Una partícula de masa m se mueve en un potencial

${\displaystyle V\left(r\right)=-\lambda \frac{{\hslash }^2}{2m}\frac{\delta \left(r-r_0\right)}{r^2}}$

El signo de λ determina si es un pozo (λ>0) o una barrera (λ<0)

Resolvemos la componente radial de la ecuación de Schrödinger en coordenadas esféricas

${\displaystyle \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\left\{\frac{2m}{{\hslash }^2}\left(E-V(r)\right)-\frac{l\left(l+1\right)}{r^2}\right\}R=0}$

Para r>r₀ y para r<r₀, V(r)=0

${\displaystyle \begin{array}{l}\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\left\{\frac{2m}{{\hslash }^2}E-\frac{l(l+1)}{r^2}\right\}R=0\\ \frac{d^{2}R}{d{r}^2}+2r\frac{dR}{dr}-\left\{k^2{r}^2+l(l+1)\right\}R=\mathrm{0,}{k}^2=-\frac{2m}{{\hslash }^2}E\end{array}}$

La energía de los niveles E<0

La solución de esta ecuación diferencial (véase el apartado Funciones esféricas de Bessel modificadas i_n y k_n)

${\displaystyle R_l\left(r\right)=A{i}_l\left(kr\right)+B{k}_l\left(kr\right)}$

• Para r<r₀ descartamos el segundo miembro ya que k_l(x)→∞ cuando x→0

• Para r>r₀ descartamos el primer miembro ya que i_l(x)→∞ cuando x→∞

La solución de la ecuación diferencial es

${\displaystyle \begin{array}{l}{R}_l^I\left(r\right)=A{i}_l\left(kr\right),r<r_0\\ R_l^{II}\left(r\right)=B{k}_l\left(kr\right),r>r_0\end{array}}$

• La función es continua en r₀

${\displaystyle \begin{array}{l}A{i}_l\left(k{r}_0\right)=B{k}_l\left(k{r}_0\right)\\ \frac{B}{A}=\frac{i_l\left(k{r}_0\right)}{k_l\left(k{r}_0\right)}\end{array}}$

• La derivada primera no es continua

Integramos la ecuación diferencial entre los límites r₀-ε a r₀+ε donde ε→0

${\displaystyle \begin{array}{l}\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)+\left\{\frac{2m}{{\hslash }^2}\left(E-V(r)\right)-\frac{l(l+1)}{r^2}\right\}R=0\\ \frac{d^{2}R}{d{r}^2}+\frac{2}{r}\frac{dR}{dr}+\left\{\frac{2m}{{\hslash }^2}\left(E+\lambda \frac{{\hslash }^2}{2m}\frac{\delta \left(r-r_0\right)}{r^2}\right)-\frac{l(l+1)}{r^2}\right\}R=0\\ {\displaystyle \underset{r_0-\epsilon }{\overset{r_0+\epsilon }{\int }}\frac{d^{2}R}{d{r}^2}dr}+{\displaystyle \underset{r_0-\epsilon }{\overset{r_0+\epsilon }{\int }}\frac{2}{r}\frac{dR}{dr}dr}+\frac{2mE}{{\hslash }^2}{\displaystyle \underset{r_0-\epsilon }{\overset{r_0+\epsilon }{\int }}Rdr}+{\displaystyle \underset{r_0-\epsilon }{\overset{r_0+\epsilon }{\int }}\lambda \frac{\delta \left(r-r_0\right)}{r^2}Rdr-l(l+1)}{\displaystyle \underset{r_0-\epsilon }{\overset{r_0+\epsilon }{\int }}\frac{R}{{\rho }^2}d\rho }=0\\ {\displaystyle \underset{{\rho }_0-\epsilon }{\overset{{\rho }_0+\epsilon }{\int }}\frac{d^{2}R}{d{r}^2}dr}+\frac{2}{r_0}{\displaystyle \underset{r_0-\epsilon }{\overset{r_0+\epsilon }{\int }}\frac{dR}{dr}}dr+\frac{2mE}{{\hslash }^2}{\displaystyle \underset{r_0-\epsilon }{\overset{r_0+\epsilon }{\int }}Rdr}+\frac{\lambda }{r_0^2}{\displaystyle \underset{r_0-\epsilon }{\overset{r_0+\epsilon }{\int }}R\delta \left(r-r_0\right)dr-\frac{l(l+1)}{r_0^2}}{\displaystyle \underset{r_0-\epsilon }{\overset{r_0+\epsilon }{\int }}Rdr}=0\\ {\frac{dR}{dr}|}_{r_0+\epsilon }-{\frac{dR}{dr}|}_{r_0-\epsilon }+\frac{2}{r_0}\left(R\left(r_0+\epsilon \right)-R\left(r_0-\epsilon \right)\right)+\frac{2mE}{{\hslash }^2}{\displaystyle \underset{r_0-\epsilon }{\overset{r_0+\epsilon }{\int }}Rdr}+\frac{\lambda }{r_0^2}R\left(r_0\right)-\frac{l(l+1)}{r_0^2}{\displaystyle \underset{r_0-\epsilon }{\overset{r_0+\epsilon }{\int }}Rdr}=0\end{array}}$

