Skip to content

Reciprocal Lattice | Physics - Wyatt's Notes

The reciprocal lattice vectors are defined by:

b1=2πa2×a3a1(a2×a3)\mathbf{b}_1 = 2\pi\frac{\mathbf{a}_2 \times \mathbf{a}_3}{\mathbf{a}_1 \cdot (\mathbf{a}_2 \times \mathbf{a}_3)}

b2=2πa3×a1a1(a2×a3)\mathbf{b}_2 = 2\pi\frac{\mathbf{a}_3 \times \mathbf{a}_1}{\mathbf{a}_1 \cdot (\mathbf{a}_2 \times \mathbf{a}_3)}

b3=2πa1×a2a1(a2×a3)\mathbf{b}_3 = 2\pi\frac{\mathbf{a}_1 \times \mathbf{a}_2}{\mathbf{a}_1 \cdot (\mathbf{a}_2 \times \mathbf{a}_3)}

Every reciprocal lattice point is at:

G=m1b1+m2b2+m3b3,miZ\mathbf{G} = m_1\mathbf{b}_1 + m_2\mathbf{b}_2 + m_3\mathbf{b}_3, \quad m_i \in \mathbb{Z}

Key property: GR=2π×integer\mathbf{G} \cdot \mathbf{R} = 2\pi \times \mathrm{integer} So eiGR=1e^{i\mathbf{G}\cdot\mathbf{R}} = 1.

Proof of key property.

GR=(m1b1+m2b2+m3b3)(n1a1+n2a2+n3a3)\mathbf{G} \cdot \mathbf{R} = (m_1\mathbf{b}_1 + m_2\mathbf{b}_2 + m_3\mathbf{b}_3) \cdot (n_1\mathbf{a}_1 + n_2\mathbf{a}_2 + n_3\mathbf{a}_3)

By the orthogonality relation biaj=2πδij\mathbf{b}_i \cdot \mathbf{a}_j = 2\pi\delta_{ij}:

GR=2π(m1n1+m2n2+m3n3)=2π×integer\mathbf{G} \cdot \mathbf{R} = 2\pi(m_1 n_1 + m_2 n_2 + m_3 n_3) = 2\pi \times \mathrm{integer}

\blacksquare

The first Brillouin zone is the Wigner-Seitz cell of the reciprocal lattice. It is the set of Points closer to the origin than to any other reciprocal lattice point.

For FCC (real space), the reciprocal lattice is BCC. The first Brillouin zone is a truncated octahedron. For BCC (real space), the reciprocal lattice is FCC.

Important volumes:

Real spaceReciprocal spaceBZ volume
SC (aa)SC (2π/a2\pi/a)(2π/a)3(2\pi/a)^3
BCC (aa)FCC (4π/a4\pi/a)2(2π/a)32(2\pi/a)^3
FCC (aa)BCC (4π/a4\pi/a)4(2π/a)34(2\pi/a)^3

VBZ=(2π)3/VcellV_{\mathrm{BZ} = (2\pi)^3 / V_{\mathrm{cell}}}.

Theorem 2.1. The reciprocal lattice vector Ghkl=hb1+kb2+lb3\mathbf{G}_{hkl} = h\mathbf{b}_1 + k\mathbf{b}_2 + l\mathbf{b}_3 Is perpendicular to the real-space planes (hkl)(hkl) And Ghkl=2π/dhkl\lvert\mathbf{G}_{hkl}\rvert = 2\pi/d_{hkl} where dhkld_{hkl} is the interplanar spacing.

Proof. The plane (hkl)(hkl) has intercepts a1/h\mathbf{a}_1/h, a2/k\mathbf{a}_2/k, a3/l\mathbf{a}_3/l. Two vectors in this plane are a2/ka1/h\mathbf{a}_2/k - \mathbf{a}_1/h and a3/la1/h\mathbf{a}_3/l - \mathbf{a}_1/h.

Ghkl(a2/ka1/h)=(2π/k)k(2π/h)h=0\mathbf{G}_{hkl} \cdot (\mathbf{a}_2/k - \mathbf{a}_1/h) = (2\pi/k)k - (2\pi/h)h = 0.

Similarly Ghkl(a3/la1/h)=0\mathbf{G}_{hkl} \cdot (\mathbf{a}_3/l - \mathbf{a}_1/h) = 0. Hence Ghkl\mathbf{G}_{hkl} is Perpendicular to the plane.

