Infinite square well with delta potential

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Homework Statement



I have infinite square well which has a potential [tex]V(x)=\frac{\hbar^2}{m}\Omega\delta(x)[/tex] in x=0, and is 0 in the interval [tex]x\in[-a,a][/tex]


Homework Equations


Schrodinger eq.


The Attempt at a Solution



I solved the time independant Schrodinger eq. by integration around x=0 by some small parameter [tex]\varepsilon[/tex]. And I got the discontinuity

[tex]u'(+0)-u'(-0)=2\Omega u(0)[/tex]

Where u(x) is the [tex]\psi(x)[/tex]. Now I also have a discontinuity in logarithmic derivative:
[tex]\frac{u'(+0)}{u(0)}-\frac{u'(-0)}{u(0)}=2\Omega[/tex].

I solved the infinite square well problem and the solutions can be divided into symmetric and antisymmetric ones:

[tex]u_n^+(x)=\frac{1}{\sqrt{a}}\cos(k_n x), n=\pm 1,\pm 3,\ldots[/tex]
[tex]u_n^-(x)=\frac{1}{\sqrt{a}}\sin(k_n x), n=\pm 2,\pm 4,\ldots[/tex]

And [tex]k_n=\frac{n\pi}{2a}[/tex].

The odd solutions have knot in x=0 and their derivative is continuous in x=0, but even ones must have this form:

[tex]u_n^+(x)=-A\sin k_n^+(x+a),\ for\quad -a\leq x<0[/tex] and
[tex]u_n^+(x)=A\sin k_n^+(x-a),\ for\quad 0<x\leq a[/tex]

Now I have to apply the derivation condition on this solutions (discontinuity one). Now I'm puzzled here. In my notes it says:
[tex]2Ak_n^+\cos(k_n^+ a)=-2\Omega A\sin(k_n^+a)[/tex]

And right hand side I got, but what about the left?!

If I derive the whole thing (the sin) I get:

[tex]u_n^+'(x)=-Ak_n^+\cos k_n^+(x+a),\ for\quad -a\leq x<0[/tex] and
[tex]u_n^+'(x)=Ak_n^+\cos k_n^+(x-a),\ for\quad 0<x\leq a[/tex]

And if I let x=0, those two would cancel :\ Or do I have to look separately from +0 and -0? I'm kinda stuck there...
 
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Answers and Replies

  • #2
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Ok, I solved the problem, I should first consider the solutions with separate constants, let's say A and B, and then apply the continuity at x=0 condition...
 

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