Paper 4, Section II, A

Applications of Quantum Mechanics | Part II, 2014

Let Λ\Lambda be a Bravais lattice in three dimensions. Define the reciprocal lattice Λ\Lambda^{*}.

State and prove Bloch's theorem for a particle moving in a potential V(x)V(\mathbf{x}) obeying

V(x+)=V(x)Λ,xR3V(\mathbf{x}+\ell)=V(\mathbf{x}) \quad \forall \ell \in \Lambda, \mathbf{x} \in \mathbb{R}^{3}

Explain what is meant by a Brillouin zone for this potential and how it is related to the reciprocal lattice.

A simple cubic lattice Λ1\Lambda_{1} is given by the set of points

Λ1={R3:=n1i^+n2j^+n3k^,n1,n2,n3Z}\Lambda_{1}=\left\{\ell \in \mathbb{R}^{3}: \ell=n_{1} \hat{\mathbf{i}}+n_{2} \hat{\mathbf{j}}+n_{3} \hat{\mathbf{k}}, n_{1}, n_{2}, n_{3} \in \mathbb{Z}\right\}

where i^,j^\hat{\mathbf{i}}, \hat{\mathbf{j}} and k^\hat{\mathbf{k}} are unit vectors parallel to the Cartesian coordinate axes in R3\mathbb{R}^{3}. A bodycentred cubic (BCC\left(\mathrm{BCC}\right. ) lattice ΛBCC\Lambda_{B C C} is obtained by adding to Λ1\Lambda_{1} the points at the centre of each cube, i.e. all points of the form

+12(i^+j^+k^),Λ1\ell+\frac{1}{2}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}), \quad \ell \in \Lambda_{1}

Show that ΛBCC\Lambda_{B C C} is Bravais with primitive vectors

a1=12(j^+k^i^)a2=12(k^+i^j^)a3=12(i^+j^k^)\begin{aligned} \mathbf{a}_{1} &=\frac{1}{2}(\hat{\mathbf{j}}+\hat{\mathbf{k}}-\hat{\mathbf{i}}) \\ \mathbf{a}_{2} &=\frac{1}{2}(\hat{\mathbf{k}}+\hat{\mathbf{i}}-\hat{\mathbf{j}}) \\ \mathbf{a}_{3} &=\frac{1}{2}(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}) \end{aligned}

Find the reciprocal lattice ΛBCC\Lambda_{B C C}^{*}. Hence find a consistent choice for the first Brillouin zone of a potential V(x)V(\mathbf{x}) obeying

V(x+)=V(x)ΛBCC,xR3 [Hint: The matrix M=12(111111111) has inverse M1=(011101110).]\begin{aligned} & V(\mathbf{x}+\ell)=V(\mathbf{x}) \quad \forall \ell \in \Lambda_{B C C}, \mathbf{x} \in \mathbb{R}^{3} \\ & \text { [Hint: The matrix } \left.M=\frac{1}{2}\left(\begin{array}{rrr}-1 & 1 & 1 \\1 & -1 & 1 \\1 & 1 & -1\end{array}\right) \text { has inverse } M^{-1}=\left(\begin{array}{lll}0 & 1 & 1 \\1 & 0 & 1 \\1 & 1 & 0\end{array}\right) .\right] \end{aligned}

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