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From relativity we have $E^2 = (cp)^2 + (mc^2)^2$ which for a photon ($m = 0$) becomes $E = cp$. From quantum mechanics we have for a photon $E = h\nu = hc/\lambda$. Thus together $$ E = \frac{hc}{\lambda} = cp. $$ If we want to generalise wave properties from photons to massive particles one assumes that the wavelength of a such particle is related to its momentum, but why not its energy? i.e. $\lambda = h/p$ instead of $\lambda = hc/E$? where $E$ is the total energy of the particle. So there seems to be two ways to generalise this.

There are two points about this:

  • Massive particles will have wavelength even if at rest, so the wave vector $\vec k$ will have no preferred direction, but what about the particle being a point source of the wave in that case ?

  • The order of magnitude, there will be a very large difference between the two generalisations, is it experiments that settled which one it is?

what is wrong with that?

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    $\begingroup$ yeah pretty much because of the two reasons you outlined in your question. Also, de Brogile wavelength is not a deep fundamental concept in quantum mechanics. The deep fundamental concept is canonical quantization, which relates coordinates and components of the momentum vector by the Fourier transform. There are other ways to generalize classical mechanics, but they all don't work, whereas Quantum Mechanics does work, that's pretty much it. $\endgroup$ Commented Dec 20, 2021 at 11:53
  • $\begingroup$ Related : About de Broglie relations, what exactly is E? Its energy of what?. $\endgroup$
    – Frobenius
    Commented Dec 20, 2021 at 13:18

2 Answers 2

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The wavelength $\lambda$ has a direction. Or, more precisely, the wave number $$\vec{k}=2\pi \begin{pmatrix}1/\lambda_x \\ 1/\lambda_y \\ 1/\lambda_z \end{pmatrix}$$ is a vector quantity (with a direction perpendicular to the wave fronts).

wave
(image from my answer to Significance of wave number?)

Therefore it makes more sense to relate the wave number $\vec{k}$ via $$\vec{p}=\frac{h}{2\pi}\vec{k}$$ to the momentum $\vec{p}$ which is also a vector quantity, and not to the energy $E$ which is a scalar quantity. It also fits nicely into special relativity where $(E/c, \vec{p})$ make up a 4-vector and $(\omega/c, \vec{k})$ make up another 4-vector. The four-vector $(\omega/c,\vec{k})$ is established by its 4-product with the 4-vector $(ct,\vec{x})$ $$\phi=-\omega t+\vec{k}\cdot\vec{x}$$ giving the phase which is a 4-scalar.

But nevertheless, in 1924 when de Broglie brought up this hypothesis, even this was still a speculative guess. Only later it was confirmed experimentally that this relation is actually true, not only for massless photons, but also for massive particles. The first experiment of this kind was the Davisson-Germer experiment (1923-1927). It involved electrons (with known momentum $\vec{p}=m\vec{v}$) scattered by the surface of a nickel crystal (with known atomic grid distance $d$). From the observed diffraction pattern and the atomic grid distance $d$ they could calculate the wavelength $\lambda$ of the electrons, and found it matched the wavelength predicted by de Broglie's $\lambda=h/p$.

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  • $\begingroup$ The first part of your answer doesn't really fit, "The wavelength λ or, more precisely, the wave number,.., is a vector". And also what part exactly of SR is that in which $(\omega/c, \vec k)$ appears ? i.e. what should it describe, a wave ? $\endgroup$
    – Physor
    Commented Dec 20, 2021 at 12:17
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    $\begingroup$ @Physor I have tried to improve the wording. $\endgroup$ Commented Dec 20, 2021 at 12:26
  • $\begingroup$ Well not better, I thought that $\lambda$ is a scalar, I think that the picture is not need $\endgroup$
    – Physor
    Commented Dec 20, 2021 at 12:26
  • $\begingroup$ But the image only inflates your answer. The most important thing above is what you mention at the end. If you know well about the history of it, how they excluded energy and took momentum, then that will be a nice answer if you expand it. Thanks! $\endgroup$
    – Physor
    Commented Dec 20, 2021 at 13:57
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    $\begingroup$ @Physor I have addressed this in my answer now. $\endgroup$ Commented Dec 20, 2021 at 14:38
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Physics is not mathematics!

Perhaps somebody could contrive a mathematical theory in which the wavelength of quantum waves is related to energy, but,

Physics is an experimental science!

The de Broglie hypothesis matches what we see in numerous experiments, so we consider it a proven part of physics. Unlike mathematicians, we test our ideas against reality.

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  • $\begingroup$ This is not a good answer, sorry $\endgroup$
    – Physor
    Commented Dec 20, 2021 at 21:10
  • $\begingroup$ @Physor What's wrong with it? Fundamentally, questions in physics must relate to experiment and observation. That's the discipline that makes physics a science. $\endgroup$
    – John Doty
    Commented Dec 20, 2021 at 21:15
  • $\begingroup$ What should I say, when you generalise something making hypothesis there might be too many ways to make hypothises, The question here is why taking one way not the other and I had my justifications above, but wondered if they are the historical ones that physicists used to take that generalisation not the other ? $\endgroup$
    – Physor
    Commented Dec 20, 2021 at 21:18
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    $\begingroup$ @Physor The historical reason that other physicists accepted the de Broglie hypothesis was that it worked in things like the Davisson–Germer experiment. Before that, the idea that electrons could be considered waves seemed daft to many, perhaps most, physicists. $\endgroup$
    – John Doty
    Commented Dec 20, 2021 at 21:28
  • $\begingroup$ Now what do you want ? $\endgroup$
    – Physor
    Commented Dec 20, 2021 at 21:29

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