Una partícula de masa m se mueve en un potencial circular

${\displaystyle V\left(\rho \right)=-\gamma \frac{{\hslash }^2}{2m}\frac{\delta \left(\rho -{\rho }_0\right)}{\rho }}$

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

Resolvemos la ecuación de Schrrödinger en coordenadas cilíndricas independiente de la altura z

${\displaystyle -\frac{{\hslash }^2}{2m}\left(\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho \frac{\partial \psi (\rho ,\phi )}{\partial \rho }\right)+\frac{1}{{\rho }^2}\frac{{\partial }^2\psi (\rho ,\phi )}{\partial {\phi }^2}\right)+V(\rho )\psi (\rho ,\phi )=E\psi (\rho ,\phi )}$

Aplicamos el procedimiento de separación de variables

${\displaystyle \begin{array}{l}\psi (\rho ,\phi )=R(\rho )F(\phi )\hfill \\ F\frac{d^{2}R}{d{\rho }^2}+\frac{F}{\rho }\frac{dR}{d\rho }+\frac{R}{{\rho }^2}\frac{d^{2}F}{d{\phi }^2}=-\frac{2m}{{\hslash }^2}\left(E-V\left(\rho \right)\right)RF\hfill \\ \left(\frac{{\rho }^2}{R}\frac{d^{2}R}{d{\rho }^2}+\frac{\rho }{R}\frac{dR}{d\rho }+\frac{2m\left(E-V\left(\rho \right)\right)}{{\hslash }^2}{\rho }^2\right)+\frac{1}{F}\frac{d^{2}F}{d{\phi }^2}=0\hfill \end{array}}$

El primer término, solamente depende de ρ, el segundo de φ, la ecuación diferencial se transforma en el sistema de dos ecuaciones diferenciales

${\displaystyle \{\begin{array}{l}\frac{{\rho }^2}{R}\frac{d^{2}R}{d{\rho }^2}+\frac{\rho }{R}\frac{dR}{d\rho }+\left({\rho }^2\frac{2m\left(E-V\left(\rho \right)\right)}{{\hslash }^2}-n_{\phi }{}^2\right)=0\\ \frac{d^{2}F}{d{\phi }^2}=-n_{\phi }{}^{2}F\end{array}}$

La ecuación angular

${\displaystyle \frac{d^{2}F}{d{\phi }^2}+n_{\phi }{}^{2}F=0}$

La solución es

${\displaystyle F(\phi )=C\exp (i{n}_{\phi }\phi ),n_{\phi }=\mathrm{0,}\pm \mathrm{1,}\pm \mathrm{2....}}$

La ecuación radial

Para ρ>ρ₀ y para ρ<ρ₀, V(ρ)=0

${\displaystyle \begin{array}{l}\frac{{\rho }^2}{R}\frac{d^{2}R}{d{\rho }^2}+\frac{\rho }{R}\frac{dR}{d\rho }+\left({\rho }^2\frac{2mE}{{\hslash }^2}-n_{\phi }^2\right)=0\\ {\rho }^2\frac{d^{2}R}{d{\rho }^2}+\rho \frac{dR}{d\rho }-\left({\rho }^2{k}^2+n_{\phi }^2\right)R=\mathrm{0,}{k}^2=-\frac{2mE}{{\hslash }^2}\end{array}}$

La energía de los niveles E<0

La solución de esta ecuación diferencial (véase el apartado Funciones de Bessel modificadas I_n y K_n)

${\displaystyle R_{n_{\phi }}\left(\rho \right)=A{I}_{n_{\phi }}\left(k\rho \right)+B{K}_{n_{\phi }}\left(k\rho \right)}$

• Para ρ<ρ₀ descartamos el segundo miembro ya que K_m(x)→∞ cuando x→0

• Para ρ>ρ₀ descartamos el primer miembro ya que I_m(x)→∞ cuando x→∞

La solución de la ecuación diferencial es

${\displaystyle \begin{array}{l}{R}_{n_{\phi }}^I\left(\rho \right)=A{I}_{n_{\phi }}\left(k\rho \right),\rho <{\rho }_0\\ R_{n_{\phi }}^{II}\left(\rho \right)=B{K}_{n_{\phi }}\left(k\rho \right),\rho >{\rho }_0\end{array}}$

