A.26 Fourier inversion theorem and Parseval

This note discusses Fourier series, Fourier integrals, and Parseval’s identity.

Consider first one-di­men­sion­al Fourier series. They apply to functions $f(x)$ that are periodic with some given period $\ell$:

\begin{displaymath}
f(x+\ell) = f(x)\quad\mbox{for all $x$}
\end{displaymath}

Such functions can be written as a “Fourier series:”
\begin{displaymath}
\fbox{$\displaystyle
f(x) = \sum_{{\rm all\ }k} f_k \fra...
...ll f(x) \frac{e^{-{\rm i}k x}}{\sqrt{\ell}} {\,\rm d}x
$} %
\end{displaymath} (A.193)

Here the $k$ values are those for which the exponentials are periodic of period $\ell$. According to the Euler formula (2.5), that means that $k\ell$ must be a whole multiple $n$ of $2\pi$, so
\begin{displaymath}
k = n \frac{2\pi}{\ell} \qquad
n = \ldots, -3, -2, -1, 0, 1, 2, 3, \ldots %
\end{displaymath} (A.194)

Note that notations for Fourier series can vary from one author to the next. The above form of the Fourier series is the prefered one for quantum mechanics. The reason is that the functions $e^{{{\rm i}}kx}$$\raisebox{.5pt}{$/$}$$\sqrt{\ell}$ form an orthonormal set:

\begin{displaymath}
\int_0^\ell \frac{e^{-{\rm i}k x}}{\sqrt{\ell}}
\frac{e^...
...e k}\\ 0\mbox{ if } k\ne{\underline k}\end{array}
\right. %
\end{displaymath} (A.195)

Quantum mechanics just loves orthonormal sets of functions. In particular, note that the above functions are momentum eigenfunctions. Just apply the linear momentum operator ${\widehat p}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar{\rm d}$$\raisebox{.5pt}{$/$}$${\rm i}{\rm d}{x}$ on them. That shows that their linear momentum is given by the de Broglie relation $p$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\hbar}k$. Here these momentum eigenfunctions are properly normalized. They would not be using different conventions.

That any (reasonable) periodic function $f(x)$ can be written as a Fourier series was already shown in {D.8}. That derivation took $\ell$ be the half-period. The formula for the coefficients $f_k$ can also be derived directly: simply multiply the expression (A.193) for $f(x)$ with $e^{{\rm i}{{\underline k}}x}$$\raisebox{.5pt}{$/$}$$\sqrt{\ell}$ for any arbitrary value of ${\underline k}$ and integrate over $x$. Because of the orthonormality (A.195), the integration produces zero for all $k$ except if $k$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\underline k}$, and then it produces $f_{\underline k}$ as required.

Note from (A.193) that if you known $f(x)$ you can find all the $f_k$. Conversely, if you know all the $f_k$, you can find $f(x)$ at every position $x$. The formulae work both ways.

But the symmetry goes even deeper than that. Consider the inner product of a pair of functions $f(x)$ and $g(x)$:

\begin{displaymath}
\int_0^\ell f^*(x) g(x) {\,\rm d}x =
\int_0^\ell
\sum_...
...k}\frac{e^{{\rm i}{\underline k}x}}{\sqrt{\ell}}
{\,\rm d}x
\end{displaymath}

Using the orthonormality property (A.195) that becomes
\begin{displaymath}
\int_0^\ell f^*(x) g(x) {\,\rm d}x = \sum_{{\rm all\ }k} f^*_k g_k
\end{displaymath} (A.196)

Now note that if you look at the coefficients $f_k$ and $g_k$ as the coefficients of infinite-​di­men­sion­al vectors, then the right hand side is just the inner product of these vectors. In short, Fourier series preserve inner products.

Therefore the equation above may be written more concisely as

\begin{displaymath}
\fbox{$\displaystyle
\big\langle f(x)\big\vert g(x)\big\rangle = \big\langle f_k\big\vert g_k\big\rangle
$} %
\end{displaymath} (A.197)

This is the so-called “Parseval identity.” Now transformations that preserve inner products are called “unitary” in mathematics. So the Parseval identity shows that the transformation from periodic functions to their Fourier coefficients is unitary.