En el límite cuando ε→0

${\displaystyle {\frac{dR}{dr}|}_{r_0^{+}}-{\frac{dR}{d\rho }|}_{{}_{r_0^{-}}}+\frac{\lambda }{r_0^2}R\left(r_0\right)=0}$

Utilizaremos las siguientes propiedades de las funciones esféricas de Bessel

${\displaystyle \begin{array}{l}\left(2n+1\right)\frac{d{i}_n\left(x\right)}{dx}=n{i}_{n-1}\left(x\right)+\left(n+1\right)i_{n+1}\left(x\right)\\ \left(2n+1\right)\frac{d{k}_n\left(x\right)}{dx}=-n{k}_{n-1}\left(x\right)-\left(n+1\right)k_{n+1}\left(x\right)\\ i_n\left(x\right)k_{n+1}\left(x\right)+k_n\left(x\right)i_{n+1}\left(x\right)=\frac{1}{x^2}\\ i_n\left(x\right)k_{n-1}\left(x\right)+k_n\left(x\right)i_{n-1}\left(x\right)=\frac{1}{x^2}\\ i_n(x)=\sqrt{\frac{\pi }{2x}}{I}_{n+\frac{1}{2}}(x),k_n(x)=\sqrt{\frac{2}{\pi x}}{K}_{n+\frac{1}{2}}(x)\end{array}}$

