Hi all!

(I wrote this article in my own language first, then i used google translator to translate it into English. Then i read whole text and checked and corrected some words)
 

 

This thinking is really based on that i try to make phase velocity to be v with assuming new de Broglie frequency and then i check whether de Broglie wave can be wobble/beating of two components since $ m\gamma v^2/h = mc^2(\gamma - 1/\gamma)/h $. And then i try to calculate phase and group velocities of such beating wave.

 

 

So, to begin with, I could say why I think something like this? I have just enough knowledge of the subject to be able to do some kind of mathematical analysis. This subject belongs to signal analysis and in deeper analysis there are some concepts of quantum mechanics that are unknown to me, such as wave packets and the Dirac equation. Therefore, my review is not very far-reaching.

The reason for my reflection is: There may be an error in the first theory of quantum mechanics, de Broglie theory. Due to the error, the phase velocity of the de Broglie matter wave becomes $ v_{phase} = c^2/v $ which is above the speed of light. The same error may also produce a velocity c for the particle in Dirac's theory, which is the first relativistic theory of quantum mechanics. (Usually this has been tried to be explained by the Zitterbewegung phenomenon.)

Ezzat Bakhoum has already noticed in his 2002 article ("Fundamental Disagreement of Wave Mechanics with Relativity", Physics essays 2002) that if the de Broglie frequency is set to $ f = \frac{m \gamma v^2}{h} $ , the phase velocity of the matter wave becomes v instead of c^2/v . The de Broglie wavelength, on the other hand, is still considered to be

$ \lambda = \frac{h}{p} $
.

Bakhoum also makes other observations that I am not very familiar with.

Unfortunately, in this article Bakhoum draws the wrong conclusion on the matter. He claims that the mass-energy equivalence would be of the form E = mv^2 instead of  E = mc^2 (in these two formulas m is now the so-called "relativistic mass"). This is a wrong conclusion.

I made another thread about "is the total energy of a particle really E=mc^2 but I finally concluded that this is indeed the case.

In the following, I will use the notation m for the rest mass of a particle, so the total energy of the particle is written as

$$ E_{tot} = m \gamma c^2 $$

, where $ \gamma $ is the Lorentz gamma function:

$$ \gamma = \frac{1}{\sqrt{1-(v/c)^2}} = \sqrt{1+(\frac{p}{mc})^2} $$

 

Well, why wasn't this de Broglie frequency error noticed ages ago?
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I don't know. I haven't found any information on what the de Broglie frequency is used for in practice. Instead, the de Broglie wavelength is used successfully in many quantum mechanics calculations and has been experimentally confirmed and very accurately. Maybe this correction doesn't work for some reason in the end of the day. It is also known that the total energy of a particle is $E_{tot} = m\gamma c^2 $ very accurately. But is it known that there is a connection between the total energy and the de Broglie frequency $ E_{tot} = hf $? If the de Broglie frequency is not needed in any significant calculation, could it still be different from $f =\frac{m\gamma c^2}{h}$ ? That's a question I can't answer myself. That's why I'm doing these calculations.

 

First
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There is the following interesting mathematical identity:

$$ m\gamma v^2/h = mc^2(\gamma - 1/\gamma)/h $$

This is a surprising coincidence. This formula resembles the formula for the oscillation frequency. It gave me the idea to think about whether a matter wave could be an oscillation instead of a wave motion with a single frequency.

When the speed of the particle is small, the de Broglie length of the matter wave $ \lambda = \frac{h}{p} $ can be very large. When the speed approaches the speed of light, the length of the matter wave can again approach zero. It might make sense to explain this by saying that the matter wave is an oscillation between two components.

De Broglie is said to have pondered the problem with the de Broglie frequency, is it affected by time dilation and slows down or does it behave like total energy? He chose the latter option. But perhaps it does both, and splits into two components?

The mathematical analysis of a "wobble/beating quantum" or "wobble/beating wave packet" is too difficult for me, so in the following I will only consider the wobble between two frequency components.

 

How did Bakhoum see the de Broglie frequency?
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Bakhoum probably thought that the new de-Broglie frequency $ f = m\gamma v^2 $ was created by just one matter wave and that the de Broglie wavelength was $ \lambda = \frac{h}{m\gamma v} $. Then the phase velocity of the wave becomes $v_{phase} = \lambda f = v $. The group velocity for a single wave is simply $ v_{group} = v $. (er...Or is it?)
 

