Bound state of 3-dimensional Dirac well

[anlon]

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.

 

[drvrm]

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

such equations we get when treating 'deuteron ' in a potential well ...they are transcedental equations and may be solved for bound state using graphs.

 

[anlon]

I know the equation is transcendental and must be solved graphically, but I'm not sure how to do that and get a value for $\alpha$. $ka$ can be its own variable, say $z$, but I don't know what kind of values I should use for $\alpha$ in setting up a graphical solution.

 

[anlon]

To confirm, I should create a graph similar to the one here:

and find the lowest value of $\alpha$ where there's only one solution (intersection of the graphs), aside from (0,0)?

 

[TSny]

Hi, anlon. You got trig functions for the general solution when r < a, but you got exponential functions for the general solution when r > a. Isn't the differential equation the same in these two regions?

 

[anlon]

The differential equation is the same, but there are two different kinds of solutions: $A\sin{(kx)} + B\cos{(kx)}$ for solutions where you want to have a "node" (wave function equal to zero) at a specific point, i.e. a place where potential is infinite, and $Ae^{-kx} + Be^{kx}$ for solutions that extend to infinity. The problem with trig solutions that extend to infinity is that they aren't normalizable; you have to be able to solve $$1 = \int_{-\infty}^{\infty} \psi^* \psi dx$$
So, for cases where the wave function only spans a finite region, such as an infinite square well (or in this case, between $0$ and $|\textbf{r}| = a$) trig solutions are normalizable, but for cases where the wave function extends to infinity, they are not.
Here, the condition $u(0) = 0$ requires a trig solution for the wave function between $0$ and $a$. For $r > a$ the wave function extends to infinity, which requires an exponential solution.

 

[TSny]

The potential is $V(r) = -\alpha \delta(r - a)$, where $r = |\vec{r}|$. For $r < a$ and for $r > a$ you have the same differential equation for $u$: $u'' = k^2 u$.

$k$ must have the same real value for $r<a$ as for $r > a$.

What is the general solution of $u'' = k^2 u$ for real $k$?

 

[anlon]

The general solution is $Ae^{-kr}+Be^{kr}$. So now I'm confused. I've always taken statements of solutions to these differential equations at face value, never bothering to work them out myself, but it seems like the solution is exponential for positive $k$ and trigonometric for negative $k$.
So for the condition $u(0) = 0$ I would need to solve using $0 = Ae^0 + Be^0 = A + B \Rightarrow A = -B$?

 

[TSny]

So for the condition $u(0) = 0$ I would need to solve using $0 = Ae^0 + Be^0 = A + B \Rightarrow A = -B$?

Yes. You can express $u$ as a hyperbolic function.

 

[anlon]

Thanks, I'll try solving it that way tomorrow and report back my results.

 

[TSny]

OK, good.

 

[anlon]

As it turns out, solving it this way ends up giving the equation $$\frac{\hbar^2 k}{m\alpha} = 1-e^{-2ka}$$ which only has a solution when $\alpha > \frac{\hbar^2}{2am}$. Thanks for the help!

Also, an error I made in the first post: $u = rR$, not $u = \frac{R}{r}$. The work is all correct, however.

 

[TSny]

OK. Your answer looks correct to me.