Talk:Neutron scattering lengths

Origin of the scattering lengths

The following description is adapted from Boualem Hammouda's (NCNR) SANS tutorial.

Consider a neutron of energy Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_i} interacting with a nucleus, which exhibits an attractive square well of depth Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -V_0} and width Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2R} ; where $V_{0}\gg E_{i}$. The Schrödinger equation is:

$\left[-{\frac {h^{2}}{8\pi ^{2}m}}\nabla ^{2}+V(r)\right]\psi =E\psi$

Outside of the square-well (Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |r|>R} ), $V(r)=0$, and so the equation is solved as simply:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{s,\mathrm{out}} = \frac{\sin(kr)}{kr} - b \frac{e^{ikr}}{r} }

where $k={\sqrt {2mE_{i}}}2\pi /h$. Inside the square-well (Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |r|<R} ), the potential is $V(r)=-V_{0}$, and the solution becomes:

$\psi _{s,\mathrm {in} }=A{\frac {\sin(qr)}{qr}}$

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle q=\sqrt{2m(E_i+V_0} 2 \pi/h} .