Analysis of Two-Wave Oscillation
-------

Let's start by setting

$$ f_2 = mc^2 \gamma $$
$$ f_1 = mc^2 1/\gamma $$

where $ \gamma $ is the Lorentz gamma function

$$ \gamma = \frac{1}{\sqrt{1-(v/c)^2}} = \sqrt{1+(\frac{p}{mc})^2} $$

and consider the following function:

$$ F(t) = cos(2\pi f_1 t) + cos(2\pi f_2 t) $$

Using trigonometry, this function can be written as

$$ F(t) = cos[(2\pi)\frac{f_2+f_1}{2}t]cos[(2\pi)\frac{f_2-f_1}{2}t] $$

And let's write this in the form

$$ F(t) = cos[(2\pi)f_{carrier}t]cos[(2\pi) f_{envelope}t] $$

Now the question is, if the function describes a quantum mechanical "wobble" - a matter wave and its quantum, then what is the energy associated with this quantum?

"In a technical sense" it could be possible that there is an energy term $ E_{QM} = m \gamma v^2 $. Which would describe "quantum mechanical energy". But is this real energy? Another option that comes to mind is that the energy of a wobble quantum is calculated in some other way than the energy of a light quantum of the corresponding frequency.

So are there two different energies? Total energy and "quantum mechanical" energy that participates in the matter wave phenomenon? I leave this question open.

It is quite easy to see, just by looking at the arguments of the above formulas, that the total energy of a particle can be expressed in terms of envelope and carrier frequencies as follows:

$$ E_{tot} = (f_{carrier}+f_{envelope})h $$

not

$$ E_{tot} = hf $$
 

The following relation would exist between the oscillation frequency and the total energy:

$$ E_{tot} = m\gamma c^2 = \frac{1}{2}(hf + \sqrt{(hf)^2 + 4mc^2} $$

$$ (hf)^2 = 2( \sqrt{m^2c^4 + E_{tot}^2} - mc^2) $$

These relations are valid if the de Broglie frequency is

$$ f = \frac{mc^2}{h}(\gamma - 1/\gamma) $$

instead of

$$ f = \frac{mc^2}{h}\gamma = \frac{E_{tot}}{h}$$

The quantum energy of a free particle is probably some kind of integral of some kind of wave packet. If the wave packet is indeed an oscillation wave, common sense tells us that its total energy cannot be E = hf , where f is the oscillation frequency, but more than this. An oscillation wave has more movement.

The "total energy of a particle with an beating quantum" corresponds to the sum of the energies of two photons with frequencies $f_{carrier} = \frac{mc^2}{2h}(\gamma + 1/\gamma) $ and $f_{envelope} = \frac{mc^2}{2h}(\gamma - 1/\gamma)$.
 

What happens at low speeds?
---

Let's now examine these frequencies $ f_{carrier} $ and $f_{envelope} $ , when the particle's speed is low.

We can make a Maclaurian series expansion of each of them, and extract the first two terms from them

$$ f_{carrier} \rightarrow \frac{mc^2}{h} + 0 $$

$$ f_{envelope} \rightarrow 0 + \frac{mv^2}{2h} $$

The formulas approach the formulas for rest energy and kinetic energy divided by h.

In this context, one can think that the carrier frequency can almost be omitted because it is very high. In this case, the envelope frequency is left. I am not sure if this corresponds to the frequency of one of the solutions of the Schrödinger equation.

The above formula seems to produce the same energy as a single light quantum, or photon, whose frequency happens to be

$$ f = \frac{m\gammac^2}{h} $$

if the particle's speed approaches the speed of light, f1 approaches zero, so the wobble disappears and only the component f2 remains. In this case, the formula E = hf is still valid.

 

Some thoughts
-----
So it seems that the formula E = hf does not hold when the matter wave is wobbling/beating. The de Broglie frequency instead of being $f_{broglie} = \frac{E_{tot}}{h}$ would be

$$ f_{broglie} = f_2 - f_1 = mc^2(\gamma - 1\gamma) $$

Note that the wobbling frequency $f = f_2-f_1 $ is actually twice as large as the Envelope frequency. The wobbling frequency f is what is observed, for example, in music when two notes are wobbling close together.

Now at this point I will present the identity from the beginning again:

$$ mc^2(\gamma - 1/gamma) = m\gamma v^2 $$

which holds for all speeds 0 < v < c.

