Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 4 / 33B / Solution

Translate the scattering solution for the atom at the origin by $\mathbf{a}$. Its incident wave becomes $e^{i\mathbf{k}\cdot (\mathbf{r}-\mathbf{a})}$; multiplying the whole solution by $e^{i\mathbf{k}\cdot \mathbf{a}}$ restores the prescribed incident wave. At large $r$,Thus the outgoing amplitude acquires the phase $e^{i(\mathbf{k}-{\mathbf{k}}^{\prime })\cdot \mathbf{a}}$, with ${\mathbf{k}}^{\prime }=k\hat{\mathbf{r}}$.

In the single-scattering approximation, amplitudes from lattice sites add coherently. If $\mathbf{Q}=\mathbf{k}-{\mathbf{k}}^{\prime }$,For a rectangular block of a Bravais lattice with $N_j$ sites along primitive vectors ${\mathbf{a}}_j$ and $N=N_1{N}_2{N}_3$, the geometric series givesEach ratio has its limiting value $N_j$ when its argument is a multiple of $2\pi$. Thus the peaks have height $N$ and widths of order $1/N_j$ in those phase variables, at $\mathbf{Q}\cdot {\mathbf{a}}_j\in 2\pi \mathbb{Z}$, exactly the reciprocal lattice vectors.

With the incident-wave normalization used here, the Born approximation isFor the stated attractive three-dimensional delta potential,independent of scattering angle. This is the formal first Born amplitude, not a claim that an unregularized three-dimensional delta interaction defines an exact scattering Hamiltonian.

The 60-degree scattering angle is twice the Bragg angle, so $\theta =30^{\circ }$. First-order elastic Bragg scattering condition gives $2d\sin \theta =\lambda$, henceHigher orders would give $d=n\lambda$; the smallest, usual first-order spacing is the likely answer.