The Poisson Summation Formula

The Poisson summation formula (PSF) is one of the deepest and most useful identities in analysis, and its statement is almost shockingly simple: for a well-behaved function $f$ , summing $f$ over the integers gives exactly the same number as summing its Fourier transform $\hat f$ over the integers. A sum in "position space" equals a sum in "frequency space." That single equality braids together the two great objects of harmonic analysis — the Fourier series of a periodic function and the Fourier transform of a function on the line — and reveals them as two faces of one structure. What makes PSF a workhorse rather than a curiosity is the reciprocity built into the Fourier transform. Narrowing $f$ in direct space broadens $\hat f$ in reciprocal space, so a sum that crawls toward its value on one side may converge at breathtaking speed on the other. That is why the formula is the hidden engine behind Ewald summation in crystalline electrostatics, the sampling theorem and aliasing in signal processing, the reciprocal-lattice description of Bragg diffraction, and the modular transformation of the Jacobi theta function that Riemann used to continue the zeta function. The cleanest modern viewpoint is distributional: the Dirac comb $\mathrm{III}(x)=\sum_n\delta(x-n)$ is its own Fourier transform, and PSF is just that self-duality paired against a test function. This route makes the hypotheses honest — one needs enough decay and smoothness, e.g. a Schwartz function — and explains exactly how the naive identity fails for functions that decay too slowly or jump at a lattice point.

Key takeaways

For a sufficiently nice $f$ , the Poisson summation formula states $\sum_{n\in\mathbb{Z}} f(n) = \sum_{k\in\mathbb{Z}} \hat f(k)$ : a lattice sum of $f$ equals a dual-lattice sum of its Fourier transform.
Equivalently, the Dirac comb (the Shah distribution) is its own Fourier transform, $\widehat{\mathrm{III}}=\mathrm{III}$ — the whole formula in one line.
The mechanism is periodization: wrapping $f$ around the circle, $F(x)=\sum_n f(x+n)$ , produces a 1-periodic function whose Fourier coefficients are exactly $\hat f$ sampled at the integers; reading $F$ at $x=0$ is PSF.
Reciprocity makes it powerful: with spacing $a$ in direct space the dual spacing is $1/a$ , so a slowly converging series (small $a$ , or small $t$ for a Gaussian) becomes an exponentially fast one — the basis of Ewald summation and theta acceleration.
It requires genuine decay and smoothness; at a jump discontinuity on the lattice the sum must be read with the midpoint rule $\tfrac12[f(n^+)+f(n^-)]$ , and conventions must be tracked carefully.
Key idea: periodizing a function and Fourier-expanding the result equates a lattice sum of $f$ with a dual-lattice sum of $\hat f$ ; equivalently, the Dirac comb is its own Fourier transform.

Where this lives today

Poisson summation is not a museum piece — it runs, invisibly, inside a great deal of working computation and measurement. Every molecular-dynamics simulation using particle-mesh Ewald (PME) is running PSF in reverse: the $1/r$ Coulomb lattice sum, hopelessly slow in direct space, is split into a short-range part summed over neighbours and a smooth part summed over the reciprocal lattice, where PSF makes it converge exponentially — this is how simulations of proteins in water compute electrostatics at all. In crystallography the formula is the reciprocal lattice: a periodic electron density is the unit cell convolved with a Dirac comb, so its diffraction pattern is nonzero only at reciprocal-lattice points, and the Bragg peaks a synchrotron records are literally the dual comb. In signal processing PSF is the exact bookkeeping of aliasing: sampling at spacing $a$ periodizes the spectrum with period $1/a$ , and the Nyquist criterion (band-limit below $1/2a$ ) is precisely the condition that the periodized copies do not overlap — the guarantee behind every digital audio and imaging pipeline. And in analytic number theory the theta transformation remains a live tool: modern high-precision algorithms for the Riemann zeta function and $L$ -functions rely on the same $t\leftrightarrow1/t$ acceleration used in Worked Example 2, keeping a two-centuries-old identity at the center of ongoing research on the primes.

