Delta Function Well

**bugatti79** · December 3rd 2010, 11:45 AM

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

**Ackbeet** · December 3rd 2010, 12:35 PM

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...

**Ackbeet** · December 3rd 2010, 12:49 PM

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

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

**HallsofIvy** · December 4th 2010, 04:53 AM

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
$\displaytype\lim_{\epsilon\to 0} \int_{-\epsilon}^\epsilon \psi(x)dx= \phi(0)- \phi(0)= 0$ .

**bugatti79** · December 4th 2010, 05:29 AM

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

**Ackbeet** · December 4th 2010, 06:51 AM

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.

**bugatti79** · December 4th 2010, 07:21 AM

Originally Posted by Ackbeet

A few comments:

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).

**CaptainBlack** · December 4th 2010, 07:46 AM

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

**bugatti79** · December 4th 2010, 07:55 AM

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!

**Ackbeet** · December 6th 2010, 04:48 AM

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

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