For the spacing: the plane through the origin has equation Ghklr=0\mathbf{G}_{hkl} \cdot \mathbf{r} = 0. The next parallel plane is Ghklr=2π\mathbf{G}_{hkl} \cdot \mathbf{r} = 2\pi (since eiGr=1e^{i\mathbf{G}\cdot\mathbf{r}} = 1 On all lattice planes). The distance from the origin to this plane is dhkl=2π/Ghkld_{hkl} = 2\pi/\lvert\mathbf{G}_{hkl}\rvert. \blacksquare

Worked Example: First Brillouin Zone of the 2D Square Lattice

For a 2D square lattice with primitive vectors a1=ax^\mathbf{a}_1 = a\hat{x}, a2=ay^\mathbf{a}_2 = a\hat{y}:

b1=2πax^,b2=2πay^\mathbf{b}_1 = \frac{2\pi}{a}\hat{x}, \quad \mathbf{b}_2 = \frac{2\pi}{a}\hat{y}

The nearest reciprocal lattice points to the origin are at (±2π/a,0)(\pm 2\pi/a,\, 0) and (0,±2π/a)(0,\, \pm 2\pi/a). Their perpendicular bisectors are the lines kx=±π/ak_x = \pm\pi/a and ky=±π/ak_y = \pm\pi/a.

The next-nearest points are at (±2π/a,±2π/a)(\pm 2\pi/a,\, \pm 2\pi/a). Their perpendicular bisectors are kx+ky=±2π/ak_x + k_y = \pm 2\pi/a and kxky=±2π/ak_x - k_y = \pm 2\pi/a.

The first Brillouin zone is bounded by the four nearest-neighbour bisectors and is a square with Vertices at (±π/a,±π/a)(\pm\pi/a,\, \pm\pi/a) and area (2π/a)2(2\pi/a)^2.

The second Brillouin zone is the region between the first zone and the bisectors of the next-nearest Neighbours.

Worked Example: First Brillouin Zone of the 2D Hexagonal Lattice

For a 2D hexagonal lattice with a1=ax^\mathbf{a}_1 = a\hat{x} a2=a(x^/2+3y^/2)\mathbf{a}_2 = a(\hat{x}/2 + \sqrt{3}\hat{y}/2):

b1=2πa(x^y^3),b2=4πa3y^\mathbf{b}_1 = \frac{2\pi}{a}\left(\hat{x} - \frac{\hat{y}}{\sqrt{3}}\right), \quad \mathbf{b}_2 = \frac{4\pi}{a\sqrt{3}}\hat{y}

The six nearest reciprocal lattice points form a regular hexagon. The perpendicular bisectors of The six nearest-neighbour vectors form a regular hexagon centred at the origin --- the first Brillouin zone.

High-symmetry points: Γ\Gamma (centre), MM (midpoint of edge), KK (corner).

The area of the BZ equals (2π)2/Acell(2\pi)^2/A_{\mathrm{cell}} where Acell=a23/2A_{\mathrm{cell} = a^2\sqrt{3}/2}.

The Ewald sphere provides a geometric criterion for when diffraction occurs. Given an incident Wave vector k\mathbf{k} (with k=2π/λ\lvert\mathbf{k}\rvert = 2\pi/\lambda) and the reciprocal lattice:

  1. Draw k\mathbf{k} terminating at the origin of reciprocal space.
  2. Construct a sphere of radius k=2π/λk = 2\pi/\lambda centred at the start of k\mathbf{k}.
  3. Diffraction occurs for every reciprocal lattice point G\mathbf{G} that lies on the sphere, since then k"=k+G\mathbf{k}" = \mathbf{k} + \mathbf{G} also has k=k\lvert\mathbf{k}'\rvert = k (elastic scattering condition).

Implications:

  • For a fixed wavelength and a single crystal, very few reciprocal lattice points lie on the Ewald sphere. The crystal must be rotated to bring different points onto the sphere.
  • As λ\lambda decreases (shorter wavelength), the Ewald sphere radius increases and more points satisfy the condition.
  • For λ0\lambda \to 0 (e.g., electron diffraction), the Ewald sphere is effectively flat, and the Laue condition reduces to a planar section through reciprocal space.
Worked Example: Ewald Sphere for Aluminium

Aluminium is FCC with a=0.405a = 0.405 nm. The reciprocal lattice is BCC with conventional cubic Constant 4π/a=3.10×10104\pi/a = 3.10 \times 10^{10} m1^{-1}.

Using Cu KαK_\alpha radiation (λ=0.15406\lambda = 0.15406 nm), the Ewald sphere radius is k=2π/λ=4.08×1010k = 2\pi/\lambda = 4.08 \times 10^{10} m1^{-1}.

The shortest reciprocal lattice vector has magnitude Gmin=2π/(a2)=1.10×1010G_{\min} = 2\pi/(a\sqrt{2}) = 1.10 \times 10^{10} m1^{-1} (the (111) reflection of FCC).

Since k=4.08×1010k = 4.08 \times 10^{10} m1>Gmin^{-1} \gt G_{\min}The (111) point can lie on the Ewald sphere When the crystal is appropriately oriented. The maximum accessible GG is Gmax=2k=8.16×1010G_{\max} = 2k = 8.16 \times 10^{10} M1^{-1}Which allows access to many reflections.

