The calculation The two appendices to the scattering calculation
The scattering-formula calculation rests on two results, derived here in full: the general solution of the Helmholtz equation and an integral that appears twice in the text.
Appendix 1. The Helmholtz equation. The equation we encountered in the text was of the type:
∇ 2 ψ + k 2 ψ = f ( r ⃗ ) \nabla^2\psi+k^2\psi=f(\vec{r}) ∇ 2 ψ + k 2 ψ = f ( r ) with the condition that ψ → e i k z \psi\to e^{ikz} ψ → e ik z z → − ∞ z\to-\infty z → − ∞
To solve this problem we first consider the homogeneous equation with the actual boundary condition:
{ ∇ 2 ψ + k 2 ψ = 0 with ψ ⟶ z → − ∞ e i k z ⇔ ψ = e i k z \begin{aligned}
\left\{\begin{gathered}
\nabla^2\psi+k^2\psi=0 \\
\text{with }\psi\underset{z\to-\infty}{\longrightarrow}e^{ikz}
\end{gathered}\right. & \Leftrightarrow \\
\psi & =e^{ikz}
\end{aligned} { ∇ 2 ψ + k 2 ψ = 0 with ψ z → − ∞ ⟶ e ik z ψ ⇔ = e ik z Now we solve the complete equation with the “null” boundary condition: { ∇ 2 ψ + k 2 ψ = f ( r ‾ ) ψ ( r ‾ ) → 0 as 1 r \left\{\begin{gathered}
\nabla^2\psi+k^2\psi=f(\overline{r}) \\
\psi(\overline{r})\to0\ \text{as}\ \frac{1}{r}
\end{gathered}\right. ⎩ ⎨ ⎧ ∇ 2 ψ + k 2 ψ = f ( r ) ψ ( r ) → 0 as r 1
We determine the solution when the source term is the Dirac function:
∇ 2 ψ + k 2 ψ = δ ( r ⃗ − r ⃗ ′ ) \nabla^2\psi+k^2\psi=\delta(\vec{r}-\vec{r}{\:}') ∇ 2 ψ + k 2 ψ = δ ( r − r ′ ) In this case the solution, as we will verify later, is
ψ = − e i k ∣ r ‾ − r ‾ ′ ∣ / ( 4 π ∣ r ‾ − r ‾ ′ ∣ ) \psi=-e^{ik|\overline{r}-{\overline{r}}'|}/(4\pi|\overline{r}-{\overline{r}}'|) ψ = − e ik ∣ r − r ′ ∣ / ( 4 π ∣ r − r ′ ∣ ) Using the linearity of the equation, from the fact that:
f ( r ‾ ) = ∭ All Space f ( r ‾ ′ ) δ ( r ‾ − r ‾ ′ ) d Ω ′ f(\overline{r})=\iiint_{\text{All Space}}f(\overline{r}{\:}')\delta(\overline{r}-\overline{r}{\:}')\:d\Omega' f ( r ) = ∭ All Space f ( r ′ ) δ ( r − r ′ ) d Ω ′ we can conclude that:
ψ = − 1 4 π ∭ All Space f ( r ⃗ ′ ) e i k ∣ r ⃗ − r ⃗ ′ ∣ ∣ r ⃗ − r ⃗ ′ ∣ d Ω ′ \psi=-\frac{1}{4\pi}\iiint_{\text{All Space}}f(\vec{r}{\:}')\frac{e^{ik|\vec{r}-\vec{r}{\:}'|}}{|\vec{r}-\vec{r}{\:}'|}d\Omega' ψ = − 4 π 1 ∭ All Space f ( r ′ ) ∣ r − r ′ ∣ e ik ∣ r − r ′ ∣ d Ω ′ Adding this solution to the one found for the homogeneous equation with the actual initial condition, we obtain the formula applied in the text:
ψ = e i k z − 1 4 π ∭ All Space f ( r ‾ ′ ) e i k ∣ r ‾ − r ‾ ′ ∣ ∣ r ‾ − r ‾ ′ ∣ d Ω ′ \psi=e^{ikz}-\frac{1}{4\pi}\iiint_{\text{All Space}}f({\overline{r}}')\frac{e^{ik|\overline{r}-{\overline{r}}'|}}{|\overline{r}-{\overline{r}}'|}d\Omega' ψ = e ik z − 4 π 1 ∭ All Space f ( r ′ ) ∣ r − r ′ ∣ e ik ∣ r − r ′ ∣ d Ω ′ Now we verify that the function ψ = − e i k ∣ r ⃗ − r ⃗ ′ ∣ / ( 4 π ∣ r ⃗ − r ⃗ ′ ∣ ) \psi=-e^{ik|\vec{r}-\vec{r}{\:}'|}/(4\pi|\vec{r}-\vec{r}{\:}'|) ψ = − e ik ∣ r − r ′ ∣ / ( 4 π ∣ r − r ′ ∣ ) ∇ 2 ψ + k 2 ψ = δ ( r ⃗ − r ⃗ ′ ) \nabla^2\psi+k^2\psi=\delta(\vec{r}-\vec{r}{\:}') ∇ 2 ψ + k 2 ψ = δ ( r − r ′ )
First we prove that the equation is satisfied for every r ‾ ≠ r ‾ ′ \overline{r}\neq{\overline{r}}' r = r ′
for r ‾ ≠ r ‾ ′ \overline{r}\neq{\overline{r}}' r = r ′ δ ( r ⃗ − r ⃗ ′ ) = 0 \delta(\vec{r}-{\vec{r}}')=0 δ ( r − r ′ ) = 0 ∇ 2 ψ + k 2 ψ = 0 \nabla^2\psi+k^2\psi=0 ∇ 2 ψ + k 2 ψ = 0 r ‾ ′ {\overline{r}}' r ′
∇ 2 ψ + k 2 ψ = 0 ⇔ ∇ 2 ( − 1 4 π e i k ∣ r ⃗ − r ⃗ ′ ∣ ∣ r ⃗ − r ⃗ ′ ∣ ) + k 2 ( − 1 4 π e i k ∣ r ⃗ − r ⃗ ′ ∣ ∣ r ⃗ − r ⃗ ′ ∣ ) = 0 ⇔ \begin{aligned}
\nabla^2\psi+k^2\psi & =0\Leftrightarrow \\
\nabla^2\left(-\frac{1}{4\pi}\frac{e^{ik|\vec{r}-{\vec{r}}'|}}{|\vec{r}-{\vec{r}}'|}\right)+k^2\left(-\frac{1}{4\pi}\frac{e^{ik|\vec{r}-{\vec{r}}'|}}{|\vec{r}-{\vec{r}}'|}\right) & =0\Leftrightarrow
\end{aligned} ∇ 2 ψ + k 2 ψ ∇ 2 ( − 4 π 1 ∣ r − r ′ ∣ e ik ∣ r − r ′ ∣ ) + k 2 ( − 4 π 1 ∣ r − r ′ ∣ e ik ∣ r − r ′ ∣ ) = 0 ⇔ = 0 ⇔ substituting the formula for the Laplacian and simplifying the common terms
⇔ 1 r 2 d d r ( r 2 d d r e i k r r ) + k 2 e i k r r = 0 ⇔ 1 r 2 d d r ( r 2 ( i k e i k r r − e i k r r 2 ) ) + k 2 e i k r r = 0 ⇔ 1 r 2 d d r ( i k r e i k r − e i k r ) + k 2 e i k r r = 0 ⇔ 1 r 2 ( i k e i k r − r k 2 e i k r − i k e i k r ) + k 2 e i k r r = 0 ⇔ − 1 r 2 r k 2 e i k r + k 2 e i k r r = 0 Verified . \begin{aligned}
\Leftrightarrow\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\frac{e^{ikr}}{r}\right)+k^2\frac{e^{ikr}}{r} & =0\Leftrightarrow \\
\frac{1}{r^2}\frac{d}{dr}\left(r^2\left(ik\frac{e^{ikr}}{r}-\frac{e^{ikr}}{r^2}\right)\right)+k^2\frac{e^{ikr}}{r} & =0\Leftrightarrow \\
\frac{1}{r^2}\frac{d}{dr}\left(ikre^{ikr}-e^{ikr}\right)+k^2\frac{e^{ikr}}{r} & =0\Leftrightarrow \\