The physical picture

Place a copy of $f$ at every integer and add them up: $F(x)=\sum_n f(x+n)$ . The result cannot help being 1-periodic — shifting $x$ by one integer merely relabels which copy is which. A 1-periodic function contains only pure frequencies that are whole numbers, and the amplitude sitting on frequency $k$ turns out to be exactly $\hat f(k)$ , the original transform sampled at the integers. Reading this periodic function at the single point $x=0$ collapses both descriptions onto each other and delivers the Poisson formula: the sum of the copies' heights at the origin equals the sum of the transform samples. The first figure makes "wrap and add" the whole story.

Figure

Periodizing f places a copy at every integer and adds them; the sum F(x)=Σₙ f(x+n) is automatically 1-periodic, so it carries only integer frequencies k with amplitudes f̂(k). Reading F at the single point x=0 gives the Poisson summation formula Σₙ f(n) = Σₖ f̂(k).

The reason PSF earns its keep is reciprocity: the Fourier transform trades width in $x$ for narrowness in $\xi$ . A Gaussian of width $\sigma$ becomes a Gaussian of width $1/\sigma$ , so squeezing the direct-space samples together spreads the dual-space samples apart. When the direct sum is agonizingly slow — many terms of nearly equal size — the dual sum is exponentially fast, often a single term to machine precision. The second figure shows this in hard numbers for the Jacobi theta function $\theta(t)=\sum_n e^{-\pi t n^2}$ at the small value $t=0.01$ : the direct series needs dozens of terms to reach ten digits, while the transformed series pins the answer $\theta(0.01)=10$ to over a hundred digits with one term.

Figure

Reciprocity in hard numbers for θ(t)=Σₙ e^(−πtn²) at t=0.01. The direct terms e^(−πtn²) form a broad picket fence: about 57 terms are needed for ten-digit accuracy. The dual terms t^(−1/2) e^(−πk²/t) = 10 e^(−100πk²) collapse to a single spike of height 10; the next term is only 10·e^(−100π) ≈ 4×10⁻¹³⁶, so one dual term gives the exact value θ(0.01)=10 to 135 digits.

The core equations

The formula's headline statement uses the symmetric convention for the Fourier transform, the one in which the exponent already carries its own factor of $2\pi$ . In this convention the integers are their own dual lattice, so PSF is completely factor-free: the sum over $\mathbb{Z}$ on the left equals the sum over $\mathbb{Z}$ on the right, with no covolume, no $2\pi$ , nothing.

\sum_{n\in\mathbb{Z}} f(n)=\sum_{k\in\mathbb{Z}} \hat{f}(k),\qquad \hat f(\xi)=\int_{-\infty}^{\infty} f(x)\,e^{-2\pi i x\xi}\,\mathrm dx \qquad (1)

where: $f$ — a sufficiently nice function (e.g. Schwartz), summed over the integers; $\hat f$ — its Fourier transform with the $e^{-2\pi i x\xi}$ kernel; $n,k$ — integer indices running over the direct and dual lattices $\mathbb{Z}$ . Read it as: the total of $f$ over the integer grid equals the total of $\hat f$ over the same grid. The engine underneath is the periodization identity. Wrapping $f$ around the unit circle gives a 1-periodic function $F$ , and expanding $F$ in a Fourier series shows its coefficients are precisely $\hat f$ evaluated at the integers. PSF is then just the $x=0$ slice of this equation.

F(x)\equiv\sum_{n\in\mathbb{Z}}f(x+n)=\sum_{k\in\mathbb{Z}}\hat f(k)\,e^{2\pi i k x} \qquad (2)

where: $F$ — the 1-periodic periodization of $f$ ; $x$ — the continuous variable (PSF is the case $x=0$ ); the coefficients $\hat f(k)$ are the transform sampled at integers. Read it as: the periodized function and its Fourier series are two names for the same object. The most compact encoding of the whole formula is distributional. The Dirac comb — the Shah distribution — is a sum of unit deltas on the integer lattice, and its Fourier transform is again a Dirac comb. That self-duality is Poisson summation, dressed as an operator identity rather than a numerical one.

