3.II.33B

Applications of Quantum Mechanics
Part II, 2005

Let {l}\{\mathbf{l}\} be the set of lattice vectors of some lattice. Define the reciprocal lattice. What is meant by a Bravais lattice?

Let i,j,k\mathbf{i}, \mathbf{j}, \mathbf{k} be mutually orthogonal unit vectors. A crystal has identical atoms at positions given by the vectors

a[n1i+n2j+n3k],a[(n1+12)i+(n2+12)j+n3k]a[(n1+12)i+j+(n3+12)k],a[n1i+(n2+12)j+(n3+12)k]\begin{array}{ll} a\left[n_{1} \mathbf{i}+n_{2} \mathbf{j}+n_{3} \mathbf{k}\right], & a\left[\left(n_{1}+\frac{1}{2}\right) \mathbf{i}+\left(n_{2}+\frac{1}{2}\right) \mathbf{j}+n_{3} \mathbf{k}\right] \\ a\left[\left(n_{1}+\frac{1}{2}\right) \mathbf{i}+\mathbf{j}+\left(n_{3}+\frac{1}{2}\right) \mathbf{k}\right], & a\left[n_{1} \mathbf{i}+\left(n_{2}+\frac{1}{2}\right) \mathbf{j}+\left(n_{3}+\frac{1}{2}\right) \mathbf{k}\right] \end{array}

where (n1,n2,n3)\left(n_{1}, n_{2}, n_{3}\right) are arbitrary integers and aa is a constant. Show that these vectors define a Bravais lattice with basis vectors

a1=a12(j+k),a2=a12(i+k),a3=a12(i+j)\mathbf{a}_{1}=a \frac{1}{2}(\mathbf{j}+\mathbf{k}), \quad \mathbf{a}_{2}=a \frac{1}{2}(\mathbf{i}+\mathbf{k}), \quad \mathbf{a}_{3}=a \frac{1}{2}(\mathbf{i}+\mathbf{j})

Verify that a basis for the reciprocal lattice is

b1=2πa(j+ki),b2=2πa(i+kj),b3=2πa(i+jk)\mathbf{b}_{1}=\frac{2 \pi}{a}(\mathbf{j}+\mathbf{k}-\mathbf{i}), \quad \mathbf{b}_{2}=\frac{2 \pi}{a}(\mathbf{i}+\mathbf{k}-\mathbf{j}), \quad \mathbf{b}_{3}=\frac{2 \pi}{a}(\mathbf{i}+\mathbf{j}-\mathbf{k})

In Bragg scattering, an incoming plane wave of wave-vector k\mathbf{k} is scattered to an outgoing wave of wave-vector k\mathbf{k}^{\prime}. Explain why k=k+g\mathbf{k}^{\prime}=\mathbf{k}+\mathbf{g} for some reciprocal lattice vector g. Given that θ\theta is the scattering angle, show that

sin12θ=g2k.\sin \frac{1}{2} \theta=\frac{|\mathbf{g}|}{2|\mathbf{k}|} .

For the above lattice, explain why you would expect scattering through angles θ1\theta_{1} and θ2\theta_{2} such that

sin12θ1sin12θ2=32\frac{\sin \frac{1}{2} \theta_{1}}{\sin \frac{1}{2} \theta_{2}}=\frac{\sqrt{3}}{2}