\frac{1}{r^2}\left(ike^{ikr}-rk^2e^{ikr}-ike^{ikr}\right)+k^2\frac{e^{ikr}}{r} & =0\Leftrightarrow \\
-\frac{1}{r^2}rk^2e^{ikr}+k^2\frac{e^{ikr}}{r} & =0 \\
\text{Verified}. &
\end{aligned} ⇔ r 2 1 d r d ( r 2 d r d r e ik r ) + k 2 r e ik r r 2 1 d r d ( r 2 ( ik r e ik r − r 2 e ik r ) ) + k 2 r e ik r r 2 1 d r d ( ik r e ik r − e ik r ) + k 2 r e ik r r 2 1 ( ik e ik r − r k 2 e ik r − ik e ik r ) + k 2 r e ik r − r 2 1 r k 2 e ik r + k 2 r e ik r Verified . = 0 ⇔ = 0 ⇔ = 0 ⇔ = 0 ⇔ = 0 As a second point we prove that the property of the Dirac function holds:
∭ ( ∇ 2 ψ + k 2 ψ ) d Ω ′ = 1 \iiint\left(\nabla^2\psi+k^2\psi\right)d\Omega'=1 ∭ ( ∇ 2 ψ + k 2 ψ ) d Ω ′ = 1 Substituting ψ = − e i k ∣ r ⃗ − r ⃗ ′ ∣ / ( 4 π ∣ r ⃗ − r ⃗ ′ ∣ ) \psi=-e^{ik|\vec{r}-\vec{r}{\:}'|}/(4\pi|\vec{r}-\vec{r}{\:}'|) ψ = − e ik ∣ r − r ′ ∣ / ( 4 π ∣ r − r ′ ∣ )
∭ Sphere centred at r ‾ ( ∇ 2 ( − 1 4 π e i k ∣ r ‾ − r ‾ ′ ∣ ∣ r ‾ − r ‾ ′ ∣ ) + k 2 ( − 1 4 π e i k ∣ r ‾ − r ‾ ′ ∣ ∣ r ‾ − r ‾ ′ ∣ ) ) d Ω ′ = \iiint_{\text{Sphere centred at }\overline{r}}\left(\nabla^2\left(-\frac{1}{4\pi}\frac{e^{ik|\overline{r}-{\overline{r}}'|}}{|\overline{r}-{\overline{r}}'|}\right)+k^2\left(-\frac{1}{4\pi}\frac{e^{ik|\overline{r}-{\overline{r}}'|}}{|\overline{r}-{\overline{r}}'|}\right)\right)d\Omega'= ∭ Sphere centred at r ( ∇ 2 ( − 4 π 1 ∣ r − r ′ ∣ e ik ∣ r − r ′ ∣ ) + k 2 ( − 4 π 1 ∣ r − r ′ ∣ e ik ∣ r − r ′ ∣ ) ) d Ω ′ = choosing a system of spherical coordinates centred at r ‾ \overline{r} r
= − 1 4 π ∭ Sphere ( ∇ 2 e i k r ′ r ′ + k 2 e i k r ′ r ′ ) d Ω ′ = − 1 4 π ∭ Sphere ( ∇ 2 e i k r ′ r ′ + k 2 e i k r ′ r ′ ) d Ω ′ = − 1 4 π ∭ Sphere ( ∇ ⋅ ∇ e i k r ′ r ′ + k 2 e i k r ′ r ′ ) d Ω ′ = \begin{aligned}
& \\
& =-\frac{1}{4\pi}\iiint_{\text{Sphere}}\left(\nabla^2\frac{e^{ikr'}}{r'}+k^2\frac{e^{ikr'}}{r'}\right)d\Omega' \\
& =-\frac{1}{4\pi}\iiint_{\text{Sphere}}\left(\nabla^2\frac{e^{ikr'}}{r'}+k^2\frac{e^{ikr'}}{r'}\right)d\Omega' \\
& =-\frac{1}{4\pi}\iiint_{\text{Sphere}}\left(\nabla\cdot\nabla\frac{e^{ikr'}}{r'}+k^2\frac{e^{ikr'}}{r'}\right)d\Omega'=
\end{aligned} = − 4 π 1 ∭ Sphere ( ∇ 2 r ′ e ik r ′ + k 2 r ′ e ik r ′ ) d Ω ′ = − 4 π 1 ∭ Sphere ( ∇ 2 r ′ e ik r ′ + k 2 r ′ e ik r ′ ) d Ω ′ = − 4 π 1 ∭ Sphere ( ∇ ⋅ ∇ r ′ e ik r ′ + k 2 r ′ e ik r ′ ) d Ω ′ = applying the divergence theorem we have
