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As simple as in the title.. I would like to know also some mathematics about it!

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It cannot. This is because energy and momentum are not both conserved if a free charged particle (say, an electron) emits a photon. It needs interaction with at least a second charged particle in order to do so (as in Bremsstrahlung). The mathematic involved is that of the energy of a photon $E=\hbar \omega$, energy of a particle $E^2 = m^2 c^4 + p^2 c^2$, momentum of a photon $p = \hbar \omega /c$ and simple trigonometry and basic algebra, very much as in the classical version of Compton scattering.

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    $\begingroup$ Note that if the particle has internal structure, this argument can fail. For example, atomic nuclei emit gamma rays. This is because they have more than one internal state with different energies. $\endgroup$
    – user4552
    Commented Oct 20, 2013 at 19:58
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    $\begingroup$ That is certainly true. Nice addition to the discussion. $\endgroup$ Commented Oct 20, 2013 at 22:14
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    $\begingroup$ @BenCrowell In the context of the shell model, one can argue that the gamma is emitted by a nucleon falling to an orbital of lower energy, i.e. by a bound particle. $\endgroup$
    – user154997
    Commented Jul 17, 2017 at 2:45
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    $\begingroup$ I know this was a long time ago, but I am just wondering why we have to consider the conservation of energy and momentum for the particle alone, and not the particle + photon after the hypothetical emission happens. $\endgroup$
    – newtothis
    Commented Jun 26, 2021 at 8:40
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    $\begingroup$ @newtothis As the answer points out, it is precisely considering the energy and momentum (already given) of the photon + particle system that makes it impossible. So they not only are being considered, they must be considered. $\endgroup$ Commented Jul 1, 2021 at 6:40
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Rather than thinking in terms of conservation mathematically we can just think of the energy.

Consider the emission case (if we can prove this is impossible, then the absorption case is also impossible if we reverse time).

Let the free particle be at rest - if it is not we can transform to a frame where it is. Then if it emits a photon, the photon's momentum must be opposite the final momentum of the particle to conserve the zero momentum initial state.

But now the electron is moving, so has more energy than its rest energy ($E^2 = p^2 + m^2$ in natural units). The photon also has energy, not that this matters. Key point: the final energy is greater than the initial energy $\implies$ forbidden.

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Let's assume it could then $\frac{1}{2} mv^2=\frac{hc}{\lambda}$ also $P=mv=\frac{h}{\lambda}$ by replacing $mv$ in first equation with$\frac{h}{\lambda}$ we get $\frac{1}{2} \frac{h}{\lambda}v=\frac{hc}{\lambda}$ now $v=2c$ which is of course impossible.

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  • $\begingroup$ 1 year later i am unable to understand what i did $\endgroup$
    – Ha'Penny
    Commented Apr 4, 2023 at 5:06
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I see the consensus above is NO, and also same in this thread Free electron can't absorb a photon

Is this an accepted 'law' of physics?? Because I need to respectfully offer a different solution:

If you are mixing particles and photons you need to include relativism. From the example in the link above (Closed discussion) it is shown that there are no valid solutions to conservation of momentum and energy, thus electron cannot absorb a photon.

There is a solution.

To simplify, we observe the electron initially stationary from the same rest frame, thereby setting $p_1=0$ in the equation:

$p_1+p_ν=p_2,(1)$

where p1=0 and p2 is the momentum of the electron after the interaction and pv is the photon momentum, will then be:

$p_ν=p_2,(2)$

which (2) is incorrect because you are now observing the electron no longer from the same frame, so it must be written as:

$p_ν=γp_2,(3)$

where γ is the Lorentz factor, but (3) is not really helpful yet.

Conservation of energy gives us (still starting with p1=0)

$m_ec^2+p_νc=γm_ec^2,(4)$

, where the right-hand side of (4) is the total energy of the particle. Then subtracting $mc^2$ both sides

$p_νc=m_ec^2(γ-1),(5)$

and the right hand side of (5) is exactly the relativistic kinetic energy of the particle. So not only can the particle absorb a photon, the absorbed photon adds to kinetic energy and will thus remain absorbed until the kinetic energy changes!

The above can be extended to any frame of reference and still holds true.

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  • $\begingroup$ Most of the time, the absorbed photon is emitted again, this time as a photon with less energy, if it previously hit a photon moving in the same direction. Or with higher energy if it slows down the electron (laser cooling). $\endgroup$ Commented Oct 18, 2023 at 17:26
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No. Photon does not change electron state ever.

Atom or molecule can have excited or no-excited state - depending how far electron is from nucleus.

According to nowadays view on the Space, it is not the emptiness. Space is heavily filled-in with waves (dark energy) and debris (like quarks, nutrino and other unknown yet particles - dark mass).

Nucleus makes a disturbance in the space and because of it creates spherical or more complex "wave boxes" (named shells or orbitals) around itself; for more than one boson in nucleus it can be cube-like-shape or some other wave combinations

Electron orbital in hydrogen is the gap between two spherical wave bumps of the space around nucleus; wave nature of electron makes orbital conductive and electrically atom does not radiate -- nucleus proton and electron shell are static to each other. Speed of electron is really does not matter and it makes random jerky moves inside the orbital. Keep in mind, orbital can have very different shapes. In molecule it is a tube between two nuclei.

Photon makes a disturbance in the atom adding electrical energy as a short pulse.

This energy pushes electron and proton apart; it is not always enough energy to push electron absolutely out of atom; it may simply push it from the lower orbital to the higher.

So the answer is: only atom, molecule or two particles with different electrical properties can emit/absorb photon. Photon is electromagnetic pulse and IS NOT a particle.

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    $\begingroup$ New age-ish gibberish. $\endgroup$
    – user154997
    Commented Jul 17, 2017 at 2:31

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