A particle of mass m is placed in a finite spherical well:

$\mathbf{V}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\{\begin{array}{l}-V_0,r\le a;\\ 0,r>a;\end{array}$

Find the ground state, by solving the radial equation with$\mathcal{l}\mathbf{=}\mathbf{0}$. Show that there is no bound state if ${\mathit{V}}_{\mathbf{0}}{\mathit{a}}^{\mathbf{2}}\mathbf{<}{\mathbf{\pi }}^{\mathbf{2}}{\mathcal{k}}^{\mathbf{2}}\mathbf{/}\mathbf{8}\mathit{m}$.

A particle of mass m is placed in a finite spherical well, where the potential is given by:

$\mathrm{V}\left(\mathrm{r}\right)=\{\begin{array}{l}-{\mathrm{V}}_0,\mathrm{r}\le \mathrm{a};\\ 0,\mathrm{r}>\mathrm{a};\end{array}$

We need to solve the radial equation:

$-\frac{{\mathrm{h}}^2}{2\mathrm{m}}\frac{{\mathrm{d}}^2\mathrm{u}}{{\mathrm{dr}}^2}+(\mathrm{V}+\frac{{\mathrm{h}}^2}{2\mathrm{m}}\frac{\mathrm{l}(\mathrm{l}+1)}{{\mathrm{r}}^2})\mathrm{u}=\mathrm{Eu}.$

With l=0, and for r<a, we get:

$$\begin{gathered} -\frac{{\mathrm{h}}^2}{2\mathrm{m}}\frac{{\mathrm{d}}^2\mathrm{u}}{{\mathrm{dr}}^2}-{\mathrm{V}}_0\mathrm{u}=\mathrm{Eu} \\ \frac{{\mathrm{d}}^2\mathrm{u}}{{\mathrm{dr}}^2}\to =-\frac{2\mathrm{m}}{{\mathrm{h}}^2}({\mathrm{V}}_0+\mathrm{E})\mathrm{u}. \\ \frac{{\mathrm{d}}^2\mathrm{u}}{{\mathrm{dr}}^2}=-{\mathrm{\mu }}^2\mathrm{\upsilon }\left(1\right) \end{gathered}$$

Whererole="math" localid="1658141017857" $\mathrm{\mu }=\sqrt{\frac{2\mathrm{m}({\mathrm{v}}_0+\mathrm{E})}{{\mathrm{h}}^2}}$

Equation (1) has a solution of:

$u\left(r\right)=C\sin \left(\mu r\right)+D\cos \left(\mu r\right)\left(2\right)$

$\mathrm{z}=\mathrm{\tau \tau }/2,\mathrm{z}=\mathrm{\tau \tau }$The actual radial function is$u\left(r\right)/r$we must eliminate the cosine term to keep the radial function finite at r=0.

Therefore:$\mathrm{u}\left(\mathrm{r}\right)=\mathrm{Csin}\left(\mathrm{\mu }\right(\mathrm{r})$ (2)

For r>a, the equation is:

$\frac{d^{2}u}{d{r}^2}=k^{2}u$ (3)

Where:$k=\sqrt{-\frac{2mE}{h^2}}$

This equation has a general solution:

$u\left(r\right)=A{e}^{kr}+B{e}^{kr}$

The function must be at infinity y, so we must set A=0, to get:

$u\left(r\right)=B{e}^{-kr}$ (4)

We have one boundary at r=a, the function u(r) and its first derivative to be continuous at the boundary. These conditions gives us:

$$\begin{gathered} C\sin \left(\mu a\right)=B{e}^{-ka} \\ \mu C\cos \left(\mu a\right)=-KB{e}^{-ka} \end{gathered}$$

By dividing these two equations we can eliminate the exponentials, so we get:

$-\frac{\mathrm{\mu }}{\mathrm{k}}=\tan \left(\mathrm{\mu a}\right)$

Introduce the following two variables,

$$\begin{gathered} \mathrm{z}=-\mathrm{\mu a}, \\ {\mathrm{z}}_0=\frac{\mathrm{a}}{\mathrm{h}}\sqrt{2{\mathrm{mV}}_0} \end{gathered}$$

Then$\mathrm{ka}=\sqrt{{\mathrm{z}}_0^2-{\mathrm{z}}^2}$ and the equation to solve is:

$\tan \left(z\right)=\frac{-1}{\sqrt{z_0^2/z^2-1}}$

This equation must be solved numerically or graphically, it cannot be solved analytically. In the following python code this equation solved using the two methods for$z_0=8$, simply the code used to draw$-1/\sqrt{z_0^2/z^2-1},\tan \left(z\right)$ , then find the intersection points.

There is no solution if$z_0\le \frac{\mathrm{\pi }}{2}$ , which is to say ifrole="math" localid="1658141915735" ${\mathrm{V}}_0{\mathrm{a}}^2/{\mathrm{h}}^2<{\mathrm{\tau \tau }}^2/4,{\mathrm{V}}_0{\mathrm{a}}^2<{\mathrm{\tau \tau }}^2{\mathrm{h}}^2/8\mathrm{m}$

The ground state energy occurs somewhere between$z=\tau \tau /2,z=\tau \tau$

$$\begin{gathered} z_0<\frac{\mathrm{\pi }}{2} \\ \frac{a}{h}\sqrt{2m{V}_0}<\frac{\mathrm{\pi }}{2} \end{gathered}$$

Therefore,$V_0{a}^2<\frac{h^2\tau \tau }{8m}$ .