\mathrm{III}(x)=\sum_{n\in\mathbb{Z}}\delta(x-n),\qquad \widehat{\mathrm{III}}=\mathrm{III} \qquad (3)

where: $\mathrm{III}$ — the Dirac comb / Shah distribution, a train of unit deltas at the integers; $\delta$ — the Dirac delta. Read it as: the comb is a fixed point of the Fourier transform. Pairing both sides against a test function $f$ recovers equation (1) instantly. On a general lattice the symmetry acquires a geometric factor. If $\Lambda=B\mathbb{Z}^d$ is the lattice generated by a basis matrix $B$ , its dual (reciprocal) lattice is $\Lambda^*=B^{-\mathsf{T}}\mathbb{Z}^d$ , and the covolume $V=|\det B|$ of the fundamental cell appears as $1/V$ on the dual side.

\sum_{x\in\Lambda}f(x)=\frac{1}{|\det B|}\sum_{y\in\Lambda^*}\hat f(y) \qquad (4)

where: $\Lambda$ — the direct lattice $B\mathbb{Z}^d$ ; $\Lambda^*$ — the dual (reciprocal) lattice $B^{-\mathsf{T}}\mathbb{Z}^d$ ; $|\det B|=V$ — the covolume of the unit cell. Read it as: a dense lattice in direct space has a sparse dual, weighted by the inverse cell volume — the exact statement that Bragg peaks live at reciprocal-lattice points. The one-dimensional special case that does all the everyday work is the scaled form. Sampling at spacing $a$ in direct space forces spacing $1/a$ in dual space, together with an overall prefactor $1/a$ — the numerical dial that trades a slow sum for a fast one.

\sum_{n\in\mathbb{Z}}f(na)=\frac{1}{a}\sum_{k\in\mathbb{Z}}\hat f\!\left(\frac{k}{a}\right) \qquad (5)

where: $a$ — the direct-space sampling spacing; $k/a$ — the dual-space sampling at spacing $1/a$ ; $1/a$ — the amplitude prefactor from the dilation rule. Read it as: shrink the grid by $a$ and the dual grid stretches by $1/a$ . Feeding a Gaussian through the scaled formula yields the crown jewel: the modular transformation of the Jacobi theta function, which turns a slow series for small $t$ into a fast one and is the analytic seed of the Riemann zeta functional equation.

\theta(t)=\sum_{n\in\mathbb{Z}}e^{-\pi t n^2}=\frac{1}{\sqrt t}\sum_{k\in\mathbb{Z}}e^{-\pi k^2/t}=t^{-1/2}\,\theta(1/t),\qquad t>0 \qquad (6)

where: $t$ — a positive parameter; small $t$ makes the direct sum decay slowly while the dual sum $e^{-\pi k^2/t}$ decays furiously. Read it as: the theta function relates its value at $t$ to its value at $1/t$ — a self-similarity across the fixed point $t=1$ .

Derivation APoisson summation from periodization and Fourier series

The classical route builds the formula in three honest moves: periodize $f$ to make a function on the circle, expand that periodic function in a Fourier series, and evaluate at the origin. The only analytic input is that the periodization converges to a continuous function whose Fourier series converges pointwise at $x=0$ , which holds comfortably for any Schwartz function because Schwartz decay makes the wrapped sum absolutely and uniformly convergent. The payoff is that the two great tools of Fourier analysis — series for periodic functions, transforms for functions on the line — are shown to be the same tool viewed from two sides.

Step 1. Form the periodization $F(x)=\sum_{n\in\mathbb{Z}}f(x+n)$ . For $f$ Schwartz this sum converges absolutely and uniformly, so $F$ is a well-defined smooth function, and shifting $x\mapsto x+1$ merely reindexes the sum, giving $F(x+1)=F(x)$ . Why: wrapping the line onto a circle of circumference one is what converts a transform problem into a series problem — periodicity is manufactured, not assumed.

Step 2. Because $F$ is 1-periodic and smooth it has a convergent Fourier series $F(x)=\sum_k c_k e^{2\pi i k x}$ with $c_k=\int_0^1 F(x)e^{-2\pi i k x}\,\mathrm dx$ . Substitute the definition of $F$ , swap sum and integral (justified by uniform convergence), and set $u=x+n$ in the $n$ -th term. Why: the coefficient integral over one period, once the sum is unfolded, tiles the individual translated pieces back into a single integral over all of $\mathbb{R}$ .

