## Table of Contents

## Compton Shift

The Compton effect demonstrates the particle-like properties of light. When X-rays collide with an electron, they scatter with a change in wavelength (the Compton shift) that depends only on the angle of scattering, $\theta$, and not on the intensity of the X-ray. This phenomenon supports the particle theory of light, emphasizing its dual nature.

## Initial Setup Of Compton Shift Equation Derivation

Using the above diagram, derive the Compton Shift Equation.

## Derivation

### Conservation Of Momentum

The initial momentum of the photon is given by $p$ (before collision), and after collision, the photon’s momentum is $p’$ at an angle $\theta$ to the original direction. The electron, initially at rest, gains momentum $P_e$ at an angle $\phi$. Applying the conservation of momentum in two dimensions (along and perpendicular to the initial photon direction), we get:

**Along the initial direction (Horizontal in this case):**

$$\begin{aligned}p \, &= p’ \, \text{cos} \, \theta + P_{e} \, \text{cos} \, \phi \\ P_{e} \, \text{cos} \, \phi \, &= p-p’ \, \text{cos} \, \theta \end{aligned} $$

**Perpendicular to the initial direction (Vertical in this case):**

$$\begin{aligned} 0 \, &= p’ \, \text{sin} \, \theta \, – P_{e} \, \text{sin} \, \phi \\ P_{e} \, \text{sin} \, \phi \, &= p’ \, \text{sin} \, \theta \end{aligned}$$

### Deriving Intermediate Equation – Equation 1

Next, we combine these momentum conservation equations and use the trigonometric identity $\sin^2 \phi + \cos^2 \phi = 1$,

$$\begin{aligned} \sin^2 \phi + \cos^2 \phi &= 1 \\ P_{e}^{2}\sin^2 \phi + P_{e}^{2}\cos^{2} \phi &= P_{e}^{2} \\ P^{2}_{e} \, &= \left( p-p’ \, \text{cos} \, \theta \right)^{2} + p’^{2} \, \text{sin}^{2} \, \theta \\ P^{2}_{e} \, &= p^{2} + p’^{2}-2pp’ \, \text{cos} \, \theta \end{aligned}$$

$$ P^{2}_{e} \, = p^{2} + p’^{2}-2pp’ \, \text{cos} \, \theta \tag{1}$$

### Conservation Of Energy

Conservation of energy for the system gives:

$$ E + E_{o} \, = E’ + E_{e} \tag{2}$$

Where $E$ and $E’$ are the energies of the photon before and after the collision, $E_0$ is the rest energy of the electron, and $E_e$ is the energy of the recoiled electron.

Next, we note that:

The relativistic energy-momentum relation for the photon is given by:

$$E = pc$$

The relativistic energy-momentum relation for the electron is given by:

$$E^{2} = m^{2} c^{4} + p^{2} c^{2}$$

We use the two relativistic energy-momentum relations in Equation 2. We will now have:

$$\begin{aligned} pc + mc^{2} \, &= p’c + \left(m^{2} c^{4} + P^{2}_{e} c^{2} \right)^{\frac{1}{2}} \\ \left(p-p’ + mc \right)^{2} \, &= m^{2}c^{2} + P^{2}_{e} \end{aligned}$$

### Final Steps

Using **equation 1** to get rid of $P^{2}_{e}$ in the above equation, we obtain

$$\require{cancel} \begin{aligned} \left(p-p’ + mc \right)^{2} \, &= m^{2}c^{2} + \left(p^{2} + p’^{2} – 2pp’ \, \text{cos} \, \theta \right) \\ \cancel{p^{2}} + \cancel{p’^{2}} + \cancel{m^{2}c^{2}} + 2 \left( – pp’ + pmc-p’mc \right) \, &= \cancel{m^{2}c^{2}} + \cancel{p^{2}} + \cancel{p’^{2}}-2pp’ \, \text{cos} \, \theta \\ pmc-p’mc \, &= pp’ \left( 1-\text{cos} \, \theta \right) \\ \frac{mc}{p’}-\frac{mc}{p} \, &= 1-\text{cos} \, \theta \\ \frac{mc \lambda’}{h}-\frac{mc \lambda}{h} \, &= 1-\text{cos} \, \theta \\ \lambda’-\lambda \, &= \frac{h}{mc} \left( 1-\text{cos} \, \theta \right) \end{aligned}$$

**We’re done!**

This equation highlights the quantum nature of light, showing that the change in wavelength ($\lambda’ – \lambda$) depends only on the scattering angle $\theta$ and the Planck constant $h$, with $m$ being the electron’s rest mass and $c$ the speed of light. This result is a cornerstone in quantum mechanics, demonstrating the quantized interaction between electromagnetic radiation and matter.