An atom in a harmonic cubic crystal is displaced from its equilibrium lattice site by…

PHYSICS · Multi-part · Fermi University Physics · Condensed Matter, Materials & Soft Matter · Crystal structure, bonding and diffraction · Lattice dynamics and thermal properties of solids

Solid State Physics, Ch Diffraction as measurement: form factors, thermal motion and refinement

Problem

An atom in a harmonic cubic crystal is displaced from its equilibrium lattice site R\mathbf{R} by a thermal displacement u=(ux,uy,uz)\mathbf{u}=(u_x,u_y,u_z), a Gaussian random variable of zero mean. Cubic symmetry makes the three components independent (no cross-correlation, ⟨uiuj⟩=0\langle u_iu_j\rangle=0 for i≠ji\ne j) and equal in variance, ⟨ux2⟩=⟨uy2⟩=⟨uz2⟩\langle u_x^2\rangle=\langle u_y^2\rangle=\langle u_z^2\rangle. Consider a scattering vector of fixed magnitude QQ pointing in an arbitrary direction relative to the cubic axes, Q=Q(sin⁡θcos⁡ϕ, sin⁡θsin⁡ϕ, cos⁡θ)\mathbf{Q}=Q(\sin\theta\cos\phi,\ \sin\theta\sin\phi,\ \cos\theta). You may use ⟨eix⟩=e−⟨x2⟩/2\langle e^{ix}\rangle=e^{-\langle x^2\rangle/2} for any Gaussian variable xx of zero mean. ((a)) Show that ⟨(Q⋅u)2⟩=⟨ux2⟩Q2\langle(\mathbf{Q}\cdot\mathbf{u})^2\rangle=\langle u_x^2\rangle Q^2, independent of θ\theta and ϕ\phi. ((b)) Starting from the scattered amplitude of a lattice of displaced identical atoms, A(Q)=f∑ReiQ⋅(R+uR)A(\mathbf{Q})=f\sum_{\mathbf{R}}e^{i\mathbf{Q}\cdot(\mathbf{R}+\mathbf{u}_{\mathbf{R}})}, derive the thermally averaged Bragg amplitude and show that the Bragg intensity is reduced by a factor e−2We^{-2W} with 2W=⟨ux2⟩Q22W=\langle u_x^2\rangle Q^2, for any direction of Q\mathbf{Q}.

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