(Assume potential is radially symmetric) [ − ℏ 2 2 m ∇ 2 + V ( r ) ] ψ ( r ) = E ψ ( r ) {\displaystyle \left[-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+V(r)\right]\psi (r)=E\psi (r)}
If we put
then
∇ = ∂ r r ^ + 1 r ∂ θ θ ^ + 1 r sin θ ∂ ϕ ϕ ^ {\displaystyle \nabla =\partial _{r}{\hat {r}}+{\frac {1}{r}}\partial _{\theta }{\hat {\theta }}+{\frac {1}{r\sin \theta }}\partial _{\phi }{\hat {\phi }}}
( ∇ 2 + k 2 ) e i k r r = − 4 π δ 3 ( r → ) {\displaystyle \left(\nabla ^{2}+k^{2}\right){\frac {e^{ikr}}{r}}=-4\pi \delta ^{3}({\vec {r}})}
G ( r → ) = ∫ d 3 p ( 2 π ) 3 e − i p → ⋅ r → 1 k 2 − p 2 = − 1 4 π r e i k r {\displaystyle G({\vec {r}})=\int {\frac {d^{3}p}{(2\pi )^{3}}}e^{-i{\vec {p}}\cdot {\vec {r}}}{\frac {1}{k^{2}-p^{2}}}=-{\frac {1}{4\pi r}}e^{ikr}}
E = | p → | 2 2 m = ℏ 2 2 m | k → | 2 = 1 2 m | v → | 2 {\displaystyle E={\frac {|{\vec {p}}|^{2}}{2m}}={\frac {\hbar ^{2}}{2m}}|{\vec {k}}|^{2}={\frac {1}{2}}m|{\vec {v}}|^{2}}
ψ k → ( r → ) = e i k → ⋅ r → + f ( k , θ , ϕ ) e i k r r {\displaystyle \psi _{\vec {k}}({\vec {r}})=e^{i{\vec {k}}\cdot {\vec {r}}}+f(k,\theta ,\phi ){\frac {e^{ikr}}{r}}}
d σ d Ω = | f ( k , θ , ϕ ) | 2 {\displaystyle {\frac {d\sigma }{d\Omega }}=|f(k,\theta ,\phi )|^{2}}
∫ d Ω σ ( Ω ) = ∫ 0 2 π d ϕ ∫ 0 π sin θ d θ | f ( θ , ϕ ) | 2 {\displaystyle \int d\Omega \,\sigma (\Omega )=\int _{0}^{2\pi }d\phi \int _{0}^{\pi }\sin \theta d\theta |f(\theta ,\phi )|^{2}}
![{\displaystyle \left[-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+V(r)\right]\psi (r)=E\psi (r)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d1686f50dbe58c065a013a0ff19c35a4975229a2)
![{\displaystyle \left[\nabla ^{2}+k^{2}-U(r)\right]\psi (r)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b86676c0a7495bb4af36c8d2c0d6fa05a1e632ae)