• La función es continua en ρ₀

${\displaystyle \begin{array}{l}A{I}_{n_{\phi }}\left(k{\rho }_0\right)=B{K}_{n_{\phi }}\left(k{\rho }_0\right)\\ \frac{B}{A}=\frac{I_{n_{\phi }}\left(k{\rho }_0\right)}{K_{n_{\phi }}\left(k{\rho }_0\right)}\end{array}}$

• La derivada primera no es continua

Integramos la ecuación diferencial entre los límites ρ₀-ε a ρ₀+ε donde ε→0

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

Utilizaremos las siguientes propiedades de las funciones de Bessel

${\displaystyle \begin{array}{l}\frac{d{I}_n\left(x\right)}{dx}=\frac{I_{n+1}\left(x\right)+I_{n-1}\left(x\right)}{2}\\ \frac{d{K}_n\left(x\right)}{dx}=-\frac{K_{n+1}\left(x\right)+K_{n-1}\left(x\right)}{2}\\ I_n\left(x\right)K_{n+1}\left(x\right)+K_n\left(x\right)I_{n+1}\left(x\right)=\frac{1}{x}\\ I_n\left(x\right)K_{n-1}\left(x\right)+K_n\left(x\right)I_{n-1}\left(x\right)=\frac{1}{x}\end{array}}$

El resultado es una ecuación transcendente

${\displaystyle \begin{array}{l}{\frac{d{R}_{n_{\phi }}^{II}}{d\rho }|}_{\rho ={\rho }_0^{+}}-{\frac{d{R}_{n_{\phi }}^I}{d\rho }|}_{\rho ={\rho }_0^{-}}=-\gamma \frac{R_{n_{\phi }}}{{\rho }_0}\\ B{\frac{d{K}_{n_{\phi }}\left(k\rho \right)}{d\rho }|}_{\rho ={\rho }_0^{+}}-A{\frac{d{I}_{n_{\phi }}\left(k\rho \right)}{d\rho }|}_{\rho ={\rho }_0^{-}}=-\gamma A\frac{I_{n_{\phi }}\left(k{\rho }_0\right)}{{\rho }_0}\\ -A\frac{I_{n_{\phi }}\left(k{\rho }_0\right)}{K_{n_{\phi }}\left(k{\rho }_0\right)}k\frac{K_{n_{\phi }+1}\left(k{\rho }_0\right)+K_{n_{\phi }-1}\left(k{\rho }_0\right)}{2}-Ak\frac{I_{n_{\phi }+1}\left(k{\rho }_0\right)+I_{n_{\phi }-1}\left(k{\rho }_0\right)}{2}=-\gamma A\frac{I_{n_{\phi }}\left(k{\rho }_0\right)}{{\rho }_0}\\ -\frac{I_{n_{\phi }}\left(k{\rho }_0\right)}{K_{n_{\phi }}\left(k{\rho }_0\right)}\frac{K_{n_{\phi }+1}\left(k{\rho }_0\right)+K_{n_{\phi }-1}\left(k{\rho }_0\right)}{2}-\frac{I_{n_{\phi }+1}\left(k{\rho }_0\right)+I_{n_{\phi }-1}\left(k{\rho }_0\right)}{2}=-2\gamma \frac{I_{n_{\phi }}\left(k{\rho }_0\right)}{{\rho }_0}\\ I_{n_{\phi }}\left(k{\rho }_0\right)\left(K_{n_{\phi }+1}\left(k{\rho }_0\right)+K_{n_{\phi }-1}\left(k{\rho }_0\right)\right)+K_{n_{\phi }}\left(k{\rho }_0\right)\left(I_{n_{\phi }+1}\left(k{\rho }_0\right)+I_{n_{\phi }-1}\left(k{\rho }_0\right)\right)=2\frac{\gamma }{k}\frac{I_{n_{\phi }}\left(k{\rho }_0\right)K_{n_{\phi }}\left(k{\rho }_0\right)}{{\rho }_0}\\ \left(I_{n_{\phi }}\left(k{\rho }_0\right)K_{n_{\phi }+1}\left(k{\rho }_0\right)+K_{n_{\phi }}\left(k{\rho }_0\right)I_{n_{\phi }+1}\left(k{\rho }_0\right)\right)+\left(I_{n_{\phi }}\left(k{\rho }_0\right)K_{n_{\phi }-1}\left(k{\rho }_0\right)+K_{n_{\phi }}\left(k{\rho }_0\right)I_{n_{\phi }-1}\left(k{\rho }_0\right)\right)=2\frac{\gamma }{k}\frac{I_{n_{\phi }}\left(k{\rho }_0\right)K_{n_{\phi }}\left(k{\rho }_0\right)}{{\rho }_0}\\ \frac{1}{k{\rho }_0}+\frac{1}{k{\rho }_0}=2\frac{\gamma }{k}\frac{I_{n_{\phi }}\left(k{\rho }_0\right)K_{n_{\phi }}\left(k{\rho }_0\right)}{{\rho }_0}\\ \gamma I_{n_{\phi }}\left(k{\rho }_0\right)K_{n_{\phi }}\left(k{\rho }_0\right)-1=0\end{array}}$

