Solution

Statement

$14.5.24.$ Prove that a free electron can neither absorb nor emit a photon.

Solution

To prove this, let us write the laws of conservation of energy and momentum for the "electron-photon" system and show that they are satisfied only for a non-existent photon with frequency $\nu = 0$.

1. Impossibility of absorption:
Let us consider the process in a reference frame where the free electron is initially at rest.

• Before absorption: the energy of the electron is $E_{e1} = mc^2$, its momentum is $p_{e1} = 0$. The energy of the incident photon is $E_{\gamma} = h\nu$, its momentum is $p_{\gamma} = \frac{h\nu}{c}$.
• After absorption: the photon disappears. According to the law of conservation of momentum, the entire momentum is transferred to the electron, so $p_{e2} = \frac{h\nu}{c}$. The total energy of the electron becomes $E_{e2} = \sqrt{p_{e2}^2 c^2 + m^2 c^4}$.

Let us write the law of conservation of energy:
$$E_{e1} + E_{\gamma} = E_{e2}$$
$$mc^2 + h\nu = \sqrt{\left(\frac{h\nu}{c}\right)^2 c^2 + m^2 c^4}$$
$$mc^2 + h\nu = \sqrt{(h\nu)^2 + m^2 c^4}$$
Squaring both sides of the equation:
$$(mc^2)^2 + 2mc^2 h\nu + (h\nu)^2 = (h\nu)^2 + m^2 c^4$$
After canceling out identical terms, we get:
$$2mc^2 h\nu = 0$$
Since the electron mass $m \neq 0$, the speed of light $c \neq 0$, and Planck's constant $h \neq 0$, this equality holds true only if **$\nu = 0$**. Photons with zero frequency do not exist; therefore, complete absorption is impossible.

2. Impossibility of emission:
Suppose a free electron at rest emits a photon with frequency $\nu$.

• Before emission: the energy of the system is equal to the rest energy of the electron $E_1 = mc^2$, and the total momentum is zero.
• After emission: a photon appears with momentum $p_{\gamma} = \frac{h\nu}{c}$. To keep the total momentum zero, the electron must acquire a recoil momentum $p_e = -\frac{h\nu}{c}$.

The law of conservation of energy is:
$$mc^2 = \sqrt{\left(-\frac{h\nu}{c}\right)^2 c^2 + m^2 c^4} + h\nu$$
$$mc^2 = \sqrt{(h\nu)^2 + m^2 c^4} + h\nu$$
Since $h\nu > 0$, the right side of the equation is always strictly greater than $mc^2$. The equality is possible only if **$\nu = 0$** (meaning no emission occurs).