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 $\ell$:

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

Such func­tions can be writ­ten as a “Fourier se­ries:”

$$\fbox{$\displaystyle f(x) = \sum_{{\rm all }k} f_k \frac{... ...\ell f(x) \frac{e^{-{\rm i}k x}}{\sqrt{\ell}} { \rm d}x $} %$$ (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

$$k = n \frac{2\pi}{\ell} \qquad n = \ldots, -3, -2, -1, 0, 1, 2, 3, \ldots %$$ (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:

$$\int_0^\ell \frac{e^{-{\rm i}k x}}{\sqrt{\ell}} \frac{e^{{... ...line k}\ 0\mbox{ if } k\ne{\underline k}\end{array} \right. %$$ (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 $g(x)$:

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

Us­ing the or­tho­nor­mal­ity prop­erty (A.195) that be­comes

$$\int_0^\ell f^*(x) g(x) { \rm d}x = \sum_{{\rm all }k} f^*_k g_k$$ (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

$$\fbox{$\displaystyle {\left\langle f(x)\right.\hspace{-\nu... ...pace{-\nulldelimiterspace}}{\left\vert g_k\right\rangle} $} %$$ (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:

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

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

$$\Delta k = \frac{2\pi}{\ell}$$

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

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

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:

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

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

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

$$\int_{-\infty}^\infty f^*(x) g(x){ \rm d}x = \int_{-\inft... ... g(k){ \rm d}k = \int_{-\infty}^\infty f^*(p) g(p){ \rm d}p$$

or in short

$$\fbox{$\displaystyle {\left\langle f(x)\right.\hspace{-\nu... ...ace{-\nulldelimiterspace}}{\left\vert g(p)\right\rangle} $} %$$ (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:

$$f(x+\ell_x,y,z) = f(x,y,z) \quad f(x,y+\ell_y,z) = f(x,y,z) \quad f(x,y,z+\ell_z) = f(x,y,z)$$

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,

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

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

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

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

The Fourier se­ries for a three-di­men­sion­al pe­ri­odic func­tion is

$$\fbox{$\displaystyle f({\skew0\vec r}) = \sum_{{\rm all }... ...kew0\vec r}}}{\sqrt{{\cal V}}} { \rm d}^3{\skew0\vec r} $} %$$ (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:

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

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

$$\fbox{$\displaystyle {\left\langle f({\skew0\vec r})\right... ...nulldelimiterspace}}{\left\vert g_{\vec k}\right\rangle} $} %$$ (A.202)

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

For in­fi­nite space

$$\fbox{$\displaystyle f({\skew0\vec r}) = \frac{1}{\sqrt{2\... ...i}{\vec k}\cdot{\skew0\vec r}} { \rm d}^3{\skew0\vec r} $} %$$ (A.203)

$$\fbox{$\displaystyle f({\skew0\vec r}) = \frac{1}{\sqrt{2\... ...c p}\cdot{\skew0\vec r}/\hbar} { \rm d}^3{\skew0\vec r} $} %$$ (A.204)

$$\fbox{$\displaystyle {\left\langle f({\skew0\vec r})\right... ...imiterspace}}{\left\vert g({\skew0\vec p})\right\rangle} $} %$$ (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

$$\begin{eqnarray*} & \displaystyle {\left\langle\vec f({\skew0\vec r})\right.\hs... ...ce{-\nulldelimiterspace}}{\left\vert{g_{\vec k}}_z\right\rangle} \end{eqnarray*}$$

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.)