# Bound state of 3-dimensional Dirac well

1. Nov 21, 2016

### anlon

1. The problem statement, all variables and given/known data
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.

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

3. 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.

Last edited: Nov 21, 2016
2. Nov 21, 2016

### drvrm

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

3. Nov 21, 2016

### 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.

4. Nov 22, 2016

### 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)?

5. Nov 22, 2016

### 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?

6. Nov 22, 2016

### 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.

7. Nov 22, 2016

### 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 $k$?

8. Nov 23, 2016

### 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 $0 = Ae^0 + Be^0 = A + B \Rightarrow A = -B$?

9. Nov 23, 2016

### TSny

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

10. Nov 23, 2016

### anlon

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

11. Nov 23, 2016

### TSny

OK, good.

12. Nov 24, 2016

### 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.

13. Nov 24, 2016