The de Broglie frequency can therefore also be written in the form

$$ f_{broglie} = mc^2(\gamma - 1/\gamma)/h = m\gamma v^2 /h = pv/h $$

So de Broglie may have postulated the wrong de Broglie frequency in his theory. The de Broglie wavelength postulate is correct instead and the wavelength remains the same, which is known to be

$$ \lambda = \frac{h}{p} = \frac{h}{m\gamma v} $$

It seems that the de Broglie wavelength postulate is independent of the de Broglie frequency postulate. It does not tell whether there is wobble in the matter wave.
 

Phase and Group Velocities of the beating Wave
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Now we could calculate the phase and group velocities of the new de Broglie wave.

The oscillation frequency formula can be written in the following form:

$$ F(x,t) = 2cos[kx-wt]cos[\Delta k x - \Delta wt] $$

and the following formulas apply to the phase and group velocities of these two waves:

$$ v_{phase} = \frac{w}{k} $$

$$ v_{group} = \frac{\Delta w}{\Delta k} $$

The angular frequency and change in angular frequency are then:

$$ \omega = \pi (\gamma + 1/\gamma)(\frac{mc^2}{h}) $$

$$ \Delta \omega = \pi (\gamma - 1/\gamma)(\frac{mc^2}{h}) $$

In order to obtain the phase velocity of the oscillation wave as v, $\Delta k$ must be

$$ \Delta k = \frac{1}{2}\frac{2\pi m}{h}\gamma v $$

,then the wavelength is

$$ \Delta \lambda = 2\pi/\Delta k =2 (\frac{h}{m \gamma v}) $$

At this point, it must be remembered that the observed oscillation frequency is twice the actual frequency, with the same logic the observed oscillation wavelength is half of this. (or is it?)

The wavenumber of the carrier wave depends on the phase velocity. The following relation exists between the phase velocity and the wavenumber of the carrier wave:

$$ v_{phase} = \frac{w}{k} = \frac{1}{k}(\frac{2\pi mc^2}{h})(\gamma + 1/\gamma) $$

is the carrier wave velocity, or phase velocity, v , or is it c or some velocity in between?

But it remains open, how do the actual frequencies, wavenumber and the above-mentioned velocities behave in a wave packet?

 

Application to the Schrödinger equation - some thoughts
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How does such a beating wave of matter fit into the Schrödinger equation? The equation in question is a postulate, or initial assumption, of non-relativistic quantum mechanics.

Thinking in this way with common sense, the complex wave function that comes as a solution to the equation probably remains the same otherwise, but it adds a very high-frequency carrier component, whose frequency is

$$ f = \frac{mc^2}{h} $$

. The carrier frequency is precisely this expression when the speeds are very small compared to the speed of light. What the wave number of the carrier wave would be remains open in this reflection.

Such a "second-order" carrier wave would perhaps produce its own interference pattern in a double-slit experiment, but this is very difficult to distinguish in practice, because the carrier frequency is very high. This frequency is of the same order as the frequency of the Zitterbewegung phenomenon. This phenomenon was needed to explain why the speed of the Dirac particle is the speed of light. According to the phenomenon, the particle moves from its position at a very fast frequency so that its average speed becomes v.
 

The end.

 

Edited November 19, 20241 yr by caracal
problems with latex math

46 minutes ago, caracal said:

The reason for my reflection is: There may be an error in the first theory of quantum mechanics, de Broglie theory. Due to the error, the phase velocity of the de Broglie matter wave becomes $ v_{phase} = c^2/v $ which is above the speed of light. The same error may also produce a velocity c for the particle in Dirac's theory, which is the first relativistic theory of quantum mechanics. (Usually this has been tried to be explained by the Zitterbewegung phenomenon.)

No there is no error in De Broglie's theory.

But you are not wrong either.

DB theory was just too simple.

In fact Sommerfield used relativistic version to obtaina good match with the observed fine structure of hydrogen.

Matter-wave interference confirms the deBroglie equation.

Hi, Can you give an example of such experiment where de Broglie frequency is measured in practice? Is it confirmed experimentally in some experiment or calculation? De Broglie wavelength is confirmed for sure, i know that.  Double slit experiment for example only needs formula of de Broglie wavelength in order to calculate the interference pattern behind the slit.  It does not need any information on de Broglie frequency.

Double slit experiment model only needs this formula

$$ \lambda = \frac{h}{p} $$

And then you just need the equation for constructive or destructive interference.

