The Theory of X-ray Diffraction — Bragg’s Law, the Reciprocal Lattice, the Structure Factor, and What Determines the Peak Positions

— From the interplanar spacing d_hkl to Bragg’s law, the reciprocal lattice, the Laue condition, the structure factor and the systematic absences, without skipping intermediate steps (for second- and third-year undergraduates)

A powder X-ray diffraction (XRD) pattern is a simple picture: 2θ on the horizontal axis, intensity on the vertical axis, and just a few bars standing up. Yet this picture contains three kinds of information, cleanly separated — the peak positions carry the shape and size of the lattice, the peak heights carry the contents of the box (which elements sit where), and the peak widths carry the crystallite size and the strain. In lectures this division of labor tends to be handed down as “that’s just how it is,” but in fact it can all be proved properly just by writing out sums of the exponential function exp to the very end. The mathematics we use is only the inner and cross products of vectors, the sum of a geometric series, and $e^{i\pi n}={(-1)}^n$.

1. Introduction — What Can X-ray Diffraction Tell Us?
2. Miller Indices and the Interplanar Spacing d_hkl
3. Derivation of Bragg’s Law
4. The Reciprocal Lattice — Why Think in Reciprocal Space?
5. The Laue Condition — Saying the Same Thing as Bragg’s Law
6. The Ewald Sphere
7. The Structure Factor F_hkl
8. Deriving the Systematic Absences by Hand
9. Why Do the Peak Positions Coincide for the Same Bravais Lattice? [The Core]
10. Other Factors That Determine the Intensity
11. The Scherrer Equation and the Peak Width
12. From the Powder Method to Rietveld Analysis
13. For Further Study
14. Exercises and Solutions
15. Appendix A. What the Simulator Shows

1. Introduction — What Can X-ray Diffraction Tell Us?

A crystal is a space completely filled, without gaps, by boxes of the same shape (unit cells). This period is only 0.2–1 nm, that is, about 2–10 Å. To “see” it, we have to shine on it waves whose wavelength is about the same as this period. The wavelength of visible light is 400–800 nm, about 1000 times too long to sense the period of a crystal. The wavelength of Cu Kα₁ radiation, the one most often used in the laboratory, is λ = 1.5405929 Å, which is of exactly the same order as the distance between atoms. That is why we use X rays.

X rays are scattered not by the nuclei but by the electrons (Thomson scattering). Each individual electron emits scattered waves in almost all directions, although with a direction-dependent strength (the polarization factor; §10.2), but in a crystal the scatterers are arranged regularly, so only in particular directions do the phases of the scattered waves line up exactly and reinforce each other. In all other directions the enormous number of waves cancel one another out to zero. What tells us these “aligned directions” is Bragg’s law, or equivalently the Laue condition.

There is a simulator that pairs with this explanation. If you get tired of following the equations, try moving the sliders in xrd-simulator-en.html yourself. At the end of each chapter we point to the corresponding tab. How to read the simulator screen, the meaning of the controls, and the approximations used in the calculations are collected in Appendix A.

2. Miller Indices and the Interplanar Spacing d_hkl

2.1 Definition of the Miller indices

To specify the orientation of a plane in a crystal, crystallography uses the Miller indices. When a plane cuts the three crystal axes at the points ${\mathbf{a}}_1/h$, ${\mathbf{a}}_2/k$, ${\mathbf{a}}_3/l$, we write that plane as (h k l). In other words, the indices are the reciprocals of the intercepts, listed in order. If h = 0, this means “the plane cuts the ${\mathbf{a}}_1$ axis at infinity,” that is, the plane is parallel to that axis. A negative index is written with an overbar, like $\bar{1}$.

2.2 The interplanar spacing — everything follows from a single equation

The symbol (h k l) stands not for a single plane but for a family of equally spaced parallel planes. Their spacing is the interplanar spacing (d-spacing) d_hkl. Textbooks list a different expression for each of the seven crystal systems, but they are all merely specializations of the following single equation. First we construct the metric tensor of the real lattice $\mathbf{G}$ (a symmetric 3×3 matrix).

Its inverse ${\mathbf{G}}^{*}={\mathbf{G}}^{-1}$ is called the reciprocal metric tensor. Then, for any crystal system, the interplanar spacing is given by the following equation.

If you write a program, this one equation is enough — there is no need to write separate expressions for each of the 7 crystal systems, and no bugs creep in. Treat the individual expressions below as “for calculating by hand” and “for reading off properties from the shape of the expression.”

Written in explicit form, the triclinic case is as follows. V is the volume of the unit cell.

2.3 Cubic crystals — the form used most often in this material

For cubic crystals the expression becomes dramatically simpler.

N is called the index sum. In a cubic crystal, reflections with the same N always fall on the same 2θ. For example, (300) and (221) both have N = 9, so in a powder pattern they are observed as a single peak — this is the first “accidental degeneracy” that occurs in cubic crystals (§10.3).

→ In xrd-simulator-en.html, tab ① Bragg’s condition, choose a substance and a plane (hkl) and see how, the larger N is, the more finely the planes are packed and the smaller d_hkl becomes (the numbers for three substances are in the table of §3.5).

3. Bragg’s Law — Derivation of 2d sinθ = nλ

3.1 Counting the path difference

Suppose that parallel planes with spacing d are stacked up. An X ray strikes the planes at an angle θ to them, and we consider the direction in which it leaves at the same angle θ (the direction of specular reflection). Compare the wave “reflected” by the upper plane with the wave “reflected” by the plane one below: the lower wave travels a longer path by the extra distance it has gone down. Let M be the scattering point on the upper plane and B the scattering point on the plane one below, and let A and C be the feet of the perpendiculars dropped from M onto the incident and diffracted rays through B, respectively. Since MB = d and the angle between MB and the incident direction is 90° − θ, the right triangles MAB and MCB give $\overline{AB}=\overline{BC}=d\sin \theta$ at once. The path difference is therefore

When this path difference is an integer multiple of the wavelength, the waves from the two planes are in phase and reinforce each other.