That is quite important for quantum mechanics. For example, assume that $f(x)$ is a wave function of a particle stuck on a ring of circumference $\ell$. Wave functions should be normalized, so:

\begin{displaymath}
\int_0^\ell f^*(x) f(x) {\,\rm d}x = 1 = \sum_{{\rm all\ }k} f^*_k f_k
\end{displaymath}

According to the Born interpretation, the left hand side says that the probability of finding the particle is 1, certainty, if you look at every position on the ring. But according to the orthodox interpretation of quantum mechanics, $f^*_kf_k$ in the right hand side gives the probability of finding the particle with momentum $p$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\hbar}k$. The fact that the total sum is 1 means physically that it is certain that the particle will be found with some momentum.

So far, only periodic functions have been covered. But functions in infinite space can be handled by taking the period $\ell$ infinite. To do that, note from (A.194) that the $k$ values of the Fourier series are spaced apart over a distance

\begin{displaymath}
\Delta k = \frac{2\pi}{\ell}
\end{displaymath}

In the limit $\ell\to\infty$, $\Delta{k}$ becomes an infinitesimal increment ${\rm d}{k}$, and the sums become integrals. Now in quantum mechanics it is convenient to replace the coefficients $f_k$ by a new function $f(k)$ that is defined so that

\begin{displaymath}
f_k =\sqrt{\Delta k} f(k) \quad\Longrightarrow\quad
\vert f_k\vert^2 = \vert f(k)\vert^2 \Delta k
\end{displaymath}

The reason that this is convenient is that $\vert f_k\vert^2$ gives the probability for wave number $k$. Then for a function $f(k)$ that is defined as above, $\vert f(k)\vert^2$ gives the probability per unit $k$-range.

If the above definition and $\sqrt{\ell}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\sqrt{2\pi}$$\raisebox{.5pt}{$/$}$$\Delta{k}$ are substituted into the Fourier series expressions (A.193), in the limit $\ell\to\infty$ it gives the “Fourier integral” formulae:

\begin{displaymath}
\fbox{$\displaystyle
f(x) = \frac{1}{\sqrt{2\pi}} \int_{...
... \int_{-\infty}^\infty f(x) e^{-{\rm i}k x} {\,\rm d}x
$} %
\end{displaymath} (A.198)

In books on mathematics you will usually find function $f(k)$ indicated as $\widehat{f}(k)$, to clarify that it is a completely different function than $f(x)$. Unfortunately, the hat is already used for something much more important in quantum mechanics. So in quantum mechanics you will have to look at the argument, $x$ or $k$, to know which function it really is.

Of course, in quantum mechanics you are often more interested in the momentum than in the wave number. So it is often convenient to define a new function $f(p)$ so that $\vert f(p)\vert^2$ gives the probability per unit momentum range rather than unit wave number range. Because $p$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\hbar}k$, the needed rescaling of $f(k)$ is by a factor $\sqrt{\hbar}$. That gives

\begin{displaymath}
\fbox{$\displaystyle
f(x) = \frac{1}{\sqrt{2\pi\hbar}}
...
..._{-\infty}^\infty f(x) e^{-{\rm i}px/\hbar} {\,\rm d}x
$} %
\end{displaymath} (A.199)

Using similar substitutions as for the Fourier series, the Parseval identity (A.197) becomes

\begin{displaymath}
\int_{-\infty}^\infty f^*(x) g(x){\,\rm d}x =
\int_{-\in...
...(k){\,\rm d}k =
\int_{-\infty}^\infty f^*(p) g(p){\,\rm d}p
\end{displaymath}

or in short
\begin{displaymath}
\fbox{$\displaystyle
\big\langle f(x)\big\vert g(x)\big\...
...ig\rangle = \big\langle f(p)\big\vert g(p)\big\rangle
$} %
\end{displaymath} (A.200)

This identity is sometimes called the “Plancherel theorem,” after a later mathematician who generalized its applicability. The bottom line is that Fourier integral transforms too conserve inner products.

So far, this was all one-di­men­sion­al. However, the extension to three dimensions is straightforward. The first case to be considered is that there is periodicity in each Cartesian direction:

\begin{displaymath}
f(x+\ell_x,y,z) = f(x,y,z) \quad
f(x,y+\ell_y,z) = f(x,y,z) \quad
f(x,y,z+\ell_z) = f(x,y,z)
\end{displaymath}

In quantum mechanics, this would typically correspond to the wave function of a particle stuck in a periodic box of dimensions $\ell_x$ $\times$ $\ell_y$ $\times$ $\ell_z$. When the particle leaves such a box through one side, it reenters it at the same time through the opposite side.

