Matter waves (split from Photon is massless why?)

[KJW]

2 minutes ago, martillo said:

Which formula do you apply to obtain v_p with λ and f?

v_p = λ f

 

2 minutes ago, martillo said:

By the way, seems f is not measurable...

I don't know about that. Earlier, I mentioned time crystals, the time-based analogue of ordinary crystals, but nobody seemed to have noticed. This at least allows frequency to be measured in principle even if not currently in practice.

 

[joigus]

27 minutes ago, KJW said:

I tend to agree with this. But ultimately, I was simply justifying the statement that phase velocity can be obtained from the measurements of wavelength and frequency.

Again. How? Be explicit, please. 

That would contradict gauge invariance, which we know to be exact. In a nutshell, gauge invariance tells us that quantum states don't have "a" phase, meaning an unambiguous local phase. Let alone a phase velocity.

The phase velocity of ripples on a pond can be measured.

The phase of the wave function cannot.

You can measure interference patterns. But those have nothing to do with phase velocity.

You talk about frequency and wavelength as if there was just one frequency and one wavelength for matter waves. It is precisely because all "realistic" matter waves package many frequencies and wavelenths that the phase velocity is rendered all but meaningless in QM. Only the group velocity makes physical sense.

I've just had a déja vu.

Didn't I say something like this before? It's either right or wrong. What's sure not to be is irrelevant to the question at hand.

[KJW]

30 minutes ago, joigus said:

Again. How? Be explicit, please. 

v_p = λ f

 

30 minutes ago, joigus said:

That would contradict gauge invariance, which we know to be exact. In a nutshell, gauge invariance tells us that quantum states don't have "a" phase, meaning an unambiguous local phase. Let alone a phase velocity.

Are you sure about that? After all, differences in phase are observable without contradicting gauge invariance.

 

30 minutes ago, joigus said:

You talk about frequency and wavelength as if there was just one frequency and one wavelength for matter waves. It is precisely because all "realistic" matter waves package many frequencies and wavelenths that the phase velocity is rendered all but meaningless in QM. Only the group velocity makes physical sense.

I see two meanings to "wave" in QM. Firstly, there is "wave" as in "wavefunction", a general term referring to any waveform. But there is also "wave" as in "wave-particle duality" which I see as referring specifically to a sinusoidal waveform. In other words, a particle with definite momentum, and therefore definite wavelength. But also definite energy, and therefore definite frequency. Phase velocity might not be meaningful for all wavefunctions, but that does not make it meaningless for all wavefunctions.

 

Edited November 18, 2023 by KJW

[martillo]

35 minutes ago, KJW said:

v_p = λ f

Then f = v_p/λ

Now λ is measurable (λ = h/p) and v_p is measurable from v_p = E/p. I conclude f is measurable...

[joigus]

27 minutes ago, KJW said:

v_p = λ f

I'm afraid we have different understandings of what "explicit" means, at least for this case.

"Explicit" as in "He gave me very explicit directions on how to get there." (Taken from Oxford's dictionary).

37 minutes ago, KJW said:

Are you sure about that?

Yes, I'm sure about that.

https://www.ippp.dur.ac.uk/~krauss/Lectures/QuarksLeptons/QED/GaugeInvariance_2.html

Read under the heading "Local Gauge Invariance". IOW: You can re-define the local phase at will as long as you accompany such change by the corresponding gradient in the gauge field. What is "the" phase now?

No wonder phase velocity is not an observable.

You can concoct situations in which phase and group velocity are essentially the same. I'd venture to say that for those cases you can "measure" the phase velocity. What you're doing (secretly) is, of course, measuring the group velocity, and using the theory to deduce a value of the phase velocity consistent with it and with the choice of quantum state (or "ray") that you have obtained in that particular gauge.

46 minutes ago, KJW said:

I see two meanings to "wave" in QM. Firstly, there is "wave" as in "wavefunction", a general term referring to any waveform. But there is also "wave" as in "wave-particle duality" which I see as referring specifically to a sinusoidal waveform. In other words, a particle with definite momentum, and therefore definite wavelength. But also definite energy, and therefore definite frequency. Phase velocity might not be meaningful for all wavefunctions, but that does not make it meaningless for all wavefunctions.

No. You got this totally wrong. Quantum waves are not sinusoidal. For starters, they are not real functions. They typically go like complex exponentials.

37 minutes ago, martillo said:

Then f = v_p/λ

Now λ is measurable (λ = h/p) and v_p is measurable from v_p = E/p. I conclude f is measurable...

