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Is the de Broglie (matter) wavelength $\lambda=\frac{h}{p}$ of a photon equal to the electromagnetic wavelength of the radiation? I guess yes, but how come that photons have both a matter wave and an electromagnetic wave but other particles not and that matter and electromagnetic wavelength miraculously match?

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Hi user24298, and welcome to Physics Stack Exchange! On this site, it's best to ask questions one at a time. I've edited out everything but your first question, and once you get answers to that, it will help you to formulate your second question so you can post it separately. –  David Z May 10 '13 at 6:12
You may also want to read this. –  David Z May 10 '13 at 6:13
"photons have both a matter wave and an electromagnetic wave" What does this mean? –  Ben Crowell Aug 8 '13 at 20:20
How the electromagnetic wave is built up is written up here : motls.blogspot.com/2011/11/… . It is no simple and needs a level of mathematical knowledge. –  anna v Jan 5 at 14:00

2 Answers 2

1) Yes. The photon's matter wave is actually its electromagnetic wave.

2) Photon emission is not such a kind of process when you get some particles emitted in some interval of time, and you can assume some emission moment within that interval to every particular particle. No, it is a quantum process. The system's intermediate states are not "some particles are emitted, and some are not emitted yet", but "system has some intemediate probability to be in the initial state, and some probability to be in the final state". Every photon is the subject to that accumulating probability. So the most we can say for every photon is that it is emitted in the same interval of time.

The frequency, polarisaion, direction, spatial distribution and all such characteristics of each photon would be the same as those of the electromagnetic wave. (Some advanced details are omitted.)

3) The same way as you may interpret some particular waveform in different bases, like a function of time or a set of sinewave weights, you can interpret photons in different bases. The non-basic waveform is then understood as a quantum superposition of states that belong to the chosen basis.

4) The question is based on the wrong assumption, see the answer to the question 2.

P.S. You deleted your 4 questions, leaving just one, but I hope my answers would still help you.

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The speed of the wave = wavelength x frequency, say the frequency is 6.0 x 10^14 Hz calculate the wavelength.

NOw use the de Broglie wavelength, Wavelength = h/MV

use E=hv to get the energy than,

Use E=MC^2 to get the mass

Plug that back in to the de Broglie eq you get the same answer

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This is unclear; at the moment, I don't think it answers the question well. –  HDE 226868 Dec 2 at 22:37

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