Questions about deBroglie (matter) waves

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The discussion centers on de Broglie waves, which are fundamentally a concept of quantum mechanics, specifically related to wave-particle duality. Participants explore the justification for de Broglie's hypothesis that particles with mass should exhibit wave-like properties, akin to photons. Key points include the relationship between frequency and total energy, the visualization of de Broglie waves in space-time diagrams, and the experimental verification of de Broglie's predictions through phenomena like Bragg and Fresnel diffraction. The conversation also touches on the implications of the Pauli Exclusion Principle and the complexities of combining quantum mechanics with general relativity.

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JDoolin
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I'd like to ask some questions about deBroglie waves.

(1) are deBroglie waves a topic under classical physics, or does it belong under quantum mechanics, or relativity?

(2) The justification for deBroglie expecting matter waves is "well, photons have frequency and wavelength, so it only makes sense that particles with mass should have frequency and wavelength, too." I have trouble believing that someone would be awarded a PhD, let alone a Nobel Prize for such an insipid analogy. Does anyone have any more detail on his reasoning?

(3) The wikipedia article alternates between saying the frequency is proportional to the kinetic energy (in the text) and the total energy (in the equations.) I am not sure I trust either idea. But on reading the wikipedia discussion page, I believe deBroglie originally meant for the Total Energy to be used.

However, if the (at rest) frequency is proportional to the rest mass energy, and the (at rest) wavelength approaches infinity. This can't be visualized. On the other hand, if you presume that the particle cannot be at rest (cf. Heisenberg's uncertainty principle), the wavelength is perhaps constrained by the container in which the particle is located?

(4) I am particularly curious about whether a space-time diagram of such a wave could be produced. The stationary particle would consist of a straight line. But would the waves, oscillating up and down form some surfaces in space-time.

I have an exercise from MTW's "Gravitation" Exercise 2.1 which gives
\psi = e^{i \phi}=exp[\vec k \cdot\vec x-\omega t]​
as the "quantum mechanical properties of a de Broglie wave"

The problem is, as given, this equation is a nonlocalized wave; extending throughout space with the same amplitude! An actual particle should have a finite extension in space. Does de Broglie's explanation have some implicit boundary conditions?

(5) The experimental verification of de Broglie's hypothesis (Bragg diffraction, Fresnel diffraction, etc.) all seem to verify that the wavelength is in line with that predicted by deBroglie, but I don't see how these experiments verify the frequency is as deBroglie predicted.
 
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All good questions!
JDoolin said:
(1) are deBroglie waves a topic under classical physics, or does it belong under quantum mechanics, or relativity?
Quantum Mechanics; any time you're talking about the wave-particle duality (or related concepts) you're in QM. Another god rule of thumb: anytime an equation involves an 'h' or 'h-bar'---you're also in the QM realm (or beyond).

JDoolin said:
(2) The justification for deBroglie expecting matter waves is "well, photons have frequency and wavelength, so it only makes sense that particles with mass should have frequency and wavelength, too." ...
That's almost right. The key argument is, 'well, photons act as both particles and waves, so maybe particles with mass will as-well.' Its beauty does, exactly, lie in its incredible simplicity. Its significance for a PhD and nobel prize lies in the repercussions and paradigm-shift that followed from it.

JDoolin said:
(3) The wikipedia article alternates between saying the frequency is proportional to the kinetic energy (in the text) and the total energy (in the equations.) I am not sure I trust either idea. But on reading the wikipedia discussion page, I believe deBroglie originally meant for the Total Energy to be used.
I believe it should technically be the kinetic energy. Often people leave it implied, and even more often they will be proportional to each-other, and the distinction doesn't matter too much for qualitative purposes.

JDoolin said:
(4) I am particularly curious about whether a space-time diagram of such a wave could be produced. ... \psi = e^{i \phi}=exp[\vec k \cdot\vec x-\omega t] ...
The problem is, as given, this equation is a nonlocalized wave; extending throughout space with the same amplitude!
Keep in mind that QM and 'space-time' (i.e. general relativity) have yet to be adequately combined. A perfect wave does extent through all of space-time, and thus couldn't be draw (very well at least) in a space-time diagram. A wave-packet would generally just be drawn as a particle on a space-time diagram.