Niveles de energía

Dado el parámetro γ y el radio ρ₀ resolvemos la ecuación transcendente

${\displaystyle \gamma I_{n_{\phi }}\left(x\right)K_{n\phi }\left(x\right)-1=\mathrm{0,}x=k{\rho }_0}$

para calcular la energía de los niveles

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

gamma=5.5;
for n=0:2
    f=@(x) gamma*besseli(n,x)*besselk(n,x)-1;
    x0=fzero(f,3);
    disp(-x0^2)
end

   -7.8694
   -6.5922
   -3.3790

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

Los mismos valores se obtienen para n_φ negativa

gamma=5.5;
for n=0:-1:-2
    f=@(x) gamma*besseli(n,x)*besselk(n,x)-1;
    x0=fzero(f,3);
    disp(-x0^2)
end

Funciones de onda

${\displaystyle {\psi }_{n_{\phi }}(\rho ,\phi )=C{\text{'}}_{n_{\phi }}\left\{\begin{array}{l}{I}_{n_{\phi }}\left(k\rho \right),\rho \le {\rho }_0\\ \frac{I_{n_{\phi }}\left(k{\rho }_0\right)}{K_{n_{\phi }}\left(k{\rho }_0\right)}{K}_{n_{\phi }}\left(k\rho \right),\rho >{\rho }_0\end{array}\right\}\exp \left(i{n}_{\phi }\phi \right)}$

Otra forma equivalente, más adecuda es

${\displaystyle {\psi }_{n_{\phi }}(\rho ,\phi )=C_{n_{\phi }}\left\{\begin{array}{l}\frac{I_{n_{\phi }}\left(k\rho \right)}{I_{n_{\phi }}\left(k{\rho }_0\right)},\rho \le {\rho }_0\\ \frac{K_{n_{\phi }}\left(k\rho \right)}{K_{n_{\phi }}\left(k{\rho }_0\right)},\rho >{\rho }_0\end{array}\right\}\exp \left(i{n}_{\phi }\phi \right)}$

El cociente entre los coeficientes se mantiene

${\displaystyle \frac{B}{A}=\frac{\frac{1}{K_{n_{\phi }}\left(k{\rho }_0\right)}}{\frac{1}{I_{n_{\phi }}\left(k{\rho }_0\right)}}=\frac{I_{n_{\phi }}\left(k{\rho }_0\right)}{K_{n_{\phi }}\left(k{\rho }_0\right)}}$