= − 1 4 π [ ∬ Sphere ∇ e i k r ′ r ′ ⋅ r ^ ′ d S ′ + ∭ Sphere ( k 2 e i k r ′ r ′ ) d Ω ′ ] =-\frac{1}{4\pi}\left[\iint_{\text{Sphere}}\nabla\frac{e^{ikr'}}{r'}\cdot{\hat{r}}'\:dS'+\iiint_{\text{Sphere}}\left(k^2\frac{e^{ikr'}}{r'}\right)d\Omega'\right] = − 4 π 1 [ ∬ Sphere ∇ r ′ e ik r ′ ⋅ r ^ ′ d S ′ + ∭ Sphere ( k 2 r ′ e ik r ′ ) d Ω ′ ] We first carry out the surface integral for a sphere of radius R
∯ Sphere ∇ e i k r ′ r ′ ⋅ r ^ ′ d S ′ = ∯ Sphere ( i k e i k r ′ r ′ − e i k r ′ r ′ 2 ) d S ′ = 4 π R 2 ( i k e i k R R − e i k R R 2 ) = 4 π ( i k R e i k R − e i k R ) \begin{aligned}
\oiint_{\text{Sphere}}\nabla\frac{e^{ikr'}}{r'}\cdot{\hat{r}}'\:dS' & =\oiint_{\text{Sphere}}\left(ik\frac{e^{ikr'}}{r'}-\frac{e^{ikr'}}{r^{'2}}\right)dS' \\
& =4\pi R^2\left(ik\frac{e^{ikR}}{R}-\frac{e^{ikR}}{R^2}\right) \\
& =4\pi\left(ikRe^{ikR}-e^{ikR}\right)
\end{aligned} ∬ Sphere ∇ r ′ e ik r ′ ⋅ r ^ ′ d S ′ = ∬ Sphere ( ik r ′ e ik r ′ − r ′ 2 e ik r ′ ) d S ′ = 4 π R 2 ( ik R e ik R − R 2 e ik R ) = 4 π ( ik R e ik R − e ik R ) Now we carry out the volume integral for the same sphere of radius R
∭ Sphere ( k 2 e i k r ′ r ′ ) d Ω ′ = ∫ 0 2 π d φ ∫ 0 π sin ϑ d ϑ ∫ 0 R k 2 e i k r ′ r ′ r ′ 2 d r ′ = 4 π ∫ 0 R k 2 e i k r ′ r ′ d r ′ = \begin{aligned}
\iiint_{\text{Sphere}}\left(k^2\frac{e^{ikr'}}{r'}\right)d\Omega' & =\int_0^{2\pi}d\varphi\int_0^\pi\sin\vartheta\:d\vartheta\int_0^Rk^2\frac{e^{ikr'}}{r'}r^{'2}\:dr' \\
& =4\pi\int_0^Rk^2e^{ikr'}r'\:dr'=
\end{aligned} ∭ Sphere ( k 2 r ′ e ik r ′ ) d Ω ′ = ∫ 0 2 π d φ ∫ 0 π sin ϑ d ϑ ∫ 0 R k 2 r ′ e ik r ′ r ′ 2 d r ′ = 4 π ∫ 0 R k 2 e ik r ′ r ′ d r ′ = integrating by parts we have
4 π ( k 2 e i k r ′ r ′ i k ∣ 0 R − ∫ 0 R k 2 e i k r ′ i k d r ′ ) = 4 π ( − i k e i k R R + e i k R − 1 ) 4\pi\left({\frac{k^2e^{ikr'}r'}{ik}|}_0^R-\int_0^Rk^2\frac{e^{ikr'}}{ik}\:dr'\right)=4\pi\left(-ike^{ikR}R+e^{ikR}-1\right) 4 π ( ik k 2 e ik r ′ r ′ ∣ 0 R − ∫ 0 R k 2 ik e ik r ′ d r ′ ) = 4 π ( − ik e ik R R + e ik R − 1 ) Adding the results of the two integrals we have
− 1 4 π [ ∬ Sphere ∇ e i k r ′ r ′ ⋅ r ^ ′ d S ′ + ∭ Sphere ( k 2 e i k r ′ r ′ ) d Ω ′ ] = − 1 4 π [ 4 π ( i k R e i k R − e i k R ) + 4 π ( − i k e i k R R + e i k R − 1 ) ] = − [ i k R e i k R − e i k R − i k e i k R R + e i k R − 1 ] = 1 \begin{aligned}