El resultado es una ecuación transcendente

${\displaystyle \begin{array}{l}{\frac{d{R}_l^{II}}{dr}|}_{\rho =r_0^{+}}-{\frac{d{R}_l^I}{dr}|}_{r=r_0^{-}}=-\frac{\lambda }{r_0^2}{R}_l\left(r_0\right)\\ B{\frac{d{k}_l\left(kr\right)}{dr}|}_{r=r_0^{+}}-A{\frac{d{i}_l\left(kr\right)}{dr}|}_{r=r_0^{-}}=-\lambda A\frac{i_l\left(k{r}_0\right)}{r_0^2}\\ -A\frac{i_l\left(k{r}_0\right)}{k_l\left(k{r}_0\right)}k\frac{l{k}_{l-1}\left(k{r}_0\right)+\left(l+1\right)k_{l+1}\left(k{r}_0\right)}{\left(2l+1\right)}-Ak\frac{l{i}_{l-1}\left(k{r}_0\right)+\left(l+1\right)i_{l+1}\left(k{r}_0\right)}{\left(2l+1\right)}=-\lambda A\frac{i_l\left(k{r}_0\right)}{r_0^2}\\ k\frac{l\left(i_l\left(k{r}_0\right)k_{l-1}\left(k{r}_0\right)+i_{l-1}\left(k{r}_0\right)k_l\left(k{r}_0\right)\right)+\left(l+1\right)\left(i_l\left(k{r}_0\right)k_{l+1}\left(k{r}_0\right)+k_l\left(k{r}_0\right)i_{l+1}\left(k{r}_0\right)\right)}{\left(2l+1\right)k_l\left(k{r}_0\right)}=\lambda \frac{i_l\left(k{r}_0\right)}{r_0^2}\\ k\left(\frac{l}{k^2{r}_0^2}+\frac{l+1}{k^2{r}_0^2}\right)=\lambda \left(2l+1\right)\frac{i_l\left(k{r}_0\right)k_l\left(k{r}_0\right)}{r_0^2}\\ \frac{1}{k}=\lambda i_l\left(k{r}_0\right)k_l\left(k{r}_0\right)\\ k\lambda i_l\left(k{r}_0\right)k_l\left(k{r}_0\right)-1=0\\ k\lambda \sqrt{\frac{\pi }{2k{r}_0}}{I}_{l+\frac{1}{2}}\left(k{r}_0\right)\sqrt{\frac{2}{\pi k{r}_0}}{K}_{l+\frac{1}{2}}\left(k{r}_0\right)-1=0\\ \frac{\lambda }{r_0}{I}_{l+\frac{1}{2}}\left(k{r}_0\right)K_{l+\frac{1}{2}}\left(k{r}_0\right)-1=0\end{array}}$

Niveles de energía

Dado el parámetro λ y el radio r₀ resolvemos la ecuación transcendente

${\displaystyle \frac{\lambda }{r_0}{I}_{l+\frac{1}{2}}\left(x\right)K_{l+\frac{1}{2}}\left(x\right)-1=\mathrm{0,}x=k{r}_0}$

para calcular la energía de los niveles

${\displaystyle E=-\frac{{\hslash }^2{k}^2}{2m}=-\frac{{\hslash }^2}{2m{r}_0^2}{x}^2}$

gamma=6;  %lambda/r0
for l=0:2
    f=@(x) gamma*besseli(l+1/2,x)*besselk(l+1/2,x)-1;
    x0=fzero(f,3);
    disp(-x0^2)
end

   -8.9548
   -6.6740
   -2.5769

La energía de los niveles se expresa en unidades $\frac{{\hslash }^2}{2m{r}_0^2}$

Funciones de onda

Combinando la componente radial y angular las funciones de onda

${\displaystyle {\psi }_{l,m}(r,\theta ,\phi )=C{\text{'}}_l\left\{\begin{array}{l}{i}_l\left(kr\right),r\le r_0\\ \frac{i_l\left(k{r}_0\right)}{k_l\left(k{r}_0\right)}{k}_l\left(kr\right),r>r_0\end{array}\right\}{Y}_{l,m}\left(\theta ,\phi \right)}$

Otra forma equivalente, más adecuda, es

${\displaystyle {\psi }_{l,m}(r,\theta ,\phi )=C_l\left\{\begin{array}{l}\frac{i_l\left(kr\right)}{i_l\left(k{r}_0\right)},r\le r_0\\ \frac{k_l\left(kr\right)}{k_l\left(k{r}_0\right)},r>r_0\end{array}\right\}{Y}_{l,m}\left(\theta ,\phi \right)}$

Calculamos la constante C_l de modo que

${\displaystyle {\displaystyle \underset{0}{\overset{2\pi }{\int }}{{\displaystyle \underset{0}{\overset{\pi }{\int }}{\displaystyle \underset{0}{\overset{\infty }{\int }}\left|{\psi }_{l,m}(r,\theta ,\phi )\right|}}}^2{r}^2·\sin \theta ·d\phi }·d\theta ·dr=1}$

