Falsification Of Heinsenberg's Uncertainty Principle?

[DRZion]

That's right. Just like the dcay of radioactive nuclei.

Yes. Radioactive nuclei decay through a probabilistic process. But there is no need for the UP to explain it. From what I understood in AlphaNumberic's post, it is due to tunneling, which is based on probability.

The explanation for the Casimir effect is just theoretical, if I'm not mistaken. It is similar to Hawking Radiation:

I'm still dubious about all this. I strongly disagree with Einstein's theory of space-time for example; this is just a hypothetical construct which has no evidence in reality. Particles have been detected in reality, yet the 'groupthink' of the last hundred years has blocked this avenue of thinking and research. Why can't radiating particles be the cause of the gravity force?

Hawking radiation is purely theoretical, with no detectable evidence, at least with today's technology. One would have to get close to a black hole and then do some kind of observation there in order to detect something. Supposedly it takes 10e100 years for a black hole to evaporate. The current age of the universe is ~1.4e11 , or about 90 orders of magnitude younger.

In quantum physics, everything can be treated as a particle. You can define vibrational waves as particles (phonons), because they have a wave characteristic. The point of the Large Hadron Collider is to define the 'particles' responsible for gravity. . . is my understanding correct here?

Once again, does anyone know about 'graininess of wavelength'? I believe this also has to do with Hawking radiation. Someone pointed out that Hawking radiation would be too weak and thus it would fall somewhere between the quanta of 0 and 1 of wavelength.

 

[AlphaNumeric]

Yes. Radioactive nuclei decay through a probabilistic process. But there is no need for the UP to explain it. From what I understood in AlphaNumberic's post, it is due to tunneling, which is based on probability.

Tunneling is related to the UP in some way.

Suppose you have a ball sitting at the bottom of a frictionless valley and you kick it. In order to get out of the valley you must give the ball more kinetic energy than is required to overcome the gravitational forces the ball experiences. If you don't you can be certain the ball will roll back down into the valley. The equivalent quantum mechanical system doesn't require you to kick the atom. You trap the atom in a 'magnetic valley' whose potential is much much larger than the thermal energy of the atom. Classically it could not escape. Quantum mechanically it'll escape at some point, when you aren't looking. Why? Because the uncertainty principle allows it to have enough energy to get out of the trap, if only for a second.

 

[kurros]

Not untrue. However, I keep my delusions in check. Here I am trying to prove a point.

Lasers have a good ability to keep wavelength close to monochromatic. Doesn't this mean that we know very certainly the wavelength/energy? And since it is a beam, we also know direction, time, etc. Surely, this breaches the UP!

Nope, knowing things about direction and time is similar information to wavelength and energy, all are really telling you about momentum. The direction part gets you the full momentum vector rather than just it's magnitude. I don't really know what you mean by saying we know the time actually. We know the time rate of change of position (the momentum), except these are massless particles so $$ \mathbf{p}\neq md\mathbf{x}/dt $$. But you get the idea.

In a laser you have very little idea about where a particular photon is at any time but you can be pretty sure about it's energy.

In quantum physics, everything can be treated as a particle. You can define vibrational waves as particles (phonons), because they have a wave characteristic. The point of the Large Hadron Collider is to define the 'particles' responsible for gravity. . . is my understanding correct here?

Phonons are quantum particles because they come in quanta, i.e. the lattice vibrations are quantised. As for the LHC, one of it's many goals is to find the Higgs boson, which gives the others particles mass in a sense. It is not responsible for gravity, although it doesn't seem unreasonable to me that a connection may be found in the future.

Once again, does anyone know about 'graininess of wavelength'? I believe this also has to do with Hawking radiation. Someone pointed out that Hawking radiation would be too weak and thus it would fall somewhere between the quanta of 0 and 1 of wavelength.

There is no 'graininess of wavelength' in general. In many systems, like the particle in a box, you are only allowed to have certain wavelengths, so wavelength is quantised in those situations, but in free space there is no such restriction and you can have whatever wavelength you like. There may be some quantum gravity limit on such things (i.e. so you can't have a particle with wavelength so short that it's energy becomes mega-high and folds spacetime in on itself or something) but who knows. I don't know anything about the Hawking radiation part though.

 

[DRZion]

Tunneling is related to the UP in some way.

Suppose you have a ball sitting at the bottom of a frictionless valley and you kick it. In order to get out of the valley you must give the ball more kinetic energy than is required to overcome the gravitational forces the ball experiences. If you don't you can be certain the ball will roll back down into the valley. The equivalent quantum mechanical system doesn't require you to kick the atom. You trap the atom in a 'magnetic valley' whose potential is much much larger than the thermal energy of the atom. Classically it could not escape. Quantum mechanically it'll escape at some point, when you aren't looking. Why? Because the uncertainty principle allows it to have enough energy to get out of the trap, if only for a second.