Calculamos la constante C_nφ de modo que

${\displaystyle \begin{array}{l}{{\displaystyle \underset{0}{\overset{\infty }{\int }}{\displaystyle \underset{0}{\overset{2\pi }{\int }}\left|{\psi }_{n_{\phi }}(\rho ,\phi )\right|}}}^2\rho ·d\rho ·d\phi =1\\ C_{n_{\phi }}^2{\displaystyle \underset{0}{\overset{\infty }{\int }}{\displaystyle \underset{0}{\overset{2\pi }{\int }}{R}_{n_{\phi }}^2\left(\rho \right)\exp \left(i{n}_{\phi }\phi \right)}}\exp \left(-i{n}_{\phi }\phi \right)\rho ·d\rho ·d\phi =2\pi C_{n_{\phi }}^2{\displaystyle \underset{0}{\overset{\infty }{\int }}{R}_{n_{\phi }}^2\left(\rho \right)\rho ·d\rho }\\ C_{n_{\phi }}^2\left\{\frac{2\pi }{I_{n_{\phi }}^2\left(k{\rho }_0\right)}{\displaystyle \underset{0}{\overset{{\rho }_0}{\int }}{I}_{n_{\phi }}^2\left(k\rho \right)}d\rho +\frac{2\pi }{K_{n_{\phi }}^2\left(k{\rho }_0\right)}{\displaystyle \underset{{\rho }_0}{\overset{\infty }{\int }}{K}_{n_{\phi }}^2\left(k\rho \right)d\rho }\right\}=1\end{array}}$

Calculamos las integrales partiendo de la ecuación diferencial radial

${\displaystyle \begin{array}{l}{z}^2\frac{d^{2}R}{d{z}^2}+z\frac{dR}{dz}-\left(z^2+n_{\phi }^2\right)R=\mathrm{0,}z=\rho k\\ \frac{d}{dz}{\left(z\frac{dR}{dz}\right)}^2=\left(z^2+n_{\phi }^2\right)\frac{d}{dz}{R}^2\\ {\displaystyle \underset{0}{\overset{z_0}{\int }}\frac{d}{dz}{\left(z\frac{d{I}_{n_{\phi }}(z)}{dz}\right)}^{2}dz={\displaystyle \underset{0}{\overset{z_0}{\int }}\left(z^2+n_{\phi }^2\right)\frac{d}{dz}{I}_{n_{\phi }}^2(z)·dz}},z_0={\rho }_{0}k\end{array}}$

Integramos por partes el segundo miembro

${\displaystyle \{\begin{array}{l}u=z^2+n_{\phi }^2,du=2z·dz\\ dv=\frac{d}{dz}{I}_{n_{\phi }}^2(z)·dz,v=I_{n_{\phi }}^2(z)\end{array}}$

El resultado es

${\displaystyle \begin{array}{l}{\left(z_0{\frac{d{I}_{n_{\phi }}(z)}{dz}|}_{z=z_0}\right)}^2={\left(z^2+n_{\phi }^2\right)I_{n_{\phi }}^2(z)|}_0^{z_0}-2{\displaystyle \underset{0}{\overset{z_0}{\int }}z{I}_{n_{\phi }}^2(z)·dz}\\ {\left(z_0{\frac{d{I}_{n_{\phi }}(z)}{dz}|}_{z=z_0}\right)}^2=\left(z_0^2+n_{\phi }^2\right)I_{n_{\phi }}^2(z_0)-n_{\phi }^2{I}_{n_{\phi }}^2(0)-2{\displaystyle \underset{0}{\overset{z_0}{\int }}z{I}_{n_{\phi }}^2(z)·dz}\\ {\displaystyle \underset{0}{\overset{z_0}{\int }}z{I}_{n_{\phi }}^2(z)·dz}=\frac{\left(z_0^2+n_{\phi }^2\right)I_{n_{\phi }}^2(z_0)-{\left(z_0{\frac{d{I}_{n_{\phi }}(z)}{dz}|}_{z=z_0}\right)}^2}{2}\end{array}}$