La componente angular ya está normalizada, lo hacemos con la componente radial

${\displaystyle \begin{array}{l}{C}_l^2{\displaystyle \underset{0}{\overset{\infty }{\int }}{R}_l^2\left(r\right)r^2·dr=}1\\ C_l^2\left\{\frac{1}{i_l^2\left(k{r}_0\right)}{\displaystyle \underset{0}{\overset{r_0}{\int }}{r}^2{i}_l^2\left(kr\right)}dr+\frac{1}{k_l^2\left(k{r}_0\right)}{\displaystyle \underset{r_0}{\overset{\infty }{\int }}{r}^2{k}_l^2\left(kr\right)dr}\right\}=1\end{array}}$

Calculamos el primer témino

${\displaystyle {\displaystyle \underset{0}{\overset{r_0}{\int }}{r}^2{i}_l^2\left(kr\right)}dr={\displaystyle \underset{0}{\overset{r_0}{\int }}{r}^2\frac{\pi }{2kr}{I}_{l+\frac{1}{2}}^2\left(kr\right)}dr=\frac{\pi }{2k}{\displaystyle \underset{0}{\overset{r_0}{\int }}r{I}_{l+\frac{1}{2}}^2\left(kr\right)}dr}$

En la página titulada Un potencial delta de Dirac en forma de anillo, hemos obtenido el resultado

${\displaystyle {\displaystyle \underset{0}{\overset{{\rho }_0}{\int }}\rho I_{n_{\phi }}^2\left(k\rho \right)·d\rho =}\frac{\left({\left(k{\rho }_0\right)}^2+n_{\phi }^2\right)I_{n_{\phi }}^2\left(k{\rho }_0\right)-{\left({\rho }_0{\frac{d{I}_{n_{\phi }}\left(k\rho \right)}{d\rho }|}_{\rho ={\rho }_0}\right)}^2}{2k^2}}$

De forma similar, cambiando el nombre de las variables que intervienen

${\displaystyle {\displaystyle \underset{0}{\overset{r_0}{\int }}r{I}_{l+\frac{1}{2}}^2\left(kr\right)·dr=}\frac{\left({\left(k{r}_0\right)}^2+{\left(l+\frac{1}{2}\right)}^2\right)I_{l+\frac{1}{2}}^2\left(k{r}_0\right)-{\left(r_0{\frac{d{I}_{l+\frac{1}{2}}\left(kr\right)}{dr}|}_{r=r_0}\right)}^2}{2k^2}}$

Seguimos los mismos pasos para evaluar el segundo término

${\displaystyle {\displaystyle \underset{0}{\overset{\infty }{\int }}{r}^2{k}_l^2\left(kr\right)}dr={\displaystyle \underset{0}{\overset{r_0}{\int }}{r}^2\frac{2}{\pi kr}{K}_{l+\frac{1}{2}}^2\left(kr\right)}dr=\frac{2}{\pi k}{\displaystyle \underset{0}{\overset{r_0}{\int }}r{K}_{l+\frac{1}{2}}^2\left(kr\right)}dr}$

En la página titulada Un potencial delta de Dirac en forma de anillo, hemos obtenido el resultado

${\displaystyle {\displaystyle \underset{{\rho }_0}{\overset{\infty }{\int }}\rho K_{n_{\phi }}^2\left(k\rho \right)·d\rho =}\frac{-\left({\left(k{\rho }_0\right)}^2+n_{\phi }^2\right)K_{n_{\phi }}^2\left(k{\rho }_0\right)+{\left({\rho }_0{\frac{d{K}_{n_{\phi }}\left(k\rho \right)}{d\rho }|}_{\rho ={\rho }_0}\right)}^2}{2k^2}}$

De forma similar

${\displaystyle {\displaystyle \underset{r_0}{\overset{\infty }{\int }}r{K}_{l+\frac{1}{2}}^2\left(kr\right)·dr=}\frac{-\left({\left(k{r}_0\right)}^2+{\left(l+\frac{1}{2}\right)}^2\right)K_{l+\frac{1}{2}}^2\left(k{r}_0\right)+{\left(r_0{\frac{d{K}_{l+\frac{1}{2}}\left(kr\right)}{dr}|}_{r=r_0}\right)}^2}{2k^2}}$