Step 3. Since $e^{-2\pi i k n}=1$ for integer $k,n$ , the translated intervals reassemble exactly, giving $c_k=\int_{-\infty}^{\infty}f(u)e^{-2\pi i k u}\,\mathrm du=\hat f(k)$ . The periodization's Fourier coefficients are therefore the transform sampled at the integers, which is equation (2). Why: this is the crux — the integer phase factor is what lets the pieces of the periodized integral glue into the full-line Fourier transform.

Step 4. Evaluate both representations at $x=0$ : the direct side is $F(0)=\sum_n f(n)$ and the series side is $\sum_k c_k=\sum_k \hat f(k)$ , so $\sum_n f(n)=\sum_k \hat f(k)$ . If instead $f$ jumps at an integer, the Fourier series at that point converges to the average of the one-sided limits, giving the midpoint correction $\tfrac12[f(n^+)+f(n^-)]$ . Why: the single evaluation point $x=0$ is where the two sums are forced to agree, and the Dirichlet-averaging caveat is exactly why decay and smoothness are not optional. *The coefficients $c_k=\hat f(k)$ are standard, and the identity $\sum_n 1/(n^2+1)=\pi\coth\pi=3.15335$ confirms the result numerically.*

Go deeper · Derivation B · 1 minThe Dirac comb is its own Fourier transform

Derivation BThe Dirac comb is its own Fourier transform

The distributional proof compresses the entire formula into a single operator identity, $\widehat{\mathrm{III}}=\mathrm{III}$ , and then reads Poisson summation off by pairing the comb against a test function. Working in the space of tempered distributions is what makes this rigorous: the comb is not a function but a perfectly legitimate tempered distribution, and the Fourier transform is defined on that space by duality, so every manipulation below is licensed. The advantage over the classical route is that it isolates why the formula is true — it is the self-duality of a single object — and simultaneously pins down the exact function class for which it holds.

Step 1. Treat the comb $\mathrm{III}(x)=\sum_n\delta(x-n)$ as a 1-periodic tempered distribution; like any such object it has a distributional Fourier series $\mathrm{III}(x)=\sum_k a_k e^{2\pi i k x}$ with constant coefficients $a_k$ . Why: periodicity guarantees a Fourier-series representation even when the object is a train of spikes rather than an honest function.

Step 2. Compute the coefficients over one period, $a_k=\int_{-1/2}^{1/2}\delta(x)\,e^{-2\pi i k x}\,\mathrm dx=e^{0}=1$ , since only the delta at $n=0$ lies inside $[-\tfrac12,\tfrac12]$ . Hence $\mathrm{III}(x)=\sum_k e^{2\pi i k x}$ . Why: the sifting property of the delta collapses each coefficient integral to the value of the exponential at the origin, which is one for every $k$ .

Step 3. Fourier-transform term by term. The transform of a pure frequency is a shifted delta, $\widehat{e^{2\pi i k x}}=\delta(\xi-k)$ , so $\widehat{\mathrm{III}}(\xi)=\sum_k\delta(\xi-k)=\mathrm{III}(\xi)$ . The comb is a fixed point of the Fourier transform. Why: summing the transforms of the individual frequencies rebuilds a comb of deltas at the integers — the self-duality (3).

Step 4. Pair both sides against a Schwartz test function $f$ using the defining relation $\langle \hat T,f\rangle=\langle T,\hat f\rangle$ . Then $\langle\widehat{\mathrm{III}},f\rangle=\langle\mathrm{III},f\rangle=\sum_n f(n)$ while $\langle\mathrm{III},\hat f\rangle=\sum_k \hat f(k)$ , and because $\widehat{\mathrm{III}}=\mathrm{III}$ these are equal: $\sum_n f(n)=\sum_k \hat f(k)$ . Why: the transform's duality moves the hat from the comb onto the test function, and the comb's self-duality forces the two pairings — hence the two sums — to coincide. *The key results $a_k=1$ and $\widehat{e^{2\pi i k x}}=\delta(\xi-k)$ are standard tempered-distribution computations.*

