91影视

The following equation is given:

$\frac{1}{R}\frac{d}{dr}\left(r^2\frac{dR}{dr}\right)鈭�\frac{2M{r}^2}{{\mathrm{鈩�}}^2}[V\left(r\right)鈭�E]=l(l+1)$

A linear partial differential equation that governs the wave function of a quantum-mechanical system is known as Schr枚dinger equation.

Consider the time independent Schr枚dinger equation.

$\frac{1}{R}\frac{鈭�}{鈭�r}\left(r^2\frac{鈭�R}{鈭�r}\right)+\frac{2M{r}^2}{{\mathrm{鈩�}}^2}E=l(l+1)$ 鈥︹��. (1)

Substitute ${尾}^2=\frac{2ME}{{\mathrm{鈩�}}^2}$in the radial part of the equation (1).

$\frac{鈭�}{鈭�r}\left(r^2\frac{鈭�R}{鈭�r}\right)+{\left(尾r\right)}^{2}R=l(l+1)R$

$\frac{鈭�}{鈭�r}\left(r^2\frac{鈭�R}{鈭�r}\right)+[{\left(尾r\right)}^2鈭�l(l+1)]R=0$ 鈥�.. (2)

Substitute the following values in the above equation.

$\begin{array}{l}l=n\\ x=尾r\\ y=R\end{array}$

Put the values in equation (2) and solve the partial derivatives.

$\begin{array}{c}\frac{鈭�}{鈭�r}=\frac{鈭�}{鈭�x}\frac{鈭�x}{鈭�r}\\ =尾\frac{鈭�}{鈭�x}\end{array}$

$\begin{array}{l}{r}^2=\frac{1}{{尾}^2}{x}^2\\ \frac{鈭�}{鈭�r}\left(r^2\frac{鈭�R}{鈭�r}\right)=尾\frac{鈭�}{鈭�x}\frac{1}{{尾}^2}{x}^2尾\frac{鈭�y}{鈭�x}\end{array}$

Simplify further.

$\begin{array}{l}尾\frac{鈭�}{鈭�x}\frac{1}{{尾}^2}{x}^2尾\frac{鈭�y}{鈭�x}=2x\frac{鈭�y}{鈭�x}+x^2\frac{{鈭�}^{2}y}{鈭�x^2}\\ x^2\frac{{鈭�}^{2}y}{鈭�x^2}+2x\frac{鈭�y}{鈭�x}+[x^2鈭�n(n+1)]y=0\\ \frac{{鈭�}^{2}y}{鈭�x^2}+\frac{2}{x}\frac{鈭�y}{鈭�x}+[1鈭�\frac{n(n+1)}{x^2}]y=0\end{array}$

The differential equation is of the form,

$y^{\text{'}\text{'}}+\frac{1鈭�2a}{x}{y}^{\text{'}}+[{\left(bc{x}^{c鈭�1}\right)}^2+\frac{a^2鈭�p^2{c}^2}{x^2}]y=0$

Find the values of the parameters $a,b,c$and $p$.

$\begin{array}{l}{\left(bc\right)}^2=1\\ 2(c鈭�1)=0\\ c=1\end{array}$

Solve further.

$\begin{array}{l}1鈭�2a=2\\ a=鈭�\frac{1}{2}\\ a^2鈭�p^2{c}^2=鈭�n(n+1)\end{array}$

$\begin{array}{c}p=(n+\frac{1}{2})\\ =\frac{2n+1}{2}\end{array}$

Use the Spherical Bessel function for the determination of Eigen-functions.

$\begin{array}{c}{y}_n\left(\frac{1}{尾}x\right)=y_n\left(\frac{1}{尾}x\right)+j_n\left(\frac{1}{尾}x\right)\\ y_n\left(\frac{1}{尾}x\right)=\sqrt{\frac{蟺}{2\frac{1}{尾}x}}{Y}_{\left(\frac{2n+1}{2}\right)}\left(\frac{1}{尾}x\right)\\ j_n\left(\frac{1}{尾}x\right)=\sqrt{\frac{蟺}{2\frac{1}{尾}x}}{J}_{\left(\frac{2n+1}{2}\right)}\left(\frac{1}{尾}x\right)\end{array}$

Determine Eigen-energy.

$\begin{array}{l}{尾}^2{r}^2鈭�n(n+1)=0\\ \frac{2M{r}^2}{{\mathrm{鈩�}}^2}{E}_n=n(n+1)\\ E_n=\frac{{\mathrm{鈩�}}^2}{2M}\frac{n(n+1)}{r^2}\end{array}$

Hence the Eigen-functions are:

$\begin{array}{c}{y}_n\left(\frac{1}{尾}x\right)=y_n\left(\frac{1}{尾}x\right)+j_n\left(\frac{1}{尾}x\right)\\ y_n\left(\frac{1}{尾}x\right)=\sqrt{\frac{蟺}{2\frac{1}{尾}x}}{Y}_{\left(\frac{2n+1}{2}\right)}\left(\frac{1}{尾}x\right)\\ j_n\left(\frac{1}{尾}x\right)=\sqrt{\frac{蟺}{2\frac{1}{尾}x}}{J}_{\left(\frac{2n+1}{2}\right)}\left(\frac{1}{尾}x\right)\end{array}$

And Eigen-energy is$E_n=\frac{{\mathrm{鈩�}}^2}{2M}\frac{n(n+1)}{r^2}$.