Monday 10 February 2020

homework and exercises - What intermediate steps of the Dirac Delta Function and Fourier Series am I missing in finding a solution to the Kronig-Penney Model?




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 $\sum^N_{n=0}\cos(Gna)=N$.





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?





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.



Answer




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:


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


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...