# Why do matter waves show refraction? / Why does wavelength change when a particle enters a medium?

I know why electromagnetic waves show refraction but matter waves?

I found this in section 2-5 in “An Introduction to Quantum Physics” by French and Taylor while reading about Davison and germer experiment.

• A wave is a wave. What properties of matter waves compared to electromagnetic waves would make them behave differently according to you? May 1, 2020 at 19:15
• electromagnetic waves are composed of electric and magnetic fields but there is no such things for matter waves. Due to electric field the speed electromagnetic waves slows down in a medium but how do I explain change in wavelength of matter wave inside a crystal May 1, 2020 at 19:28
• @QuantumApple: Matter waves do not have electric & magnetic fields oscillating perpendicular to each other this makes a difference that matter waves cann't show polarization phenomenon. Am I right? May 1, 2020 at 19:28
• Yes ........... May 1, 2020 at 19:30
• Diffraction effects are not related to polarization though (or am I mistaken?). As for change of wavelength, this can be due for instance to going through a potential step. Say you go from $U(x<0) = 0$ to $U(x>0) = U_0$. The dispersion relation writes $E = \frac{\hbar^2 k^2}{2m} + U$ for both parts, so going from $U = 0$ to $U > 0$ changes $k$ and thus the wavelength. May 1, 2020 at 19:36

While it is true that matter wave and electromagnetic waves have different properties (for instance, Maxwell equations describe the evolution of electric and magnetic fields, which are vector fields, whereas matter waves are scalar, so only EM waves can exhibit polarization effects), they still share the same universal properties of waves. One of them is that waves diffract.

This comes from the fact that the most basic solutions to the wave equation in both cases is a plane wave:

$$\phi(\overrightarrow{r}, t) = \exp \left(i\left(\overrightarrow{k}\cdot \overrightarrow{r} - \omega t\right)\right)$$

These go in a straight line, at constant velocity as long as the medium is homogeneous. However, plane waves are objects that are infinitely spread both in time and space, which is not physical. In real life, you would need to describe any finite-sized wavepacket/beam by an infinite combination of these plane waves, which will all evolve slightly differently. This is what causes diffraction. Note that this is not dependent on the exact nature of the waves (scalar or vector), nor the dispersion relation between $$\omega$$ and $$k$$ (linear for EM waves, quadratic for matter waves without a potential).

As for what causes a change in wavelength when you go into a medium, you have to look at the dispersion relation. For EM waves, it writes:

$$\left| \overrightarrow{k} \right| = n(\omega) \frac{\omega}{c},$$

where the optical index $$n(\omega)$$ depends on the way the medium responds to the EM field.

We have a similar dispersion relation for matter waves. Indeed, the Schrödinger equation can be written as:

$$E \psi = \frac{\hbar^2 k^2}{2m} \psi + U(\overrightarrow{r}) \psi$$

Assuming $$U(\overrightarrow{r})$$ is locally constant and equal to $$U_0 < E$$, the solutions are plane waves:

$$\psi(\overrightarrow{r}, t) = \exp \left(i\left(\overrightarrow{k}\cdot \overrightarrow{r} - \omega t\right)\right), \quad \mathrm{with} \, \left| \overrightarrow{k} \right| = \sqrt{\frac{2m \omega}{\hbar}- \frac{2m U}{\hbar^2}}.$$

As before, if we move from a region of space from a certain uniform $$U$$ to a region of space with a different $$U$$ (for instance, a potential step $$U(x<0) = 0$$ to $$U(x > 0) = U_0 < E$$), the wavelength must change accordingly.

Note that in real life, $$U(\overrightarrow{r})$$ is rarely uniform, so the previous description is not exactly true. For an electron in a crystal for instance, $$U(\overrightarrow{r})$$ is periodic, which will lead to diffraction just in the same way as a periodic optical index $$n(\overrightarrow{r})$$ would lead to light diffraction.

If matter waves do not refract, diffract or extinguish in a medium this implies that the medium is transparent to matter. That would be a vacuum.