El resultado final es

${\displaystyle \begin{array}{l}{C}_l^2\left\{\frac{1}{\frac{\pi }{2k{r}_0}{I}_{l+\frac{1}{2}}^2\left(k{r}_0\right)}\frac{\pi }{2k}{\displaystyle \underset{0}{\overset{r_0}{\int }}r{I}_{l+\frac{1}{2}}^2\left(kr\right)}dr+\frac{1}{\frac{2}{\pi k{r}_0}{K}_{l+\frac{1}{2}}^2\left(k{r}_0\right)}\frac{2}{\pi k}{\displaystyle \underset{0}{\overset{r_0}{\int }}r{K}_{l+\frac{1}{2}}^2\left(kr\right)}dr\right\}=1\\ C_l^2\left\{\frac{r_0}{I_{l+\frac{1}{2}}^2\left(k{r}_0\right)}{\displaystyle \underset{0}{\overset{r_0}{\int }}r{I}_{l+\frac{1}{2}}^2\left(kr\right)}dr+\frac{r_0}{K_{l+\frac{1}{2}}^2\left(k{r}_0\right)}{\displaystyle \underset{0}{\overset{r_0}{\int }}r{K}_{l+\frac{1}{2}}^2\left(kr\right)}dr\right\}=1\\ C_l^2\left\{\frac{r_0}{I_{l+\frac{1}{2}}^2\left(k{r}_0\right)}\frac{\left({\left(k{r}_0\right)}^2+{\left(l+\frac{1}{2}\right)}^2\right)I_{l+\frac{1}{2}}^2\left(k{r}_0\right)-{\left(r_0{\frac{d{I}_{l+\frac{1}{2}}\left(kr\right)}{dr}|}_{r=r_0}\right)}^2}{2k^2}+\frac{r_0}{K_{l+\frac{1}{2}}^2\left(k{r}_0\right)}\frac{-\left({\left(k{r}_0\right)}^2+{\left(l+\frac{1}{2}\right)}^2\right)K_{l+\frac{1}{2}}^2\left(k{r}_0\right)+{\left(r_0{\frac{d{K}_{l+\frac{1}{2}}\left(kr\right)}{dr}|}_{r=r_0}\right)}^2}{2k^2}\right\}=1\\ C_l^2{r}_0\left\{\frac{1}{I_{l+\frac{1}{2}}^2\left(k{r}_0\right)}\frac{-{\left(r_0{\frac{d{I}_{l+\frac{1}{2}}\left(kr\right)}{dr}|}_{r=r_0}\right)}^2}{2k^2}+\frac{1}{K_{l+\frac{1}{2}}^2\left(k{r}_0\right)}\frac{{\left(r_0{\frac{d{K}_{l+\frac{1}{2}}\left(kr\right)}{dr}|}_{r=r_0}\right)}^2}{2k^2}\right\}=1\\ C_l^2\frac{r_0^3}{2k^2}\left\{{\left(\frac{{\frac{d{K}_{l+\frac{1}{2}}\left(kr\right)}{dr}|}_{r=r_0}}{K_{l+\frac{1}{2}}\left(k{r}_0\right)}\right)}^2-{\left(\frac{{\frac{d{I}_{l+\frac{1}{2}}\left(kr\right)}{dr}|}_{r=r_0}}{I_{l+\frac{1}{2}}\left(k{r}_0\right)}\right)}^2\right\}=1\\ C_l^2\frac{r_0^3}{2k^2}\left\{{\left(\frac{-\frac{k}{2}\left(K_{l-\frac{1}{2}}\left(k{r}_0\right)+K_{l+\frac{3}{2}}\left(k{r}_0\right)\right)}{K_{l+\frac{1}{2}}\left(k{r}_0\right)}\right)}^2-{\left(\frac{\frac{k}{2}\left(I_{l-\frac{1}{2}}\left(k{r}_0\right)+I_{l+\frac{3}{2}}\left(k{r}_0\right)\right)}{I_{l+\frac{1}{2}}\left(k{r}_0\right)}\right)}^2\right\}=1\\ C_l^2\frac{r_0^3}{2^3}\left\{\left(\frac{K_{l-\frac{1}{2}}\left(k{r}_0\right)+K_{l+\frac{3}{2}}\left(k{r}_0\right)}{K_{l+\frac{1}{2}}\left(k{r}_0\right)}\right)-{\left(\frac{I_{l-\frac{1}{2}}\left(k{r}_0\right)+I_{l+\frac{3}{2}}\left(k{r}_0\right)}{I_{l+\frac{1}{2}}\left(k{r}_0\right)}\right)}^2\right\}=1\end{array}}$

