Demostración

Para ver que efectivamente la forma integral propuesta $$ R(r)={\cal N}_{k,l} (k r)^{l+1}\int_{-1}^{1}\frac{e^{{\bf i} k r u}}{2}(1+u)^{l+\frac{{\bf i}}{2 k r_0}}(1-u)^{l-\frac{{\bf i}}{2 k r_0}} du $$

es solución de la ecuación radial, calculamos la derivada de $F(k r)=\frac{(k r)^{-(l+1)}}{{\cal N}}R_{kl}(k r)$

\begin{eqnarray*} \frac{\partial F( k r)}{\partial r} &=& {\bf i}k \int_{-1}^{1} u \frac{e^{{\bf i} k r u}}{2} (1+u)^{l+\frac{{\bf i}}{2 k r_0}}(1-u)^{l-\frac{{\bf i}}{2 k r_0}}du \;\;\mbox{Integrando por partes}\\ &=& {\bf i}k \int_{-1}^{1} \frac{1-u^2}{2} \frac{d }{d u}\left(\frac{e^{{\bf i} k r u}}{2} (1+u)^{l+\frac{{\bf i}}{2 k r_0}}(1-u)^{l-\frac{{\bf i}}{2 k r_0}}\right)du \\ &=& -k^2 r \int_{-1}^{1} \frac{1-u^2}{2} \frac{e^{{\bf i} k r u}}{2 } (1+u)^{l+\frac{{\bf i}}{2 k r_0}}(1-u)^{l-\frac{{\bf i}}{2 k r_0}}du\\ && - \int_{-1}^{1} ( \frac{1}{2 r_0} + l {\bf i} k u )\frac{e^{{\bf i} k r u}}{2 } (1+u)^{l+\frac{{\bf i}}{2 k r_0}}(1-u)^{l-\frac{{\bf i}}{2 k r_0}}du \;\;\mbox{reagrupando términos}\\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% &=& -(\frac{k^2 r}{2}+\frac{1}{2 r_0}) \int_{-1}^{1} \frac{e^{{\bf i} k r u}}{2 } (1+u)^{l+\frac{{\bf i}}{2 k r_0}}(1-u)^{l-\frac{{\bf i}}{2 k r_0}}du -\frac{r}{2} \int_{-1}^{1} ({\bf i} k u)^2 \frac{e^{{\bf i} k r u}}{2 } (1+u)^{l+\frac{{\bf i}}{2 k r_0}}(1-u)^{l-\frac{{\bf i}}{2 k r_0}}du\\ && - l \int_{-1}^{1} ({\bf i} k u) {\bf i} \frac{e^{{\bf i} k r u}}{2 } (1+u)^{l+\frac{{\bf i}}{2 k r_0}}(1-u)^{l-\frac{{\bf i}}{2 k r_0}}du \;\;\;\mbox{\'o, en t\'erminos de }F(kr) \mbox{ y sus derivadas}\\ %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% &=& -(\frac{k^2 r}{2}+\frac{1}{2 r_0}) F(k r) -\frac{r}{2} \frac{\partial^2}{\partial r^2}F(kr) - l \frac{\partial}{\partial r}F(kr) \end{eqnarray*}

Pasando todo al primer miembro encontramos que $F(kr)$ satisface la ecuación diferencial $$ \frac{\partial^2}{\partial r^2}F(kr)

