D.8 Completeness of Fourier modes

The purpose of this note is to show completeness of the “Fourier modes”

$$\ldots, \frac{e^{-3{\rm i}x}}{\sqrt{2\pi}}, \frac{e^{-... ...}{\sqrt{2\pi}}, \frac{e^{3{\rm i}x}}{\sqrt{2\pi}}, \ldots$$

for describing functions that are periodic of period $2\pi$. It is to be shown that “all” these functions can be written as combin­ations of the Fourier modes above. Assume that $f(x)$ is any reasonable smooth function that repeats itself after a distance $2\pi$, so that $f(x+2\pi)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $f(x)$. Then you can always write it in the form

$$f(x) = \ldots + c_{-2} \frac{e^{-2{\rm i}x}}{\sqrt{2\pi}... ...t{2\pi}} + c_3 \frac{e^{3{\rm i}x}}{\sqrt{2\pi}} + \ldots$$

or

$$f(x) = \sum_{k=-\infty}^\infty c_k \frac{e^{k{\rm i}x}}{\sqrt{2\pi}}$$

for short. Such a represen­tation of a periodic function is called a “Fourier series.” The coefficients $c_k$ are called “Fourier coefficients.” The factors 1$\raisebox{.5pt}{$/$}$$\sqrt{2\pi}$ can be absorbed in the definition of the Fourier coefficients, if you want.

Because of the Euler formula, the set of exponential Fourier modes above is completely equivalent to the set of real Fourier modes

$$\frac{1}{\sqrt{2\pi}}, \frac{\cos(x)}{\sqrt{\pi}}, \frac... ...{\cos(3x)}{\sqrt{\pi}}, \frac{\sin(3x)}{\sqrt{\pi}}, \ldots$$

so that $2\pi$-periodic functions may just as well be written as

$$f(x) = a_0 \frac{1}{\sqrt{2\pi}} + \sum_{k=1}^\infty a_k... ...{\pi}} + \sum_{k=1}^\infty b_k \frac{\sin(kx)}{\sqrt{\pi}}.$$

The extension to functions that are periodic of some other period than $2\pi$ is a trivial matter of rescaling $x$. For a period $2\ell$, with $\ell$ any half period, the exponential Fourier modes take the more general form

$$\ldots, \frac{e^{-k_2{\rm i}x}}{\sqrt{2\ell}}, \frac{e... ..._2 = \frac{2\pi}{\ell},\; k_3 = \frac{3\pi}{\ell}, \ldots$$

and similarly the real version of them becomes

$$\frac{1}{\sqrt{2\ell}}, \frac{\cos(k_1x)}{\sqrt{\ell}}, ... ...k_3x)}{\sqrt{\ell}}, \frac{\sin(k_3x)}{\sqrt{\ell}}, \ldots$$

See [39, p. 141] for detailed formulae.

Often, the functions of interest are not periodic, but are required to be zero at the ends of the interval on which they are defined. Those functions can be handled too, by extending them to a periodic function. For example, if the functions $f(x)$ relevant to a problem are defined only for 0 $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ $\ell$ and must satisfy $f(0)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $f(\ell)$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, then extend them to the range $\vphantom0\raisebox{1.5pt}{$-$}$$\ell$ $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ 0 by setting $f(x)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $-f(-x)$ and take the range $\vphantom0\raisebox{1.5pt}{$-$}$$\ell$ $\raisebox{-.3pt}{$\leqslant$}$ $x$ $\raisebox{-.3pt}{$\leqslant$}$ $\ell$ to be the period of a $2\ell$-periodic function. It may be noted that for such a function, the cosines disappear in the real Fourier series represen­tation, leaving only the sines. Similar extensions can be used for functions that satisfy symmetry or zero-derivative boundary conditions at the ends of the interval on which they are defined. See again [39, p. 141] for more detailed formulae.

If the half period $\ell$ becomes infinite, the spacing between the discrete $k$ values becomes zero and the sum over discrete $k$ values turns into an integral over continuous $k$ values. This is exactly what happens in quantum mechanics for the eigen­functions of linear momentum. The represen­tation is now no longer called a Fourier series, but a “Fourier integral.” And the Fourier coefficients $c_k$ are now called the “Fourier transform” $F(k)$. The complete­ness of the eigen­functions is now called Fourier’s integral theorem or inversion theorem. See [39, pp. 190-191] for more.

The basic complete­ness proof is a rather messy mathematical derivation, so read the rest of this note at your own risk. The fact that the Fourier modes are orthogonal and normalized was the subject of various exercises in chapter 2.6 and will be taken for granted here. See the solution manual for the details. What this note wants to show is that any arbitrary periodic function $f$ of period $2\pi$ that has continuous first and second order derivatives can be written as

$$f(x) = \sum_{k=-\infty}^{k=\infty} c_k \frac{e^{k{\rm i}x}}{\sqrt{2\pi}},$$

in other words, as a combin­ation of the set of Fourier modes.

