1
$\begingroup$

Intro

We're looking at the Kronig-Penney model in class and one of the conundrums is related to the Kronig-Penney potential for a chain of $N$ atoms. I'm supposed to squeeze out some expression for the Fourier components, $U_G$, but I don't end up with the right expression, although I think I did the right stuff.

The following is given for the delta potential function, \begin{equation} \tag 1 U(x)=A\sum\limits_{n=-\infty}^\infty \delta(x-na), \end{equation} of which I have to show that the Fourier components are given by \begin{equation} \tag 2 U_G=\frac{A}{a}, \qquad G\in\mathbb{Z}, \end{equation} using the hint that $\sum^N_{n=0}\cos(Gna)=N$.


Now I've done the following:

I'll start with the generalized form of $U_G$ (I am not sure whether this expression is correct), \begin{equation} \tag 1 U_G=\frac1a\int_a U(x)e^{-iGx}\mathrm{d}x, \end{equation} in which I can substitute my potential function $U(x)$ ($U(x)=A\sum\limits_{n=-\infty}^\infty \delta(x-na)$), to find \begin{equation} \tag 2 U_G=\frac1a\int_a A\sum\limits_{n=-\infty}^\infty \delta(x-na)e^{-iGx}\mathrm{d}x. \end{equation} $A$ is just a constant and the sum and integral signs can be swapped, so \begin{equation} \tag 3 U_G=\frac{A}a\sum\limits_{n=-\infty}^\infty \int_a\delta(x-na)e^{-iGx}\mathrm{d}x. \end{equation} Now by the definition of the delta function, \begin{equation}\tag 4 \int_a\delta(x-na)e^{-iGx}\mathrm{d}x=\int_a\delta(x-na)f(x)\mathrm{d}x=f(na)=e^{-iGna} \end{equation} Back to our original equation, \begin{align} \tag 5 U_G&=\frac{A}a\sum\limits_{n=-N}^Ne^{-iGna}\\ \tag 6 &=\frac{2A}a\sum\limits_{n=0}^N\frac{e^{-iGna}+e^{iGna}}2\\ \tag 7 &=\frac{2A}{a}\sum\limits_{n=0}^N\cos(Gna)\\ \tag 8 &=\frac{2A}{a}N. \end{align} ...which doesn't add up?


Wikipedia

The wikipedia article on this derivation simply skips a few steps but I'm having some trouble filling those in: \begin{align} \tag 9 U_G&=\frac1a\int_{-a/2}^{a/2}\mathrm{d}x U(x)e^{-iGx}\\ \tag {10} &=\frac1a\int_{-a/2}^{a/2}\mathrm{d}x \sum^\infty_{n=-\infty}A\cdot \delta(x-na)e^{-iGx}\\ \tag {11} &=\frac A a \end{align}

See this link for more info.

$\endgroup$
  • 1
    $\begingroup$ about the wikipedia page: your problem is with the last step of the derivation? If so note that the integral (I assumed a typo about the lower integration limit) $$ \int_{-a/2}^{a/2} dx \,\,\delta(x-na)e^{-ikx} $$ is zero unless $na$ is in the interval $(-a/2,a/2)$. This rules out any $n\neq 0$ hence the result $\endgroup$ – glS Feb 7 '15 at 11:25
  • $\begingroup$ @glance And it's $e^{-ikx}$ for $n=0$? $\endgroup$ – user55789 Feb 7 '15 at 16:31
  • $\begingroup$ for $n=0$ you have $$ \int_{-a/2}^{a/2} dx \,\, \delta(x) e^{-ikx} = 1,$$ from the defining properties of the delta function $\endgroup$ – glS Feb 7 '15 at 16:40
  • $\begingroup$ @glance I can only find material claiming that $\int_{-\infty}^\infty dx\delta(x)=1$?! The exponential factor is completely new to me here. $\endgroup$ – user55789 Feb 8 '15 at 12:30
  • $\begingroup$ @glance And if I use the other rule, $\int dx\delta(x-y)f(x)=f(y)$, I end up with $\int dx\delta(x-na)e^{-iGx}=e^{-iGna}$, which gives me the wrong result (see my updated question). $\endgroup$ – user55789 Feb 8 '15 at 12:32
2
$\begingroup$

From this step,

$$U_G=\frac1a\int_{-a/2}^{a/2}\mathrm{d}x U(x)e^{-iGx}=\frac1a\int_{-a/2}^{a/2}\mathrm{d}x \sum^\infty_{n=-\infty}A\cdot \delta(x-na)e^{-iGx}$$

note that the summation is an impulse train with spacing $a$. Since the integral is from $-\frac{a}{2}$ to $\frac{a}{2}$, just the impulse at $x = 0$ is integrated over so only the $n=0$ term in the summation contributes to the integral:

$$U_G = \frac1a\int_{-a/2}^{a/2}\mathrm{d}x \sum^\infty_{n=-\infty}A\cdot \delta(x-na)e^{-iGx} = \frac1a\int_{-a/2}^{a/2}\mathrm{d}xA\cdot \delta(x)e^{-iGx} = \frac{Ae^{-iG0}}{a} = \frac{A}{a}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.