JDoolin said:
(5) The experimental verification of de Broglie's hypothesis (Bragg diffraction, Fresnel diffraction, etc.) all seem to verify that the wavelength is in line with that predicted by deBroglie, but I don't see how these experiments verify the frequency is as deBroglie predicted.
The wavelength and frequency are directly related to each other---establishing one defines the other.
 
JDoolin said:
(3) The wikipedia article alternates between saying the frequency is proportional to the kinetic energy (in the text) and the total energy (in the equations.) I am not sure I trust either idea. But on reading the wikipedia discussion page, I believe deBroglie originally meant for the Total Energy to be used.

zhermes said:
I believe it should technically be the kinetic energy. Often people leave it implied, and even more often they will be proportional to each-other, and the distinction doesn't matter too much for qualitative purposes.

The frequency goes with the total (relativistic) energy, i.e. rest-energy plus kinetic energy:

E = E_0 + K = \frac {m_0 c^2} {\sqrt {1 - v^2 / c^c}}
 
jtbell said:
The frequency goes with the total (relativistic) energy, i.e. rest-energy plus kinetic energy:

E = E_0 + K = \frac {m_0 c^2} {\sqrt {1 - v^2 / c^c}}
Woops, right, definitely. I was stuck on thinking about it NOT including any potential energy terms... sorry!
 
"Matter waves" don't really exist, de Broglie was awarded the nobel for proposing wave-particle duality or more precisely "for the discovery of the wave nature of electrons"( nobelprize.org), which was confirmed by electron diffraction experiments. de Broglie had the wrong idea about what the waves represented, but Schrödinger was able to use his brilliantly original idea to create a full wave mechanics compatible with Heisenberg's matrix mechanics.

As you probably know, the wave had to be interpreted as a probability wave, and described by a fundamentally complex wave equation preventing any possibility of a real-valued quantity being associated with the wave.

A translation of de Broglie's phd thesis ("On the theory of quanta") and many other historical documents can be found here

You can decide for yourself if it was just a trivial (and lucky) guess or brilliantly inspired :smile:
 
unusualname said:
As you probably know, the wave had to be interpreted as a probability wave, and described by a fundamentally complex wave equation preventing any possibility of a real-valued quantity being associated with the wave.

This statement may be too strong:https://www.physicsforums.com/showpost.php?p=3008318&postcount=11
 
Hi again.

I thought I'd mention a couple of things. Since the frequency has to do with total energy (rest mass energy plus kinetic energy), while the wavelength is just the momentum, it means the wavelength and frequency are not proportional, and are not related by a simple formula. (depending on your idea of simple.)

For instance, if a particle had zero velocity, it's frequency would be proportional to its rest mass, while it's wavelength would be zero. However, quantum mechanics (Pauli Exclusion Principle) says no two fermion particles can occupy the same quantum state. I don't know if this question is particularly clear, but is this related to reasoning related to deBroglie wavelengths, saying no particle can have a zero wavelength?

I guess I've never heard comprehendible reasoning behind Pauli Exclusion Principle, so that's why I'm speculating they are related. If I recall correctly, the closest I ever heard to an explanation of Pauli had something to do with statistics and the moment generating function, but I didn't know enough about statistics for it to make sense. (probably still don't)
 
http://physics.about.com/od/quantumphysics/a/dbhypothesis.htm

Here is another reference that explicitly uses kinetic energy, instead of total energy for the deBroglie frequency.
For slow moving massive particles, the difference is drastic.
For massive particles at rest, the difference is infinite.
I notice that Wikipedia gives E=hw for matter waves and E=hw/2 for the zero point energy of a harmonic oscillator.
(where h=h-bar and w = omega)
Is this just a coincidence,
or is the zero point energy related to the deBroglie frequency?
 
A standing wave has an equation like:
y=sin(2 \pi \frac{x}{\lambda})sin(2 \pi \frac{t}{T})

While a traveling wave has an equation like:
y=sin(2 \pi (\frac{x}{\lambda}-\frac{t}{T}))

And a deBroglie wave has an equation somewhere in between?
 