Go deeper · Derivation C · 1 minThe theta transformation from a Gaussian

Derivation CThe theta transformation from a Gaussian

Poisson summation becomes a concrete miracle the moment a Gaussian is fed through the scaled form, because the Gaussian is the one function whose Fourier transform is another Gaussian. The result is the modular transformation $\theta(t)=t^{-1/2}\theta(1/t)$ of the Jacobi theta function, an identity that is both a spectacular convergence accelerator for numerics and the analytic seed from which Riemann grew the functional equation of the zeta function. The derivation needs only the Gaussian transform and one substitution into equation (1).

Step 1. Take $f(x)=e^{-\pi t x^2}$ . In the symmetric convention its Fourier transform is the Gaussian $\hat f(\xi)=t^{-1/2}e^{-\pi\xi^2/t}$ — the width inverts and a prefactor $t^{-1/2}$ appears. Why: the self-Fourier character of the Gaussian is precisely what will let both sides of Poisson summation be theta series, closing the identity onto itself.

Step 2. Substitute into $\sum_n f(n)=\sum_k\hat f(k)$ to get $\sum_n e^{-\pi t n^2}=t^{-1/2}\sum_k e^{-\pi k^2/t}$ . Both sides are theta series; the left has decay rate set by $t$ , the right by $1/t$ . Why: the single substitution converts an abstract summation identity into a transformation law relating the function at $t$ and at $1/t$ .

Step 3. Recognize the two sides as $\theta(t)$ and $t^{-1/2}\theta(1/t)$ , giving $\theta(t)=t^{-1/2}\theta(1/t)$ , with the fixed point $t=1$ where $\theta(1)$ is self-consistent. For small $t$ the direct series decays slowly, but the dual series decays like $e^{-\pi k^2/t}$ , so a single term is essentially exact — for $t=0.01$ one dual term fixes $\theta(0.01)=10$ to over a hundred digits. Why: this reciprocity is the practical heart of PSF, and analytically continued it delivers $\xi(s)=\xi(1-s)$ for the completed zeta function. *This is the standard Jacobi transformation. Numerically, $\theta(0.01)=10.000$ : the direct sum needs ~57 terms for ten digits, while the $k=0$ dual term already gives it, with the next term being $10\,e^{-100\pi}\approx 4\times10^{-136}$ .*

Limiting cases

How the formula specializes, and where care is mandatory:

Case	Result	Meaning
Gaussian self-duality	$\sum_{n}e^{-\pi t n^2}=t^{-1/2}\sum_{k}e^{-\pi k^2/t}$	The theta modular relation; both sides converge, but for small $t$ the dual side is exponentially faster — the acceleration engine.
General spacing $a$	$\sum_{n}f(na)=\tfrac{1}{a}\sum_{k}\hat f\!\left(\tfrac{k}{a}\right)$	Direct spacing $a$ forces dual spacing $1/a$ and prefactor $1/a$ ; as $a\to0$ the left sum times $a$ becomes $\int f=\hat f(0)$ , the Riemann-sum limit.
Slow decay or a jump	$f\notin$ Schwartz $\Rightarrow$ midpoint rule	At a lattice-point discontinuity the periodized series converges to $\tfrac12[f(n^+)+f(n^-)]$ ; without decay the identity can fail pointwise.
Angular-frequency convention	$\sum_n f(n)=\tfrac{1}{2\pi}\sum_k\tilde f(2\pi k)$	Using $\tilde f(k)=\int f\,e^{-ikx}\,\mathrm dx$ moves the dual sampling to $2\pi k$ and introduces $1/2\pi$ ; keep conventions consistent.
High-dimensional lattice	$\sum_{x\in\Lambda}f(x)=\tfrac1V\sum_{y\in\Lambda^*}\hat f(y)$	The covolume $V=\|\det B\|$ weights the dual sum — the inverse-cell-volume scaling of reciprocal space in crystallography.

