# Solving quantum radial equation for infinite potential spherical annulus for $l=0$

There is a mass $m$ in a potential such that

$$V(r) = \left\{ \begin{array}{lr} 0, & a \leq r \leq b\\ \infty, & \text{everywhere else} \end{array} \right.$$ I am looking to find the solution $u(r)$ to the radial equation $$-\frac{\hbar^2}{2m} u^{\prime \prime}(r) + \bigg( V(r) + \frac{\hbar^2}{2m} \frac{l(l+1)}{r^2}\bigg) u(r) = E\, u(r)$$ in the case $l=0$.

$\textbf{Progress so far}$

Looking at the region with zero potential, and letting $l=0$, I define $$k \equiv \frac{\sqrt{2mE}}{{\hbar}}$$ so that I have the second order differential equation $$u^{\prime \prime}(r) = - k^2 u(r)$$ which has the solution $$u(r) = A \sin(kr) + B \cos( kr)$$ The potential is such that the boundary conditions are $$u(a) = u(b) = 0$$ $$A \sin(ka) + B \cos(ka) = A \sin(kb) + B \cos(kb)$$ It seems that, since, for instance, $\sin(kb) = \sin(ka)$, $$kb = ka + 2\pi n$$But this must be wrong because, for instance, $$\sin\bigg (\frac{2 \pi n a}{b - a} \bigg ) \neq 0 .$$ Could someone tell me where I went wrong in finding $k?$

Your radial equation is wrongly formulated. It looks as an equation for flat 1D space, lacking a first-derivative term, arising from spherical coordinates Jacobian.

Instead of postulating the equation, as you seem to have done, let's derive it. I'll use units such that $\frac{\hbar^2}{2m}=1$, so that the equations look simpler, you should be able to reproduce it with your units.

So, the Schrödinger equation for a particle in a spherically symmetric potential $V(r)$ with $r=|\vec x|$ is:

$$-\nabla^2\Psi(\vec x)+V(r)\Psi(\vec x)=E\Psi(\vec x)$$

As your potential is spherically symmetric, we can make use of this symmetry and switch to spherical coordinates. Our Laplacian would look like:

$$\nabla^2 f=\frac1{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial f}{\partial r}\right)+\frac1{r^2\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial f}{\partial \theta}\right)+\frac1{r^2\sin^2\theta}\frac{\partial^2 f}{\partial\phi^2}$$

With such a Laplacian, our Schrödinger equation is separable, so we can look for solution in the form of $\Psi(r,\theta,\phi)=u(r)v(\theta,\phi)$.

Substituting it into new formula for Laplacian, we have:

$$\nabla^2\Psi(r,\theta,\phi)=v(\theta,\phi)\frac1{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u(r)}{\partial r}\right)+u(r)\frac1{r^2\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial v(\theta,\phi)}{\partial \theta}\right)+\\+u(r)\frac1{r^2\sin^2\theta}\frac{\partial^2 v(\theta,\phi)}{\partial\phi^2}$$

Or, denoting radial part of Laplacian with $-\hat R$ and angle part with $-\hat L^2/r^2$, we have:

$$\nabla^2\Psi(r,\theta,\phi)=-v(\theta,\phi)\hat R u(r)-u(r)\frac{\hat L^2v(\theta,\phi)}{r^2}$$

Now we can write our Schrödinger equation as:

$$v(\theta,\phi)\hat R u(r)+u(r)\frac{\hat L^2v(\theta,\phi)}{r^2}+V(r)u(r)v(\theta,\phi)=Eu(r)v(\theta,\phi)$$

Multiply both sides by $\frac{r^2}{u(r)v(\theta,\phi)}$ and rearrange the terms:

$$-\frac{r^2\hat Ru(r)}{u(r)}-V(r)+E=\frac{\hat L^2v(\theta,\phi)}{v(\theta,\phi)}$$

Now we have separated radial variables from angle ones, so we introduce a separation constant, which we'll write as $l(l+1)$. It's an eigenvalue of $\hat L^2$ operator (and eigenfunctions are spherical harmonics). Now, multiplying everything by $u(r)/r^2$, we have:

$$\hat Ru(r)+V(r)u(r)+\frac{l(l+1)}{r^2}u(r)=Eu(r),$$

or, finally, writing out expression for $\hat R$,

$$-\frac1{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial u(r)}{\partial r}\right)+V(r)u(r)+\frac{l(l+1)}{r^2}u(r)=Eu(r).$$

This is the equation you should be trying to solve (up to units). Now its solution should be in terms of spherical Bessel functions.

Why could we conclude from $\sin\left(\frac{2 \pi n a}{b-a}\right)\neq0$ that the value of $k$ is incorrect? On the contrary, I think the conclusion is partly correct. In fact, We can rewrite $A\sin(k x)+B\cos(k x)=C\sin(k x + \delta)$ to simplify the boundary condition. Thus it is evident that $k= \frac{n \pi}{b-a}$.

• Yes, the solution you are looking for is $$u(x)=C\,\sin\left(\frac{n\pi(x-a)}{b-a}\right).$$ – pressure Jan 16 '14 at 8:58