A spherical infinite well has potential energy

$\mathit{U}\left(r\right)\mathbf{=}\{\begin{array}{l}0r<a\\ +\infty r>a\end{array}$

Since this is a central force, we may use the Schrodinger equation in the form (7-30)-that is, just before the specific hydrogen atom potential energy is inserted. Show that the following is a solution

$\mathit{R}\left(r\right)\mathbf{=}\frac{\mathbf{A}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{b}\mathbf{r}}{\mathbf{r}}$

Now apply the appropriate boundary conditions. and in so doing, find the allowed angular momenta and energies for solutions of this form.

The equation 7.30 is given by,

$-\frac{{\sout{h}}^2}{2m}\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d}{dr}\right)R\left(r\right)+\frac{{\sout{h}}^{2}l\left(l+l\right)}{2m{r}^2}R\left(r\right)+U\left(r\right)R\left(r\right)-ER\left(r\right)=0$

The potential energy,$U=\left\{\begin{array}{ll}0& r<a\\ +\infty & r>a\end{array}\right.$

Consider the equation 7.30,

$\mathbf{-}\frac{{\mathbf{h}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\frac{\mathbf{1}}{{\mathbf{r}}^{\mathbf{2}}}\frac{\mathbf{d}}{\mathbf{d}\mathbf{r}}\left(r^2\frac{d}{dr}\right)\mathit{R}\left(r\right)\mathbf{+}\frac{{\mathbf{h}}^{\mathbf{2}}\mathbf{l}\left(l+l\right)}{\mathbf{2}\mathbf{m}{\mathbf{r}}^{\mathbf{2}}}\mathit{R}\left(r\right)\mathbf{+}\mathit{U}\left(r\right)\mathbf{+}\mathit{R}\left(r\right)\mathbf{-}\mathit{E}\mathit{R}\left(r\right)\mathbf{=}\mathbf{0}$

Since the potential energy is 0 for $r<a$. Therefore, the equation can be written as,

role="math" localid="1660024455548" $$\begin{gathered} -\frac{{\mathrm{h}}^2}{2\mathrm{m}}\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{d}}{\mathrm{dr}}\right)R\left(\mathrm{r}\right)+\frac{{\mathrm{h}}^2\mathrm{l}(\mathrm{l}+\mathrm{l})}{2{\mathrm{mr}}^2}R\left(\mathrm{r}\right)+U\left(\mathrm{r}\right)+R\left(\mathrm{r}\right)-ER\left(\mathrm{r}\right)=0 \\ -\frac{{\mathrm{h}}^2}{2\mathrm{m}}\frac{1}{{\mathrm{r}}^2}\frac{\mathrm{d}}{\mathrm{dr}}\left({\mathrm{r}}^2\frac{\mathrm{d}}{\mathrm{dr}}\right)R\left(\mathrm{r}\right)+\frac{{\mathrm{h}}^2\mathrm{l}(\mathrm{l}+\mathrm{l})}{2{\mathrm{mr}}^2}R\left(\mathrm{r}\right)-ER\left(\mathrm{r}\right)=0....\left(i\right) \end{gathered}$$

The term of the above equation $\frac{d}{dr}\left(r^2\frac{d}{dr}R\left(r\right)\right)$can be calculated as,

$$\begin{gathered} \frac{d}{dr}\left(r^2\frac{d}{dr}R\left(r\right)\right)=\frac{d}{dr}\left(r^2\frac{d}{dr}\frac{A\sin br}{r}\right) \\ \frac{d}{dr}\left(r^2\frac{d}{dr}R\left(r\right)\right)=A\frac{d}{dr}\left(r^2\frac{d}{dr}{r}^{-1}\sin br\right) \\ \frac{d}{dr}\left(r^2\frac{d}{dr}R\left(r\right)\right)=A\frac{d}{dr}\left(br\cos br-\sin br\right) \\ \frac{d}{dr}\left(r^2\frac{d}{dr}R\left(r\right)\right)=-A{b}^{2}r\sin br \end{gathered}$$

Solve further as,

$$\begin{gathered} \frac{d}{dr}\left(r^2\frac{d}{dr}R\left(r\right)\right)=-b^2{r}^2\frac{A\sin br}{r} \\ \frac{d}{dr}\left(r^2\frac{d}{dr}R\left(r\right)\right)=-b^2{r}^{2}R\left(r\right) \end{gathered}$$

substitute $b^2{r}^{2}R\left(r\right)$for $\frac{d}{dr}\left(r^2\frac{d}{dr}R\left(r\right)\right)$into equation (i),

$$\begin{gathered} -\frac{{\sout{h}}^2}{2m}\frac{1}{r^2}\left(-b^2{r}^{2}R\left(r\right)\right)+\frac{{\sout{h}}^{2}l\left(l+l\right)}{2m{r}^2}R\left(r\right)-ER\left(r\right)=0 \\ \left[\frac{{\sout{h}}^2}{2m}\frac{1}{r^2}{b}^2{r}^2+\frac{{\sout{h}}^{2}l\left(l+l\right)}{2m{r}^2}-E\right]R\left(r\right)=0 \\ \left[\frac{{\sout{h}}^2{b}^2}{2m}+\frac{{\sout{h}}^{2}l\left(l+l\right)}{2m{r}^2}-E\right]R\left(r\right)=0 \end{gathered}$$

Since the R(r) doesn’t equal zero. So, L.H.S is equal to zero if $l=0,-1$.

$E=\frac{{\sout{h}}^2{b}^2}{2m}$ ……. (ii)

By restricting the value of the $l$, the expression for the angular momentum is $L=\sout{h}\sqrt{l\left(l+l\right)}=0$.

By imposition the boundary condition, at$r=a,R\left(a\right)=0$

$$\begin{gathered} R\left(0\right)=\frac{A\sin ba}{r} \\ 0=\frac{A}{a}\sin ab \\ 0=\sin ab \\ n\mathrm{\pi }=\mathrm{ab} \end{gathered}$$

Solve further as,

$b=\frac{n\mathrm{\pi }}{a}$ Where $n=0,1,2,....$

Substitute $\frac{n\mathrm{\pi }}{a}$for b into equation (ii).

$$\begin{gathered} E=\frac{{\sout{h}}^2}{2m}{\left(\frac{n\mathrm{\pi }}{a}\right)}^2 \\ E=\frac{{\sout{h}}^2{n}^2{\mathrm{\pi }}^2}{2m{a}^2} \end{gathered}$$

Hence it is proved that R(r) is the solution of the Schrodinger equation, angular momentum is L = 0 and energy is$E=\frac{{\sout{h}}^2{n}^2{\mathrm{\pi }}^2}{2m{a}^2}$ .