Common trapThe formula is not an unconditional identity

It is tempting to write $\sum_n f(n)=\sum_k\hat f(k)$ for any function in sight, but the equality genuinely requires both decay and smoothness, and it is convention-dependent. Two failure modes bite in practice. First, slow decay: a function that only falls off like $1/x^2$ can leave the periodized Fourier series conditionally convergent or divergent, so the naive equality gives a wrong or ambiguous number. Second, a jump exactly on the lattice: if $f$ is discontinuous at an integer, the periodized series converges to the average of the one-sided limits, so you must replace $f(n)$ by $\tfrac12[f(n^+)+f(n^-)]$ — forget this and you are off by a finite amount. And always fix your Fourier convention first: the clean factor-free form holds only in the symmetric ( $2\pi$ -in-the-exponent) convention; the angular-frequency convention sprinkles factors of $2\pi$ and changes the dual spacing to $2\pi k$ .

Worked examples

1 · Summing 1/(n²+1) in closed form (fully worked)

Evaluate the infinite sum $\sum_{n=-\infty}^{\infty}\dfrac{1}{n^2+1}$ using Poisson summation.

Step 1. Choose $f(x)=1/(x^2+1)$ , so the target is $\sum_n f(n)$ . Its Fourier transform in the symmetric convention is the two-sided exponential $\hat f(\xi)=\pi\,e^{-2\pi|\xi|}$ (the standard Lorentzian-to-exponential pair, here with unit width). This decays exponentially, so the dual sum will be a geometric series — the whole point.

Step 2. Apply PSF, $\sum_n f(n)=\sum_k\hat f(k)$ , and split off $k=0$ : $\sum_k \pi e^{-2\pi|k|}=\pi\big[1+2\sum_{k=1}^{\infty}e^{-2\pi k}\big]$ . The tail is a geometric series with ratio $e^{-2\pi}=0.00186744$ , summing to $e^{-2\pi}/(1-e^{-2\pi})=0.00187093$ .

Step 3. Assemble: $\pi\big[1+2(0.00187093)\big]=\pi\cdot 1.00374187=\mathbf{3.15335}$ . In closed form the geometric series is exactly $(1+e^{-2\pi})/(1-e^{-2\pi})=\coth\pi$ , so the result is the classical identity $\sum_{n}\frac{1}{n^2+1}=\pi\coth\pi$ .

Step 4. Sanity-check the direct side: $1+2\big(\tfrac12+\tfrac15+\tfrac1{10}+\tfrac1{17}+\cdots\big)=3.15335$ , matching term for term but converging only like $1/n^2$ — dozens of terms for a few digits, versus the two-term dual computation above. The sum equals π coth π = 3.15335, and PSF turned an algebraically slow series into a one-line geometric sum.

2 · Accelerating the theta function (fully worked)

Compute $\theta(0.01)=\sum_{n}e^{-\pi(0.01)n^2}$ to high accuracy, and count how many terms each side needs.

Step 1. Direct series. With $t=0.01$ the general term is $e^{-0.0314159\,n^2}$ , which falls off slowly: $n=0$ gives $1$ , $n=\pm5$ gives $0.456$ , $n=\pm10$ gives $0.043$ , $n=\pm20$ gives $3.4\times10^{-6}$ . The term first drops below $10^{-10}$ at $n=28$ , so reaching ten-digit accuracy needs about 57 terms. Summing them gives $\theta(0.01)=\mathbf{10.000000\ldots}$ .

Step 2. Dual series. Equation (6) gives $\theta(0.01)=t^{-1/2}\sum_k e^{-\pi k^2/t}=10\sum_k e^{-100\pi k^2}$ . The $k=0$ term is exactly $10$ ; the $k=\pm1$ terms are $10\,e^{-100\pi}=10\times3.65\times10^{-137}=3.65\times10^{-136}$ .

Step 3. So a single dual term delivers $\theta(0.01)=10$ correct to about 135 digits, while the direct series needed dozens for ten. The reciprocity $t\leftrightarrow 1/t$ has moved essentially all the "work" onto the fast-decaying side. One dual term beats ~57 direct terms; the exact value is θ(0.01)=10 to 135 digits. This is exactly the trick Ewald summation uses for lattice Coulomb sums and that theta-based algorithms use to evaluate the Riemann zeta function.