3.2 Why is it enough to consider only the “neighboring plane”?

This is where many students get stuck. A crystal contains tens of millions of planes, yet why may we write the condition using only two adjacent ones? The answer is simple. If the path difference between adjacent planes is nλ, then for planes j apart it is $jn\lambda$, which is also an integer multiple of the wavelength. In other words, if neighbors are in step, everything is automatically in step. So it is enough to write the condition for adjacent planes only.

3.3 Absorbing n into d — the crystallographic convention

The n in Eq. (6) can in fact be eliminated by changing the way we write it. Dividing both sides by n gives

Now, as Eq. (4) shows, $d_{hkl}/n=d_{nh,nk,nl}$ holds (for a cubic crystal, $a/\sqrt{n^2(h^2+k^2+l^2)}=d_{hkl}/n$. In a general crystal system the same thing happens because 1/d² is a quadratic form in h, k and l). Crystallography therefore adopts the convention of always fixing n = 1 and instead multiplying the indices by n.

In this convention, the second-order reflection of the (111) plane is read as the first-order reflection of (222), the third-order reflection of the (100) plane as the first-order reflection of (300), and so on. Indices that could be reduced by a common factor (such as (200), (222), (400)) are also treated as independent reflections. These unreduced indices are called Laue indices. From here on, this page uses the convention of Eq. (7).

3.4 There is a lower limit to the observable d

Solving Eq. (7) for sinθ gives $\sin \theta =\lambda /(2d)$, but a solution exists only if sinθ ≤ 1. Therefore

A plane spacing finer than half the wavelength can never be seen at that wavelength. This is the limit of resolution. With Cu Kα₁, planes finer than λ/2 = 0.77030 Å cannot be observed by any means. To see finer structure, use the short-wavelength Mo Kα₁ or Ag Kα₁.

3.5 Checking with three substances — d_hkl is fixed by the indices, and the peak height by the contents

Let us apply Eqs. (4) and (7) to the three cubic crystals that can be selected in simulator ①. Cesium chloride CsCl (Pm-3m, a = 4.209055 Å) and strontium titanate SrTiO₃ (Pm-3m, perovskite type, a = 3.945130 Å) are simple cubic P, and sodium chloride NaCl (Fm-3m, rock-salt type, a = 5.691694 Å) is face-centered cubic F. The atomic positions (fractional coordinates in the conventional unit cell) are as follows.

• CsCl: Cs (0, 0, 0), Cl (1/2, 1/2, 1/2).
• SrTiO₃: Sr (0, 0, 0), Ti (1/2, 1/2, 1/2), O (1/2, 1/2, 0), (0, 1/2, 1/2), (1/2, 0, 1/2).
• NaCl: Na (0, 0, 0), (1/2, 1/2, 0), (1/2, 0, 1/2), (0, 1/2, 1/2), Cl (0, 0, 1/2), (1/2, 0, 0), (0, 1/2, 0), (1/2, 1/2, 1/2).

The numbers in the table are rounded outputs computed from the structure data above (Reference 14) by the Mochizuki Research Group’s XRD calculation script (pymatgen’s XRDCalculator, Reference 3). The wavelength is Cu Kα₁ = 1.5405929 Å and the range is 2θ ≤ 120°; “Intensity” is the relative intensity with the strongest line in that range set to 100 (it includes the multiplicity m_hkl and the Lorentz-polarization factor but not the temperature factor; §10). The NaCl (110) line had no row in the output, so d and 2θ were calculated with Eqs. (4) and (7), and, since the structure factor is 0, its intensity is written as “0 (absent).” Reflections whose intensity in the output is 10⁻²⁹ or less (double-precision rounding error) are also written as “0 (absent).” The lattice constants are first-principles values, so they deviate slightly from the measured 2θ (see the box in Appendix A.6).

There are two things to read from the table. The first is that the interplanar spacing d_hkl is determined only by the lattice constant a and the indices (hkl). Within the same substance, the larger N = h² + k² + l² is, the smaller d is and the higher the angle at which 2θ appears. Changing the substance changes the ratios of d only through the ratio of a (in every column d₁₀₀ : d₁₁₀ : d₁₁₁ = 1 : 1/√2 : 1/√3). (200) and (222) are the second-order reflections of (100) and (111), respectively, rewritten in the convention of §3.3, and their d is exactly half.

The second is that whether a peak appears, and how high it is, is determined not by d but by the contents of the unit cell (§7, §8). For CsCl and SrTiO₃, which are simple cubic P, all ten reflections appear, whereas for NaCl, which is face-centered cubic F, the reflections (100), (110), (210), (211) and (310), whose indices h, k, l are of mixed parity, vanish (Eq. (33)). The contents also affect the heights. A rough estimate in terms of the numbers of electrons goes as follows.

• CsCl: for reflections with h+k+l odd, the waves scattered by Cs and Cl are in antiphase and F = f_Cs − f_Cl. Even so, the difference between Cs (55 electrons) and Cl (17 electrons) is large, so (100) stands out clearly at 36.5.
• SrTiO₃: for (100), F = f_Sr − f_Ti − f_O (38 − 22 − 8 = 8 electrons’ worth) almost cancels, so it is weak at 3.9, whereas for (110), F = f_Sr + f_Ti − f_O (52 electrons’ worth), and it becomes the strongest line.
• NaCl: (111) and (311), with all-odd indices, are weak with F = 4(f_Na − f_Cl), while (200), (220) and (222), with all-even indices, are strong with F = 4(f_Na + f_Cl) (see also the box in §9.4 and Exercise 9).

→ In xrd-simulator-en.html, tab ① Bragg’s condition, choose a substance and (hkl), vary θ, and watch the moment when the two waves come into phase and how the waves from the atomic layers reinforce or cancel one another. For Cu Kα₁, the 2θ_B and the relative intensity read off in ① are the same as the values in the table above (how to read the screen is explained in Appendix A.2).

4. The Reciprocal Lattice — Why Think in Reciprocal Space?

Bragg’s law is a view in which the “planes” play the leading role. But planes are awkward to count (which plane is (100) and which is (200), and so on), and they quickly become unmanageable in three dimensions. The idea of the reciprocal lattice is therefore to represent each whole family of planes by a single point. We replace “the family of parallel planes (hkl), extending to infinity” with “a single vector perpendicular to the planes, with a length proportional to 1/d_hkl.” Then the diffraction condition can be written as a single vector addition (§5).

