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## Homework Statement

Consider the double delts-function potential

[tex]

V(x)=-\alpha[\delta(x+a)+\delta(x-a)]

[/tex]

How many bound states does this possess? Find the allowed energies for

[tex]\alpha=\frac{\hbar^{2}}{ma^{2}}[/tex]and[tex]\alpha=\frac{\hbar^{2}}{4ma^{2}}[/tex]

## Homework Equations

## The Attempt at a Solution

I divided the region into three parts x<-a(Region 1) ; -a<x<+a(Region 2) ; x>+a(Region 3)

Since we consider bound states, E<0 and solving the SE yields

[tex]Ae^{kx} (x<-a)[/tex]

[tex]Be^{kx}+Ce^{-kx} (-a<x<a)[/tex]

[tex]De^{-kx}(x>a)[/tex]

where [tex]k=\frac{\sqrt{-2mE}}{\hbar}[/tex]

Continuity at x=-a and x=+a respectively give

[tex]A-B=Ce^{2ka}.....(1)[/tex]

[tex]D-C=Be^{2ka}.....(2)[/tex]

For this infinite potentials at points x=-a and x=+a,

[tex]\Delta(\frac{d\psi}{dx})=-\frac{2m\alpha}{\hbar^{2}}\psi(\underline{+}a)[/tex]

So these give two more BC

[tex]A(1-\frac{2m\alpha}{k\hbar^{2}})=B-Ce^{2ka}...(3)[/tex]

[tex]D(1-\frac{2m\alpha}{k\hbar^{2}})=C-De^{2ka}...(4)[/tex]

So I tried to solve these (eqns 1 to 4)and what I got was A=D and B=C

and taking A/B ratios from 1 and 3, i get [tex]ke^{4ka}=\frac{2m\alpha}{\hbar^{2}}[/tex]

And I am not able to solve this equation for k..i'd be grateful for any help :)

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