  • #10
Read De Broglie's nobel lecture, it explains how he thought of his waves

http://nobelprize.org/nobel_prizes/physics/laureates/1929/broglie-lecture.pdf
 
  • #11
Hi,
I realize that this thread is a little old and some of the participants may not be around, but I was wondering if anybody might be able to clarify the basis of the deBroglie wavelength, which ties in with some of the questions raised about whether the energy is related to kinetic or total energy. If I start with the Compton wavelength, it seems to fall out of combining Einstein’s and Planck’s energy equations, i.e.
[1] E=mc^2 = hf = \frac {hc}{\lambda}
[2] \lambda = \frac {h}{mc} = \frac {h}{\rho}
In the context of a photon, we have the wavelength being inversely proportional to momentum of the photon, where [m] presumably is representative of its kinetic mass. If I directly apply this logic to a particle, the momentum would also be a function of its kinetic mass [m], which is presumably related to its rest mass via gamma, i.e.
[3] \lambda = \frac {h}{\rho} = \frac {h}{mv} = \frac {h}{\gamma m_0 v}
This seems reflective of the deBroglie wavelength of a particle, where the wavelength is inversely proportional to velocity [v]. However, if I start with the relativistic energy equation, http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/releng.html" , I appear to end up with a different interpretation:
[4] E^2 = m_0^2c^4 + \rho^2c^2 = h^2f^2 = \frac {h^2c^2}{\lambda^2}
[5] \lambda^2 = \frac {h^2}{m_0^2c^2 + p^2} = \frac {h^2}{m_0^2c^2 + \gamma m_0^2v^2}
On this basis, the previous form of the deBroglie equation is lost and only when the velocity [v] approaches [c] would the kinetic component swamp the rest component and revert to the standard form of the deBroglie equation. However, presumably time dilation might also be a factor at these speeds? I would appreciate any clarification of this issue on offer. Thanks
 
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  • #12
Hi mysearch,

Probably the most simple way to derive the equations for a photon is to start with the general relativistic equations for energy and momentum of a particle with mass

E = \frac {m_0 c^2}{\sqrt{1 - \beta^2}}
\rho = \frac {m_0 v}{\sqrt{1 - \beta^2}}

Eliminate m to get

\rho = \frac {Ev}{c^2}

De Broglie assumes \ \ \ \ \ \ hf = E \ \ \ \ \ \ where the phase velocity of the wave packet is \ \ \ \ \ \ v_{phase} = \frac {c^2}{v} = \frac {c}{\beta}

Then \ \ \ \ \ \ \frac {hf}{v_{phase}} = \frac {Ev}{c^2} = \rho

For a photon \ \ \ v = c \ \ \ so that \ \ \ v_{phase} = \frac {c^2}{c} = c

Whence solving for energy and momentum

E = hf \ \ \ \ \ \ \rho = \frac {hf}{c}
 
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  • #13
Hi PhilDSP,

Thanks for the feedback. By way of background, I have just read Manjit Kumar’s book ‘Quantum’ which is an interesting account of the historical developments leading to quantum theory that has renewed my interest in this subject. As such, I have only just started to look into the underlying science, plodding my way through blackbody radiation, photoelectric effect, Bohr atom, Compton effect and now deBroglie, i.e. so still have a way to go before Heisenberg, Schrödinger, QED, QCD, QFT and the rest. Anyway, while I agree with your equations, they still seem to lead to some interesting questions, e.g.
v_{phase} = \frac {c^2}{v} = \frac {c}{\beta}
The implication of this equation appears to be that the phase velocity vp is always greater than c. As I understand it, de Broglie thought of his matter waves in terms of a pilot wave that guided the particle, but not too sure on this point as yet. Other sources also introduce the concept of phase and group velocity, where it is the group velocity vg that represents the velocity of the observed particle. However, I only understand these concepts in terms of wave superposition, e.g. standing waves, which required 2 underlying physical waves to exist. It is not clear to me, yet, how quantum theory describes these waves, maybe Schrödinger wave equations will enlighten me when I get to them. Returning to my post #11, I have a problem with equation [1] when applied to matter waves because there is an assumption that:
c=f \lambda not v=f \lambda
If you apply the latter then deBroglie’s equation would become:
E=mc^2 = hf = \frac {hv}{\lambda}
\lambda = \frac {hv}{\gamma m_0 c^2}
While this reflects the phase velocity vp=c^2/v, it is the group velocity vg implied by deBroglie. The denominator also aligns to relativistic energy, which then returns to the issue raised in post #11 regarding equation 5.

