Poisson Summation Formula

The Poisson summation formula relates samples of a function on a real-space lattice to samples of its Fourier transform on the reciprocal lattice. It is the mathematical bridge behind many “sum over modes versus sum over images” identities in quantum mechanics.

In one dimension, with lattice spacing $a>0$, the formula says

$$\sum_{n\in\mathbb Z} f(na) = \frac{1}{a} \sum_{m\in\mathbb Z} F \left( \frac{2\pi m}{a} \right),$$

where the Fourier transform convention is

$$F(k) = \int_{-\infty}^{\infty} f(x)e^{-ikx}\,dx.$$

The formula is exact for sufficiently nice functions, such as Schwartz functions, and extends distributionally to many important kernels.

Why It Matters in Quantum Mechanics

Poisson summation appears whenever a calculation has both periodicity and Fourier analysis:

• finite periodic boxes versus continuum limits;
• position-space image sums versus momentum-space mode sums;
• reciprocal lattices in crystals;
• diffraction peaks from periodic arrays;
• theta-function identities for Gaussian sums;
• semiclassical trace formulas and periodic-orbit calculations;
• numerical aliasing and sampling diagnostics.

The formula is not merely a trick for evaluating sums. It says that a lattice in one representation becomes a reciprocal lattice in the conjugate representation.

One-Dimensional Statement

Using the ordinary $k$-space convention

$$F(k) = \int_{-\infty}^{\infty} f(x)e^{-ikx}\,dx, \qquad f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(k)e^{ikx}\,dk,$$

the Poisson summation formula for spacing $a$ is

$$\sum_{n\in\mathbb Z} f(na) = \frac{1}{a} \sum_{m\in\mathbb Z} F(G_m), \qquad G_m=\frac{2\pi m}{a}.$$

The numbers $G_m$ are the reciprocal-lattice wave numbers because

$$e^{iG_m na} = e^{2\pi i mn} = 1$$

for all integers $m,n$.

Dirac Comb Form

The distributional heart of the formula is the Dirac comb identity

$$\sum_{n\in\mathbb Z} \delta(x-na) = \frac{1}{a} \sum_{m\in\mathbb Z} e^{iG_m x}, \qquad G_m=\frac{2\pi m}{a}.$$

Pairing both sides with a test function $f$ gives

$$\sum_{n\in\mathbb Z} f(na) = \frac{1}{a} \sum_{m\in\mathbb Z} \int_{-\infty}^{\infty} f(x)e^{iG_m x}\,dx.$$

Since $F(k)=\int f(x)e^{-ikx}\,dx$, the right-hand side is

$$\frac{1}{a} \sum_{m\in\mathbb Z} F(-G_m).$$

As $m$ ranges over all integers, $-G_m$ ranges over the same reciprocal lattice, so this is the same as the standard formula. This distributional form also explains why the formula belongs with Delta Function and Distributions.

Derivation from a Periodic Sum

Define the periodic image sum

$$P(x) = \sum_{n\in\mathbb Z} f(x+na).$$

It satisfies $P(x+a)=P(x)$. Therefore it has a Fourier series

$$P(x) = \sum_{m\in\mathbb Z} c_m e^{iG_m x}, \qquad G_m=\frac{2\pi m}{a}.$$

The coefficients are

$$c_m = \frac{1}{a} \int_0^a P(x)e^{-iG_m x}\,dx.$$

Substituting the image sum and unfolding the integral over all translated cells gives

$$\begin{aligned} c_m &= \frac{1}{a} \sum_{n\in\mathbb Z} \int_0^a f(x+na)e^{-iG_m x}\,dx \\ &= \frac{1}{a} \int_{-\infty}^{\infty} f(u)e^{-iG_m u}\,du \\ &= \frac{1}{a}F(G_m). \end{aligned}$$

Thus

$$P(x) = \frac{1}{a} \sum_{m\in\mathbb Z} F(G_m)e^{iG_m x}.$$

Setting $x=0$ gives the Poisson summation formula.

Gaussian Example

Let

$$f(x)=e^{-\pi t x^2}, \qquad t>0.$$

With the convention used here,

$$F(k) = \frac{1}{\sqrt t} e^{-k^2/(4\pi t)}.$$

For $a=1$, Poisson summation gives

$$\sum_{n\in\mathbb Z} e^{-\pi t n^2} = \frac{1}{\sqrt t} \sum_{m\in\mathbb Z} e^{-\pi m^2/t}.$$

This identity is the modular transformation of the basic theta-function sum. In physics it relates a narrow real-space Gaussian sum to a broad reciprocal-space Gaussian sum, a pattern that reappears in finite-temperature and periodic-boundary calculations.