No. I can write A=XY/Z, the relation be totally right, and neither A, nor X, Y or Z be measurable.

In fact, what you both are saying contradicts the principles of mainstream quantum mechanics --except for free particles.

If the particle is interacting, the so-called Hamiltonian has a potential energy term, and the momentum (the inverse wavelength) does not commute with it, so the principles say no, you can't measure both at the same time. They're called "incompatible."

You both are confusing a component (among infinitely many) of the quantum state with the whole quantum state.

More on this:

Quote

It is quite possible for the phase velocity of a perfectly monochromatic wave of light, assuming such a thing exists, to exceed the value of c, because it conveys no information. In fact, the concept of a perfectly monochromatic beam of light is similar to the idea of a "free photon", in the sense that neither of them has any physical significance or content, because a photon must be emitted and absorbed, just as a beam of light cannot be infinite in extent and duration, but must always have a beginning and an end, which introduces a range of spectral components to the signal. Any modulation will propagate at the group velocity, which, in dispersive media, is always less than c.

https://www.mathpages.com/home/kmath210/kmath210.htm

And so on, and so on, and so on...

[studiot]

1 hour ago, joigus said:

a beam of light cannot be infinite in extent and duration, but must always have a beginning and an end, which introduces a range of spectral components to the signal.

So many forget this fact that in our mathematical idealisations functions like sinx , solutions to the linear wave equation, and many more  have no beginning and no end. They extend to infinity in all directions.

+1

But functions and waves in the real world have a beginning and an end.

So we must employ artificial mathematical devices to suppress beyond initial and end conditions.

Sometime we have to go further and match curvatures at initial and end points as well.

Edited November 18, 2023 by studiot

[joigus]

55 minutes ago, studiot said:

So many forget this fact that in our mathematical idealisations functions like sinx , solutions to the linear wave equation, and many more  have no beginning and no end. They extend to infinity in all directions.

+1

But functions and waves in the real world have a beginning and an end.

So we must employ artificial mathematical devices to suppress beyond initial and end conditions.

Sometime we have to go further and match curvatures at initial and end points as well.

Yes, thank you for your appreciation of this point. That's what I meant when I said,

1 hour ago, joigus said:

You both are confusing a component (among infinitely many) of the quantum state with the whole quantum state.

The very moment you trim down a quantum state from the idealisation of an infinite monochromatic wave to a short pulse, you are introducing infinitely many frequencies, and you need the group velocity to describe its motion. No other velocity makes any sense. In the relativistic case, it's actually superluminal, as I proved from simple relativistic-dynamical and quantum-dynamical constraints.

Edited November 19, 2023 by joigus
minor correction

[martillo]

1 hour ago, joigus said:

In fact, what you both are saying contradicts the principles of mainstream quantum mechanics --except for free particles.

If the particle is interacting, the so-called Hamiltonian has a potential energy term, and the momentum (the inverse wavelength) does not commute with it, so the principles say no, you can't measure both at the same time. They're called "incompatible."

You both are confusing a component (among infinitely many) of the quantum state with the whole quantum state.

Could I say then that the phase is not measurable but it is determinable? I mean we cannot experimentally measure it but we could determine it. Could it has sense this way? "Matter waves" could be just abstract things in our mind but we should be able to know about their properties like their wavelength, their group velocity and why not their phase velocity? 

"Particles", "waves", everything is just models in our minds but why we cannot have good models with well determined properties? Isn't it the main purpose of Physics to elaborate good models for everything? We should be able to know all the properties of those models even if they were just abstractions in our minds...

Edited November 19, 2023 by martillo

[joigus]

9 hours ago, martillo said:

Could I say then that the phase is not measurable but it is determinable? I mean we cannot experimentally measure it but we could determine it. Could it has sense this way? "Matter waves" could be just abstract things in our mind but we should be able to know about their properties like their wavelength, their group velocity and why not their phase velocity? 

"Particles", "waves", everything is just models in our minds but why we cannot have good models with well determined properties? Isn't it the main purpose of Physics to elaborate good models for everything? We should be able to know all the properties of those models even if they were just abstractions in our minds...

I think it's probably always fair to say that theories are models in our minds. But not just any models. They are under the obligation to explain and predict the world. And that's a very stringent constriction.

As to the phase velocity, the wave function, and its local phase, all indications are that there are part of a "factor" of the world we can move through using mathematics, but quite invisible for us. You seem to feel some discomfort because of this fact, that there are quantities we seem to be forced to talk about but cannot see. But that's part of physics since the inception of QM.