Many Thanks
 
  • #14
The wikipedia article on matter waves (http://en.wikipedia.org/wiki/Matter_wave#The_de_Broglie_Relations ) clarifies some of these points -- E is kinetic, and phase velocity >= c, although I think there are some non-vacuum exceptions to the latter.

In my limited experience the wavelength calculation was always based on momentum, hence the use of kinetic E. (Also, the case of 0 momentum yields infinite, not 0 wavelength.)
 
  • #15
Hi,
Thanks for the link to wikipedia, I had read this article and might have also triggered the originator of this thread to ask for clarification between kinetic and total energy in this case:
JDoolin said:
(3) The wikipedia article alternates between saying the frequency is proportional to the kinetic energy (in the text) and the total energy (in the equations.) I am not sure I trust either idea. But on reading the wikipedia discussion page, I believe deBroglie originally meant for the Total Energy to be used.
zhermes said:
I believe it should technically be the kinetic energy. Often people leave it implied, and even more often they will be proportional to each-other, and the distinction doesn't matter too much for qualitative purposes.
jtbell said:
The frequency goes with the total (relativistic) energy, i.e. rest-energy plus kinetic energy.
JDoolin said:
Since the frequency has to do with total energy (rest mass energy plus kinetic energy), while the wavelength is just the momentum, it means the wavelength and frequency are not proportional, and are not related by a simple formula…..For instance, if a particle had zero velocity, it's frequency would be proportional to its rest mass, while it's wavelength would be zero.
I think the statement highlighted in bold is incorrect and is, as you pointed out, infinite based on the accepted deBroglie equation, where velocity is in the denominator. However, I think the point made about the relationship between frequency and wavelength is a good one. Based on equating Einstein’s and Planck’s energy equation there appears to be a direct relationship between frequency and mass. If mass increases due to gamma at relativistic speed then it would seem that so must frequency, i.e. this frequency of a particle goes to infinity as the particle approaches c. However, the logic of the deBroglie wavelength seems to conform to equations [1-3] in post #11, but as pointed out in post #13, this logic appears to be predicated on frequency*wavelength=c, i.e. this wave would have to propagate at c, irrespective of the velocity of particle. However, if you come at this from a wave perspective you appear end with the idea of 2 waves, presumably in superposition, which then define two velocities, i.e. phase and group:
v_p = \frac {\omega}{k} = \frac {E}{p}
v_p = \frac {d \omega}{dk} = \frac {dE}{dp}
As PhilDSP showed in post #12, the phase velocity vp of a particle is always greater than c, while the group velocity reflects the kinetic velocity v<c. See Wikipedia:

Phase velocity: http://en.wikipedia.org/wiki/Phase_velocity
Group velocity: http://en.wikipedia.org/wiki/Group_velocity

What is unclear to me, at this stage, is the nature of these waves. I understand superposition in terms of 2 physical waves that combine to produce a standing wave. In this context, the standing wave is more of an interference pattern, but which contrary to its name, can propagate with a group velocity vg. However, this wave would result due to the inference of 2, or more, waves each having a specific phase velocity. However, I also have only limited knowledge on these matters, which is why I am raising these questions, while still trying to read further into the issues. As JDoolin pointed out, there appears to be quite a bit of conflicting information on this subject. Any way, appreciate any insights on offer.
 
  • #16
The phase velocity is always greater than or equal to c as KWillets pointed out. It has a reciprocal relationship with group velocity around c where both are equal (at c) for photons.

The phase velocity is not regarded as being something physical. It is the speed and direction of the plane in which the phase is constant and unvarying. It really has nothing to do with superposition other than very generally a wave packet can be analyzed as consisting of a collection of sinusoidal waves (as is done when a Fourier transform is applied).

As for your use of m_0 in those equations, that may only be applied to a particle with mass. If you want to generate equations for the photon you must eliminate the mass term in some way because the photon (in SR) is defined as having no rest mass. Otherwise I think your equations are correct.

If you are starting from the relativistic energy or momentum equations as de Broglie did, then you've already applied, in principle, the Lorentz transformation so that the subsequent equations assume the observer's frame of reference.
 
  • #17
Just noticed post 11 assumes the phase velocity is c in the wavelength/frequency relation [1]. That's incorrect; it's dispersive.

