📂Quantum Mechanics

Compton scattering

Formula

Let $\lambda$ be the wavelength of the incident light and $\lambda^{\prime}$ be the wavelength of the scattered photon. The following equation then holds true:

$$ \lambda^{\prime} -\lambda = \frac{h}{m_{e}c}(1-\cos\theta) $$

Here, $h$ is Planck’s constant, $m_{e}$ is the mass of the electron, $c$ is the speed of light, and $\theta$ is the scattering angle. In terms of energy, we have:

$$ \cos \theta=1-\frac{m_{e}c^{2}(E-E^{\prime})}{E^{\prime}E} $$

Explanation

Compton scattering¹ refers to the phenomenon where X-rays scatter when they encounter electrons, resulting in both the X-rays and the electrons being deflected. When this happens, the wavelength of the scattered X-ray increases, which in terms of energy means that its energy decreases. This serves as evidence that X-rays, or light, possess particle-like properties.

Since $ \lambda ^{\prime}-\lambda=\dfrac{h}{m_{e} c}(1-\cos\theta) > 0$ holds true, the wavelength of light increases after the collision. This matches well with experimental results and supports the particle nature of light.

Derivation

Strategy: Use the conservation of momentum and conservation of energy principles to derive the result.

Let $\mathbf{p}_\gamma$ be the momentum of the photon before the collision, $\mathbf{p}_{e}$ be the momentum of the electron before the collision, $\mathbf{p}_\gamma^{\prime}$ be the momentum of the photon after the collision, and $\mathbf{p}_{e}^{\prime}$ be the momentum of the electron after the collision.

Part 1. Conservation of Momentum

Since there is no information about the electron after the collision, let’s solve for $\mathbf{p}_{e}^{\prime}$.

$$ \mathbf{p}_{\gamma}+\mathbf{p}_{e}=\mathbf{p}_{\gamma}^{\prime}+\mathbf{p}_{e}^{\prime} $$

Assuming the electron is initially at rest, we have $\mathbf{p}_{e}=0$.

$$ \mathbf{p}_{\gamma}+\mathbf{p}_{e}-\mathbf{p}_{\gamma}^{\prime}=\mathbf{p}_{e}^{\prime} $$

Since the rest mass of the photon is $0$, we have $ p_\gamma=\dfrac{E}{c}=\dfrac{h\nu}{c}$. Substituting this results in

$$ \frac{h^2\nu^2}{c^2}+\frac{h^2{\nu^{\prime}}^{2}}{c^2}-\frac{2h^2\nu\nu^{\prime}}{c^2}\cos\theta=(p_{e}^{\prime})^2\tag{1} $$

Part 2. Conservation of Energy

Now let $E_\gamma$ be the energy of the photon before the collision, $E_{\gamma}^{\prime}$ be the energy of the photon after the collision, $E_{e}$ be the energy of the electron before the collision, and $E_{e}^{\prime}$ be the energy of the electron after the collision. Then,

$$ \begin{align*} && E_{\gamma}^{\prime}+E_{e}^{\prime} &= E_\gamma+E_{e} \\ \implies && E_{e}^{\prime} &= E_\gamma+E_{e}-E_{\gamma}^{\prime} \\ \implies && (E_{e}^{\prime})^2 &= (E_\gamma+E_{e}-E_{\gamma}^{\prime})^2 \\ \implies && (E_{e}^{\prime})^2 &= (E_\gamma)^2+(E_{e})^2+(E_{\gamma}^{\prime})^2+2E_\gamma E_{e}-2E_\gamma E_{\gamma}^{\prime}-2E_{e}E_{\gamma}^{\prime} \end{align*} $$

The energy of the photon is $E=h\nu$, and the relativistic energy is $ E=\sqrt{(mc^2)^2+p^2c^2}$. Reformulating this yields

$$ h^2\nu^2+m_{e}^2c^4+h^2{\nu^{\prime}}^{2}+2h\nu m_{e}c^2-2h^2\nu\nu^{\prime}-2m_{e}c^2h\nu^{\prime}=m_{e}c^4+(p_{e}^{\prime})^2c^2 $$

Rearranging for $(p_{e}^{\prime})^2$, we get

$$ \frac{h^2\nu^2}{c^2} +\frac{h^2{\nu^{\prime}}^{2}}{c^2} +2m_{e} h(\nu-\nu^{\prime}) -\frac{2h^2\nu\nu^{\prime}}{c^2}=(p_{e}^{\prime})^2 \tag{2} $$

Part 3.

Using $(1)$ and $(2)$, we have

$$ \begin{align*} && \frac{h^2\nu^2}{c^2}+\frac{h^2{\nu^{\prime}}^{2}}{c^2}-\frac{2h^2\nu\nu^{\prime}}{c^2}\cos\theta&=\ \frac{h^2\nu^2}{c^2} +\frac{h^2{\nu^{\prime}}^{2}}{c^2} +2m_{e} h(\nu-\nu^{\prime}) -\frac{2h^2\nu\nu^{\prime}}{c^2} \\ \implies && -\frac{2h^2\nu\nu^{\prime}}{c^2}\cos\theta& =2m_{e} h(\nu-\nu^{\prime}) -\frac{2h^2\nu\nu^{\prime}}{c^2} \\ \implies && 2m_{e} h(\nu-\nu^{\prime})&=\ \frac{2h^2\nu\nu^{\prime}}{c^2}(1-\cos\theta) \\ \implies && (\nu-\nu^{\prime})&=\ \frac{h}{m_{e}}\frac{\nu\nu^{\prime}}{c^2}(1-\cos\theta) \end{align*} $$

Dividing both sides by $\nu\nu^{\prime}$ and multiplying by c,

$$ \frac{c}{\nu^{\prime}}-\frac{c}{\nu}=\frac{h}{m_{e}}\frac{1}{c}(1-\cos\theta) $$

Since $\displaystyle \lambda=\frac{c}{\nu}$ holds, we get

$$ \lambda ^{\prime}-\lambda=\frac{h}{m_{e} c}(1-\cos\theta) $$

Due to $E=h\nu=\dfrac{ hc }{ \lambda }$, simplifying the above equation results in

$$ \cos \theta=1-\frac{m_{e}c^{2}(E-E^{\prime})}{E^{\prime}E} $$

■