refraction - How are neutron refractive indices calculated?

I have been informed of the fact that neutrons refract similarly to light in accordance to Snell's Law. How would one calculate the refraction index for a neutron?

Answer

We use the Fermi pseudopotential.

The interaction of a free neutron with a free nucleus can be summarized by the effective scattering length of the interaction. For simple probabalistic reasons (explained nicely by Golub, Richardson, and Lamoreaux) most neutron-nucleus scattering lengths are positive, and so we can think of a nucleus as "a thing that pushes neutrons away from itself a little."

The pseudopotential is just the sum of all the tiny repulsions from all the individual nuclei. For neutron mass $m$, scattering length $b$ from a nucleus at $\vec r_i$, the psuedopotential which reproduces the scattering length without any neutron-nucleus details is

$$V_i = \frac{2\pi\hbar^2}{m}b\cdot\delta^{(3)}(\vec r - \vec r_i).$$ You find yourself in the domain of neutron optics when the neutron wavelength $\lambda$ is long relative to the spacing between atoms. In that case the neutron doesn't distinguish between nearby scatters, and the pseudopotential becomes $$V_F = \frac{2\pi\hbar^2}{m}bN$$ where $N$ is the number density of the scatterers.

A neutron which passes from vacuum ($V_F=0$) to some medium with $V_F\neq0$ will effectively see a one-dimensional step potential, with some probabilities for reflection and transmission. The transmitted neutron will generally ($V_F>0$) have a slower speed than the incident/reflected neutrons. I'll let you work out that the index of refraction with momentum $\vec p = \hbar \vec k$ is

$$n \approx 1-\frac{2\pi Nb}{k^2}$$