But maybe it was always this way. We cannot directly see energy, or angular momentum. Think about it.

[martillo]

2 hours ago, joigus said:

I think it's probably always fair to say that theories are models in our minds. But not just any models. They are under the obligation to explain and predict the world. And that's a very stringent constriction.

As to the phase velocity, the wave function, and its local phase, all indications are that there are part of a "factor" of the world we can move through using mathematics, but quite invisible for us. You seem to feel some discomfort because of this fact, that there are quantities we seem to be forced to talk about but cannot see. But that's part of physics since the inception of QM.

But maybe it was always this way. We cannot directly see energy, or angular momentum. Think about it.

I still don't get it.

We cannot see energy or momentum but we can determine their values. Not observable, not directly measurable but determinable. Why not the same for the phase velocity? If the energy E and the momentum p can be determined the quantity E/p is determinable and so v_p = E/p is determinable. The uncertainty principle puts a limit in the precision of the determination of course but this does not mean is not determinable. It would be determinable under an appropriated degree of certainty.

[joigus]

1 minute ago, martillo said:

I still don't get it.

We cannot see energy or momentum but we can determine their values. Not observable, not directly measurable but determinable. Why not the same for the phase velocity? If the energy E and the momentum p can be determined the quantity E/p is determinable and so v_p = E/p is determinable. The uncertainty principle puts a limit in the precision of the determination of course but this does not mean is not determinable. It would be determinable under an appropriated degree of certainty.

You were waxing philosophical, "everything is an abstraction", "it's only in our minds". So Ok, I guess in a way you're right. Ultimately, the only thing we see, the only thing we measure, is positions of measuring scales, numbers flashing in LED lights, and the like. So everything is position, colour, time, what may have you. But there's always a theoretical layer beneath that. My point was that there are concepts that seem to demand you to consider them from the theory, although they cannot be measured, directly or indirectly. Example: phase velocity of a quantum wave.

[studiot]

14 hours ago, KJW said:

I see two meanings to "wave" in QM. Firstly, there is "wave" as in "wavefunction", a general term referring to any waveform. But there is also "wave" as in "wave-particle duality" which I see as referring specifically to a sinusoidal waveform. In other words, a particle with definite momentum, and therefore definite wavelength. But also definite energy, and therefore definite frequency. Phase velocity might not be meaningful for all wavefunctions, but that does not make it meaningless for all wavefunctions.

 

13 hours ago, joigus said:

No. You got this totally wrong. Quantum waves are not sinusoidal. For starters, they are not real functions. They typically go like complex exponentials.

 

Actually I agree with the spirit of KJW's paragraph, allbeit the letter is verey loosly worded and certainly the restriction to sinusoidal waves is inappropriate.

With this correction the distinction he makes is relevant to 'matter waves'  - the subject of this thread.

It is really also part of the larger relevant question 'what is a wave ?' and the confusion that has accrued over the last century.

 

It is also true that the physical dimensions (units) for the quantum wave function are weird in that they vary according to the number of spatial dimensions you are working in.

12 hours ago, martillo said:

Could I say then that the phase is not measurable but it is determinable? I mean we cannot experimentally measure it but we could determine it.

Not necessarily.

Applying 'Quantum Uncertainty' to matter waves leads to some interesting results.

 

I am preparing a 'road map' for you so that you can see the development of De Broglies original idea over the last century, where it has gon in and out of mainstream fashion and is currently coming back into fashion at the (not so early) beginning of the 21st century, being revived by such bodies as NASA and CERN.

[martillo]

10 minutes ago, joigus said:

My point was that there are concepts that seem to demand you to consider them from the theory, although they cannot be measured, directly or indirectly. Example: phase velocity of a quantum wave.

What I don't get is that you have E, you have p so, why can't you have E/p. What does the quotient E/p means to you? Why to not just associate v_p = E/p? I don't get it.

 

13 minutes ago, studiot said:

Not necessarily.

Applying 'Quantum Uncertainty' to matter waves leads to some interesting results.

As I already said to @joigus "quantum uncertainty" just puts a limit in the precision on the determination of some quantity. It does not means some quantity would be not determinable. Different things.

[studiot]

17 minutes ago, martillo said:

As I already said to @joigus "quantum uncertainty" just puts a limit in the precision on the determination of some quantity. It does not means some quantity would be not determinable. Different things.