4.1 Definition of the reciprocal-lattice vectors

From the primitive vectors of the real lattice ${\mathbf{a}}_1$, ${\mathbf{a}}_2$, ${\mathbf{a}}_3$ and the volume of the unit cell $V={\mathbf{a}}_1·({\mathbf{a}}_2\times {\mathbf{a}}_3)$ we construct the following three vectors.

The cross product ${\mathbf{a}}_2\times {\mathbf{a}}_3$ is perpendicular to both ${\mathbf{a}}_2$ and ${\mathbf{a}}_3$, so ${\mathbf{b}}_1·{\mathbf{a}}_2={\mathbf{b}}_1·{\mathbf{a}}_3=0$ holds. On the other hand, ${\mathbf{a}}_1·{\mathbf{b}}_1=2\pi ·{\mathbf{a}}_1·({\mathbf{a}}_2\times {\mathbf{a}}_3)/V=2\pi$ — because the scalar triple product is exactly V. In summary,

This duality relation is the tool for all the calculations below; we use nothing else. And we define the reciprocal-lattice vector corresponding to (hkl) as

As h, k, l run over all sets of integers, the tip of ${\mathbf{G}}_{hkl}$ forms a lattice in space. This is the reciprocal lattice.

4.2 ${\mathbf{G}}_{hkl}$ is perpendicular to the (hkl) planes

By the definition in §2.1, the (hkl) plane cuts the three axes at ${\mathbf{a}}_1/h$, ${\mathbf{a}}_2/k$, ${\mathbf{a}}_3/l$. Let us take two independent vectors lying in this plane. Connecting the intercepts, we obtain

Let us calculate the inner products of these with ${\mathbf{G}}_{hkl}$ using Eq. (10) alone.

By the same calculation, ${\mathbf{G}}_{hkl}·\mathbf{v}=2\pi -2\pi =0$ also holds. ${\mathbf{G}}_{hkl}$ is orthogonal to both of the two independent in-plane vectors, so it is perpendicular to the plane itself. □

4.3 Proof that $|{\mathbf{G}}_{hkl}|=2\pi /d_{hkl}$

d_hkl is the distance between “the plane through the origin” and “the plane next to it.” The neighboring plane passes through the point ${\mathbf{a}}_1/h$. By §4.2 the unit normal of the planes is $\hat{\mathbf{n}}=\mathbf{G}/|\mathbf{G}|$, so projecting the vector from the origin to that point onto the normal direction gives the distance.

All we used along the way was the single calculation ${\mathbf{a}}_1·{\mathbf{G}}_{hkl}=2\pi h$. Dividing by h makes the h disappear, and the result is the same whichever of h, k, l we choose — this is the beautiful part. When h = 0, choose one of h, k, l that is not zero (for example, if k ≠ 0, use ${\mathbf{a}}_2/k$) and carry out exactly the same calculation, which leads to the same conclusion $d=2\pi /|\mathbf{G}|$ (if h = k = l = 0, the (hkl) plane itself is not defined).

Now “a family of planes” and “a single reciprocal-lattice point” correspond one to one. The distance of the reciprocal-lattice point (hkl) from the origin is directly proportional to the reciprocal of the interplanar spacing (1/d_hkl in the crystallographic convention, 2π times that in the physics convention of this page) — this is all there is to the reciprocal lattice. Indeed, Eq. (2) is nothing but the square of the length of the reciprocal-lattice vector in the crystallographic convention $|h{\mathbf{a}}_1^{*}+k{\mathbf{a}}_2^{*}+l{\mathbf{a}}_3^{*}{|}^2$ (= ${|{\mathbf{G}}_{hkl}|}^2/{(2\pi )}^2$) written out in components.

→ In xrd-simulator-en.html, tab ② Reciprocal lattice and Ewald sphere, see how the planes of the real lattice and the reciprocal-lattice points move in step.

5. The Laue Condition — Saying the Same Thing as Bragg’s Law

5.1 Summing the scattering amplitudes over all atoms

Let the wave vector of the incident X ray be $\mathbf{k}$ ($|\mathbf{k}|=2\pi /\lambda$), and that after scattering be ${\mathbf{k}}^{\prime }$. The wave from a scatterer at position $\mathbf{r}$ is shifted in phase, relative to the scatterer at the origin, by $-\Delta \mathbf{k}·\mathbf{r}$. Here $\Delta \mathbf{k}\equiv {\mathbf{k}}^{\prime }-\mathbf{k}$ is called the scattering vector. The scattering amplitude from the whole crystal is

In a crystal the position of an atom can be decomposed into “lattice translation $\mathbf{R}$ + position within the unit cell,” so the sum also splits in two. The sum over lattice translations is a geometric series, and when there are $N$ lattice points (this N is not the index sum of Eq. (4))

This is because, for $\Delta \mathbf{k}=\mathbf{G}$, Eq. (10) together with $\mathbf{R}=n_1{\mathbf{a}}_1+n_2{\mathbf{a}}_2+n_3{\mathbf{a}}_3$ gives $\mathbf{G}·\mathbf{R}=2\pi (h{n}_1+k{n}_2+l{n}_3)$ = 2π × (an integer), so every exponential becomes 1 and the N terms simply add up. Otherwise the phases go round and round, and the enormous number of terms cancel one another.

Let us confirm this “cancellation” with the geometric series. Taking out just one direction and writing $x\equiv \Delta \mathbf{k}·{\mathbf{a}}_1$, the sum of the $N$ terms is $\sum _{n=0}^{N-1}{e}^{-inx}=\frac{e^{-iNx}-1}{e^{-ix}-1}$ and its squared absolute value (= a quantity proportional to the intensity) is ${\sin }^2(Nx/2)/{\sin }^2(x/2)$. The value reaches N² only where x = 2π × (an integer), and away from there it drops to a size negligible compared with N². Writing this sum out to the end gives Eq. (42) of §11.1 (the Laue interference function). See there for the exact form and for the peak width that follows from it (= the Scherrer equation). If the crystal is sufficiently large (N → ∞), this sum is effectively zero. Therefore

5.2 Deriving Bragg’s law from the Laue condition

X-ray scattering is elastic, so the energy does not change and $|{\mathbf{k}}^{\prime }|=|\mathbf{k}|$. We write Eq. (17) as ${\mathbf{k}}^{\prime }=\mathbf{k}+\mathbf{G}$ and square both sides.

