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De Broglie postulated that because nature loves symmetry particles should also behave like waves.just as electromagnetic waves also behave as particles.

The wavelength of the particle is $$\lambda = h/p$$

Now the problem I encountered following this line of reasoning was.

i) We know that if photons are waves these are our good old electromagnetic wave in Electric and magnetic field(Electromagnet field) but what kind of wave say an electron or proton is and in which field?

ii) The equation we have given above was derived for photons which are massless and travel at speed of light (two very unique characters which are not true in general for a particle at all) how can we assume the equation for a general particle which have mass and can't travel at speed of light is similar to photon equation

neither $$E=hc/\lambda$$ or $$E=pc$$ follows for such a particle

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  • $\begingroup$ In both cases it comes down to photons but one photon does not make a wave. It takes billions of coherent photons to resemble a light wave. i) A single photon has a frequency that completes one oscillation over a certain distance. For instance a green photon will complete one cycle every 500 nm as it travels along at the speed of light. For convenience we call that a wavelength but it’s not really a wave. ii) General particles also resemble waves as the accelerated electrons emit billions of coherent photons, which have frequencies completing oscillations in distances we call wavelengths. $\endgroup$ Jan 25, 2020 at 16:57

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At least as I understand it, de Broglie more or less had the idea, but as with many things in physics, it wasn't put on rigorous mathematical footing right away. What matters ultimately in science is what nature does, you can speculate all day until you're blue in the face, but at the end of the day you have to check against nature. And wouldn't you know it, at about the same time an experiment was done to check whether de Broglie was right, not only that matter behaves as a wave, but also that his wavelength was right. https://en.wikipedia.org/wiki/Davisson%E2%80%93Germer_experiment The experiment was essentially firing electrons with a specific momentum at a nickel target and measuring the interference pattern caused by scattering in the nickel. It's essentially a more technical version of the Young double slit experiment, which Akira Tonamura performed the electron version of in 1989 https://www.youtube.com/watch?v=jvO0P5-SMxk . If you read up on the Young experiment, you will see a formula $d\sin{\theta} = m\lambda$. Using this, you can determine the wavelength by measuring the deflection angle of each of the fringes if you know the distance between the two slits that the wave passes through.

I hope this helps, cheers!

Edit in response to comment: The fields that particles belong to are described in Quantum Field Theory. If you want an introduction to ideas involved, I found Griffiths' Introduction to Elementary Particle Physics to be helpful. There really is an electron field, but in the context of QFT, we usually only calculate how much particles scatter off each other, not so much how they freely propagate in space. To properly understand what's going on, you also need to understand how photons can be a thing if light is a wave: https://en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_field

As for the second part, $E=pc$ is a special case of $E^2=(mc^2)^2 + (pc)^2$, which $E=mc^2$ is a special case of as well, for massive particles in their rest frame. This comes from special relativity, which already existed in 1924 when de Broglie posited his idea. Also, in actuality $E=hf$, it just so happens that since the frequency of light is $f=\frac{c}{\lambda}$. The equation for light that he used was not for energy, it was the one for momentum $p=\frac{h}{\lambda}$. He essentially said that perhaps this would work for massive particles too, using $pc=\sqrt{E^2-(mc^2)^2}$. If you're doing this with electrons in a cathode ray tube, you can get the energy of the electrons by adding the mass energy to the kinetic energy given to the electrons by accelerating through the electron gun.

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    $\begingroup$ Well, thanks for the comment but it does not provide any clue for what I have asked :( $\endgroup$ Jan 25, 2020 at 15:33
  • $\begingroup$ I apologize, I wasn't initially sure what you were asking about due to wording, but I looked closer and I think that I addressed what you were asking. $\endgroup$
    – Liam Clink
    Jan 25, 2020 at 15:59
  • $\begingroup$ For ii) part that is what I was saying since m is not equal to zero we can not get E=pc which was fundamental for the wavelength of photon to come out as h/p. Is there a particular proof that $$\lambda =h/p$$ holds for any particle. $\endgroup$ Jan 25, 2020 at 16:16
  • $\begingroup$ To my knowledge there is no such proof. It was hypothesized, since it works for light it might work for matter, and it turns out (by experiment) that it does. $\endgroup$
    – Liam Clink
    Jan 25, 2020 at 17:54
  • $\begingroup$ Physics is about constructing mathematical models of reality and then experimentally confirming or falsifying them. It isn’t about “proof”. The de Broglie relations for massive particles are not logical consequences of those for massless particles. Instead we observe that the same relations between energy and frequency, and momentum and wavelength, hold for both. In hindsight, this seems natural because one can consider the limit $m\to 0$ and it is nice that nothing weird happens. $\endgroup$
    – G. Smith
    Jan 25, 2020 at 22:33

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