I want to prove that the complex exponentials $(\phi_n(x)=\frac{1}{\sqrt{a}}e^{i\frac{2\pi nx}{a}})_{n=0, \pm 1, \pm2,...}$ form a complete orthonormal system for the space of square-integrable functions on the interval $[0,a]$. I have shown orthonormality, which wasn't too hard. Now I want to show completeness. I believe the easiest way to show completeness is to show the completeness relation: $\sum_{n=-\infty}^{\infty}\phi_n^*(x) \phi_n(x')= \delta(x-x'). $ (see https://ocw.mit.edu/courses/physics/8-05-quantum-physics-ii-fall-2013/lecture-notes/MIT8_05F13_Chap_01.pdf for an explantion of this relation). The last time I wanted to show that some function was equal to the dirac delta function, I integrated it's product with an appropriate test function $f(x)$, and proved that it acted just like the dirac delta. In this case, this would look like so:$\int \sum_{-\infty}^{\infty}\phi_n^*(x) \phi_n(x') \cdot f(x)dx= \int \delta(x-x')\cdot f(x) dx = f(x'). $ So I want to show (assuming my approach works) $\int \sum_{n=-\infty}^{\infty}\phi_n^*(x) \phi_n(x') \cdot f(x)dx= \int\sum_{n=-\infty}^{\infty}\frac{1}{\sqrt{a}}e^{-i\frac{2\pi nx}{a}} \frac{1}{\sqrt{a}}e^{i\frac{2\pi nx'}{a}}f(x) dx = f(x') $. First of all is this a good approach? And second, how would you proceed from here?

  • $\begingroup$ Do you already know that the result is true for $a=1$? $\endgroup$ – user159517 May 1 '18 at 8:58
  • $\begingroup$ No I don't. I'm still stuck with the sum of integrals even in that special case. Is that special case significantly easier? $\endgroup$ – ghthorpe May 1 '18 at 9:02
  • $\begingroup$ @user159517 The only idea I have right now is identifying a fourier transform of $f(x)$ to get something like $\hat{f}(n)$ and then taking a fourier series of that... Is that at all a reasonable approach? Or am I completely off? $\endgroup$ – ghthorpe May 1 '18 at 9:19
  • $\begingroup$ No, not really, but I bet you're already familiar with that case becase it is shown in Fourier theory (although you might not be familiar with in the form it is formulated here). Once you know this is true the general case can be deduced with a simple substitution. $\endgroup$ – user159517 May 1 '18 at 9:19
  • $\begingroup$ @user159517 Could you give me a hint as to how to proceed (with the special or the general case)? $\endgroup$ – ghthorpe May 1 '18 at 9:20

I'll try to reduce it to standard Fourier theory. We begin by substituting $y= x/a$ to obtain

\begin{align}&\int_{0}^{a}\sum_{n=-\infty}^{\infty}\frac{1}{\sqrt{a}}e^{-i\frac{2\pi nx}{a}} \frac{1}{\sqrt{a}}e^{i\frac{2\pi nx'}{a}}f(x) dx = \sum_{n=0}^{\infty}e^{i\frac{2\pi nx'}{a}}\int_{0}^{1} e^{-i2\pi ny}f(ay)~\mathrm dy \\ &= \sum_{n=0}^{\infty} e^{i\frac{2\pi n x'}{a}} \hat{f_a}(n). \end{align} According to standard fourier theory, this is the Fourier series of the funtion $f_a(x) := f(ax)$ evaluated at the point $x'/a$. Accordingly, we find $$\sum_{n=0}^{\infty} e^{i\frac{2\pi n x'}{a}} \hat{f_a}(n) = f(x').$$

  • $\begingroup$ thanks. I had suspected a solution of that flavor, but what confused me (and still does) is that here I'm using continuous Fourier transformation in one direction and discrete Fourier series in the other. Is this mathematically legitimate? Also is there a physical interpretation for this? $\endgroup$ – ghthorpe May 1 '18 at 13:59
  • $\begingroup$ Are we really using continuous Fourier transformation? It's possible to interpret it that way by setting $f = 0$ outside $[0,a]$. But I wouldn't see it that way. We want to show completeness on $[0,a]$, so we can only look at functions defined on $[0,a]$ and points $x,x' \in [0,a]$. Mathematically, all the steps are fine, perhaps with the exception that the last equality does not hold pointwise but in $L^2(0,a)$. $\endgroup$ – user159517 May 1 '18 at 15:24

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