# Determining bound states for delta function potential

I'm working on a problem out of Griffith's Intro to QM 2nd Ed. and it's asking to find the bound states for for the potential $V(x)=-\alpha[\delta(x+a)+\delta(x-a)]$ This is what I'm doing so far:

$$\mbox{for x\lt-a:}\hspace{1cm}\psi=Ae^{\kappa a}\\ \mbox{for -a\lt x\lt a}\hspace{1cm}\psi=Be^{-\kappa x}+Ce^{\kappa x}\\ \mbox{and for x\gt a:}\hspace{1cm}\psi=De^{-\kappa a}$$

However, this is what the solution reads: $$\mbox{for x\lt a:}\hspace{1cm}\psi=Ae^{-\kappa a}\\ \mbox{for -a\lt x\lt a}\hspace{1cm}\psi=B(e^{\kappa x}+e^{-\kappa x})\\ \mbox{and for x\lt-a:}\hspace{1cm}\psi=Ae^{\kappa a}$$

Can someone explain what I'm doing wrong - why they are getting the coefficients they are - and what I should be doing? Maybe go over the general way of approaching these kind of problems? I'm also wondering why they don't have the case for $x>a$?

${\bf New}$ ${\bf Question}$

After this I am trying to figure out what $\Delta\left(\frac{d\psi}{dx}\right)$ is. Integrating the potential part of SWE and taking the limit as $\epsilon$ approaches $\pm a$I get: $$\Delta\left(\frac{d\psi}{dx}\right)=-\frac{2m}{\hbar ^2}\left[\alpha\psi(a)+\alpha\psi(-a)\right]$$ but the solution reads $\Delta\left(\frac{d\psi}{dx}\right)=-\frac{2m\alpha}{\hbar^2}\psi(a)$ I think I'm close but not sure how they got their result. Do you see what I'm doing? Can you correct where I made any mistakes?

• There's a typo. The first line should read $x>a$. But what was your solution? Your start is ok. Where did you go from there? – garyp Oct 14 '14 at 2:07

Can someone explain what I'm doing wrong

You're not doing anything wrong but you haven't taken the next logical step.

The potential is an even function so there (may) be both an even and an odd solution (proving this is also a problem in this chapter of the textbook).

For an even solution, and using your coefficients, we require

$$A = D$$

and

$$B = C$$

It turns out that there is always an even solution but there may not be an odd solution when $a$ is small enough.

• after this I am trying to figure out what $\Delta\left(\frac{d\psi}{dx}\right)$ is. Integrating the potential part of SWE and taking the limit as $\epsilon$ approaches $\pm a$I get: $$\Delta\left(\frac{d\psi}{dx}\right)=-\frac{2m}{\hbar ^2}\left[\alpha\psi(a)+\alpha\psi(-a)\right]$$ but the solution reads $\Delta\left(\frac{d\psi}{dx}\right)=-\frac{2m\alpha}{\hbar^2}\psi(a)$ I think I'm close but not sure how they got their result. Do you see what I'm doing? Can you correct where I made any mistakes? – Logan Oct 14 '14 at 23:20

You have started with the correct functions, but you need to demand that the wave function is continuous.

So, if we call the three solutions $\psi_\text{I}(x)$, $\psi_\text{II}(x)$ and $\psi_\text{III}(x)$ from left to right in the domain $x$, we need to impose

$$\psi_\text{I}(-a) = \psi_\text{II}(-a)$$

$$\psi_\text{II}(a) = \psi_\text{III}(a)$$

Doing this you should get the result (I didn't check it myself).

• The delta function makes it impossible to make the first derivative continuous. – garyp Oct 14 '14 at 2:08
• Yes, right, I was thinking in a square well. That's the reason why we get two constants $A$ and $B$. I will correct it right now. – dpravos Oct 14 '14 at 2:11