# Two Dimensional Self-Reciprocal BravaisLattice

I've been reading Quantum States of Atoms Molecules and Solids by Morrison et al. for a condensed matter course. They make the claim that all 2D Bravais lattices are self-reciprocal, but I'm having difficulty picturing that or preferably rigorously proving it.

I know I could write the lattice vector as $$R = k_1 \vec{a} + k_2 \vec{b}$$ and then the corresponding reciprocal lattice would be $$n_1 \vec{A} + n_2 \vec{B}$$ where the vectors satisfy $$2\pi \delta_{ij}$$.

I don't actually know to calculate the reciprocal lattice for only 2D though? They gave an example for the square lattice where they introduced a orthongonal third vector, but I would like to do it more general than this.

Does anyone have a nice proof for why this is true? Thanks in advance.

• – d_b Mar 30 '19 at 20:37
• This doesn't answer my question about why the new form preserves the geometry though. That's why I posted a new question – Andrew Hardy Mar 31 '19 at 17:56
• Sorry, I'm not sure what you mean by "the new form preserves the geometry" – d_b Mar 31 '19 at 20:42

Let $$\mathbf{a}_1$$, $$\mathbf{a}_2$$ be the lattice vectors a two-dimensional Bravais lattice, let \begin{align} R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \end{align} be the orthogonal matrix rotating by $$\pi/2$$ in two dimensions, and let $$\epsilon^{ij}$$ be the antisymmetric symbol in two dimensions. From Emilio Pisanty's answer here, the reciprocal lattice vectors for a two-dimensional Bravais lattice have components \begin{align} b_{1i} &= \frac{1}{\epsilon^{kl} a_{1k}a_{2l}}R_{ij}a_{2j}\\ b_{2i} &= -\frac{1}{\epsilon^{kl} a_{1k}a_{2l}}R_{ij}a_{2j}.\\ \end{align} in Einstein notation. We observe directly that the $$\mathbf{b}_i$$ are scalar multiples of the original lattice vectors (rotated by $$\pi/2$$).