I'm just now reading http://nobelprize.org/nobel_prizes/physics/laureates/1929/broglie-lecture.pdf with the explanation of these relations. He derives frequency without wavelength when discussing mass-energy, and this frequency varies relativistically with velocity in his discussion of momentum.
 
  • #18
Hi,
Many thanks for the useful insights, it is giving me something to mull over. If possible I would like to outline some of the maths and physical interpretation that I am trying to resolve in my mind. Bear in mind that I am not proposing a solution only trying to understand the accepted position within quantum theory.
KWillets said:
Just noticed post 11 assumes the phase velocity is c in the wavelength/frequency relation [1]. That's incorrect; it's dispersive.
I agree with your comment, although I would like to clarify whether the dispersion being described is in terms of frequency dispersion, e.g. frequency diffraction through a prism, or velocity dispersion due to the index of the medium.

How are these concepts being applied to a photon or electron wave propagating through a vacuum, which is non-dispersive in the case of light, i.e. is the phase or group wave affected?

The point being made with equation [1-3] in post #11 and then followed up in #13 was that the velocity of any wave is defined by its frequency*wavelength. In this respect, I described a standing wave, i.e. v=0, as a superposition or interference pattern of 2 waves, because I am assuming ‘real’ waves have to have a velocity. This assumption does not stop a standing or superposition wave from having a group velocity, which reflects the frequency and wavelength of the superposition wave envelope. One of the key issues I am trying to understand is the definition of momentum used in the various equations in this thread and in deBroglie’s lecture paper cited:
\lambda = \frac {h}{p}
For a photon this appears to be unambiguous, i.e. follows from Einstein’s and Planck’s energy equations:
E=mc^2=hf
In the case of a photon, the propagation velocity appears to be unambiguously defined by c, such that we can substitute for frequency, arriving at what appears to be the Compton wavelength of a photon:
\lambda = \frac {h}{mc}= \frac{h}{p}
Just substituting p=mv to get the deBroglie wavelength seemed suspect, because the previous logic was predicated on c not v. So what about some other wave, which may not propagate at c? I tried to highlight this issue in post #13 by citing the following equations:
E=mc^2 = hf = \frac {hv}{\lambda}
\lambda = \frac {hv}{\gamma m_0 c^2}
If we equate the previous equations for wavelength we get:
\frac{h}{p}= \frac {hv}{\gamma m_0 c^2}
p= \frac {\gamma m_0 c^2}{v}
Where the term (c^2/v) conforms to the definition for the phase velocity, as shown in #12. We can also establish the phase velocity via another route:
v_p=f \lambda
v_p=\frac {E}{h} \frac {h}{p} = \frac {E}{p}
If we substitute the relativistic energy expression touched on in post #11, we get another form of expression for the phase velocity of a photon and a particle:
v_p=\frac{ \sqrt{m_0^2c^4 + p^2c^2}}{p}
v_{photon}=\frac{pc}{p}=c
v_{particle}=\frac{mc^2}{mv}=\frac {c^2}{v}
Now deBroglie touches on some of these relationships on p.249, but in his definition it seems that he is using the phase velocity to justify the wavelength = h/p. While I have not read or checked the entire paper carefully, I not sure about this relationship, especially if I take the following quote at face value:
PhilDSP said:
The phase velocity is not regarded as being something physical. It is the speed and direction of the plane in which the phase is constant and unvarying. It really has nothing to do with superposition other than very generally a wave packet can be analyzed as consisting of a collection of sinusoidal waves (as is done when a Fourier transform is applied).
We can also establish a framework for the idea of group velocity using a standard wave equation and the assumption of kinetic energy:
v_p= \frac {dE}{dp}
E=\frac {mv^2}{2}= \frac {p^2}{m}
v_{g(photon)}= \frac {p}{m} = \frac {mc}{m} = c
v_{g(particle)}= \frac {p}{m} = \frac {mv}{m} = v
So, at one level, we appear to see the use of the idea of both phase and group waves, but it is unclear to me how to physically interpret some of the ideas being outlined. So my questions, not assertions, are really about trying to understand the physical interpretation of the waves being discussed.

What substance is a ‘point’ particle, like an electron, made of?
Does the electron have some underlying physical wave structure?
If so, where can I find a coherent description?

Sorry, this post is a bit long, but it would be really useful to get a better perspective of what appears to be a fundamental issue before moving on to other ideas in quantum theory. Thanks.
 
