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?

  • $\begingroup$ A duplicate of a question asked a week ago. While that one is unanswered, a comment there points the way to various papers on the topic. $\endgroup$ – Jon Custer Nov 20 '15 at 23:59
  • $\begingroup$ Are you truly interested in calculation qua calculation, or is it sufficient to say how they are determined? Because the optical index of refraction for real materials is determined experimentally, and efforts to calculate them from first principles are used to validate the models on which the calculations are based. And I imagine the situation is the same in neutron refraction. $\endgroup$ – dmckee --- ex-moderator kitten Nov 21 '15 at 0:07

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


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