91影视

The spherical motion is $Y_{l,m}(胃,蠒)$and potential energy is$V\left(r\right)=鈭�e^2/r$.

An Eigen function of a linear operator D defined on some function space is any non-zero function f in that space that, when acted on by D, is only multiplied by some scaling factor known as an eigenvalue.

Rewrite the time-independent Schrodinger-radial Equation's portion as follows.

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

Make the following substitution:

$\begin{array}{c}x=\frac{2r}{伪}\\ y=rR\\ =\frac{伪}{2}xR\end{array}$

Determine the value of $\frac{d}{dr}$.

$\begin{array}{c}\frac{d}{dr}=\frac{d}{dx}\frac{dx}{dr}\\ =\frac{2}{伪}\frac{d}{dx}\end{array}$

Define the value of $r^2\frac{dR\left(r\right)}{dr}$as below.

$\begin{array}{c}{r}^2\frac{dR\left(r\right)}{dr}=\frac{{伪}^2}{4}{x}^2\frac{2}{伪}\frac{d}{dx}\left(\frac{2}{伪x}y\left(x\right)\right)\\ =\frac{伪}{2}{x}^2(鈭�\frac{2}{伪x^2}y\left(x\right)+\frac{2}{伪x}{y}^{\text{'}}\left(x\right))\\ =鈭�y\left(x\right)+x{y}^{\text{'}}\left(x\right)\end{array}$

Put this value in$\frac{d}{dr}$.

$\begin{array}{c}\frac{d}{dr}\left(r^2\frac{dR\left(r\right)}{dr}\right)=\frac{2}{伪}\frac{d}{dx}(鈭�y\left(x\right)+x{y}^{\text{'}}\left(x\right))\\ =\frac{2}{伪}(鈭�y^{\text{'}}\left(x\right)+y^{\text{'}}\left(x\right)+x{y}^{\text{'}\text{'}}\left(x\right))\\ =\frac{2}{伪}x{y}^{\text{'}\text{'}}\left(x\right)\end{array}$

Make another substitution.

$\begin{array}{l}{伪}^2=鈭�\frac{2ME}{{\mathrm{鈩�}}^2}\text{鈥�}\\ 位=\frac{M{e}^2}{{\mathrm{鈩�}}^2}伪\end{array}$

Find the value of equation

$\begin{array}{l}\frac{1}{\frac{2y\left(x\right)}{伪x}}\left(\frac{2}{伪}x{y}^{\text{'}\text{'}}\left(x\right)\right)鈭�\frac{{伪}^2}{4}{x}^2(鈭�\frac{4位}{{伪}^{2}x}+{伪}^2)=l(l+1)\\ \frac{1}{y\left(x\right)}{x}^2{y}^{\text{'}\text{'}}\left(x\right)+位x鈭�\frac{{伪}^4}{4}{x}^2=l(l+1)\end{array}$

$y^{\text{'}\text{'}}\left(x\right)+(\frac{位}{x}鈭�\frac{{伪}^4}{4}鈭�\frac{l(l+1)}{x^2})y\left(x\right)=0$ 鈥�.. (2)

Consider the following equation .

$y\left(x\right)=x^{(l+1)}{e}^{鈭�\frac{x}{2}}v\left(x\right)$

Find the first and second derivative.

$\begin{array}{l}{y}^{\text{'}}\left(x\right)=(l+1)x^l{e}^{鈭�\frac{x}{2}}v\left(x\right)+x^{l+1}(鈭�\frac{1}{2}{e}^{鈭�\frac{x}{2}}v\left(x\right)+e^{鈭�\frac{x}{2}}{v}^{\text{'}}\left(x\right))\\ y^{\text{'}\text{'}}\left(x\right)=x^l{e}^{鈭�\frac{x}{2}}\left(\begin{array}{l}\frac{l(l+1)}{x}v\left(x\right)鈭�\frac{1}{2}(l+1)v\left(x\right)+(l+1)v^{\text{'}}\left(x\right)鈭�\frac{1}{2}(l+1)v\left(x\right)\\ +(l+1)v^{\text{'}}\left(x\right)+\frac{1}{4}xv\left(x\right)鈭�v^{\text{'}}\left(x\right)x+x{v}^{\text{'}\text{'}}\left(x\right)\end{array}\right)\end{array}$

Put the derivatives value in equation (2).

$x{v}^{\text{'}\text{'}}\left(x\right)+(2l+2鈭�x)v^{\text{'}}\left(x\right)+(\frac{(1鈭�{伪}^4)}{4}x+位鈭�(l+1))v\left(x\right)=0$

The Laguerre-Differential Equation, whose solutions are the Laguerre-Polynomials of the algebraic form $L_{位鈭�l鈭�1}^{2l+1}\left(x\right)$, can be used to identify the derived Differential Equation for$v\left(x\right)$.

The dependency of the higher Bohr-orbits are accounted for by $\frac{(1鈭�{伪}^4)}{4}x$. As a result, $伪=n^{\text{'}}a$, with being the value of for the first Bohr orbit.

$\begin{array}{l}鈭�\frac{2ME}{{\mathrm{鈩�}}^2}=n^{\text{'}2}{a}^2\\ E_{n^{\text{'}}}=鈭�\frac{{\mathrm{鈩�}}^2}{2M}{n}^{\text{'}2}{a}^2\\ y\left(x\right)=x^{l+1}{e}^{鈭�\frac{x}{2}}{L}_{位=n^{\text{'}}鈭�l鈭�1}^{2l+1}\left(x\right)\\ R_{n^{\text{'}}}\left(r\right)=\frac{1}{r}{\left(\frac{2r}{n^{\text{'}}a}\right)}^{l+1}{e}^{鈭�\frac{r}{n^{\text{'}}a}}{L}_{n^{\text{'}}鈭�l鈭�1}^{2l+1}\left(\frac{2r}{n^{\text{'}}a}\right)\end{array}$