# Why can't a free electron absorb or emit photon?

## Homework Statement

In an article, I am informed that a free electron can neither absorb nor emit a photon. I just want to know why?! Please help me.

## The Attempt at a Solution

I've no idea at all. Photon is the medium boson carrying electromagnetic force, so the relationship between electron and photon is interesting. The above process is said to bef orbidden, and I think some conservation laws must be violated!

Related Advanced Physics Homework Help News on Phys.org
look at the conduction electrons in a metal and how they interact with light.

Hello, Zeus!!! I'm so glad to come across your problem. Nice and attractive!
It is essential in special relativity that, a free electron can neither absorb nor emit photons. Hints are given in details below.

1. Absorbing of photon is forbidden.

If a free electron could absorb a photon, then, according to conservation of energy and momentum,
$$\hbar \omega +mc^2 = \sqrt{p^2c^2 + m^2c^4} (Eq1)$$
$$\hbar k = p (Eq2)$$
where $\omega$ and $k$ are the frequency and wavenumber of the photon, respectively, $m$ the electron's rest mass, $p$ the momentum of the electron after absorbing the photon. Eq2 leads to
$$p^2=\hbar^2 k^2 = \hbar^2 \frac{\omega^2}{c^2} (Eq3)$$
Insert Eq3 into Eq2, and square of the left side is
$$(\hbar \omega +mc^2 )^2 =\hbar^2 \omega^2 + m^2c^4+2mc^2\hbar \omega$$
when square of the right side is
$$\hbar^2\omega^2+m^2c^4$$
So, if Eq1 holds, $\hbar \omega=0$. There is no photon carrying vanishing energy. Hence, absorbing of a photon by a free electron is forbidden.

2. Emitting of a photon is Forbidden
Suppose the intial (before photon emitting) and final (after photon emitting) 4-momentum of the electron are separately $p_\mu$, $p'_\mu$:
$$p_\mu=(0,0,0,imc), \qquad p'_\mu = \left[p, i\frac{mc}{\sqrt{1-\frac{v^2}{c^2}}} \right] (Eq4)$$
According to conservation of energy and momentum:
$$p_\mu=q_\mu+p'_\mu (Eq5) (p_\mu - p'_\mu)^2 =q_\mu^2 (Eq6)$$
where $q_\mu$ refers to the 4-momentum of the photon. From Eq5 and Eq6, we have
$$(p_\mu - p'_\mu)^2 = p_\mu^2 -2p_\mu p'_\mu+p'_\mu^2 =-m^2c^2- 2p_\mu p'_\mu -\frac{mc^2}{1-\frac{v^2}{c^2}} (Eq7)$$
Recall that
$$p_\mu p'_\mu = -\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}} (Eq8)$$
Insert Eq8 into Eq7, we have
$$p_\mu p'_\mu = -m^2 c^2 \left[ 1-\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \right] (Eq9)$$
Yet $q_\mu^2=0$, so, $p_\mu=p'_\mu$, to ensure Eq5 and Eq6 hold. Thus the state of motion of the electron is not disturbed, with no momentum transported to the photon. Hence, emitting of a photon by the free electron is forbidden.

berkeman
Mentor

## Homework Statement

In an article, I am informed that a free electron can neither absorb nor emit a photon. I just want to know why?! Please help me.

## The Attempt at a Solution

I've no idea at all. Photon is the medium boson carrying electromagnetic force, so the relationship between electron and photon is interesting. The above process is said to bef orbidden, and I think some conservation laws must be violated!
Sounds like the article is wrong. Or else they are placing constraints on the electrons that are not met in FELs...

http://en.wikipedia.org/wiki/Free_electron_laser

.

sylas
Sounds like the article is wrong. Or else they are placing constraints on the electrons that are not met in FELs...

http://en.wikipedia.org/wiki/Free_electron_laser
The electrons in a Free-electron laser aren't actually free in the sense used for the question. They are tightly controlled by an oscillating magnetic field, so that they emit coherent radiation.

An electron can't absorb a photon all by itself, because you can't get conservation of energy and momentum with the electron alone.

Cheers -- sylas

I would like to draw your attention to https://www.physicsforums.com/showthread.php?t=308657 , on which another interesting problem on the relationship between photon and electron is discussed. This problem is ----why can an electron - positron pair not be created by a photon in free space（by sheelbe999 ）? Here also, mathematical formulation are given in details as a hint.

sylas