Spherical Well

Let's consider spherical well potentials,

$V(\mathbf {r} )={\begin{cases}V_{0},&0\leq r<a\\0,&r>a\end{cases}}$

The Schrödinger equations for these two regions can be written by

$\left({\frac {-\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {\hbar ^{2}l(l+1)}{2mr^{2}}}-V_{0}\right)u_{l}(r)=Eu_{l}(r)$

for $0\leq r<a\!$ and

$\left({\frac {-\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial r^{2}}}+{\frac {\hbar ^{2}l(l+1)}{2mr^{2}}}\right)u_{l}(r)=Eu_{l}(r)$

for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r>a \! } .

The general solutions are

${\begin{aligned}&{\frac {u_{\ell }(r)}{r}}=C_{\ell }j_{\ell }(k'r),&r\leq d,\\&{\frac {u_{\ell }(r)}{r}}=A_{\ell }j_{\ell }(kr)+Bn_{\ell }(kr),&r>d,\end{aligned}}$

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k' = \sqrt{\frac{2m(E+V_0)}{\hbar^2}} \!} and $k={\sqrt {\frac {2mE}{\hbar }}}\!$.

For the $l=0\!$ term, the centrifugal barrier drops out and the equations become the following

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{cases} 0\leq r< a & (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial r^2}-V_0)u_0(r)=Eu_0(r)\\ r>a & (\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial r^2})u_0(r)=Eu_0(r) \end{cases} }

The generalized solutions are

${\begin{cases}0\leq r<a&u_{0}(r)=Ae^{ikr}+Be^{-ikr}\\r>a&u_{0}(r)=Ce^{ik'r}+De^{-ik'r}\end{cases}}$

Using the boundary condition, $u(r=0)=0\!$, we find that $A=-B\!$. The second equation can then be reduced to sinusoidal function where $\alpha =2iA\!$.

$u_{0}(r)=2iA\sin(kr)=\alpha \sin(kr)=\alpha \sin \left({\frac {r}{\hbar }}{\sqrt {2m(E+V_{0})}}\right)$

for $r>a\!$, we know that $D=0\!$ since as $r\!$ approaches infinity, the wavefunction does not go to zero.

$u_{0}(r)=Ce^{ik'r}+De^{-ik'r}=Ce^{-{\frac {r}{\hbar }}{\sqrt {-2mE}}}$

Matching the conditions that at $r=a\!$, the wavefunctions and their derivatives must be continuous which results in 2 equations

$\alpha \sin \left({\frac {a}{\hbar }}{\sqrt {2m(E+V_{0})}}\right)=Ce^{-{\frac {a}{\hbar }}{\sqrt {-2mE}}}$

$\alpha {\frac {\sqrt {2m(E+V_{0})}}{\hbar }}\cos \left({\frac {a}{\hbar }}{\sqrt {2m(E+V_{0})}}\right)=-{\frac {\sqrt {-2mE}}{\hbar }}Ce^{-{\frac {a}{\hbar }}{\sqrt {-2mE}}}$

Dividing the above equations, we find

$-\cot \left({\sqrt {{\frac {2m}{\hbar ^{2}}}(V_{0}-|E|)a^{2}}}\right)={\frac {\sqrt {\frac {2m|E|}{\hbar ^{2}}}}{\sqrt {\frac {2m(V_{0}-|E|)}{\hbar ^{2}}}}}$, which is the solution for the odd state in 1D square well.

Solving for $V_{0}\!$, we know that there is no bound state for

$V_{0}<{\frac {\pi ^{2}\hbar ^{2}}{8ma^{2}}}$.