The left-hand side is equal to ${|\mathbf{k}|}^2$ so subtracting it gives

Here, $\mathbf{G}$ is the normal to the (hkl) planes (§4.2), so if the angle between the incident beam and the planes is θ, $\mathbf{k}$ makes an angle of 90° + θ with the normal direction (because the incident beam is heading “down into” the planes). Therefore $\mathbf{k}·\mathbf{G}=|\mathbf{k}||\mathbf{G}|\cos (90°+\theta )=-|\mathbf{k}||\mathbf{G}|\sin \theta$ holds. Substituting this into Eq. (18) gives

Finally, substituting $|\mathbf{k}|=2\pi /\lambda$ and, from Eq. (14), $|\mathbf{G}|=2\pi /d$, we obtain

This is exactly Eq. (7), that is, Bragg’s law in the convention of n = 1. □ We have seen that the Laue condition and Bragg’s law say the same thing in different words. The nth-order reflection corresponds to ${\mathbf{G}}_{nh,nk,nl}=n{\mathbf{G}}_{hkl}$, another reciprocal-lattice point — the Laue condition is already in the form where n has been absorbed into the indices from the start.

5.3 A bonus — the diffraction condition is exactly the Brillouin-zone boundary

Let us rewrite Eq. (18). Dividing both sides by 4 gives

This is the condition that “the tip of $\mathbf{k}$ lies on the perpendicular bisector plane (Bragg plane) of the segment joining the origin and ±$\mathbf{G}$.” (If $\mathbf{G}$ is a reciprocal-lattice point, so is −$\mathbf{G}$, so the distinction of ± is not essential.) The smallest region enclosed by the perpendicular bisector planes of the reciprocal-lattice points is called the first Brillouin zone. That is,

→ In xrd-simulator-en.html, tab ② Reciprocal lattice and Ewald sphere, see how $\mathbf{k}$, ${\mathbf{k}}^{\prime }$ and $\mathbf{G}$ form a triangle.

6. The Ewald Sphere — Constructing the Condition for Diffraction

6.1 Steps of the construction

The standard way to draw this is in the crystallographic convention (without 2π, with radius 1/λ), so we explain it that way. If you want to draw it in the physics convention, multiply all the lengths by 2π.

1. Draw the reciprocal lattice on paper. Call the origin O.
2. Along the direction of the incident beam, draw a vector of length 1/λ whose end point is at O, and call its starting point C (C corresponds to the position of the crystal).
3. Draw a sphere centered at C with radius 1/λ. This is the Ewald sphere (reflection sphere). By construction, this sphere always passes through O.
4. When a reciprocal-lattice point P lies on the surface of the Ewald sphere, the reflection (hkl) is observed. The direction of the diffracted beam is C → P.

6.2 Why this is the same as Bragg’s law — one triangle is enough

O and P both lie on the sphere with center C and radius 1/λ. So the triangle C–O–P is isosceles. The chord OP has length $|\mathbf{g}|=1/d$. Writing half of the apex angle ∠OCP as θ, the right triangle obtained by cutting the isosceles triangle in half gives

This one line is all there is to the Ewald sphere. And from this triangle we can also read off the relation $s=\sin \theta /\lambda =1/(2d)=|\mathbf{g}|/2$. This s becomes the argument of the atomic scattering factor in the next chapter.

6.3 In a single crystal almost no reflections are visible

Suppose that we shine X rays of a fixed wavelength on a fixed single crystal. The reciprocal-lattice points are discrete points scattered through space, while the Ewald sphere is a two-dimensional surface. The probability that a point happens to lie on that surface is almost zero. Therefore

• Laue method: vary λ continuously (white X rays). The Ewald sphere becomes a “shell” of spheres with different radii.
• Rotating-crystal method and four-circle diffractometer: rotate the crystal to bring reciprocal-lattice points onto the sphere.

6.4 Why does a powder show everything at once?

A powder sample is a collection of microcrystals pointing in every orientation. Consider a reflection (hkl). Since the microcrystals are oriented in all directions, the reciprocal-lattice vector ${\mathbf{g}}_{hkl}$ can also point in any direction. That is, in reciprocal space it is not “a single point” but a spherical shell centered at the origin O with radius 1/d_hkl.

This spherical shell (center O, radius 1/d) and the Ewald sphere (center C, radius 1/λ) always intersect in a circle if they intersect at all. Two spheres intersect when the distance between their centers lies between the sum and the difference of the two radii. Here the distance between the centers equals the radius of the Ewald sphere, 1/λ, so the sum condition 1/λ ≤ 1/d + 1/λ holds automatically. What remains is the difference condition |1/d − 1/λ| ≤ 1/λ. It always holds if 1/d < 1/λ, and if 1/d ≥ 1/λ it reads 1/d − 1/λ ≤ 1/λ, so, altogether,

— this is exactly Eq. (8) of §3.4. Therefore

→ In xrd-simulator-en.html, tab ② Reciprocal lattice and Ewald sphere, see how a peak appears at the moment a reciprocal-lattice point crosses the sphere.

7. The Structure Factor F_hkl — What Determines the Peak Heights

7.1 The atomic scattering factor f

X rays are scattered by the electrons in an atom. One atom has Z electrons spread over a radius of about 1 Å, so there are phase differences among the waves scattered by the individual electrons as well. The atomic scattering factor (also called the atomic form factor) f is “the scattering amplitude from the whole electron cloud of that atom divided by the scattering amplitude from one free electron,” and it is the Fourier transform of the electron density $n(\mathbf{r})$. (Here the sign of the exponent is taken as + to match the usual crystallographic convention. For a spherically symmetric n(r), f is real, so the value is the same whichever sign is used. See also the note in §7.2.)