Physics of the portrait

Poisson summation carries a working mathematician's name onto a formula that has since become indispensable to physicists — the reciprocal lattice, the sampling theorem, and Ewald's method are all its children. The portrait belongs to the man who first wrote the identity down in his studies of heat and infinite series.

Origins · 1 minSiméon Denis Poisson and how this was discovered

Siméon Denis Poisson

1781–1840 · French mathematical physics and analysis

Siméon Denis Poisson was one of the towering figures of French mathematical physics in the generation after Laplace and Lagrange, and his name is attached to an astonishing spread of results — the Poisson equation of potential theory, the Poisson distribution, the Poisson bracket of analytical mechanics, and the summation formula treated here. A student of Laplace and Lagrange at the École Polytechnique, where he became a professor at twenty-one, Poisson worked at the interface of analysis and physics his whole career, on electricity and magnetism, elasticity, celestial mechanics, and above all the theory of heat. It was in that last setting, alongside his running rivalry with Fourier over the foundations of harmonic analysis in the 1810s and 1820s, that the summation formula first appeared: a tool for converting a slowly convergent series over a lattice into a rapidly convergent series over its dual. Poisson would not have phrased it in terms of Fourier transforms and dual lattices — that language came later — but the essential reciprocity, that a sum and its transformed counterpart are equal, is his. The distributional reading as the self-duality of the Dirac comb waited more than a century for Laurent Schwartz's theory of distributions to make the train of deltas rigorous, but the identity itself is a direct descendant of Poisson's work on heat and series — which is why his name now sits over the reciprocal lattice of every crystallographer and the sampling theorem of every signal engineer.

History

1807

— Fourier presents his memoir on the propagation of heat, founding harmonic analysis

1827

— Poisson derives the summation formula in his work on heat conduction and series

1829

— Jacobi's Fundamenta Nova establishes the theta-function transformation θ(t)=t⁻¹ᐟ²θ(1/t)

1859

— Riemann uses the theta relation to prove the functional equation of ζ(s)

1950

— Schwartz's theory of distributions makes the comb identity III-hat = III rigorous

Going further

Texts. Stone & Goldbart, Mathematics for Physics, ch. on Fourier transforms and distributions · Arfken & Weber, Mathematical Methods for Physicists, Fourier-transform chapter · Hassani, Mathematical Physics, distributions and Fourier analysis · Stein & Shakarchi, Fourier Analysis: An Introduction, ch. 5 · Grafakos, Classical Fourier Analysis, ch. 3 · Byron & Fuller, Mathematics of Classical and Quantum Physics, transform methods. The key relations used throughout this article are: (1) the periodization coefficients $c_k=\hat f(k)$ ; (2) the comb coefficients $a_k=1$ and $\widehat{e^{2\pi ikx}}=\delta(\xi-k)$ , hence $\widehat{\mathrm{III}}=\mathrm{III}$ ; (3) the Gaussian pair $\widehat{e^{-\pi tx^2}}=t^{-1/2}e^{-\pi\xi^2/t}$ and the theta relation $\theta(t)=t^{-1/2}\theta(1/t)$ ; (4) the dilation rule $\widehat{f(a\cdot)}=\tfrac1a\hat f(\cdot/a)$ ; (5) the closed forms $\sum_n 1/(n^2+1)=\pi\coth\pi=3.15335$ and $\theta(0.01)=10$ .

Prerequisites. Fourier Series and Periodic Functions · The Fourier Transform · The Dirac Delta Function · Convergence of Fourier Series. Leads to. Tempered Distributions and Fourier Transforms · Theory of Distributions · Parseval's and Plancherel's Theorems · The Convolution Theorem.

Ask the tutor: Want me to walk through applying Poisson summation to your own series — picking $f$ , computing $\hat f$ , and choosing the side that converges fastest?

Is the Poisson summation formula the same as Parseval's theorem?

