The Reciprocal Lattice

Introduction

Atoms occupy positions in ordinary, or direct, space. Diffraction spots are easier to understand in a related space whose axes measure inverse distance. This reciprocal lattice is not a second physical crystal; it is a mathematical map of the crystal's periodicities. A wide spacing between real-space planes gives a short reciprocal vector, while tightly spaced planes give a long one. Its nodes index which scattering directions can interfere constructively throughout an ideal crystal.

Core explanation

Let the primitive direct vectors be a, b and c, with cell volume V = a · (b × c). Using a crystallographic convention without factors of 2π, define a = (b × c)/V, b = (c × a)/V and c = (a × b)/V. These satisfy a · a = 1, a · b = 0, and the corresponding cyclic relations. Every reciprocal-lattice vector can be written H hkl = h a + k b + l c , with integer Miller indices h, k and l. The units of these vectors are inverse length. In a physics wavevector convention, one instead defines G hkl = 2πH hkl; this distinction must be checked before using a formula from another text.

The vector H hkl is normal to the family of (hkl) planes. For the first-order plane family described by those indices, its magnitude is 1/d hkl in the crystallographic convention, where d hkl is the interplanar spacing. The IUCr Online Dictionary of Crystallography states the node, plane-normal and spacing relationship. For a cubic direct cell with edge length a, a , b and c are orthogonal and have magnitude 1/a. Hence 1/d hkl² = (h² + k² + l²)/a², or d hkl = a/√(h²+k²+l²). That simple formula does not apply unchanged to noncubic cells; use the reciprocal metric for oblique axes.

Why does this mathematical lattice govern diffraction? Radiation scattered by atoms in cells separated by direct translation R accumulates a relative phase. Constructive interference from all equivalent cells requires that phase to be an integer multiple of 2π for every R. With the physical scattering vector q = k out − k in, this is q · R = 2π integer, satisfied when q = G hkl = 2πH hkl. This is the Laue condition. The geometric Ewald construction draws a sphere representing the allowed elastic wavevectors; a reciprocal-lattice node lying on its surface gives a possible reflection for the crystal orientation and incident wavelength. The IUCr educational treatment develops this relation from phase matching.

Meeting the node condition tells where a reflection can occur, not how bright it will be. Atoms within the basis scatter with different phases and amplitudes. Their sum is the structure factor , and it can cancel for some indexed reflections even when the translation condition is geometrically met. Centring, glide planes and screw axes can lead to systematic absences. Distinguishing reciprocal-lattice geometry from structure-factor intensity is vital when interpreting a powder pattern: peak positions primarily report cell geometry, while intensities also encode the basis, scattering factors, thermal motion and texture.

Finite crystals and disorder broaden ideal reciprocal points. A small crystallite gives broader diffraction peaks; microstrain can also broaden peaks. These effects do not make the reciprocal-lattice construction useless. Instead, the ideal nodes provide reference positions around which real scattering intensity is distributed. In electron-band theory, k-space uses the same inverse-length geometry to classify electronic Bloch states, although k values are continuous within a large crystal's allowed mesh and are often reduced to a Brillouin zone.

Step-by-step reasoning

1. State whether reciprocal vectors include 2π. 2. Obtain a , b and c from the direct-cell vectors. 3. Form H hkl with the chosen integer indices. 4. Use its direction for the plane normal and its magnitude for reciprocal plane spacing. 5. For diffraction, apply q = 2πH hkl and then ask whether the basis gives nonzero intensity.

Visual explanation

Draw parallel real-space planes a distance d apart at the left. To the right draw an arrow perpendicular to them whose length is 1/d, ending at a reciprocal-space node. A second set of more closely spaced real-space planes gets a longer arrow. Add an Ewald sphere passing through one node but missing another to show why only certain reflections appear at a fixed crystal orientation.

Real-world analogy

A musical beat pattern can be described either by the distance between successive beats or by the number of beats per second. The reciprocal lattice resembles the frequency description of spatial repetition. Shorter real-space intervals mean larger spatial frequencies. Diffraction selects spatial frequencies that match the scattering geometry.

Real-world example

If a cubic solid has lattice parameter a = 0.400 nm, its (100) plane spacing is 0.400 nm and its (110) spacing is 0.400/√2 = 0.283 nm. Their reciprocal-vector magnitudes in the no-2π convention are 2.50 nm⁻¹ and 3.54 nm⁻¹ respectively. Thus the (110) reciprocal node lies farther from the origin than the (100) node, consistent with the smaller plane spacing.

Why?

Why is the reciprocal vector perpendicular to the planes? Moving within a plane should not change the phase label h x + k y + l z of that plane. The gradient of that label points perpendicular to its equal-phase surfaces, and it is precisely the reciprocal-vector direction.

Common misconception

"A reciprocal-lattice node is a real atom." It is a vector representing a spatial periodicity. Several different atomic bases can share the same translation lattice and therefore node positions, while producing very different diffraction intensities.

Worked example

Question: A cubic crystal has a = 0.300 nm. Find the spacing and magnitude of the crystallographic reciprocal vector for (111), then compare it with (100).

Reasoning: For cubic axes, d 111 = a/√(1²+1²+1²) = 0.300/√3 = 0.173 nm. Therefore H 111 = 1/d 111 = 5.77 nm⁻¹. For (100), d 100 = a = 0.300 nm and H 100 = 3.33 nm⁻¹. The (111) planes are closer, so their reciprocal vector is longer. If a physics convention is used, multiply both vector magnitudes by 2π, but do not multiply the plane spacings.

Answer: d 111 ≈ 0.173 nm and H 111 ≈ 5.77 nm⁻¹, versus 0.300 nm and 3.33 nm⁻¹ for (100).

Quick check

1. What are the units of a reciprocal-lattice vector? Answer: Inverse length. If a plane spacing doubles, the magnitude of its first-order crystallographic reciprocal vector halves.

Exam focus

Write the convention before applying H = 1/d or G = 2π/d. Distinguish diffraction position from diffraction intensity. For cubic systems use d hkl = a/√(h²+k²+l²), while avoiding that shortcut for oblique cells.

Advanced insight

The reciprocal basis is determined by the inverse transpose of the matrix whose columns are the direct basis vectors. That algebra explains why changing the chosen direct unit cell also changes the coordinates of reciprocal nodes. It does not change the physical scattering vectors. Crystallographic index transformations therefore require careful paired transformations of real and reciprocal bases.

Summary

Reciprocal vectors encode the normal direction and inverse spacing of plane families. Their integer nodes specify phase-matched diffraction directions, while the atomic basis determines intensity through the structure factor. The crystallographic reciprocal basis uses 1/d; the physics wavevector basis includes 2π. Checking that convention prevents many quantitative errors.

Practice questions

1. What does H hkl point normal to? Answer: The family of direct-lattice planes labelled (hkl). 2. Does a permitted reciprocal node guarantee a bright diffraction peak? Answer: No; the structure factor can cancel or weaken it. 3. How is q related to H in the convention used here? Answer: q = 2πH for an allowed elastic reflection. 4. For a cubic cell, which has larger reciprocal-vector magnitude, (100) or (200)? Answer: (200) is twice as large, corresponding to half the plane spacing under the indexed-family convention.