If the electron cloud is spherically symmetric, f is independent of the direction of $\mathbf{g}$ and is a function only of $s=|\mathbf{g}|/2=\sin \theta /\lambda$. It is enough to remember just three properties.

• $f(0)=Z$ — in forward scattering (θ = 0) all the electrons scatter in phase, so the total number of electrons comes out as is.
• f decreases monotonically as s increases — because at higher angles the phase shifts within the electron cloud matter more. This is the main reason why high-angle peaks become weak.
• The heavier the atom, the larger f — which is why hydrogen is hardly visible to X rays (and why neutron diffraction is used as a complement).

When actually computing numbers, we use a table in which f(s) is approximated by four Gaussian functions. This material and the simulator adopt the following expression and the same coefficients as pymatgen.

(The unit of s is Å⁻¹. Note that f(0) = Z holds exactly.)

7.2 Derivation of the structure factor F_hkl

Write the position of the j-th atom in the unit cell, in fractional coordinates, as ${\mathbf{r}}_j=x_j{\mathbf{a}}_1+y_j{\mathbf{a}}_2+z_j{\mathbf{a}}_3$ (0 ≤ x, y, z < 1). The phase difference relative to the atom at the origin is ${\phi }_j=\Delta \mathbf{k}·{\mathbf{r}}_j=\mathbf{G}·{\mathbf{r}}_j$ (since we consider positions where the Laue condition is satisfied, $\Delta \mathbf{k}=\mathbf{G}$). Substituting Eq. (11) and the above ${\mathbf{r}}_j$ into this, and using ${\mathbf{a}}_i·{\mathbf{b}}_j=2\pi {\delta }_{ij}$ from Eq. (10), all the cross terms vanish, giving

This one line is the source of every systematic absence that follows. It is the heart of the derivation; note that the lattice constants a, b, c and the angles appear nowhere in it. The scattering amplitude from one unit cell is therefore

This is called the structure factor. The sum runs over all the atoms in the unit cell. The observed intensity is

7.3 Separation into a lattice part and a basis part — the separation theorem

Note: The name “separation theorem” is peculiar to this page. It is not standard terminology, so a search will not find it, and it will not be understood if you use it in other courses. The same content is stated as Theorem 3.3 in Chapter 3 of the Mochizuki Research Group’s Introduction to Solid State Physics (in Japanese).

Here begins the core of this material. A crystal is made of “Bravais lattice + basis.” If we use the conventional cell, the lattice includes the centering translations. Let us write the set of these as $\{{\mathbf{t}}_1,\dots ,{\mathbf{t}}_M\}$ (M = 1 for P, 2 for I and C, 3 for R (hexagonal axes), and 4 for F). Let the collection of atoms attached to one lattice point (the basis) be $\{{\mathbf{r}}_1,\dots ,{\mathbf{r}}_n\}$. Then, by the very definition of the Bravais lattice and the basis, the positions of all the atoms in the unit cell are

and this representation is unique. Moreover, the basis attached to every ${\mathbf{t}}_m$ is exactly the same — because that is the very definition of a Bravais lattice. Substitute this into Eq. (26). By Eq. (25), $\mathbf{G}·{\mathbf{r}}_j=2\pi (h{x}_j+k{y}_j+l{z}_j)$, so 2π is already contained in $\mathbf{G}·\mathbf{r}$ — below we write the exponent as $i\mathbf{G}·\mathbf{r}$ (do not multiply by 2π twice). Since the exponent is a sum, the exponential function splits into a product of exponentials.

The step from the second line to the third is the key. The summand of $\sum _m$ does not contain j, and the summand of $\sum _j$ does not contain m, so the double sum simply factorizes into a product ($\sum \sum AB=(\sum A)(\sum B)$). We therefore write

Here the fractional coordinates of ${\mathbf{t}}_m$ are denoted by $(u_m,v_m,w_m)$. The scattering power has been split completely into the product of a “lattice factor” and a “contents factor.” Two conclusions follow immediately.

→ In xrd-simulator-en.html, tab ③ Structure factor and systematic absences, let us see how the vectors cancel one another on the complex plane.

8. Deriving the Systematic Absences by Hand

8.1 The simple lattice P

The only centering translation is (0, 0, 0) (M = 1), so

It is always 1. There are no absences from the lattice — the Bravais-lattice side allows every (hkl). (There are separate absences that come from the basis. See ② of §8.5 and §8.6.) For example, in the CsCl-type structure (a = 4.123 Å), no (hkl) vanishes — 100, 110, 111, 200, 210, 211, 220, 300/221, 310, 311, … and so on (N = 7 is skipped not because of a systematic absence but because there is no set of integers with h²+k²+l² = 7. N = 15, 23, 28 and so on do not appear for the same reason).

8.2 The body-centered lattice I

There are two centering translations, (0, 0, 0) and (½, ½, ½) (M = 2).

Rule: the reflection is absent when h + k + l is odd. Physically, the planes made by the body-centered atoms lie exactly halfway between the original planes, so the waves scattered from them differ in phase by π (antiphase) from the waves from the corner atoms, and the two cancel. If h + k + l is even, then N = h² + k² + l² is also even (and vice versa), so in bcc only the reflections with even N appear: 110, 200, 211, 220, 310, 222, 321, 400, …

8.3 The face-centered lattice F

There are four centering translations: (0,0,0), (0,½,½), (½,0,½) and (½,½,0) (M = 4).

We write out all four cases. Permuting which of the indices are even and which are odd is symmetric, so these cases are exhaustive.

Rule: the reflection is absent when the parities of h, k, l are mixed (allowed only when they are all the same). The allowed values of N are 3(111), 4(200), 8(220), 11(311), 12(222), 16(400), 19(331), 20(420), 24(422), 27(511/333), 32(440), 35(531), 36(442/600), …, and N = 1, 2, 5, 6, 7, 9, 10 do not appear (7 alone is missing not because of an absence rule of the F lattice but because no such set of indices exists in the first place).

8.4 The base-centered lattice C

For the C lattice, in which the ab face is centered, the centering translations are (0,0,0) and (½, ½, 0).