  • #19
mysearch said:
So, at one level, we appear to see the use of the idea of both phase and group waves, but it is unclear to me how to physically interpret some of the ideas being outlined.

What substance is a ‘point’ particle, like an electron, made of?
Does the electron have some underlying physical wave structure?

There is no generally accepted interpretation, beyond the fact that we can use this wave function to predict the probabilities of outcomes of experiments. There are a number of possible interpretations which all (so far) predict the same results for experiments, so we cannot distinguish among them experimentally. People argue about it a lot, sometimes with an intensity that approaches religious fervor. :rolleyes:
 
  • #20
I just want to say I'm glad you guys are discussing this again. I haven't really spent any time thinking about it since my last post in this thread.

Is there a relationship between the Schrödinger Wave equation, and the DeBroglie wave? Is the DeBroglie wave the solution to some differential equation, like the Schrödinger Wave?
 
  • #21
mysearch said:
How are these concepts being applied to a photon or electron wave propagating through a vacuum, which is non-dispersive in the case of light, i.e. is the phase or group wave affected?

I've never heard the terms "phase wave" or "group wave" and don't think they're proper and accepted concepts if you have heard them used. There is a wave or a wave packet that can be described as having particular characteristics such as a group velocity or a phase velocity. They merely help describe the wave and how it either changes in time and position or doesn't change. These essentials of Wave Mechanics go back to Lord Raleigh over a hundred years ago and in the 60's were more thoroughly investigated by L. Brillouin.

You might think instead of a wave as consisting of separate harmonics - a sine wave at 1 Hz plus a sine wave at 2 Hz plus a sine wave at 3 Hz, for instance. Combined, because the principle of superposition means their amplitudes add linearly, they produce a new wave with a distinct and fixed pattern. The pattern repeats itself at the interval synchronized with the 1 Hz wave component.

If no dispersion occurs within the medium they are passing then all frequencies remain unaltered - 1 Hz plus 2 Hz plus 3 Hz. In that case, group velocity equals phase velocity. This happens with EM waves in a vacuum - no dispersion.

If dispersion occurs the frequencies will be altered. Possibly .9 Hz plus 1.7 Hz plus 2.5 Hz for example. Group velocity should correlate with the transition time of the lowest frequency harmonic (.9 Hz). You can see that the other harmonics are no longer synchronized in phase. The phase velocity will be much higher than the group velocity in this case. This happens with EM waves in the vicinity of charges (plasmas, gasses, liquids and solid substances)

mysearch said:
What substance is a ‘point’ particle, like an electron, made of?
Does the electron have some underlying physical wave structure?
If so, where can I find a coherent description?

To offer another take on this, it's not resolved yet what the structure of an electron definitely is or whether waves produce the structure or are rather a result of it. A point particle is of course a mathematical abstraction that radically simplifies very many problems. It signifies that the energy and forces around the point are spherically symmetric and centered at the point - not necessarily that they have no spatial extension. Some types of diffraction experiments on electrons fail to find a definite diameter at the power levels tested.

One hint at the structure of an electron was given by Paul Dirac. But to follow his reasoning and mathematical description requires a fairly heavy and deep study of generally lesser known aspects of QM.
 
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  • #22
The purpose of this post is mainly to log some issue for future reference, as I need to consider other aspects of the quantum model, which might then give me a better perspective of the de Broglie wavelength. My main issue, at this stage, is what appears to be the somewhat circular logic that led de Broglie to derived his equation in 1924, prior to the first experimental verification in 1927 via an electron scattering/diffraction experiment – http://www.qudev.ethz.ch/phys4/PHYS4_lecture02v1_2page.pdf" . On this basis, de Broglie’s equation seems to be verified in the form:
\lambda = \frac {h}{mv} = \frac {h}{p}
However, if you initially generalise the wave propagation to v, not c, using the energy relationship as a starting point, it seems difficult to arrive at the de Broglie equation, e.g.
v=f \lambda
E=mc^2=hf= \frac {hv}{\lambda}
\lambda = \frac {hv}{mc^2}
In this context, the velocity in question appears to be the phase velocity, not the group velocity, which has been shown to be related to the kinetic energy of the particle. However, if we assume the relationship wavelength=h/p, as defined by deBroglie, we arrive at the following definition of the phase velocity of a particle, which exceeds c. However, it has been suggested that this should only be seen as a mathematical, not a physical, interpretation.
v_p=f \lambda= \frac {E}{h} \frac {h}{p} = \frac {E}{p} = \frac {mc^2}{mv} = \frac {c^2}{v}
However, the various experiments appear to have verified that the deBroglie equation is correct, although in this case p=mv corresponds to the kinetic energy, where v now appears to be linked to the idea of group velocity. In this context, it would appear that a particle is actually being described in terms of finite wave packet, which can be mathematically constructed in the form of many ‘harmonic’ waves in superposition, all presumably propgating faster than c, at least, mathematically. Clearly, whether any of these waves have any physical interpretation may be debated, but quantum theory seems to be satisfied that when described in terms of probability, it can justify the position of a particle within the inference patterns observed. As such, physics seems to have a means of prediction, but not necessarily a means of physically describing the nature of the wave-particle duality. I would appreciate correction of any of the points raised. Thanks
 