But that would be incorrect.

Quantum uncertainty leads to the conclusion that certain quantites cannot be determined (or known by any means) either by calculation or by measurement.

It is more than just a limit on meassurement, and of quite a different nature from clasical uncertainty as set out so ably in

Robinson and Whittaker

The Calculus of Observations.

Edited November 19, 2023 by studiot

[martillo]

51 minutes ago, studiot said:

I am preparing a 'road map' for you so that you can see the development of De Broglies original idea over the last century, where it has gon in and out of mainstream fashion and is currently coming back into fashion at the (not so early) beginning of the 21st century, being revived by such bodies as NASA and CERN.

I was reading something about that Schrodinger found problems in the development of his equation with some things related to Relativity and so he at the end developed a non relativistic equation. The relativistic Klein-Gordon equation was developed after. Lot of new things being developed quite at the same time in those years...

 

29 minutes ago, studiot said:

But that would be incorrect.

Quantum uncertainty leads to the conclusion that certain quantites cannot be determined (or known by any means) either by calculation or by measurement.

It is more than just a limit on meassurement.

Seems I would need to enter deeper into QM to be able to understand the thing...

I just want to ask you now the same as to @joigus:

34 minutes ago, martillo said:

What I don't get is that you have E, you have p so, why can't you have E/p. What does the quotient E/p means to you? Why to not just associate v_p = E/p? I don't get it.

Edited November 19, 2023 by martillo

[joigus]

49 minutes ago, studiot said:

Actually I agree with the spirit of KJW's paragraph, allbeit the letter is verey loosly worded and certainly the restriction to sinusoidal waves is inappropriate.

With this correction the distinction he makes is relevant to 'matter waves'  - the subject of this thread.

It is really also part of the larger relevant question 'what is a wave ?' and the confusion that has accrued over the last century.

 

It is also true that the physical dimensions (units) for the quantum wave function are weird in that they vary according to the number of spatial dimensions you are working in.

Ok. Yes, thank you for pointing this out. The word "wave" is sometimes used very loosely, for these and other reasons. People also use the word "wave" for solitons, which are objects somewhere in between. I did go too far when I said "totally wrong". My apologies to @KJW.

[studiot]

42 minutes ago, martillo said:

Seems I would need to enter deeper into QM to be able to understand the thing...

I just want to ask you now the same as to @joigus:

55 minutes ago, martillo said:

What I don't get is that you have E, you have p so, why can't you have E/p. What does the quotient E/p means to you? Why to not just associate v_p = E/p? I don't get it.

 

Because in the matter wave interpretation neither p nor E are conserved for short intervals of time, as determined by the quantum uncertainty principle.

This is what I was leading up to when I said

1 hour ago, studiot said:

Applying 'Quantum Uncertainty' to matter waves leads to some interesting results.

 

[martillo]

21 hours ago, studiot said:

Just as a matter of interest I think this whol argument about direct measurement is a red herring.

Any measurement can be called non direct, it just depends upon how far you take it.

For example

If I connect my modern digital multimeter into a circuit and 'directly' measure the current, I am not really measuring the current at all. I am measuring voltage.

If I connect my old fashioned analog meter into that circuit and measure voltage, I am not really measuring voltage at all. I am measuring current.

But then I could take this argument one stage further.

I am not measuring current or voltage at all, I am measuring an illuminated pattern of lines/dots on a screen on one hand and a needle deflection on the other

and I could go on and on down this rabbit hole.

But nobody does in reality.

I have highlighted a point I am particularly interested. As you say voltmeters actually measure current. The current passes through an internal resistance and they take the voltage over that resistance to move a needle or generate a number in a digital display. Two questions:

In which way could a voltage be measured directly?

Is there any known case where the voltage has been measured directly?

I'm particularly interested about high voltage measurements.

Edited November 19, 2023 by martillo

[studiot]

29 minutes ago, martillo said:

In which way could a voltage be measured directly?

Is there any known case where the voltage has been measured directly?

Two methods spring to mind.

In the analog world electrometers and potentiometers measure voltage directly.

In the digital world the digital version of the potentiometer also measures voltage directly, hence my comment about digital multimeters.

 

[martillo]

8 minutes ago, studiot said:

Two methods spring to mind.

In the analog world electrometers and potentiometers measure voltage directly.

In the digital world the digital version of the potentiometer also measures voltage directly, hence my comment about digital multimeters.

 

Thanks. I'll look further on this subject.