory explicitly rules out particle-like behavior of the electromagnetic field!

You completely missed my point...

Anyway I don't think there is probably much point in arguing about this since if you were just referring to the classical wave theory of course I agree with you, I was more trying to confront the macroscopic wave behaviour with what we know about the microscopic behaviour of matter and fields. Certainly not trying to say here that the classical theory picture applies in the quantum realm.


To go back on topic, I think what jtbell was saying about waves, HUP and Fourier analysis was spot on. Quantization(and therefore the introducing of noncommutation /HUP) of classical fields and matter has much more to do with the microscopic wave nature (in the modern sense) of matter and fields than it is usually acknowledged. The kind of counterintuitive thing is that at the same time quantization is what allows also the quantum particle behaviour of matter and fields, the fact that unlike the classical particles we cannot know simultaneously momentum and position is then an unavoidable consequence of the quantum wave nature.

 

[Rosen]

Hello, This is my first entrance into the Physics Forum so I'll ask a naive question. If an electron wave packet collapses into a particle, wouldn't it have to have a definite position and momentum in a single instant? Particularly if we don't observe it.

 

[DrChinese]

Hello, This is my first entrance into the Physics Forum so I'll ask a naive question. If an electron wave packet collapses into a particle, wouldn't it have to have a definite position and momentum in a single instant? Particularly if we don't observe it.

Welcome to PhysicsForums, Rosen!

If you don't observe it, it doesn't collapse. But you can choose what basis you want to observe it.

The collapse of the wave function places the electron into an eigenstate on that specific basis. Let's say it is position. Then the non-commuting partner - momentum - is undefined. Same for other pairs of non-commuting properties. So no, you cannot meaningfully say it had both position and momentum at any time.

Notice that I said "non-commuting" above. Commuting properties CAN have simultaneously well-defined values. So for example: spin and momentum can both be known, but not position and momentum.

 

[Ger]

Isn't the basics of it all that we observe differences? A completely static object would have a delta of 0 for all its observables, meaning there is nothing there to observe. A constant can be added and removed from any differential or integration, even if the constant to add would be infinite: renormalization. If one is stuck with such a non-removable singularity, it might have a physical meaning or the interpretation is wrong.