# Infinite square well with delta potential

## Homework Statement

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

Schrodinger eq.

## The Attempt at a Solution

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

$$u'(+0)-u'(-0)=2\Omega u(0)$$

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

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

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

And $$k_n=\frac{n\pi}{2a}$$.

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

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

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

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

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

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

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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