First an expression for the values of the Fourier coefficients $c_k$ is needed. It can be obtained from taking the inner product $\big\langle{e}^{l{\rm i}{x}}/\sqrt{2\pi}\big\vert f(x)\big\rangle$ between a generic eigen­function $e^{l{\rm i}{x}}$$\raisebox{.5pt}{$/$}$$\sqrt{2\pi}$ and the represen­tation for function $f(x)$ above. Noting that all the inner products with the exponentials representing $f(x)$ will be zero except the one for which $k$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l$, if the Fourier represen­tation is indeed correct, the coefficients need to have the values

$$c_l = \int_{x=0}^{2\pi} \frac{e^{-l{\rm i}x}}{\sqrt{2\pi}} f(x){\,\rm d}x,$$

a requirement that was already noted by Fourier. Note that $l$ and $x$ are just names for the eigen­function number and the integr­ation variable that you can change at will. Therefore, to avoid name conflicts later, the expression will be renotated as

$$c_k = \int_{\bar x=0}^{2\pi} \frac{e^{-k{\rm i}\bar x}}{\sqrt{2\pi}} f(\bar x){\,\rm d}\bar x,$$

Now the question is: suppose you compute the Fourier coefficients $c_k$ from this expression, and use them to sum many terms of the infinite sum for $f(x)$, say from some very large negative value $\vphantom0\raisebox{1.5pt}{$-$}$$K$ for $k$ to the corre­sponding large positive value $K$; in that case, is the result you get, call it $f_K(x)$,

$$f_K(x) \equiv \sum_{k=-K}^{k=K} c_k \frac{e^{k{\rm i}x}}{\sqrt{2\pi}},$$

a valid approxi­mation to the true function $f(x)$? More specifi­cally, if you sum more and more terms (make $K$ bigger and bigger), does $f_K(x)$ reproduce the true value of $f(x)$ to any arbitrary accuracy that you may want? If it does, then the eigen­functions are capable of reproducing $f(x)$. If the eigen­functions are not complete, a definite difference between $f_K(x)$ and $f(x)$ will persist however large you make $K$. In mathematical terms, the question is whether $\lim_{K\to\infty}f_K(x)$ $\vphantom0\raisebox{1.5pt}{$=$}$ $f(x)$.

To find out, the trick is to substitute the integral for the coefficients $c_k$ into the sum and then reverse the order of integr­ation and summation to get:

$$f_K(x) = \frac{1}{2\pi} \int_{\bar x=0}^{2\pi} f(\bar x) ... ...{k=-K}^{k=K} e^{k{\rm i}(x-\bar x)}\right] {\,\rm d}\bar x.$$

The sum in the square brackets can be evaluated, because it is a geometric series with starting value $e^{-K{\rm i}(x-\bar{x})}$ and ratio of terms $e^{{\rm i}(x-\bar{x})}$. Using the formula from [39, item 21.4], multi­plying top and bottom with $e^{-{\rm i}(x-\bar{x})/2}$, and cleaning up with, what else, the Euler formula, the sum is found to equal

$$\frac{\sin\Big((K+\frac12)(x-\bar x)\Big)} {\sin\Big(\frac12(x-\bar x)\Big)}.$$

This expression is called the “Dirichlet kernel”. You now have

$$f_K(x) = \int_{\bar x=0}^{2\pi} f(\bar x) \frac{\sin\Big... ...)} {2\pi\sin\Big(\frac12(x-\bar x)\Big)} {\,\rm d}\bar x.$$

The second trick is to split the function $f(\bar{x})$ being integrated into the two parts $f(x)$ and $f(\bar{x})-f(x)$. The sum of the parts is obviously still $f(\bar{x})$, but the first part has the advantage that it is constant during the integr­ation over $\bar{x}$ and can be taken out, and the second part has the advantage that it becomes zero at $\bar{x}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $x$. You get

$$\begin{eqnarray*} \lefteqn{f_K(x) = f(x) \int_{\bar x=0}^{2\pi} \frac{\sin... ...ig)} {2\pi\sin\Big(\frac12(x-\bar x)\Big)} {\,\rm d}\bar x. \end{eqnarray*}$$

Now if you backtrack what happens in the trivial case that $f(x)$ is just a constant, you find that $f_K(x)$ is exactly equal to $f(x)$ in that case, while the second integral above is zero. That makes the first integral above equal to one. Returning to the case of general $f(x)$, since the first integral above is still one, it makes the first term in the right hand side equal to the desired $f(x)$, and the second integral is then the error in $f_K(x)$.

To manipulate this error and show that it is indeed small for large $K$, it is convenient to rename the $K$-independent part of the integrand to

$$g(\bar x) = \frac{f(\bar x)-f(x)}{2\pi\sin\Big(\frac12(x-\bar x)\Big)}$$

Using l’Hôpital's rule twice, it is seen that since by assumption $f$ has a continuous second derivative, $g$ has a continuous first derivative. So you can use one integr­ation by parts to get

$$f_K(x) = f(x) + \frac{1}{K+\frac12} \int_{\bar x=0}^{2\p... ...(K+{\textstyle\frac{1}{2}})(x-\bar x)\Big) {\,\rm d}\bar x.$$

And since the integrand of the final integral is continuous, it is bounded. That makes the error inversely propor­tional to $K+\frac12$, implying that it does indeed become arbitrarily small for large $K$. Complete­ness has been proved.

It may be noted that under the stated conditions, the convergence is uniform; there is a guaranteed minimum rate of convergence regardless of the value of $x$. This can be verified from Taylor series with remainder. Also, the more continuous derivatives the $2\pi$-periodic function $f(x)$ has, the faster the rate of convergence, and the smaller the number $2K+1$ of terms that you need to sum to get good accuracy is likely to be. For example, if $f(x)$ has three continuous derivatives, you can do another integr­ation by parts to show that the convergence is propor­tional to 1$\raisebox{.5pt}{$/$}$$(K+\frac12)^2$ rather than just 1/$(K+\frac12)$. But watch the end points: if a derivative has different values at the start and end of the period, then that derivative is not continuous, it has a jump at the ends. (Such jumps can be incorporated in the analysis, however, and have less effect than it may seem. You get a better practical estimate of the convergence rate by directly looking at the integral for the Fourier coefficients.)

The condition for $f(x)$ to have a continuous second derivative can be relaxed with more work. If you are familiar with the Lebesgue form of integr­ation, it is fairly easy to extend the result above to show that it suffices that the absolute integral of $f^2$ exists, something that will be true in quantum mechanics appli­cations.