# Delta Function Well

• Dec 3rd 2010, 11:45 AM
bugatti79
Delta Function Well
I Have a query regarding the delta function well in quantum physics.

As part of the process to determine the discontinuity of the derivative of the function psi at x =0 (see attached), the integral needs to be evaluated from -E to +E and then taking the limit E tends to 0.

What i dont understand is how the RHS of eqn goes to zero. Could anyone explain this, the rest is fine.

Thanks
• Dec 3rd 2010, 12:35 PM
Ackbeet
The integrand on the RHS is continuous, right? So you're taking the area under a continuous curve, and the width of the area is getting narrower and narrower...
• Dec 3rd 2010, 12:49 PM
Ackbeet
Another way of thinking about it is that a continuous function achieves its maximum on a closed interval. Hence,

$\displaystyle \displaystyle 0\le\left|\int_{-\epsilon}^{\epsilon}\psi(x)\,dx\right|\le\int_{-\epsilon}^{\epsilon}|\psi(x)|\,dx\le |\psi_{\max}|\int_{-\epsilon}^{\epsilon}dx=|\psi_{\max}|(2\epsilon)\to 0.$
• Dec 4th 2010, 04:53 AM
HallsofIvy
As long as $\displaystyle \psi$ has an anti-derivative (which is certainly true if it is continuous), say $\displaystyle \phi$, which is continuous, so that $\displaystyle \int_{-\epsilon}^\epsilon \psi(x)dx= \phi(\epsilon)- \phi(-\epsilon)$ and then
$\displaystyle \displaytype\lim_{\epsilon\to 0} \int_{-\epsilon}^\epsilon \psi(x)dx= \phi(0)- \phi(0)= 0$.
• Dec 4th 2010, 05:29 AM
bugatti79
Thanks guys,

I wanted to see how this goes to zero mathematically, see atached. I think it is similar to Hallsofivy definition. Is my approach look ok?

THanks
• Dec 4th 2010, 06:51 AM
Ackbeet

That's definitely overkill, but it gets the job done. You can get the last expression on the second-to-last line immediately by noticing that the integrand is even, and you're integrating over a symmetric interval.
• Dec 4th 2010, 07:21 AM
bugatti79
Quote:

Originally Posted by Ackbeet

That's definitely overkill, but it gets the job done. You can the last expression on the second-to-last line immediately by noticing that the integrand is even, and you're integrating over a symmetric interval.

I see the interval is symmetric because the limits are from -E to +E.
How is the integrand an even function? since psi (-x) not equal to psi (x).
• Dec 4th 2010, 07:46 AM
CaptainBlack
Quote:

Originally Posted by bugatti79
I see the interval is symmetric because the limits are from -E to +E.
How is the integrand an even function? since psi (-x) not equal to psi (x).

It is an even function and $\displaystyle $$\psi(-x) does equal \displaystyle$$\psi(x)$, suppose $\displaystyle $$x positive, \displaystyle \psi(x)=Be^{kx} and \displaystyle \psi(-x)=Be^{-k(-x)}=Be^{kx} CB • Dec 4th 2010, 07:55 AM bugatti79 Quote: Originally Posted by CaptainBlack It is an even function and \displaystyle$$\psi(-x)$ does equal $\displaystyle $$\psi(x), suppose \displaystyle$$x$ positive, $\displaystyle \psi(x)=Be^{kx}$ and $\displaystyle \psi(-x)=Be^{-k(-x)}=Be^{kx}$

CB

ah, ok. I was focusing on the one expression for for x>0 and substituting x and -x. Clearly I need to use the correct expression for each subsitution as you've shown. I see that now. Based on that I think you meant

$\displaystyle \psi(-x)=Be^{k(-x)}=Be^{-kx}$ and

$\displaystyle \psi(x)=Be^{-k(x)}=Be^{-kx}$

Thanks alot!
• Dec 6th 2010, 04:48 AM
Ackbeet
You're welcome for my contribution. Have a good one!