a) Check that $\mathit{A}\mathit{r}{\mathit{j}}_{\mathbf{1}}\left(kr\right)$satisfies the radial equation with $\mathit{V}\left(r\right)\mathbf{=}\mathbf{0}$and $\mathit{I}\mathbf{=}\mathbf{1}$.

(b) Determine graphically the allowed energies for the infinite spherical well, when $\mathit{I}\mathbf{=}\mathbf{1}$. Show that for large $\mathit{n}\mathbf{,}{\mathit{E}}_{\mathbf{n}\mathbf{1}}\mathbf{\approx }\left({\sout{h}}^2{\pi }^2/2{\mathrm{ma}}^2\right){\left(n+1/2\right)}^{\mathbf{2}}$. Hint: First show that ${\mathit{j}}_{\mathbf{1}}\left(x\right)\mathbf{=}\mathbf{0}\mathbf{\Rightarrow }\mathit{x}\mathbf{=}\mathbf{tan}\mathit{x}$. Plot xand$\mathbf{tan}\mathit{x}$on the same graph, and locate the points of intersection.

The Schrodinger equation is given by:

$\mathbf{-}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{u}}{\mathbf{d}{\mathbf{r}}^{\mathbf{2}}}\mathbf{+}\mathbf{[}\mathbf{V}\left(r\right)\mathbf{+}\frac{{\sout{\mathbf{h}}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\frac{\mathbf{I}\left(I+1\right)}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{]}\mathit{u}\mathbf{=}\mathit{E}\mathit{u}$

a)

Using the standard substitution $u\left(r\right)=R\left(r\right)/r$the radial part of the Schrodinger equation reads:

$-\frac{{\sout{h}}^2}{2m}\frac{d^{2}u}{d{r}^2}+[V\left(r\right)+\frac{{\sout{h}}^2}{2m}\frac{I(I+1)}{r^2}]u=Eu$

Now, if we set $I=1$and $V\left(r\right)=0$the latter expression simplifies to:

$\frac{{\sout{\mathrm{h}}}^2}{2\mathrm{m}}\frac{{\mathrm{d}}^2\mathrm{u}}{{\mathrm{dr}}^2}+\frac{{\sout{\mathrm{h}}}^2}{2\mathrm{m}}\frac{2}{r^2}u=Eu$

Next, we multiply both sides by $2m/{\sout{h}}^2$and define $k^2\equiv 2mE/{\sout{h}}^2$, so the previous equation becomes:

$-\frac{d^{2}u}{d{r}^2}+\left(\frac{2}{r^2}-k^2\right)u=0$

The spherical Bessel function must now be inserted into the final differential equation. We are aware that:

$j_1=\frac{\sin \left(kr\right)}{{\left(kr\right)}^2}-\frac{\cos \left(kr\right)}{kr}$

As a result, we must examine the following solution:

$u\left(r\right)=Ar{j}_1=A\left[\frac{\sin \left(kr\right)}{k^{2}r}-\frac{\cos \left(kr\right)}{k}\right]$

The second derivative of the $u\left(r\right)$function is required to check whether the solution satisfies the given differential equation. So, we find:

$$\begin{gathered} \frac{du}{dr}=A\left[\frac{k\cos \left(kr\right)}{k^{2}r}-\frac{\sin \left(kr\right)}{k^2{r}^2}+\frac{k\sin \left(kr\right)}{k}\right] \\ \frac{d^{2}u}{d{r}^2}=A\left[-\frac{\sin \left(kr\right)}{r}-2\frac{\cos \left(kr\right)}{k{r}^2}+2\frac{\sin \left(kr\right)}{k^2{r}^3}+k\cos \left(kr\right)\right] \end{gathered}$$

This must be equal to the term$\left(2/r^2-k^2\right)u$ in order for the differential equation to be satisfied. Therefore, we will calculate it explicitly:

$$\begin{gathered} \left(\frac{2}{r^2}-k^2\right)u=A\left(\frac{2}{r^2}-k^2\right)\left[\frac{\sin \left(kr\right)}{k^{2}r}-\frac{\cos \left(kr\right)}{k}\right] \\ =A\left[A\frac{\sin \left(kr\right)}{k^2{r}^3}-\frac{\sin \left(kr\right)}{r}-2\frac{\cos \left(kr\right)}{k{r}^2}+k\cos \left(kr\right)\right] \\ =A\left[-\frac{\sin \left(kr\right)}{r}-2\frac{\cos \left(kr\right)}{k{r}^2}+2\frac{\sin \left(kr\right)}{k^2{r}^3}+k\cos \left(kr\right)\right] \end{gathered}$$

As a result, the two terms are equal, and the proposed solution is radial differential equation-satisfying.

b)

Since, we are dealing with infinite well, our wave function must vanish at $r=a$. This implies that $u\left(r\right)$has to vanish. Therefore:

$$\begin{gathered} u\left(r\right)=0 \\ \Rightarrow A\left[\frac{\sin \left(ka\right)}{k^{2}a}-\frac{\cos \left(ka\right)}{k}\right] \\ \Rightarrow ka=\tan \left(ka\right) \end{gathered}$$

This is a transcendental equation.

By identifying intersections of two functions, we may solve it graphically.

Here, the red curve shows the tangent curve and blue curve is x-curve.

As energy increases, $k$increases and intersections occur near $ka=\left(n+\frac{1}{2}\right)\mathrm{\pi }$. As a result, the high energies are nearly equal to:

$E_n=\frac{{\sout{h}}^2{k}^2}{2m}=\frac{{\sout{h}}^2{\mathrm{\pi }}^2}{2m{a}^2}{\left(n+\frac{1}{2}\right)}^2$