0
$\begingroup$

So I'm trying to understand what appears to me to be a paradox in the Ashcroft & Mermin Solid State Physics book. In Eqn 5.8 it states $\exp(i\vec{k}\cdot\vec{R})=1$ where $\vec{R}$ is a direct lattice vector, $$\vec{R}=n_1\vec{a}_1+n_2\vec{a}_2+n_3\vec{a}_3 , $$ with $\vec{a}_i$ being a Bravais lattice vector and $n_i\in Z$ ($Z$ is the integers). Given how the reciprocal lattice is defined as $\vec{b}_i=2\pi\vec{a}_j\times\vec{a}_k$ (Eqn. 5.3) we can write any $$\vec{k}=k_1\vec{b}_1+k_2\vec{b}_2+k_3\vec{b}_3$$ where $k_i\in Z$. From this definition we can see $$\vec{k}\cdot\vec{R}=2\pi(k_1n_1+k_2n_2+k_3n_3)$$ and $$\exp(i\vec{k}\cdot\vec{R})=\left(\exp(i2\pi)\right)^{k_1n_1+k_2n_2+k_3n_3}=1^{k_1n_1+k_2n_2+k_3n_3}=1$$

Then later in Chapter 8, when he uses the Born-von Karman boundary conditions to define $\vec{k}=x_1\vec{b}_1+x_2\vec{b}_2+x_3\vec{b}_3$ (Eqn. 8.20); where $x_i=m_i/N_i;\ m_i\in Z$ (Eqn 8.26) and $N_i$ is the number of unit cells in that dimension. Clearly $x_i$ does not have to be $\in Z$, and now $\exp(i\vec{k}\cdot\vec{R})\neq 1$ for any arbitary $\vec{k}$.

The whole point I'm trying to reconcile is when working with Wannier Functions, how to treat terms such as $$\sum_k e^{i\vec{k}\cdot\vec{R}} \ \ {\rm and} \ \ \sum_R e^{i\vec{k}\cdot\vec{R}}$$ which I think can be evaluated to delta functions. When $\vec{k}$ is defined as rational multiples of the reciprocal Bravais lattice vectors it becomes obvious it's like summing the nth roots of unity, which is clear to me how they evaluate to Kronecker deltas. If $\exp(i\vec{k}\cdot\vec{R})=1$ always however, the sums would always just sum to the number of unit cells.

$\endgroup$
0
$\begingroup$

I think I figured it out. In the original formulation of the reciprocal lattice, the unit cell contains 1 primitive cell. In other words its the Born-von Karman boundary conditions for 1 unit cell, aka the Gamma point approximation, where $N_i=1$ in all directions. Obviously the sums would then evaluate to 1, which is the same as the delta function when you only have 1 choice for any index.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.