Rule: absent when h + k is odd (l is irrelevant). By the same calculation, for A-centering (bc face, (0,½,½)) the reflection is absent when k + l is odd, and for B-centering (ac face, (½,0,½)) it is absent when h + l is odd. C-centering affects only one direction, so its hallmark is that the absence rule does not depend on l. Thanks to this feature, the presence or absence of C-centering can be judged from an indexed powder pattern. The monoclinic C lattice (C2/c and so on) is the lattice that occurs most often after monoclinic P and triclinic P.

Rhombohedral (R) centering follows exactly the same procedure. In hexagonal axes the centering translations are (0,0,0), (⅔,⅓,⅓) and (⅓,⅔,⅔), three in all (M = 3), and, setting $t\equiv -h+k+l$, we have $S_{\text{lattice}}=1+{\omega }^t+{\omega }^{2t}$ (where $\omega =e^{2\pi i/3}$). The sum of the cube roots of unity is 3 when t is a multiple of 3 and 0 otherwise, so the reflection is absent when −h+k+l is not a multiple of 3 (this is exactly the same tool as the calculation with $\omega$ used for hcp in §8.6). Al₂O₃ (corundum, R-3c) and calcite have this lattice. For details, see Section 2.6 of Chapter 2 of Introduction to Solid State Physics (in Japanese).

8.5 The diamond structure — lattice and basis working together

The diamond structure consists of “Bravais lattice = cubic F” plus “basis = two atoms of the same element {(0,0,0), (¼,¼,¼)}.” Using the separation theorem (30), the lattice factor is already obtained in §8.3: it is Eq. (33) as it stands. All that remains is the basis factor.

Here we used $e^{2\pi i/4}=e^{i\pi /2}=i$. Setting $n\equiv h+k+l$, $i^n$ is determined by n mod 4 alone. For the F-lattice condition only two cases remain, “all odd” and “all even,” so we examine each. The numerical column at the end of the table below is |F_hkl|/f after multiplying by the F-lattice factor S_lattice = 4.

Let us check with Si (a = 5.43102 Å, Cu Kα₁).

8.6 hcp — absences that come from the basis alone

Hexagonal close packing (hcp) has the hexagonal-P Bravais lattice (no centering, $S_{\text{lattice}}=1$), and the basis is two atoms of the same element, {(0,0,0), (⅓, ⅔, ½)}. All the systematic absences therefore come entirely from the basis.

$\omega$ is a primitive cube root of unity, with ${\omega }^0=1$, ${\omega }^{\pm 1}=-1/2\pm i\sqrt{3}/2$. We also used $e^{2\pi i·l/2}={(-1)}^l$. Now we write out all four cases.

|F| is largest for the reflections with h + 2k a multiple of 3 and l even ((0002), $(11\bar{2}0)$, and so on) (the measured intensity is further multiplied by the multiplicity and the Lorentz–polarization factor (§10), so these are not necessarily the strongest lines. In Mg, $(10\bar{1}1)$ is the strongest line).

→ In xrd-simulator-en.html, tab ③ Structure factor and systematic absences, let us actually watch the vectors cancel one another.

9. Why Do the Peak Positions Coincide for the Same Bravais Lattice? [The Core]

9.1 The claim — from the lecture slides of the Mochizuki Research Group

This claim can be proved using only the equations built up so far. However, one condition has to be added explicitly: the lattice constants must also be the same. Without it, the claim is plainly wrong (even for the same fcc, the peaks move if a differs).

9.2 Proof

Take the relation obtained midway through §5.2, $|\mathbf{G}|=2|\mathbf{k}|\sin \theta$, solve it for sinθ, and substitute $|\mathbf{k}|=2\pi /\lambda$ into it.

Look at the right-hand side. It contains only λ (a measurement condition) and $|\mathbf{G}|$. Moreover, for $\mathbf{G}=h{\mathbf{b}}_1+k{\mathbf{b}}_2+l{\mathbf{b}}_3$, the ${\mathbf{b}}_i$ are, by Eq. (9), built from ${\mathbf{a}}_i$ alone. What determines ${\mathbf{a}}_i$ is only the type of the Bravais lattice and the lattice constants. The basis (which element sits where) does not appear anywhere.

On the other hand, which (hkl) are “allowed” is determined by whether, in Eq. (30), $S_{\text{lattice}}$ is zero or not, and, as we saw in §8, this too is determined by the centering translations, that is, by the Bravais lattice alone. And the intensity is ${|S_{\text{lattice}}{F}_{\text{basis}}|}^2$, in which the whole basis enters. Therefore

9.3 Numerical evidence — reproducing slide 1-5

Let us actually compute. We calculate powder XRD patterns keeping the lattice constants fixed at a common value and swapping only the contents of the box (Cu Kα₁, λ = 1.5405929 Å).

Group A: Bravais lattice = cubic F. Fixed at a = 5.64000 Å.

All 14 lines agree completely, to the fifth decimal place. The four structures differ in atomic arrangement and in space group, and, depending on the structure, in the kind and number of atoms (the numbers of atoms per conventional cell are 8 for the rock-salt type, 8 for the zinc-blende type, 12 for the fluorite type and 24 for SrSO₄), yet the peaks stand at the same positions without a single exception. Only the heights differ. For example, (111) is only 8.08 in the rock-salt type but 100.00 (the strongest line) in the zinc-blende type.

Group B: Bravais lattice = cubic P. Fixed at a = 4.20000 Å.

9.4 The precise statement — “the positions where peaks can appear”

Look closely at the table for Group B. The CsCl type and the anti-ReO₃ type agree in all 20 lines, but only Ag₂O is missing six lines (100, 210, 300, 320, 322, 421). Is this a counterexample to the claim? No. In the space group Pn-3m of Ag₂O, just as for hcp in §8.6, additional basis-derived absences are at work. They are of two kinds: 100, 210, 300, 320, 410 are systematic absences of the n glide itself (0kl is absent when k+l is odd, and h00 is absent when h is odd), while 221 and 322, which are superposed at the same 2θ as other reflections, and 421 on its own are absences that come from O (Wyckoff 2a) and Ag (4c) occupying special positions (if atoms were placed at general positions of Pn-3m, these three would not vanish). Either way, they follow from the same equation once the structure factor is written down.