Fijarse que I_m(0)=0 expecepto I₀(0)=1

Deshacemos el cambio de variable z=ρk

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

Seguimos los mismos pasos para evaluar la segunda integral en K_nφ

${\displaystyle \begin{array}{l}\frac{d}{dz}{\left(z\frac{dR}{dz}\right)}^2=\left(z^2+n_{\phi }^2\right)\frac{d}{dz}{R}^2\\ {\displaystyle \underset{z_0}{\overset{\infty }{\int }}\frac{d}{dz}{\left(z\frac{d{K}_{n_{\phi }}(z)}{dz}\right)}^{2}dz={\displaystyle \underset{z_0}{\overset{\infty }{\int }}\left(z^2+n_{\phi }^2\right)\frac{d}{dz}{K}_{n_{\phi }}^2(z)·dz}}\\ {\left(\infty {\frac{d{K}_{n_{\phi }}(z)}{dz}|}_{z=\infty }\right)}^2-{\left(z_0{\frac{d{K}_{n_{\phi }}(z)}{dz}|}_{z=z_0}\right)}^2={\left(z^2+n_{\phi }^2\right)K_{n_{\phi }}^2(z)|}_{z_0}^{\infty }-2{\displaystyle \underset{z_0}{\overset{\infty }{\int }}z{I}_{n_{\phi }}^2(z)·dz}\\ -{\left(z_0{\frac{d{K}_{n_{\phi }}(z)}{dz}|}_{z=z_0}\right)}^2=-\left(z_0^2+n_{\phi }^2\right)K_{n_{\phi }}^2(z_0)-2{\displaystyle \underset{z_0}{\overset{\infty }{\int }}z{I}_{n_{\phi }}^2(z)·dz}\\ {\displaystyle \underset{z_0}{\overset{\infty }{\int }}z{K}_{n_{\phi }}^2(z)·dz}=\frac{-\left(z_0^2+n_{\phi }^2\right)K_{n_{\phi }}^2(z_0)+{\left(z_0{\frac{d{K}_{n_{\phi }}(z)}{dz}|}_{z=z_0}\right)}^2}{2}\end{array}}$

Se ha utilizado el resultado

${\displaystyle \begin{array}{l}\underset{z\to \infty }{\lim }\left(z·K_m(z)\right)=0\\ \underset{z\to \infty }{\lim }\left(z\frac{d{K}_m(z)}{dz}\right)=\underset{z\to \infty }{\lim }\left(-z\frac{K_{m+1}(z)+K_{m-1}(z)}{2}\right)=0\end{array}}$

Comprobamos con MATLAB

>> m=3;
>> z=100;
>> z*(besselk(m+1,z)+besselk(m-1,z))/2
ans =   4.8963e-43
>> z*besselk(m,z)
ans =   4.8699e-43

Deshacemos el cambio de variable z=ρk

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

El resultado final es

${\displaystyle \begin{array}{l}{C}_{n_{\phi }}^2\left\{\frac{2\pi }{I_m^2\left(k{\rho }_0\right)}\frac{\left({\left(k{\rho }_0\right)}^2+n_{\phi }^2\right)I_{n_{\phi }}^2\left(k{\rho }_0\right)-{\left({\rho }_{0}k{\frac{d{I}_{n_{\phi }}\left(k\rho \right)}{d\rho }|}_{\rho ={\rho }_0}\right)}^2}{2k^2}+\frac{2\pi }{K_{n_{\phi }}^2\left(k{\rho }_0\right)}\frac{-\left({\left(k{\rho }_0\right)}^2+n_{\phi }^2\right)K_{n_{\phi }}^2\left(k{\rho }_0\right)+{\left({\rho }_0{\frac{d{K}_{n_{\phi }}\left(k\rho \right)}{d\rho }|}_{\rho ={\rho }_0}\right)}^2}{2k^2}\right\}=1\\ C_{n_{\phi }}^2\frac{\pi }{k^2}{\rho }_0^2\left\{{\left(\frac{{\frac{d{K}_{n_{\phi }}\left(k\rho \right)}{d\rho }|}_{\rho ={\rho }_0}}{K_{n_{\phi }}\left(k{\rho }_0\right)}\right)}^2-{\left(\frac{{\frac{d{I}_{n_{\phi }}\left(k\rho \right)}{d\rho }|}_{\rho ={\rho }_0}}{I_{n_{\phi }}\left(k{\rho }_0\right)}\right)}^2\right\}=1\\ C_{n_{\phi }}^2\frac{\pi }{4}{\rho }_0^2\left\{{\left(\frac{K_{n_{\phi }+1}\left(k{\rho }_0\right)+K_{n_{\phi }-1}\left(k{\rho }_0\right)}{K_{n_{\phi }}\left(k{\rho }_0\right)}\right)}^2-{\left(\frac{I_{n_{\phi }+1}\left(k{\rho }_0\right)+I_{n_{\phi }-1}\left(k{\rho }_0\right)}{I_{n_{\phi }}\left(k{\rho }_0\right)}\right)}^2\right\}=1\end{array}}$

