# Bound States in a Double Delta Function Potential [closed]

Let $V(x) = −u \delta(x) - v \delta(x − a)$ where $u, v > 0$ correspond to a potential with two $\delta$ wells. Let $v > u$. If $a$ is very large, there is certainly a bound state: the particle sits in the $\delta$-well. As $a$ decreases to a certain critical value, the bound state disappears. I need help finding that value.

My idea was: Before the bound state disappears, its energy approaches $0$. I'm trying to assume that the energy $E$ is a very small negative number, solve the Schrodinger equation, and find the suitable value of $a$, but I'm having trouble doing this.

Would someone be able to help me with this problem?

## closed as too localized by dmckee♦Nov 7 '12 at 14:16

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

• I would suggest at least finding the bound state and its energy before you try to do anything sexy with limits and such. – DJBunk Nov 5 '12 at 23:43
• Do know how to find the eigenvectors and eigenvalues in a case like this? – DJBunk Nov 6 '12 at 3:09
• I got something for you, give me a minute – Dylan O. Sabulsky Nov 6 '12 at 3:14

I'm not going to answer your exact question, but this is a good example (from an old copy of Griffith's that my loser chem bro uses [real women and men of physics use Shankar and Sakurai]

Consider the double delta-function potential $$V(x)=-\alpha[\delta(x+a)+\delta(x-a)]$$ where $a$ and $\alpha$ are positive constants. Hope this helps! -Dylan

• Hi Dylan - you shouldn't directly put in scans from a textbook, instead type out the relevant parts of the material, using block quote syntax if necessary. Could you edit this accordingly? (Perhaps someone else will be willing to do it, but don't count on that.) – David Z Nov 6 '12 at 18:29
• Hey David, I'll try to later on today. Sorry about that! – Dylan O. Sabulsky Nov 6 '12 at 22:08
• @DylanSabulsky meanwhile. – ItamarG3 Apr 28 '18 at 8:00
• This is what I see: i.stack.imgur.com/JDXOo.png – user191954 Jan 18 at 7:40