Lattices and Reciprocal Lattices

In $d$ dimensions, let $\Lambda$ be a Bravais lattice generated by primitive vectors $\mathbf a_1,\ldots,\mathbf a_d$:

$$\Lambda = \left\{ \mathbf R = \sum_{j=1}^d n_j\mathbf a_j : n_j\in\mathbb Z \right\}.$$

The reciprocal lattice is

$$\Lambda^* = \left\{ \mathbf G: e^{i\mathbf G\cdot\mathbf R}=1 \ \text{for every}\ \mathbf R\in\Lambda \right\}.$$

Equivalently, if $\mathbf b_i$ are reciprocal primitive vectors, then

$$\mathbf a_i\cdot\mathbf b_j = 2\pi\delta_{ij}.$$

Let $V_\Lambda$ be the primitive-cell volume:

$$V_\Lambda = \left\lvert \det(\mathbf a_1,\ldots,\mathbf a_d) \right\rvert.$$

For a sufficiently nice function $f:\mathbb R^d\to\mathbb C$, define

$$\widehat f(\mathbf k) = \int_{\mathbb R^d} f(\mathbf x) e^{-i\mathbf k\cdot\mathbf x} d^d x.$$

The $d$-dimensional Poisson summation formula is

$$\sum_{\mathbf R\in\Lambda} f(\mathbf R) = \frac{1}{V_\Lambda} \sum_{\mathbf G\in\Lambda^*} \widehat f(\mathbf G).$$

The associated lattice-comb identity is

$$\sum_{\mathbf R\in\Lambda} \delta(\mathbf x-\mathbf R) = \frac{1}{V_\Lambda} \sum_{\mathbf G\in\Lambda^*} e^{i\mathbf G\cdot\mathbf x}.$$

This is the distributional source of reciprocal-lattice peaks.

Periodic Boxes and Momentum Sums

For a particle in a one-dimensional periodic box of length $L$, the allowed wave numbers are

$$k_n = \frac{2\pi n}{L}.$$

Poisson summation is one way to move between:

• a sum over periodic images in position space;
• a sum over discrete momentum modes in reciprocal space.

For example, periodicizing a kernel by images,

$$K_L(x) = \sum_{n\in\mathbb Z} K(x+nL),$$

produces a Fourier series whose wave numbers are exactly $2\pi m/L$. This is the same finite-volume logic used in Periodic Functions and Fourier Series, but Poisson summation emphasizes the duality between the image lattice and the momentum lattice.

Bloch Theory Connection

For a periodic potential,

$$V(\mathbf x+\mathbf R)=V(\mathbf x), \qquad \mathbf R\in\Lambda,$$

the Fourier series contains only reciprocal-lattice wave vectors:

$$V(\mathbf x) = \sum_{\mathbf G\in\Lambda^*} V_{\mathbf G}e^{i\mathbf G\cdot\mathbf x}.$$

This is the Fourier-analysis input behind Bloch theory. A Bloch wave has the form

$$\psi_{n\mathbf k}(\mathbf x) = e^{i\mathbf k\cdot\mathbf x} u_{n\mathbf k}(\mathbf x),$$

where

$$u_{n\mathbf k}(\mathbf x+\mathbf R) = u_{n\mathbf k}(\mathbf x).$$

The periodic factor $u_{n\mathbf k}$ has reciprocal-lattice Fourier components. As a result, a periodic Hamiltonian couples plane-wave momenta that differ by reciprocal-lattice vectors:

$$\mathbf k \longleftrightarrow \mathbf k+\mathbf G.$$

Detailed band theory belongs with quantum matter, but the reciprocal-lattice algebra starts here.

Sampling and Aliasing View

Poisson summation also explains aliasing. Sampling a function in real space multiplies it by a Dirac comb. In Fourier space, multiplication becomes convolution, so the transform is replicated at reciprocal-lattice spacings. If the copies overlap, distinct Fourier components become indistinguishable after sampling.

This is the continuum version of the Nyquist warning in numerical work: a real-space grid with spacing $a$ cannot distinguish momenta that differ by $2\pi/a$ without additional information.