Resultados

Representamos la componenente radial para n_φ=0, 1 y 2, fijados γ=5.5 y ρ₀=1

gamma=5.5; %parámetro
r0=1; %radio 
hold on
colores=['k','r','g'];
for m=0:2 %nùmeros enteros
    f=@(x) gamma*besseli(m,x)*besselk(m,x)-1;
    k=fzero(f,3);%nivel de energía
    d_K=-(besselk(m+1,k*r0)+besselk(m-1,k*r0));
    d_I=besseli(m+1,k*r0)+besseli(m-1,k*r0);
    den=pi*r0^2*((d_K/besselk(m,k*r0))^2-(d_I/besseli(m,k*r0))^2)/4;
    C=sqrt(1/den);
    f=@(x) C*besseli(m,k*x)/besseli(m,k*r0);
    g=@(x) C*besselk(m,k*x)/besselk(m,k*r0);
    fplot(f,[0,r0],colores(m+1))
    fplot(g,[r0,3],colores(m+1))
end
grid on
xlabel('\rho/\rho_0')
ylabel('R_n(\rho)')
title('Componente radial')

Funciones de onda para γ=5.5 y ρ₀=1

gamma=5.5; %parámetro
r0=1; %radio 
m=0; %número entero 
f=@(x) gamma*besseli(m,x)*besselk(m,x)-1;
k=fzero(f,3);
disp(-k^2) %energía
d_K=-(besselk(m+1,k*r0)+besselk(m-1,k*r0));
d_I=besseli(m+1,k*r0)+besseli(m-1,k*r0);
den=pi*r0^2*((d_K/besselk(m,k*r0))^2-(d_I/besseli(m,k*r0))^2)/4;
C=sqrt(1/den);
hold on
rho=linspace(0,r0,50);
phi=linspace(0,2*pi, 100);
[Rho,Phi]=meshgrid(rho,phi);
X=Rho.*cos(Phi);
Y=Rho.*sin(Phi);
Z=C*cos(m*Phi).*besseli(m,k*Rho)/besseli(m,k*r0);
surfl(X,Y,Z)
rho=linspace(r0,3,50);
phi=linspace(0,2*pi, 100);
[Rho,Phi]=meshgrid(rho,phi);
X=Rho.*cos(Phi);
Y=Rho.*sin(Phi);
Z=C*cos(m*Phi).*besselk(m,k*Rho)/besselk(m,k*r0);
surfl(X,Y,Z)
shading interp
colormap(gray);
fplot3(@(t) r0*cos(t), @(t) r0*sin(t), @(t) 0*t, [0,2*pi],'r')
hold off
xlabel('X')
ylabel('Y')
zlabel('Z')
grid on
title('Función de onda')
view(-10,60)

Función de onda para n_φ=0, (en el código, m=0;), energía E=-7.8694

Función de onda para n_φ=1, (en el código, m=1;), energía E=-6.5922

La circuferencia de color rojo, señala la posición del anillo de potencial de profundidad infinita

Función de onda para n_φ=2, (en el código, m=2;), energía E=-3.3790

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