# Thread: Delta Function Well

1. ## 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

2. 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...

3. Another way of thinking about it is that a continuous function achieves its maximum on a closed interval. Hence,

$\displaystyle 0\le\left|\int_{-\epsilon}^{\epsilon}\psi(x)\,dx\right|\le\int_{-\epsilon}^{\epsilon}|\psi(x)|\,dx\le |\psi_{\max}|\int_{-\epsilon}^{\epsilon}dx=|\psi_{\max}|(2\epsilon)\to 0.$

4. As long as $\psi$ has an anti-derivative (which is certainly true if it is continuous), say $\phi$, which is continuous, so that $\int_{-\epsilon}^\epsilon \psi(x)dx= \phi(\epsilon)- \phi(-\epsilon)$ and then
$\displaytype\lim_{\epsilon\to 0} \int_{-\epsilon}^\epsilon \psi(x)dx= \phi(0)- \phi(0)= 0$.

5. 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

6. A few comments:

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.

7. 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).

8. 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 $\psi(-x)$ does equal $\psi(x)$, suppose $x$ positive, $\psi(x)=Be^{kx}$ and $\psi(-x)=Be^{-k(-x)}=Be^{kx}$

CB

9. Originally Posted by CaptainBlack
It is an even function and $\psi(-x)$ does equal $\psi(x)$, suppose $x$ positive, $\psi(x)=Be^{kx}$ and $\psi(-x)=Be^{-k(-x)}=Be^{kx}$

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

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

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

Thanks alot!

10. You're welcome for my contribution. Have a good one!