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  • #23
I'm having some problems too, and I found some vindication here: Resolving Inconsistencies in de Broglie's Relation http://www.ptep-online.com/index_files/2010/FPP-20-04.pdf . I've also been struggling with the issues noted in section 1 ("three different frequencies"). De Broglie seems to introduce factors which are Lorentz-variant, after an argument based on Lorentz-invariance and relativity of simultaneity (pg. 247).

Some of the trouble may arise from his assumption of an infinite speed of propagation in the rest frame; this defeats the idea of a wave at all, and suggests something like a clock in an SR thought experiment, which, in the rest frame, has a single value (phase) across all space, yet becomes spatially variant (wavelike) in other reference frames.
 
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  • #24
Here's the abstract to Mackinnon's paper, FWIW:

http://adsabs.harvard.edu/abs/1976AmJPh..44.1047M

De Broglie's thesis: A critical retrospective

MacKinnon, Edward
American Journal of Physics, Volume 44, Issue 11, pp. 1047-1055 (1976).
Louis de Broglie's doctoral thesis developed a concept of waves associated with material particles that was soon incorporated into wave mechanics and later supported by experimental demonstrations. De Broglie's original development, however, relied on an incorrect identification of two quite different relations: the relation between the velocity of a particle and the relation between the group velocity of a wave packet and the velocity of individual waves in the packet. A clarification of this difference and its historial significance is followed by a tentative account of the reasons why de Broglie's well-known formula proved successful, though the theory supporting it rested on a conceptual confusion. Finally, this development and clarification are related to some of the difficulties involved in the attempts top fashion a coherent interpretation of quantum theory.
 
  • #25
mysearch said:
v_p=f \lambda= \frac {E}{h} \frac {h}{p} = \frac {E}{p} = \frac {mc^2}{mv} = \frac {c^2}{v}

Yes, this seems to be the correct formulation. Please consider that when I said that the phase velocity is not considered physical that it still may be a critical parameter of a physical phenomenon. By analogy you might liken phase velocity to electric field. An electric field is not an object that is directly measurable. But based on a series of measurements of other parameters you can (mathematically) construct what the electric field should look like in a particular situation and from the behavior of the field you can derive predictions of observable parameters that are very likely to be extremely accurate. The electric field as well as phase velocity is in some sense a hidden variable in that it can't be measured, but it is an extremely important variable that dictates the outcome of observable values.

mysearch said:
In this context, it would appear that a particle is actually being described in terms of finite wave packet, which can be mathematically constructed in the form of many ‘harmonic’ waves in superposition, all presumably propgating faster than c, at least, mathematically.

A correction on this: the harmonic components of a wave packet travel at the same speed as the entire wave packet when there is no dispersion. EM wave components always propagate at c in a vacuum according to de Broglie theory (and most any other accepted theory)
 
  • #26
KWillets said:
Here's the abstract to Mackinnon's paper, FWIW:

http://adsabs.harvard.edu/abs/1976AmJPh..44.1047M

De Broglie's thesis: A critical retrospective

Thanks for the references. I did a very quick scan of MacKinnon's paper to try to get the jist of it. There are some interesting mathematical observations which bring out some paradoxes. However this is very telling (end of section IV):

Now let the same observer ride a de Broglie electron or plane with oscillating springs. In the center-of-mass system there is no distinction to be drawn between particle velocity and phase velocity. Nor is there any spread in frequencies and velocities. There is simply one system moving at constant velocity v and oscillating with one frequency v_0. In the rest system there is nothing analogous to the relation between wave and group velocity. Consequently, there is no dispersion in the rest system. A Lorentz transformation to any other system does not introduce dispersion. De Broglie's identification of the relationship between wave and group velocity is simply wrong. It rests on a conflagration of two radically different concepts.