• 2\frac{l+1}{r} \frac{\partial}{\partial r}F(kr)+(k^2+\frac{1}{r r0}) F(k r)=0 $$ pero, remplazando $F(k r)=\frac{(k r)^{-(l+1)}}{{\cal N}}R{kl}(k r)$ encontramos que \begin{eqnarray*} 0&=&\frac{\partial^2}{\partial r^2}\left(\frac{R_{kl}(k r)}{{\cal N}(kr)^{l+1}}\right)
• 2\frac{l+1}{r} \frac{\partial}{\partial r} \left(\frac{R_{kl}(k r)}{{\cal N}(kr)^{l+1}}\right) +(k^2+\frac{1}{r r0}) \left(\frac{R{kl}(k r)}{{\cal N}(kr)^{l+1}}\right)\;\;\;\mbox{ multiplicando por }-{\cal N}r^{l+1}\ &=&-(k r)^{l+1}\frac{\partial^2}{\partial r^2}\left(\frac{R_{kl}(k r)}{(kr)^{l+1}}\right)
• 2\frac{l+1}{r}(k r)^{l+1} \frac{\partial}{\partial r}\left(\frac{R_{kl}(k r)}{(kr)^{l+1}}\right) -(k^2+\frac{1}{r r0}) R{kl}(k r)\ &=& -\frac{\partial^2 R_{kl}(k r)}{\partial r^2}
• 2 \frac{l+1}{r}\frac{\partial R{kl}(k r)}{\partial r} -\frac{(l+2)(l+1)}{r^2}R{kl}(k r)
• 2\frac{(l+1)^2}{r^2}R_{kl}(k r)
• 2\frac{l+1}{r} \frac{\partial R_{kl}(k r)}{\partial r} -(k^2+\frac{1}{r r0}) R{kl}(k r)\ &=&-\frac{\partial^2 R{kl}(k r)}{\partial r^2} +\frac{((2l+2)- (l+2))(l+1)}{r^2}R{kl}(k r) -(k^2+\frac{1}{r r0}) R{kl}(k r)\ &=&-\frac{\partial^2 R{kl}(k r)}{\partial r^2} +\frac{l(l+1)}{r^2}R{kl}(k r) -\frac{R_{kl}(k r)}{r r0}-k^2 R{kl}(k r)) \ \end{eqnarray*} que es la ecuación radial.

Normalización a la $\delta^3(r)$

Para normalizar a la "delta" necesitamos encontrar la constante ${\cal N}$ tal que se cumpla que, para $r\rightarrow \infty$, $F(kr) \approx \frac{\sin(k r + \delta_{kr,l}(r))}{r}$. Para esto, necesitamos determinar el comportamiento de la forma integral en ese límite, lo que logramos vía el cambio de variables $$z= {\bf i} k r (1-u)$$ $$u= 1-z / ({\bf i } k r)$$ $$du = 1/({\bf i}k r)dz$$

Con este cambio de variables,

\begin{eqnarray} \psi(r)&=&r^l k^{l+1} 2 \Re\left[\int_{0}^{1} e^{i k r u} (1+u)^{l+i/(2 k r_0)}(1-u)^{l-i/(2k r_0)} du\right]\\ &=& r^l k^{l+1} 2 \Re\left[ e^{i k r} ({\bf i} k r)^{-1} \int_{0}^{{\bf i} k r} e^{-z} (2-z/({\bf i} k r_0))^{l+i/(2 k r_0)}(z/({\bf i}k r_0))^{l-i/(2 k r_0)} dz\right]\\ &=& r^l k^{l+1} 2 \Re\left[ e^{i k r} ({\bf i}k r)^{-l-1+i/(2 k r_0)} \int_{0}^{{\bf i} k r} e^{-z} (2-z/({\bf i} k r))^{l+i/(2 k r_0)}(z)^{l-i/(2 k r_0)} dz\right]\\ &\approx& \frac{2^{l+1}}{r} \Re\left[ {\bf i}^{-l-1} e^{i k r} (2{\bf i}k r)^{i/(2 k r_0)} \int_{0}^{{\bf i} k r} e^{-z} (z)^{l-i/(2 k r_0)} dz\right]\\ &=& \frac{2^{l+1}}{r} \Re\left[ {\bf i}^{-l-1} e^{i k r} (2{\bf i}k r)^{i/(2 k r_0)} \Gamma(l +1- \frac{{\bf i}}{2 k r_0})\right]\\ &=& \frac{2^{l+1}}{r} ({\bf i})^{{\bf i}/(2 k r_0)} \left|\Gamma(l +1- \frac{{\bf i}}{2 k r_0})\right| \Re\left[ {\bf i}^{-l-1} e^{i k r} (2 k r)^{{\bf i}/(2 k r_0)} \frac{\Gamma(l +1- \frac{{\bf i}}{2 k r_0})}{\left|\Gamma(l +1- \frac{{\bf i}}{2 k r_0})\right|}\right]\\ &=& \frac{2^{l+1}}{r} e^{-\pi/(4 k r_0)} \left|\Gamma(l +1- \frac{{\bf i}}{2 k r_0})\right| \sin(k r + \log(2 k r)/(2 k r_0) +\delta_{l,k r_0})\\ \end{eqnarray}

donde en la cuarta linea aproximamos $(2-z/(kr))^{l+{\bf i}/(2 k r_0)}$ a orden más bajo en $1/(k r)$. Esto es posible porque en la región de integración $0<|z|< kr$, y por lo tanto, su expansión en serie de potencias converge uniformemente. Los siguientes pasos resultan entonces completamente analíticos.

Soluciones numéricas para l=0

l=1

l=6