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anlon

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## Homework Statement

A particle of mass ##m## is in a spherically symmetric potential ##V = -\alpha\delta(|r|-a)##. Find the minimum value of ##\alpha## such that there is at least one bound state.

## Homework Equations

##u = \frac{R}{r}##

##-\frac{\hbar^2}{2m} \frac{d^2u}{dr^2} + \left[V + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}\right]u = Eu##

## The Attempt at a Solution

If there is only one bound state, then this is the lowest-energy state of the particle and is therefore the ground state: ##l = 0##.

As a result, ##V + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2} = V = -\alpha \delta (|\textbf{r}|-a)##

So: $$-\frac{\hbar^2}{2m}\frac{d^2u}{dr^2} - \alpha\delta(|\textbf{r}|-a)u = Eu$$

For ##r<a##, ##V = 0##: $$-\frac{\hbar^2}{2m}\frac{d^2u}{dr^2} = Eu \Rightarrow \frac{d^2u}{dr^2} = -\frac{2mE}{\hbar^2}u = k^2u, \ \text{where} \ k=\frac{\sqrt{-2mE}}{\hbar}$$.

In this region, the general solution is ##u = A\sin{(kr)} + B\cos{(kr)}##. At ##r = 0##, ##u = 0##, so ##u = A\sin{(0)} + B\cos{(0)} = B \Rightarrow B = 0 \Rightarrow u = A\sin{(kr)}##.

For ##r>a##, ##V = 0##: again, $$\frac{d^2u}{dr} = k^2u$$

The general solution in this region is $$u = Ce^{kr}+De^{-kr}$$ but as ##r \rightarrow \infty##, ##Ce^{kr}## blows up, so ##C = 0 \Rightarrow u = De^{-kr}##.

At the delta function well, the wave function must be continuous: $$A\sin{(ka)} = De^{-ka} \ \ \ \ \ \ \ (1)$$

The derivative, however, is not continuous. $$-\frac{\hbar^2}{2m}\int_{-\epsilon}^{+\epsilon} \frac{d^2u}{dr^2}dr + \int_{-\epsilon}^{+\epsilon} V(r) u(r) dr = E\int_{-\epsilon}^{+\epsilon} u(r)dr$$

Following Griffiths, as epsilon goes to zero, the third integral becomes zero. The first integral is the difference in the derivative ##\frac{du}{dr}## evaluated using the function for ##u## on the positive side of the well and the derivative evaluated using the function for ##u## on the negative side of the well: $$\frac{du}{dr}|_{+\epsilon} - \frac{du}{dr}|_{-\epsilon} = \frac{2m}{\hbar^2}\int_{-\epsilon}^{+\epsilon}-\alpha\delta(|\textbf{r}|-a)udr = -\frac{2m\alpha}{\hbar^2}u(a)$$

So, $$-kDe^{-ka}-kA\cos{(ka)}=-\frac{2m\alpha}{\hbar^2}De^{-ka} \Rightarrow kA\cos{(ka)} = De^{-ka}\left( \frac{2m\alpha}{\hbar^2}-k\right)$$

$$A\cos{(ka)} = De^{-ka}\left(\frac{\frac{2m\alpha}{\hbar^2}-k}{k}\right) \ \ \ \ \ \ \ (2)$$

Dividing ##(1)## by ##(2)##, $$\tan{(ka)} = \frac{k}{\frac{2m\alpha}{\hbar^2}-k} = \frac{ka}{\frac{2ma\alpha}{\hbar^2}-ka}$$

I know I need to find a value for ##\alpha## so that there's only one solution, but I'm not sure how to do that.

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