Writing out the structure factor, we find that all six missing lines are reflections with exactly one odd index. In the conventional cell of Ag₂O, O sits at (0,0,0) and (½,½,½), so the O contribution is $1+{(-1)}^{h+k+l}$, and it is 0 when h+k+l is odd; Ag sits at (¼,¼,¾), (¾,¼,¼), (¼,¾,¼) and (¾,¾,¾), and when exactly one index is odd the phase sum is $i-i+i-i=0$. Both vanish at the same time, which is why the reflection disappears.

9.5 Bravais lattice → reciprocal lattice → Brillouin zone

As we saw in §5.3, the diffraction condition is the same equation as the condition for the boundary of the Brillouin zone. And the Brillouin zone is “the Wigner–Seitz cell of reciprocal space,” so it is determined by the reciprocal lattice alone. The reciprocal lattice, in turn, is determined by the real lattice alone through Eq. (9). That is,

This single path is the content of the statement on slide 1-6, “The Bravais lattice determines the Brillouin zone.” The symbols Γ–X–L–W–K line up in the electronic band diagram of Si because the Bravais lattice of Si is cubic F — not because it has the diamond structure. NaCl, which is also cubic F, has a Brillouin zone of exactly the same shape (a truncated octahedron). The story of diffraction and the story of bands take place on the same stage: the reciprocal lattice.

→ In xrd-simulator-en.html, tab ④ The Bravais lattice sets the peak positions, switch the contents while keeping the lattice constants fixed, and see with your own eyes that the positions of the bars do not move.

10. Other Factors That Determine the Intensity

10.1 The overall form

C is a scale factor that applies to the whole measurement (it is easily confused with the $s=\sin \theta /\lambda$ that we have used since §2, so this page writes it as C. Note that it is often written as s in the literature).

10.2 The Lorentz factor and the polarization factor

The breakdown is as follows. Note: be careful about the constant 1/2. Multiplying the L and P below as they stand gives $LP=(1+{\cos }^{2}2\theta )/(2{\sin }^2\theta \cos \theta )$, which is exactly half of Eq. (40). In the literature (and in pymatgen’s implementation) there is a convention of absorbing this 1/2 into the scale factor C of Eq. (39) and dropping the 2 in the denominator, and Eq. (40) adopts that convention. Since only relative intensities matter, the conclusion is the same either way.

• Polarization factor $P=(1+{\cos }^{2}2\theta )/2$ — the angular dependence of Thomson scattering. For an unpolarized incident X ray, the electric-field component in the scattering plane is scattered by a factor of cos 2θ and the component perpendicular to the plane by a factor of 1. Intensity is the square of the amplitude, so on average it becomes (1 + cos² 2θ)/2.
• Lorentz factor $L=1/({\sin }^2\theta \cos \theta )$ — a geometric factor. It combines two effects: how the Ewald sphere intersects the reciprocal-lattice sphere (= the angular width over which the diffraction condition is satisfied), and the fact that the radius of the Debye–Scherrer ring changes with angle.

10.3 Multiplicity m_hkl

In a powder, symmetry-equivalent planes (for example (100), (010), (001), (−100), (0−10), (00−1)) all overlap at the same 2θ. The number of such planes is the multiplicity (multiplicity factor). You do not need to memorize how to count; the following formula gives it.

(For cubic crystals of point group m-3m. Checked by brute-force counting over the 48 signed permutations.)

10.4 Temperature factor and absorption

Atoms vibrate thermally, so their positions are not completely fixed. As a result, high-angle reflections become even weaker. This is the Debye–Waller factor (temperature factor), which is multiplied onto the f_j of each atom. $⟨u^2⟩$ is the mean-square atomic displacement.

→ In xrd-simulator-en.html, tab ⑤ Powder XRD pattern, let us compute a pattern from an actual crystal structure.

11. The Scherrer Equation and the Peak Width

11.1 The interference function — adding up a finite number of planes

In §3.2 we assumed that “the planes are stacked infinitely” and that the waves reinforce one another only when the condition is met exactly. A real crystallite is finite and has only N planes of the (hkl) family (here too N is the number of planes, not the index sum). Put the phase of the wave scattered from plane j (j = 0, 1, …, N−1) as ${\phi }_j=2\pi j\xi$, $\xi \equiv 2d\sin \theta /\lambda$ (the Bragg condition is satisfied when ξ is an integer). Since the sum of the amplitudes is a geometric series,

This is called the Laue interference function. It makes a sharp peak of height N² wherever ξ is an integer, and the larger N is, the sharper it becomes. As N → ∞ it becomes a δ function and returns to the Laue condition of §5.

11.2 Finding the width

Near $\xi =n$ (an integer), putting $\epsilon \equiv \xi -n$, for |ε| ≪ 1 we can approximate $\sin (\pi \epsilon )\approx \pi \epsilon$, which gives

This is a ${\text{sinc}}^2$-type function. The half maximum occurs at the point where $\sin (x)/x=1/\sqrt{2}$, and solving numerically gives x = 1.3915574. The full width at half maximum in ξ is therefore $\Delta \xi =2\times 1.3915574/(N\pi )=0.885893/N$. We convert this to 2θ. Differentiating $\xi =2d\sin \theta /\lambda$ with respect to θ gives $d\xi =(2d\cos \theta /\lambda )d\theta$, so, writing the crystallite thickness as $D=Nd$, and the width in 2θ as $\beta =2\Delta \theta$, we get

Here K = 0.885893 (for plate-like or columnar crystallites, with D taken as the thickness along the plane normal) — the conventional value 0.9 is a rounding of this number.

11.3 The Williamson–Hall method — separating size and strain

There are two causes of peak broadening.

1. Crystallite size: ${\beta }_{\text{size}}=K\lambda /(D\cos \theta )$ — depends only on cosθ.
2. Inhomogeneous strain: differentiating Bragg’s law gives $\Delta d/d=-\cot \theta \Delta \theta$, so, if the width of the strain distribution is $\epsilon =\Delta d/d$ (a different quantity from the $\epsilon \equiv \xi -n$ used in §11.2), we have ${\beta }_{\text{strain}}=4\epsilon \tan \theta$ — depends on tanθ.