Common Mistakes

• Mixing Fourier conventions and losing factors of $2\pi$.
• Forgetting the factor $1/a$ in one dimension or $1/V_\Lambda$ in $d$ dimensions.
• Applying the formula to slowly decaying functions without convergence or distributional care.
• Confusing the direct lattice spacing $a$ with reciprocal spacing $2\pi/a$.
• Treating reciprocal-lattice vectors as ordinary momenta rather than wave-vector shifts; physical momentum is $\hbar\mathbf k$.
• Omitting degeneracy or branch information when reducing a multidimensional lattice problem to a one-dimensional notation.

Cross-Links

• Periodic Functions and Fourier Series
• Fourier Transform
• Crystalline Symmetry Preview
• Delta Function
• Distributions
• Convolution
• Momentum Representation
• Plancherel and Parseval Theorems
• Fourier Transform Table

References

• G. B. Folland, Fourier Analysis and Its Applications, American Mathematical Society, 1992.
• E. M. Stein and R. Shakarchi, Fourier Analysis: An Introduction, Princeton University Press, 2003.
• G. B. Arfken, H. J. Weber, and F. E. Harris, Mathematical Methods for Physicists, 7th ed., Academic Press, 2013.
• N. W. Ashcroft and N. D. Mermin, Solid State Physics, Holt, Rinehart and Winston, 1976.
• C. Kittel, Introduction to Solid State Physics, 8th ed., Wiley, 2004.

Exercises

1. Starting from the periodic image sum

$$P(x) = \sum_{n\in\mathbb Z} f(x+na),$$

derive the Fourier coefficient $c_m=F(G_m)/a$.

Solution

Since $P$ is periodic with period $a$,

$$c_m = \frac{1}{a} \int_0^a P(x)e^{-iG_m x}\,dx.$$

Substitute the image sum:

$$c_m = \frac{1}{a} \sum_{n\in\mathbb Z} \int_0^a f(x+na)e^{-iG_m x}\,dx.$$

Let $u=x+na$. Since $e^{-iG_m na}=1$, the translated integrals unfold to the real line:

$$c_m = \frac{1}{a} \int_{-\infty}^{\infty} f(u)e^{-iG_m u}\,du = \frac{1}{a}F(G_m).$$

2. Use Poisson summation with $a=1$ to prove

$$\sum_{n\in\mathbb Z} e^{-\pi t n^2} = \frac{1}{\sqrt t} \sum_{m\in\mathbb Z} e^{-\pi m^2/t}, \qquad t>0.$$

Solution

For $f(x)=e^{-\pi t x^2}$,

$$F(k) = \frac{1}{\sqrt t} e^{-k^2/(4\pi t)}.$$

With $a=1$, the reciprocal points are $G_m=2\pi m$. Thus

$$F(2\pi m) = \frac{1}{\sqrt t} e^{-(2\pi m)^2/(4\pi t)} = \frac{1}{\sqrt t} e^{-\pi m^2/t}.$$

Poisson summation gives the stated identity.

3. For a one-dimensional lattice with spacing $a$, show that the reciprocal-lattice wave numbers are $G_m=2\pi m/a$.

Solution

The defining condition is

$$e^{iG_m na}=1$$

for every integer $n$. This holds exactly when $G_m a=2\pi m$ for some integer $m$. Therefore

$$G_m=\frac{2\pi m}{a}.$$

4. Explain why a periodic potential couples momenta differing by reciprocal-lattice vectors.

Solution

Write the periodic potential as

$$V(\mathbf x) = \sum_{\mathbf G} V_{\mathbf G}e^{i\mathbf G\cdot\mathbf x}.$$

Acting on a plane wave gives

$$V_{\mathbf G}e^{i\mathbf G\cdot\mathbf x} e^{i\mathbf k\cdot\mathbf x} = V_{\mathbf G} e^{i(\mathbf k+\mathbf G)\cdot\mathbf x}.$$

Thus the potential connects the plane-wave component $\mathbf k$ to $\mathbf k+\mathbf G$.

5. What happens in Fourier space when a function is sampled on a grid with spacing $a$?

Solution

Sampling multiplies the function by a real-space Dirac comb. Fourier transformation turns multiplication into convolution, and the Fourier transform of the comb is another comb with spacing $2\pi/a$. Therefore the transform is replicated at reciprocal-lattice shifts. If those copies overlap, aliasing occurs.