Representamos la componente radial de la función de onda para l=0, 1 y 2, fijados λ/r₀=6 y r₀=1

${\displaystyle R_l\left(r\right)=C_l\left\{\begin{array}{l}\frac{i_l\left(kr\right)}{i_l\left(k{r}_0\right)},r\le r_0\\ \frac{k_l\left(kr\right)}{k_l\left(k{r}_0\right)},r>r_0\end{array}\right\}=C_l\left\{\begin{array}{l}\frac{\sqrt{\frac{\pi }{2kr}}{I}_{l+\frac{1}{2}}\left(kr\right)}{\sqrt{\frac{\pi }{2k{r}_0}}{I}_{l+\frac{1}{2}}\left(k{r}_0\right)}\\ {\frac{\sqrt{\frac{2}{\pi kr}}{K}_{l+\frac{1}{2}}\left(kr\right)}{\sqrt{\frac{2}{\pi k{r}_0}}{K}_{l+\frac{1}{2}}\left(k{r}_0\right)}}_0\end{array}\right\}=C_l\left\{\begin{array}{l}\sqrt{\frac{r_0}{r}}\frac{I_{l+\frac{1}{2}}\left(kr\right)}{I_{l+\frac{1}{2}}\left(k{r}_0\right)}\\ \sqrt{\frac{r_0}{r}}\frac{K_{l+\frac{1}{2}}\left(kr\right)}{K_{l+\frac{1}{2}}\left(k{r}_0\right)}\end{array}\right\}}$

gamma=6;  %lambda/r0
r0=1; %radio 
hold on
colores=['k','r','g'];
for l=0:2 %nùmeros enteros
    f=@(x) gamma*besseli(l+1/2,x)*besselk(l+1/2,x)-1;
    k=fzero(f,3);%nivel de energía
    disp(-k^2)
    d_K=-(besselk(l+3/2,k*r0)+besselk(l-1/2,k*r0))/besselk(l+1/2,k*r0);
    d_I=(besseli(l+3/2,k*r0)+besseli(l-1/2,k*r0))/besseli(l+1/2,k*r0);
    C=sqrt(2^3/(r0^3*(d_K^2-d_I^2)));
    f=@(x) C*sqrt(r0./x).*besseli(l+1/2,k*x)/besseli(l+1/2,k*r0);
    g=@(x) C*sqrt(r0./x).*besselk(l+1/2,k*x)/besselk(l+1/2,k*r0);
    fplot(f,[0,r0],colores(l+1))
    fplot(g,[r0,3],colores(l+1))
end
hold off
grid on
xlabel('\rho/\rho_0')
ylabel('R_l(\rho)')
title('Componente radial')

Se representa en color negro l=0, en rojo, l=1 y en verde l=2

Referencias

Luis F. Castillo-Sánchez, Julio C. Gutiérrez-Vega. Quantum solutions for the delta ring and delta shell. Am. J. Phys. 93 (7), July 2025. pp. 557-565