Assuming that the two simply add as Lorentzian profiles, put $\beta ={\beta }_{\text{size}}+{\beta }_{\text{strain}}$ and multiply both sides by cosθ to obtain

Plotting 4 sinθ on the horizontal axis and β cosθ (with β in radians) on the vertical axis gives a straight line, from whose intercept the crystallite size D = Kλ/intercept and from whose slope the microstrain ε can be read off. There is also a convention of taking sinθ on the horizontal axis, in which case the slope is 4ε — state clearly on the axis of the figure which convention you use.

11.4 Peak height and area — why smaller crystallites give lower peaks

The Scherrer equation was about the width, but what happens to the height when the width changes? The answer is: “the area (integrated intensity) is constant, independent of the crystallite size D, and the height falls in inverse proportion to the width.” We look at the reason in two steps.

(1) One crystallite. The interference function of Eq. (42) has a peak height of N² and a full width at half maximum of 0.885893/N. Its area (the integral over one period) is proportional to N, as a rough estimate of height × width suggests, and moreover is given exactly by

This is because, expanding the sum of Eq. (42) as ${|A|}^2={\sum }_j{\sum }_{j^{\prime }}{e}^{2\pi i(j-j^{\prime })\xi }$ and integrating over one period, all the terms with j ≠ j′ vanish, and only the N terms with j = j′ remain, each contributing 1. In other words, the area of the diffraction from one crystallite is proportional to the number of planes it contains (in three dimensions, the number of unit cells, that is, the volume). The peak height N² is a consequence of “the waves from all the planes add in phase”; it is the product of the area becoming N times larger and the width narrowing to 1/N.

(2) The same amount of powder. Suppose we make the crystallites smaller while keeping constant the volume V_sample of the powder sample irradiated by X rays. If the volume of one crystallite is V₁, the number of crystallites is V_sample/V₁. Separate crystallites have random positions and orientations, so the phases of their scattered waves are not aligned, and it is the intensities (areas), not the amplitudes, that add. The area of the whole sample is therefore

The area is thus independent of D. On the other hand, by Eq. (43) the width broadens as β ∝ 1/D. Recalling “integral breadth = area ÷ height” from the table in §11.2, we have height = area ÷ integral breadth, so when the size-induced width dominates, the peak height decreases in proportion to D (reducing D to 1/10 lowers the height to about 1/10, too). When D becomes large and the width approaches the instrumental width, the width cannot narrow any further, so the height levels off as well.

The calculated patterns from VESTA and pymatgen are normally displayed with the strongest line scaled to 100 (§10, §12). This is convenient for identifying a substance by comparing the sequence of relative intensities, but with this normalization, the lowering of the height as D is reduced is no longer visible. The raw data (counts) that a measuring instrument outputs are not normalized, so for two samples measured under the same conditions, the peak of the one with smaller crystallites looks lower and broader. This is why simulator tab ⑥ draws the intensity on a fixed scale without normalizing it (Appendix A.7).

→ In xrd-simulator-en.html, tab ⑥ Scherrer equation, see how the peak broadens and, in step with that, becomes lower when the crystallite size is changed (the vertical axis is not normalized).

12. Real Measurements — From the Powder Method to Rietveld Analysis

12.1 Bragg–Brentano geometry

The most common geometry for a laboratory powder diffractometer is the Bragg–Brentano focusing geometry. The X-ray source, the sample and the detector are placed on a single circle (the goniometer circle), and are coupled so that when the sample is rotated by θ, the detector rotates by 2θ. In this geometry, only the planes parallel to the sample surface (those for which $\mathbf{g}$ is perpendicular to the sample surface) satisfy the diffraction condition, so it has the weakness that it is easily affected by preferred orientation (a bias in the orientation of the crystallites). In practice, the usable range of 2θ is limited on the low-angle side to about 5–10° by the direct beam and air scattering and by the irradiated width, which is set by the divergence slit, spilling over the edge of the sample (the footprint), and on the high-angle side to about 150–160° by the mechanical limits of the goniometer.

12.2 Indexing

The procedure when starting from the pattern of an unknown sample is as follows.

1. Calculate d from the 2θ of each peak (Eq. (7)).
2. Assuming a cubic crystal, compare the ratios of ${\sin }^2\theta \propto N=h^2+k^2+l^2$ among the peaks. If they form ratios of small integers, the crystal is probably cubic, and the pattern of the ratios lets you guess P / I / F (Exercise 4). However, an integer ratio is only a necessary condition: a tetragonal crystal with c/a close to 1, for example, gives the same ratios, so check it including the lines on the high-angle side. To distinguish P from I, it matters whether or not there is a line corresponding to N = 7.
3. Calculate the lattice constant a from N and d.
4. If a cubic crystal does not fit, increase the degrees of freedom — tetragonal, hexagonal, … (the job of automatic indexing programs).

12.3 Rietveld analysis

Rather than picking out the peaks one by one, the Rietveld method fits the whole pattern with a calculated one. All the parameters — lattice constants, atomic coordinates, occupancies, temperature factors, peak-profile functions, background, preferred orientation, scale factor, and so on — are varied by nonlinear least squares so that the residual with respect to the observed values, $R_{\text{wp}}$, is minimized. It is a powerful method that can refine the crystal structure itself from the pattern, but if the starting model is wrong, it either fails to converge or, even if it appears to converge, settles on a wrong structure, so the work of §8, “first narrow down the space group by the systematic absences,” is a prerequisite.

$ python3 ~/script/xrd_pattern_poscar.py -p POSCAR_NaCl_Fm-3m -s F

13. For Further Study

The Mochizuki Research Group has a solid state physics textbook for third-year undergraduates, Introduction to Solid State Physics (in Japanese) (29 chapters plus appendices). The content of this page corresponds fully to its Chapters 1, 2 and 3. That book is for third-year students, so you do not need to understand all of it now. Just remember it as a place to come back to when you get stuck on this material.

14. Exercises and Solutions