Thermal energy is not closely related to radioactive decay.
http://www.newton.dep.anl.gov/askasci/phy00/phy00543.htm
thanks, google

Nope, knowing things about direction and time is similar information to wavelength and energy, all are really telling you about momentum. The direction part gets you the full momentum vector rather than just it's magnitude. I don't really know what you mean by saying we know the time actually. We know the time rate of change of position (the momentum), except these are massless particles so $$ \mathbf{p}\neq md\mathbf{x}/dt $$. But you get the idea.

In a laser you have very little idea about where a particular photon is at any time but you can be pretty sure about it's energy.

Yes, momentum of a photon would be very very tricky to establish. Not impossible. If you shoot a laser from 100,000,000 kilometers away, you will know direction very very well.. enough to beat hbar? I don't know. However, there is no limiting factor as to how far away the laser beam can be. The solid angle of the aperture will at some point decrease below hbar.

And once you know momentum, you cannot know position very well. Am I right?

There is no 'graininess of wavelength' in general. In many systems, like the particle in a box, you are only allowed to have certain wavelengths, so wavelength is quantised in those situations, but in free space there is no such restriction and you can have whatever wavelength you like. There may be some quantum gravity limit on such things (i.e. so you can't have a particle with wavelength so short that it's energy becomes mega-high and folds spacetime in on itself or something) but who knows. I don't know anything about the Hawking radiation part though.

I wrote an email to a scientist at UCLA who studies thermal radiation.

Hi xxxxxxxx,

I am glad that you find our work interesting, and yes we are still
exploring microscopic thermal radiation. We do expect wavelength to
be quantized at some level. We haven't seen that yet in our work, but
we are not done yet.

It really makes sense, if everything else is quantized, then wavelength should be as well. For example, wavelength can only change by hbar .. or some other strictly defined quantity.

 

[kurros]

Thermal energy is not closely related to radioactive decay.

Alphanumeric was talking about a situation analagous to radioactive decay, not radioactive decay itself. Trapping a bunch of atoms in a magnetic or optical trap is a very similar situation to trapping nucleons in the nuclear potential of a nucleus. Each have a quantum chance to escape the trap without having to overcome the classical energy barrier (tunnelling). He was quite correct about the relationship between tunnelling and the UP too; I'd never really thought about it that way but it is a nice way of looking at it.

The momentum of a photon would be very very tricky to establish

Not so tricky really, since you can measure the wavelength accurately you can deduce the momentum accurately. Direction can be measured very accurately as well, diffraction experiments in electron microscopy do this kind of thing all the time. Similar experiment to yours but using a very small measuring device rather than a huge optical path length. But yes knowing this stuff screws up your knowledge of position.

As for the radiation stuff, I didn't say it wasn't quantised, I said it wasn't quantised in free space. The quantisation depends on the system. For instance in collisions between fundamental particles there is no quantisation of the wavelength of the collision products.

 

[DRZion]

Not so tricky really, since you can measure the wavelength accurately you can deduce the momentum accurately. Direction can be measured very accurately as well, diffraction experiments in electron microscopy do this kind of thing all the time. Similar experiment to yours but using a very small measuring device rather than a huge optical path length. But yes knowing this stuff screws up your knowledge of position.

Using both long optical path length and very small measuring device, couldn't we get more certainty about momentum? I think yes.
However, this still doesn't say anything about position.
Position is still fairly certain. . . You know that the particle strikes the surface of the detector, right? That is a lot more certainty than knowing NOTHING of position.
If you know EXACTLY the momentum, it should follow you know NOTHING of the position, right?

As for the radiation stuff, I didn't say it wasn't quantised, I said it wasn't quantised in free space. The quantisation depends on the system. For instance in collisions between fundamental particles there is no quantisation of the wavelength of the collision products.

Oh, I think it is very much quantized in free space. You said that phonon energies are quantized. Can you say that each photon has a discrete wavelength?

The point here is that if you define the momentum of a photon to the very smallest measuring unit, the quantum, you should have no idea about it's position. However, if the detector still picks up the photon then there is that much certainty in position.

Of course, wavelength is not the only component of momentum, which is why the experimental set-up has to be devious.



I apologize to AlphaNumeric, you weren't actually talking about radioactive decay. However, you quoted my response which was a response to a post about radioactive decay, so I anchored my response on that.

 

[kurros]

"Using both long optical path length and very small measuring device, couldn't we get more certainty about momentum? I think yes.
However, this still doesn't say anything about position.
Position is still fairly certain. . . You know that the particle strikes the surface of the detector, right? That is a lot more certainty than knowing NOTHING of position.
If you know EXACTLY the momentum, it should follow you know NOTHING of the position, right?"

Yes, but no matter how clever you try to be you aren't going to be able to beat this. I haven't really thought deeply about this particular setup just now though, sorry. If I find time later I'll think about it.

"Oh, I think it is very much quantized in free space. You said that phonon energies are quantized. Can you say that each photon has a discrete wavelength?"

In free space, yes a photon can have a definite wavelength, but you can alter it by arbitrarily small amounts so it isn't quantised. If you collide the photon with an electron at a very very shallow collision angle it will give a tiny amount of energy to the electron and have a tiny bit less energy itself (its wavelength will be a tiny bit longer). The same kind of thing is not true in an atomic system for example. You can't nudge an electron in an atomic orbital just a little to change its energy, you have to give a kick hard enough to put it into the next quantised energy level, otherwise nothing will happen.

