What is the difference between de Broglie wavelength and wavelength of electromagnetic radiation? Is there any relation between $\lambda=\dfrac{hc}{E}$ and $\lambda= \dfrac{h}{mv}$? (E stands for energy of electromagnetic radiation.)
[Physics] Difference between de Broglie wavelength and electromagnetic wavelength
quantum mechanicswave-particle-duality
Related Solutions
Yes the product $\nu \lambda$ makes sense as a velocity. Defining $E = \hslash \omega$ and $p=\hslash k$ (the Planck constant $h=2\pi \hslash$, where the $2\pi$ is injected into the $\hslash$, since physicists usually prefer to discuss the angular frequency $\omega=2\pi\nu$ and the wave vector $k=2\pi p$ rather than the frequency $\nu$ and the momentum $p$ for Fourier-transform notational convenience), you end up with $$\nu \lambda = \frac{\omega}{2\pi}\frac{2\pi}{k}=\frac{\omega}{k}$$ which is the standard definition for the phase velocity. It corresponds to the velocity of the wave component at frequency $\nu$ propagating over distance $\lambda$ per unit time.
This is always true, but for more complicated situation, a system is represented by a superposition of different waves propagating at different phase velocity. It results a wave-packet propagating, as an ensemble, at the group velocity $$v_{g}=\frac{\partial \omega \left(k\right)}{\partial k}$$ where the dispersion relation of the wave-packet is noted $\omega \left(k\right)$.
Only the group velocity has some physical clear interpretations. For instance, the phase velocity can be larger than the speed of light, but the group velocity can never been larger than the speed of light $v_{g}\leq c$, at least in vacuum.
For a photon in free space for instance, $\omega \left(k\right)=c k$ and thus it's group velocity is $c$. For a free non-relativistic particle of mass $m$, $\omega = \hslash k^2/2m$, and $v_{g}=\hslash k/m$. etc...
For a human body, you have to count all the atoms constituting the body. The individual frequency of one atom interferes with the frequencies of all the others, resulting in an almost flat dispersion relation. The group velocity is then ridiculously small. A human body does not move due to quantum effect ! You can get a basic idea of the group velocity of a human body supposing you are a free particle of mass $100 \textrm{kg}$ and wavelength $1 \textrm{m}^{-1}$ (i.e. the order of magnitude of your size is about $1 \textrm{m}$) the $\hslash$ kills $v_{g} \sim \left(10^{-36}-10^{-34}\right) \textrm{m.s}^{-1}$ !
More about that :
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.
Best Answer
Yes, there is a relationship. The base relationship is between an object's momentum and its de Broglie wavelength: $$ \lambda = \frac{h}{p}. $$ For a particle that is not moving at relativistic velocities ($v \ll c$), we have $p = mv$ and so the relationship becomes $\lambda = h/mv$. However, for a photon, its momentum is not equal to $mv$; instead $p = E/c$, where $E$ is the photon's energy. If you plug this in to the above relationship, you obtain $\lambda = hc/E$.
You may be wondering why a photon has a different relationship between its momentum and its velocity than a conventional particle does. That's probably a separate question, and one that I'm confident has been answered many times on this site; I would encourage you to search this site for "photon momentum" for answers to this.