There are now wave numbers for each direction,

\begin{displaymath}
k_x = n_x \frac{2\pi}{\ell_x} \qquad
k_y = n_y \frac{2\pi}{\ell_y} \qquad
k_z = n_z \frac{2\pi}{\ell_z}
\end{displaymath}

where $n_x$, $n_y$, and $n_z$ are whole numbers. For brevity, vector notations may be used:

\begin{displaymath}
{\skew0\vec r}\equiv x{\hat\imath}+y{\hat\jmath}+z{\hat k}...
...y y}e^{{\rm i}k_z z} = e^{{\rm i}{\vec k}\cdot{\skew0\vec r}}
\end{displaymath}

Here ${\vec k}$ is the “wave number vector.”

The Fourier series for a three-di­men­sion­al periodic function is

\begin{displaymath}
\fbox{$\displaystyle
f({\skew0\vec r}) = \sum_{{\rm all\...
...w0\vec r}}}{\sqrt{{\cal V}}} {\,\rm d}^3{\skew0\vec r}
$} %
\end{displaymath} (A.201)

Here $f({\skew0\vec r})$ is shorthand for $f(x,y,z)$ and ${\cal V}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\ell_x\ell_y\ell_z$ is the volume of the periodic box.

The above expression for $f$ may be derived by applying the one-di­men­sion­al transform in each direction in turn:

\begin{eqnarray*}
f(x,y,z) & = & \sum_{{\rm all\ }k_x} f_{k_x}(y,z)
\frac{e^...
...\ell_x}\smash{\ell_y}}}
\frac{e^{{\rm i}k_zz}}{\sqrt{\ell_z}}
\end{eqnarray*}

This is equivalent to what is given above, except for trivial changes in notation. The expression for the Fourier coefficients can be derived analogous to the one-di­men­sion­al case, integrating now over the entire periodic box.

The Parseval equality still applies

\begin{displaymath}
\fbox{$\displaystyle
\big\langle f({\skew0\vec r})\big\v...
...\big\langle f_{\vec k}\big\vert g_{\vec k}\big\rangle
$} %
\end{displaymath} (A.202)

where the left inner product integration is over the periodic box.

For infinite space

\begin{displaymath}
\fbox{$\displaystyle
f({\skew0\vec r}) = \frac{1}{\sqrt{...
...{\vec k}\cdot{\skew0\vec r}} {\,\rm d}^3{\skew0\vec r}
$} %
\end{displaymath} (A.203)


\begin{displaymath}
\fbox{$\displaystyle
f({\skew0\vec r}) = \frac{1}{\sqrt{...
...p}\cdot{\skew0\vec r}/\hbar} {\,\rm d}^3{\skew0\vec r}
$} %
\end{displaymath} (A.204)


\begin{displaymath}
\fbox{$\displaystyle
\big\langle f({\skew0\vec r})\big\v...
...{\skew0\vec p})\big\vert g({\skew0\vec p})\big\rangle
$} %
\end{displaymath} (A.205)

These expressions are all obtained completely analogously to the one-di­men­sion­al case.

Often, the function is a vector rather than a scalar. That does not make a real difference since each component transforms the same way. Just put a vector symbol over $f$ and $g$ in the above formulae. The inner products are now defined as

\begin{eqnarray*}
& \displaystyle \big\langle\vec f({\skew0\vec r})\big\vert\v...
... + \big\langle{f_{\vec k}}_z\big\vert{g_{\vec k}}_z\big\rangle
\end{eqnarray*}

For the picky, converting Fourier series into Fourier integrals only works for well-behaved functions. But to show that it also works for nasty wave functions, you can set up a limiting process in which you approximate the nasty functions increasingly accurately using well-behaved ones. Now if the well-behaved functions are converging, then their Fourier transforms are too. The inner products of the differences in functions are the same according to Parseval. And according to the abstract Lebesgue variant of the theory of integration, that is enough to ensure that the transform of the nasty function exists. This works as long as the nasty wave function is square integrable. And wave functions need to be in quantum mechanics.

But being square integrable is not a strict requirement, as you may have been told elsewhere. A lot of functions that are not square integrable have meaningful, invertible Fourier transforms. For example, functions whose square magnitude integrals are infinite, but absolute value integrals are finite can still be meaningfully transformed. That is more or less the classical version of the inversion theorem, in fact. (See D.C. Champeney, A Handbook of Fourier Theorems, for more.)