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Raze2dust
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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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