"The point here is that if you define the momentum of a photon to the very smallest measuring unit, the quantum, you should have no idea about it's position. However, if the detector still picks up the photon then there is that much certainty in position."

Hmm this is a little interesting, I'll have to think about it some more. I am imagining say a resonating cavity with a single photon in it, so the photon will be in some resonant mode of the cavity. You then stick some kind of detector in the cavity and wait for the photon to crash into it and make a click, from which you can measure its energy, and can also claim the photon was localised to the position of the detector.
Actually this is not so confusing, a photomultiplier tube is a very common detection instrument. When you look at the spike in an energy vs time graph hooked up the detector the spike will not be super sharp. The spike is widened by the uncertainty principle, i.e. you can't tell exactly when the photon hit the detector or what it's exact energy was, but you get both to a pretty good accuracy (just not below hbar/2)

 

[DRZion]

"Using both long optical path length and very small measuring device, couldn't we get more certainty about momentum? I think yes.
However, this still doesn't say anything about position.
Position is still fairly certain. . . You know that the particle strikes the surface of the detector, right? That is a lot more certainty than knowing NOTHING of position.
If you know EXACTLY the momentum, it should follow you know NOTHING of the position, right?"

Yes, but no matter how clever you try to be you aren't going to be able to beat this. I haven't really thought deeply about this particular setup just now though, sorry. If I find time later I'll think about it.

"Oh, I think it is very much quantized in free space. You said that phonon energies are quantized. Can you say that each photon has a discrete wavelength?"

In free space, yes a photon can have a definite wavelength, but you can alter it by arbitrarily small amounts so it isn't quantised. If you collide the photon with an electron at a very very shallow collision angle it will give a tiny amount of energy to the electron and have a tiny bit less energy itself (its wavelength will be a tiny bit longer). The same kind of thing is not true in an atomic system for example. You can't nudge an electron in an atomic orbital just a little to change its energy, you have to give a kick hard enough to put it into the next quantised energy level, otherwise nothing will happen.

"The point here is that if you define the momentum of a photon to the very smallest measuring unit, the quantum, you should have no idea about it's position. However, if the detector still picks up the photon then there is that much certainty in position."

Hmm this is a little interesting, I'll have to think about it some more. I am imagining say a resonating cavity with a single photon in it, so the photon will be in some resonant mode of the cavity. You then stick some kind of detector in the cavity and wait for the photon to crash into it and make a click, from which you can measure its energy, and can also claim the photon was localised to the position of the detector.
Actually this is not so confusing, a photomultiplier tube is a very common detection instrument. When you look at the spike in an energy vs time graph hooked up the detector the spike will not be super sharp. The spike is widened by the uncertainty principle, i.e. you can't tell exactly when the photon hit the detector or what it's exact energy was, but you get both to a pretty good accuracy (just not below hbar/2)

Do you even have to know time in order to know momentum? If wavelength and direction are known, this constitutes momentum.
"The energy and momentum of a photon depend only on its frequency (ν) or equivalently, its wavelength (λ)" - good ol' wikipedia
When you have both wavelength and direction, momentum is well defined.

This does not tell you the position, however. This is where time is needed. Hmm.

Point is, if you define momentum to a quantum (not sure if its possible... wavelength may be quantized, but solid angle?!?!..), the time should be completely unpredictable.
Perhaps, the fact that the limiting parameters are so strict will make the emission of such a singularly particular photon a matter of pure chance. IE, even if you put a high intensity light source [trillions of trillions of trillions of photons per second], the chance that such a photon is emitted will still be so small as to be unpredictable.

 

[kurros]

Do you even have to know time in order to know momentum? If wavelength and direction are known, this constitutes momentum.
"The energy and momentum of a photon depend only on its frequency (ν) or equivalently, its wavelength (λ)" - good ol' wikipedia
When you have both wavelength and direction, momentum is well defined.

This does not tell you the position, however. This is where time is needed. Hmm.

Ok you don't exactly need to 'know time' to know momentum, I just meant there are relationships between these things so we can figure things out about one from the other.

Point is, if you define momentum to a quantum (not sure if its possible... wavelength may be quantized, but solid angle?!?!..), the time should be completely unpredictable.

Well there isn't a quantity called a 'quantum of momentum', it is only quantised in certain systems. You can certainly have systems in which angle is quantised as well (for instance a particle in a magnetic field will have its magnetic moment quantised so that it can only point in certain directions)

Perhaps, the fact that the limiting parameters are so strict will make the emission of such a singularly particular photon a matter of pure chance. IE, even if you put a high intensity light source [trillions of trillions of trillions of photons per second], the chance that such a photon is emitted will still be so small as to be unpredictable.

I don't really know what you mean here, sorry. Perhaps you were referring to an earlier post which I have forgotten about already. My memory is pretty bad.