Yet it suggested some remarkable conclusions which were quickly verified. There may be some deep reason for this, but I have not been able to discover it.

To be blunt, the author expresses a complete lack of understanding the basic physics of the situation. The electron at rest effects the dispersion of all radiation impinging on it. It's senseless to consider the electron as being isolated. The paper seems to be a classic case of "criticizing something you don't understand" and the confusion the author attributes to de Broglie is certainly his own. Everything therein needs to taken with a grain of salt. With that in mind I'll study the paper more thoroughly as I have time.
 
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  • #27
I thought I would summarize and expand on what little I grasp so far.

JDoolin said:
...the frequency has to do with total energy (rest mass energy plus kinetic energy), while the wavelength is just the momentum, it means the wavelength and frequency are not proportional, and are not related by a simple formula. (depending on your idea of simple.)

wich mysearch confirmed:

mysearch said:
However, I think the point made about the relationship between frequency and wavelength is a good one.

I will repeat what I said before and making it more explicit (giving the relativistic momentum and total energy rather than the nonrelativistic approximation)

The frequency has to do with the total energy:
h f = \gamma m c^2

while the wavelength has to do with the momentum:

\lambda=\frac{h}{p}=\frac{h}{\gamma m v}=\frac{h}{\beta \gamma m c}

(\beta = \frac{v}{c} \;\;\; , \;\;\; \gamma = \frac{1}{\sqrt{1-\beta^2}})

the wavelength and frequency are not proportional, and are not related by a simple formula.

Now, if the phase velocity is v_p=f \lambda as mysearch said:

mysearch said:
v_p=f \lambda= \frac {E}{h} \frac {h}{p} = \frac {E}{p} = \frac {mc^2}{mv} = \frac {c^2}{v}

then the relativistic formulas yield the same final result
v_p=f \lambda= \frac {E}{h} \frac {h}{p} = \frac {E}{p} = \frac {\gamma m c^2}{\gamma m v} = \frac {c^2}{v}

Which would mean the phase velocity is independent of of the mass, and faster than the speed of light; going to infinity when v=0, and going to c when v=c.

I think it might be productive for me to go back and review the derivation of formulas for wave and group velocity, but for someone who has seen or taught these derivations, does this relationship v_p = \lambda f hold for both standing waves, plane waves, and everything in between, or is it a relationship which holds only for plane waves?
 
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  • #28
"There is simply one system moving at constant velocity v and oscillating with one frequency v0" -- I believe he's stating that there's no application of group velocity to this system, simply due to the lack of multiple frequencies to mix. Do you have access to this paper?
 
  • #29
KWillets said:
"There is simply one system moving at constant velocity v and oscillating with one frequency v0" -- I believe he's stating that there's no application of group velocity to this system, simply due to the lack of multiple frequencies to mix. Do you have access to this paper?

I interpret MacKinnon as claiming that no dispersion occurs with an electron at rest. Therefore group velocity must equal phase velocity or rather something undefined happens whereas de Broglie showed that is specifically not the case (group velocity is zero while phase velocity is infinite).

I have a copy of the paper (which must be purchased if you don't have access to a library with a subscription). I believe a very limited amount of text may be presented on a public forum under the "fair use" clause of the copyright act when it is done for purely educational reasons. Maybe the forum mentors can clarify what the limitations are. Otherwise, I'm reluctant to post any additional text from the paper.
 
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  • #30
JDoolin said:
I think it might be productive for me to go back and review the derivation of formulas for wave and group velocity, but for someone who has seen or taught these derivations, does this relationship v_p = \lambda f hold for both standing waves, plane waves, and everything in between, or is it a relationship which holds only for plane waves?

Good question. I don't have a worked out answer and the investigation is involved. I believe it holds for any wave configuration impinging on a single point. But that is not likely true for different wave configurations impinging on spatially extended structures.
 

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