No, though they are cousins. Parseval (Plancherel) says the total energy is convention-invariant: $\int|f|^2=\int|\hat f|^2$ , equating two integrals of squared magnitudes. Poisson summation equates two sums of samples, $\sum_n f(n)=\sum_k\hat f(k)$ , with no squaring. The deeper link is that both follow from unitarity of the Fourier transform — Parseval pairs $f$ with itself, PSF pairs $f$ with the Dirac comb. But you cannot derive one from the other by inspection: PSF is about lattice sampling, which is why it, not Parseval, controls aliasing and reciprocal lattices.

Why does the Dirac comb turn out to be its own Fourier transform?

Because it is the unique (up to scale) structure that is invariant under both translation by the lattice and modulation by the dual lattice, and the Fourier transform swaps those two symmetries. Concretely: the comb is 1-periodic, so it has a Fourier series; computing the coefficients with the delta's sifting property gives every coefficient equal to one, so the comb equals $\sum_k e^{2\pi i k x}$ , a sum of pure frequencies at the integers. Transforming each pure frequency gives a delta at that integer, rebuilding the comb. The self-duality is really the statement that "sampling" (multiplying by a comb) and "periodizing" (convolving with a comb) are Fourier-dual operations — the entire content of PSF.

How is Poisson summation an instance of the convolution theorem?

Periodizing $f$ is exactly convolving it with the Dirac comb: $F=f*\mathrm{III}$ , since convolving with a delta at $n$ shifts $f$ by $n$ and summing over $n$ produces $\sum_n f(x+n)$ . The convolution theorem says the transform of a convolution is the product of transforms, so $\hat F=\hat f\cdot\widehat{\mathrm{III}}=\hat f\cdot\mathrm{III}$ — that is, $\hat f$ sampled at the integers. Those samples are precisely the Fourier coefficients of the periodic $F$ , which is the periodization identity (2). So PSF is the convolution theorem applied to the one distribution, the comb, that is its own transform; the whole formula is "convolve-equals-multiply" specialized to the lattice.

Do the conventions really matter, or is that just pedantry?

They matter, and getting them wrong changes the answer by factors of $2\pi$ . In the symmetric convention $\hat f(\xi)=\int f e^{-2\pi i x\xi}\,\mathrm dx$ , the integers are self-dual and PSF is factor-free. Switch to the physicist's angular-frequency convention $\tilde f(k)=\int f e^{-ikx}\,\mathrm dx$ and two things change: the dual sampling moves from the integers to $2\pi k$ , and an overall $1/2\pi$ appears, so the formula becomes $\sum_n f(n)=\tfrac1{2\pi}\sum_k \tilde f(2\pi k)$ . The safest habit is to write your transform definition at the top of any calculation and carry the $2\pi$ 's explicitly — most "PSF gave the wrong answer" bugs are really convention mismatches.

Why does the formula fail for a function that decays only like 1/x²?

Slow decay undermines the two convergence facts the derivation quietly uses. First, the periodization $F(x)=\sum_n f(x+n)$ may converge only conditionally, so it is not obviously a nice continuous periodic function with a well-behaved Fourier series. Second, the dual sum $\sum_k\hat f(k)$ may itself converge slowly or conditionally, and rearranging or truncating it is then unsafe. For Schwartz functions both sides converge absolutely and everything is airtight; for merely $1/x^2$ -type decay you are at the boundary where the equality can still hold but only under extra hypotheses (continuity, bounded variation, careful symmetric summation). The honest statement is that PSF holds for a specific function class, and $1/x^2$ decay sits right at its ragged edge.

What is the reciprocal (dual) lattice, and why does the covolume appear?

For a lattice $\Lambda=B\mathbb{Z}^d$ generated by a basis matrix $B$ , the dual lattice $\Lambda^*=B^{-\mathsf T}\mathbb{Z}^d$ is the set of vectors whose inner product with every $\Lambda$ -vector is an integer — physically, the wavevectors that "fit" the periodicity. A dense direct lattice (small covolume $V=|\det B|$ ) has a sparse dual: their cell volumes are reciprocal. The factor $1/V$ in equation (4) is the bookkeeping of that reciprocity — the same $1/V$ that scales reciprocal space in solid-state physics. Crystallographers meet PSF as the statement that a lattice in real space diffracts into its dual lattice in momentum space, weighted by exactly this inverse cell volume.