The limiting sphere of radius 2k2k centred at the origin contains all reciprocal lattice points That can potentially be accessed by rotating the crystal. Points outside this sphere can never Satisfy the diffraction condition for the given wavelength.

Worked Example: Structure Factor of the NaCl Structure

NaCl has an FCC lattice with a two-atom basis: Na+^+ at (0,0,0)(0,0,0) and Cl^- at (a/2,0,0)(a/2, 0, 0) (or equivalently, Cl^- at (1/2,0,0)(1/2, 0, 0) in fractional coordinates).

The FCC sublattice contributes a factor SFCC=f[1+eiπ(h+k)+eiπ(h+l)+eiπ(k+l)]S_{\mathrm{FCC} = f[1 + e^{-i\pi(h+k)} + e^{-i\pi(h+l)} + e^{-i\pi(k+l)}]} Which is zero unless h,k,lh,k,l are all even or all odd.

For allowed FCC reflections, the basis factor is:

Fbasis=fNa+fCleiπhF_{\mathrm{basis} = f_{\mathrm{Na} + f_{\mathrm{Cl}\, e^{-i\pi h}}}}

When h,k,lh,k,l are all even: F=fNa+fClF = f_{\mathrm{Na} + f_{\mathrm{Cl}}}. When h,k,lh,k,l are all odd: F=fNafClF = f_{\mathrm{Na} - f_{\mathrm{Cl}}}.

The intensity IS2I \propto \lvert S\rvert^2:

  • All even: I(fNa+fCl)2I \propto (f_{\mathrm{Na} + f_{\mathrm{Cl})^2}} (strong)
  • All odd: I(fNafCl)2I \propto (f_{\mathrm{Na} - f_{\mathrm{Cl})^2}} (weak, since fNafClf_{\mathrm{Na} \approx f_{\mathrm{Cl}}} at high scattering angles where form factors converge)
flowchart TD
A[2_Reciprocal Lattice] --> B[Key Concepts]
A --> C[Core Principles]
A --> D[Practical Applications]
B --> E[Fundamental definitions]
C --> F[Design patterns]
D --> G[Real-world usage]

The reciprocal lattice is a mathematical construction that transforms the problem of crystal diffraction into simple geometry. Real-space lattice points represent where atoms sit, while reciprocal lattice points represent the allowed wavevectors for diffracted beams. When X-rays hit a crystal, they scatter strongly only when the scattering vector matches a reciprocal lattice vector, which is the Laue condition. The first Brillouin zone is the set of momentum states unique to that crystal, analogous to how a single unit cell tiles all of real space. This dual-space picture makes it easy to predict diffraction patterns without tracking individual atoms.

Mistake 1: Confusing the reciprocal lattice with the dual lattice. The reciprocal lattice is defined in terms of the real-space lattice vectors and has dimensions of inverse length. It is not the same as the dual lattice in mathematics, which is a different concept. Do not confuse the two; the reciprocal lattice is specific to crystallography.

Mistake 2: Forgetting the factor of 2π2\pi in the reciprocal lattice vectors. The reciprocal lattice vectors are defined with a factor of 2π2\pi: biaj=2πδij\mathbf{b}_i \cdot \mathbf{a}_j = 2\pi\delta_{ij}. Some conventions omit the 2π2\pi, but it is essential for the correct relationship between real and reciprocal space. Do not forget the 2π2\pi factor.

Mistake 3: Assuming that the reciprocal lattice has the same symmetry as the real lattice. The reciprocal lattice has the same symmetry as the real lattice only for cubic systems. For other crystal systems, the reciprocal lattice may have different symmetry. Do not assume that the symmetry is preserved; check the specific crystal system.

Mistake 4: Confusing the first Brillouin zone with the Wigner-Seitz cell in real space. The first Brillouin zone is the Wigner-Seitz cell of the reciprocal lattice, not the real lattice. It is the set of points in reciprocal space closer to the origin than to any other reciprocal lattice point. Do not confuse it with the Wigner-Seitz cell in real space.

Mistake 5: Forgetting that diffraction peaks occur at reciprocal lattice points. Bragg diffraction peaks occur when the scattering vector G\mathbf{G} is a reciprocal lattice vector. This is the Laue condition: GR=2π×integer\mathbf{G} \cdot \mathbf{R} = 2\pi \times \mathrm{integer}. Do not assume that diffraction can occur at arbitrary angles; it is restricted to specific directions determined by the reciprocal lattice.

  • Crystal Structures: The reciprocal lattice vectors are constructed from the real-space primitive lattice vectors using the cross-product formula.

  • Diffraction: Bragg diffraction and the Laue condition are formulated most logically in terms of reciprocal lattice vectors.

  • Lattice Vibrations and Phonons: Phonon wavevectors are defined in the Brillouin zone, which is the Wigner-Seitz cell of the reciprocal lattice.

  • Semiconductors: Electronic band structures are computed in the Brillouin zone of the reciprocal lattice.

  • Calculus

  • Linear Algebra