Integrating the time-indepdent Schrodinger equation

[wrldt]

Homework Statement

A particle of mass m is confined to move in a one-dimensional and Diract delta-function attractive potential [itex]V(x)=-\frac{\hbar^2}{m}\alpha\delta(x)[/itex] where [itex]\alpha[/itex] is positive.

Integrate eh time-independent Schrodinger equation between [itex]-\epsilon[/itex] and [itex]\epsilon[/itex]. Let [itex]\epsilon\to 0[/itex], show that the derivative of the eigenfunction [itex]\phi(x)[/itex] is discontinuous at x=0, and determine its jump at the point.

Homework Equations

[itex]\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}-V(x)\psi=E\psi[/itex]

The Attempt at a Solution

First I want to determine if I am approaching the problem correctly.
First we have:
[itex]\int\limits_{-\epsilon}^{\epsilon}-\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}+\int\limits_{-\epsilon}^{\epsilon}\frac{\hbar^2}{m}\alpha\delta(x)\psi=\int\limits_{-\epsilon}^{\epsilon}E\psi[/itex].
Then this simplifies to
[itex]\int\limits_{-\epsilon}^{\epsilon}\frac{\hbar^2}{m}\alpha\delta\psi=\int\limits_{-\epsilon}^{\epsilon}\frac{\hbar^2}{2m}\frac{\partial^2 \psi}{\partial x^2}[/itex]
Further simplifying
[itex]2\alpha\int\limits_{-\epsilon}^{\epsilon}\delta(x)\psi(x)=\frac{\partial \psi}{\partial x}\bigg|_{-\epsilon}^{\epsilon}[/itex]

The last equation shows that as $\epsilon \to 0$, the right hand side will tend to zero but the left hand side will explode. I'm wondering if my work sufficiently shows this.

The part I am stuck on is showing what its jump is at x=0.

 

[TSny]

[itex]2\alpha\int\limits_{-\epsilon}^{\epsilon}\delta(x)\psi(x)=\frac{\partial \psi}{\partial x}\bigg|_{-\epsilon}^{\epsilon}[/itex]

Your work up to here looks very good.

The last equation shows that as ##\epsilon \to 0##, the right hand side will tend to zero but the left hand side will explode. I'm wondering if my work sufficiently shows this.

Here's where you need to think some more about what you're getting. The left hand side does not explode. What does integrating over the delta function give you for any finite ##\epsilon##? Then let ##\epsilon## → 0. Similarly for the right hand side.

 

[wrldt]

[itex]2\alpha\int\limits_{-\epsilon}^{\epsilon}\delta(x)\psi(x)=\frac{\partial \psi}{\partial x}\bigg|_{-\epsilon}^{\epsilon}[/itex]

This simplies to something like [itex]2\alpha\psi(0)=\Delta\left(\frac{\partial \psi}{\partial x}\right)[/itex].

After some work I believe this simplies down to [itex]k=\alpha[/itex]

So now our [itex]E=-\frac{\hbar^2\alpha^2}{2m}[/itex].

I'm basically following the process that is going on on p. 71-72 of Introduction to Quantum Mechanics.