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.

  • $\begingroup$ See physics.stackexchange.com/a/340862/37496 $\endgroup$
    – d_b
    Mar 30, 2019 at 20:37
  • $\begingroup$ This doesn't answer my question about why the new form preserves the geometry though. That's why I posted a new question $\endgroup$ Mar 31, 2019 at 17:56
  • $\begingroup$ Sorry, I'm not sure what you mean by "the new form preserves the geometry" $\endgroup$
    – d_b
    Mar 31, 2019 at 20:42

1 Answer 1


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 $\pi/2$).


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