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## Homework Statement

We have an infinite square well potential of width 2L centered at the origin, with an attractive delta function potential V

_{0}δ(x) at the origin, with the properties

[tex]V_0<0, -V_0>\frac{\hbar^2}{mL^2}[/tex]

Determine the conditions for a negative energy bound state.

There are a few other parts to the question, but I do not have the sheet at the moment.

## Homework Equations

Schrodinger Equation

## The Attempt at a Solution

In the absence of the delta function, we get

[tex]-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}=E\psi[/tex]

This differential equation has character polynomial given by

[tex]x^2+\frac{2mE}{\hbar^2}=0, x=\pm\frac{i\sqrt{2mE}}{\hbar}[/tex]

The solution is then

[tex]\psi=Asin\left(\frac{\sqrt{2mE}}{\hbar}x\right)+Bcos\left(\frac{\sqrt{2mE}}{\hbar}x\right)[/tex]

Using the boundary condition that the wave function is zero at ±L and normalizing the wave function, I get

[tex]\psi=L^{-\frac{1}{2}}cos\left(\frac{n\pi x}{2L}\right), L^{-\frac{1}{2}}sin\left(\frac{n\pi x}{2L}\right)[/tex]

Where the cosine solution is for odd n and the sine solution is for even n. Also, the energy spectrum is given by

[tex]E_n=\frac{n^2\pi^2\hbar^2}{8mL^2}[/tex]

For any positive integer n. This is the solution for the infinite well without the delta function, the full Schrodinger equation should be

[tex]-\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}+V_0\delta(x)\psi=E\psi[/tex]

I thought about integrating both sides over the width of the well to eliminate the delta function and get

[tex]-\frac{\hbar^2}{2m}\int_{-L}^{L}\frac{\partial^2\psi}{\partial x^2}dx+V_0\psi(0)=E\int_{-L}^{L}\psi dx[/tex]

For even n this is trivial, but for odd n it seems to give me that V

_{0}=0, which is not true, nor very helpful. Obviously the bound state will most likely be the ground state n=1, so I thought it could be

[tex]E_1<-V_0, \frac{\pi^2\hbar^2}{8mL^2}<-V_0[/tex]

However, I have a feeling that I am supposed to work in the conditions imposed on V

_{0}somehow. Is the work so far on the right track, or have I missed something important? There are a few more parts to the question, but this is all I could remember without the sheet near me.