One of the first confirmation tests was the Davisson-Germer experiment.

https://en.m.wikipedia.org/wiki/Davisson–Germer_experiment

DeBroglie also ties into photoelectric effect ie quantum Tunneling such as done by Einstein for which he received a Nobel prize. (Matter wave duality is needed)

Edit the first test is a common classroom test used today so you should find plenty of example and related articles.

Edited November 20, 20241 yr by Mordred

Hi, thanks for the link i'll read it

The equation

$$ E = hf $$

would Still be valid for photon that has velocity c. I am not saying that it would be any different for photon.

And the de Broglie frequency if it is

$$ f = \frac{m\gamma v^2}{h} $$

it would become also close to

$$ f \rightarrow \frac{m\gamma c^2}{h} = \frac{E_{tot}}{h}$$

when the velocity of particle approaches light velocity.

Photoelectric effect only confirms that light is consisting of photons (or either absorbed as energy packets in metal) whose energy is E = hf. The maximum kinetic energy of electrons leaving the metal is

$$E_{kin_max} = hf - W $$

Where W is a work needed for electrons to get out from the metal.

I looked this wikipedia article, i need to read it better.

But looks like Crystal scattering experiment model also needs only equation for de Broglie wavelength for electron, not frequency and equation for constructive/destructive interference. It is similar experiment than double slit experiment.

If the electrons phase velocity and Group velocity are both v, then the equation for de Broglie frequency would be

$$ f \lambda = v $$

$$ \rightarrow f = v/\lambda = \frac{m \gamma v^2}{h} = \frac{m \gamma c^2}{h}(\gamma - 1/\gamma) $$

(This is exactly a term what bakhoum was thinking. i also was thinking could de Broglie frequency be beating, that is a difference of two frequency components. )

But as commonly what is accepted for de Broglie frequency being

$$ f = \frac{m \gamma c^2}{h} = E_{tot}/h $$

Then this equation between wavelength, frequency and velocity does not hold. The phase velocity is theoretically thought to be c^2/v while group velocity is v.

But is this only a theoretical argument?

Edited November 20, 20241 yr by caracal
question added into end

If your looking into group velocity you need a superposition of waves for example if the two waves have different momentum terms the wavepacket will continually change as it's travelling.

 

1 hour ago, Mordred said:

If your looking into group velocity you need a superposition of waves for example if the two waves have different momentum terms the wavepacket will continually change as it's travelling.

And different speeds (since this is for massive particles), so the two elements of it will separate, which is not what we observe happening.

Thankfully just using energy momentum relation greatly simplifies things hence we rarely use De-Broglie

5 hours ago, Mordred said:

Thankfully just using energy momentum relation greatly simplifies things hence we rarely use De-Broglie

As someone who worked on an atom interferometer (in grad school), I’m in that minority.

So am I when dealing with spectroscopic examinations as group velocity dispersion occurs in interstellar mediums as well as spacetime vacuum.

In many ways using DeBroglie waves is far simpler than wave vector methodologies such as via a Fourier transformation which gets incredibly tricky when dispersion occurs.

Particularly since the further you look the mean average density increases which has huge ramifications on dispersion as the medium density isn't consistent.

Extremely useful though for determining the medium properties.

For the OP De-Broglie waves becomes useful as you will see dispersion in matter waves as the waves propogate through any medium including quantum and spacetime vacuums.

It's extremely common it's that most times the mathematical treatments have already factored in the De-Broglie relations for example under QM treatments so it tends to get missed as being involved.

Here examine this article to get a better understanding of phase velocity and group velocity including dispersion.

https://www.mlsu.ac.in/econtents/784_PHASE VELOCITY AND GROUP VELOCITY.pdf

Make note of the detail that the phase velocity doesn't transport energy in terms of your opening post where phase velocity can exceed c but as no energy is transported there is no violation.

This link provides some additional details 

https://www.sciencedirect.com/topics/physics-and-astronomy/phase-velocity#:~:text=It can also be greater,not carry energy or information.

In essence keep matter velocity, phase velocity and group velocity as seperate as each has its own distinctive relations with regards to De-Broglie waves. 

Edited November 21, 20241 yr by Mordred

13 hours ago, caracal said:

 

Then this equation between wavelength, frequency and velocity does not hold. The phase velocity is theoretically thought to be c^2/v while group velocity is v.

But is this only a theoretical argument?

That should help answer this question. Hint will also help to identify the longitudinal and transverse relations of group velocity and phase velocity.

Edited November 21, 20241 yr by Mordred