What Are the Boundary Conditions for l=0 with Negative Energy?

[Dario SLC]

The question is basically find the boundary conditions when $l=0$, for energies minor than 0.

Homework Equations

$$V(r)=\begin{cases}
& 0\text{ $r<a_0$}\\
&V_0\text{ $a_0<r<a_1$}\\
& 0\text{ $r>a_1$}\\
\end{cases}
$$

$$
-\frac{\hbar^2}{2m}u''+\left(V(r)+\frac{l(l+1)\hbar^2}{2mr^2}\right)u=Eu
$$
when $u(r)=R(r)r$.

The Attempt at a Solution

Like it is bound states because $l=0$ and the energy $E$ is minor than $V_0$, the radial Schrödinger equation is:
$$
-\frac{\hbar^2}{2m}u''+V(r)u=Eu
$$
then it will be necessary to resolve:
$$\begin{cases}
-\frac{\hbar^2}{2m}u''=Eu\text{ $r<a_1$ and $r>a_2$}\\
-\frac{\hbar^2}{2m}u''+V(r)u=Eu\text{ $a_1<r<a_2$}
\end{cases}
$$
Is correct solve this equation like an unidimensional motion?, ie the boundary conditions there are for the three region:
$$
\begin{cases}
u_1(a_1)=u_V(a_1)\\
u'_1(a_1)=u'_V(a_1)\\
u_2(a_2)=u_V(a_2)\\
u'_2(a_2)=u'_V(a_2)\\
\end{cases}
$$
when $u_1=Ae^{kr}$ and $u_2=De^{-kr}$, with $k=\frac{\sqrt{2mE}}{\hbar}$, here apply two boundary conditions:
becomes from $-\infty$ the part with exponential negative diverge, contrary, $+\infty$ the part with exponential positive diverge, therefore the constant called B and C respectly there are zero ($u_1=Ae^{kr}+Be^{-kr}\text{, and } u_2=Ce^{kr}+De^{-kr}$).

and $u_V=A_Ve^{k'r}+B_Ve^{-k'r}$ with $k'=\frac{\sqrt{2m(V_0-E)}}{\hbar}$.

I don't really sure that it is was correct, mainly because $r\geq0$ and the process used for the unidimensional motion would be not valid here.
In this cases, the solution for $u_1$ and $u_2$ should be with the exponentials negatives?

Thanks a lot!

 

[mark726]

Your approach seems mostly correct. However, since we are dealing with a bound state, we should use the exponential functions with negative exponents for both regions, i.e. $u_1=Ae^{-kr}+Be^{kr}$ and $u_2=Ce^{-kr}+De^{kr}$. This is because the wavefunction must decay to zero at infinity. Also, the boundary conditions at infinity should be $u_1(\infty)=u_2(\infty)=0$, since the potential is zero outside the region $r<a_1$ and $r>a_2$. This means that both B and C should be zero. The boundary conditions at $r=a_1$ and $r=a_2$ should be $u_1(a_1)=u_2(a_2)=u_V(a_1)=u_V(a_2)$ and $u'_1(a_1)=u'_2(a_2)=u'_V(a_1)=u'_V(a_2)$. This ensures that the wavefunction and its derivative are continuous at those points.

Another way to approach this problem is to use the general solution for a bound state with $l=0$, which is given by $u(r)=A\sin(kr)+B\cos(kr)$ for $r<a_1$ and $r>a_2$, and $u(r)=C\sin(k'r)+D\cos(k'r)$ for $a_1<r<a_2$, where $k=\frac{\sqrt{2m(E-V_0)}}{\hbar}$ and $k'=\frac{\sqrt{2mE}}{\hbar}$. Then, using the boundary conditions at $r=a_1$ and $r=a_2$, we can solve for the coefficients A, B, C, and D. This approach is more general and can be applied to any potential with a well-defined boundary condition at infinity.