& -\frac{1}{4\pi}\left[\iint_{\text{Sphere}}\nabla\frac{e^{ikr'}}{r'}\cdot{\hat{r}}'\:dS'+\iiint_{\text{Sphere}}\left(k^2\frac{e^{ikr'}}{r'}\right)d\Omega'\right] \\
& \qquad=-\frac{1}{4\pi}\left[4\pi\left(ikRe^{ikR}-e^{ikR}\right)+4\pi\left(-ike^{ikR}R+e^{ikR}-1\right)\right] \\
& \qquad=-\left[ikRe^{ikR}-e^{ikR}-ike^{ikR}R+e^{ikR}-1\right] \\
& \qquad=1
\end{aligned} − 4 π 1 [ ∬ Sphere ∇ r ′ e ik r ′ ⋅ r ^ ′ d S ′ + ∭ Sphere ( k 2 r ′ e ik r ′ ) d Ω ′ ] = − 4 π 1 [ 4 π ( ik R e ik R − e ik R ) + 4 π ( − ik e ik R R + e ik R − 1 ) ] = − [ ik R e ik R − e ik R − ik e ik R R + e ik R − 1 ] = 1 As was to be shown.
Appendix 2. In the text we have an integral of the type:
∫ 0 + ∞ sin ( α r ′ ) e − β r ′ d r ′ \int_0^{+\infty}\sin\;\left(\alpha r'\right)e^{-\beta r'}\:dr' ∫ 0 + ∞ sin ( α r ′ ) e − β r ′ d r ′ We denote the integral by the symbol I . We will integrate by parts twice, thus obtaining an equation in the unknown I :
I = sin ( α r ′ ) e − β r ′ − β ∣ 0 + ∞ − ∫ 0 + ∞ α cos ( α r ′ ) e − β r ′ − β d r ′ = − α cos ( α r ′ ) e − β r ′ β 2 ∣ 0 + ∞ + ∫ 0 + ∞ − α 2 cos ( α r ′ ) e − β r ′ β 2 d r ′ = α β 2 − α 2 β 2 I \begin{aligned}
I & =\sin(\alpha r')\frac{e^{-\beta r'}}{-\beta}|_0^{+\infty}-\int_0^{+\infty}\alpha\cos(\alpha r')\frac{e^{-\beta r'}}{-\beta}dr' \\
& =-\alpha\cos(\alpha r')\frac{e^{-\beta r'}}{\beta^2}|_0^{+\infty}+\int_0^{+\infty}-\alpha^2\cos(\alpha r')\frac{e^{-\beta r'}}{\beta^2}dr' \\
& =\frac{\alpha}{\beta^2}-\frac{\alpha^2}{\beta^2}I
\end{aligned} I = sin ( α r ′ ) − β e − β r ′ ∣ 0 + ∞ − ∫ 0 + ∞ α cos ( α r ′ ) − β e − β r ′ d r ′ = − α cos ( α r ′ ) β 2 e − β r ′ ∣ 0 + ∞ + ∫ 0 + ∞ − α 2 cos ( α r ′ ) β 2 e − β r ′ d r ′ = β 2 α − β 2 α 2 I So we have the equation
I = α β 2 − α 2 β 2 I ⟺ ( 1 + α 2 β 2 ) I = α β 2 ⟺ β 2 + α 2 β 2 I = α β 2 ⟺ I = α β 2 + α 2 \begin{aligned}
I & =\frac{\alpha}{\beta^2}-\frac{\alpha^2}{\beta^2}I\Longleftrightarrow \\
\left(1+\frac{\alpha^2}{\beta^2}\right)I & =\frac{\alpha}{\beta^2}\Longleftrightarrow \\
\frac{\beta^2+\alpha^2}{\beta^2}I & =\frac{\alpha}{\beta^2}\Longleftrightarrow \\
I & =\frac{\alpha}{\beta^2+\alpha^2}
\end{aligned} I ( 1 + β 2 α 2 ) I β 2 β 2 + α 2 I I = β 2 α − β 2 α 2 I ⟺ = β 2 α ⟺ = β 2 α ⟺ = β 2 + α 2 α This is the formula applied in the text.