Why the reciprocal lattice is a lattice?

Question

I would like to know why the following ensemble :

$$\{ \overrightarrow{G} / \forall \overrightarrow{R_n}, e^{i \overrightarrow{G}.\overrightarrow{R_n}}=1\}$$

where $\overrightarrow{R_n}$ is a vector of a lattice, is a lattice ?

The justification that I have found is that it is because if $\overrightarrow{G_1}$ and $\overrightarrow{G_2}$ are in the ensemble, then $\overrightarrow{G_1}+\overrightarrow{G_2}$ is also inside but I don't understand this explanation.

Indeed for me a lattice is a set of points such as any point of the system can be written as $\overrightarrow{R}=\sum_i \alpha_i \overrightarrow{a_i}$ where $\alpha_i$ are relative integers.

Any sum of points written like this has the same form, but how is the reciprocal of this affirmation true ?

Answer 1

You can think of a lattice as a discretized vector space. In a vector space, one can define it by, say, the set described by $\Sigma a_i \mathbf{e_i}$. However, the more fundamental definition, roughly speaking, is given by a set of vectors closed under linear combinations. In the same way, the more fundamental definition of a lattice is given by the set of discrete translations closed under discrete linear transformations.

Now, to answer your question more directly, why does the closure of the set imply that we can describe the set in the form $\Sigma n_i \mathbf{e_i}$? Let us simplify to the finite dimensional case. Consider the sum $\Sigma n_i \mathbf{v_i}$, where the sum is over the entire set of possible translations. Clearly, there are relations between the $\mathbf{v_i}$ because certain combinations are linearly dependent, so we can restrict to a linearly independent subset $\mathbf{e_i}$ (aka a basis). Hence, we get exactly what you have asked for.

Can this be done for an infinite dimensional case? I'm not entirely sure, you might run into issues of lattices and limits in the general case, cause I can imagine translating along an infinite series of vectors to get a net displacement that is infinitesimally small, meaning you could ostensibly "get back" to the regular vector space. In that case, can you really even think of it as a lattice anymore? But I digress.

Answer 2

Another way of seeing it is that there is a complete symmetry in $G$, and $R$ in the equation $ e^{iG\cdot R}=1$. That is that if you can imagine $ R= \sum_i \alpha_i a_i $ for a fixed $G$ which you agree can be made into a lattice then for a fixed $R$ you can certainly imagine $G = \sum_i \beta_i b_i$. In other words all that matters is that the dot product eventually gives you a multiply of $ 2\pi $.

Answer 3

As other people have pointed out, the fact that the reciprocal of a lattice is a lattice follows by definition.

Definition. A lattice in $\mathbb R ^n$ is a subgroup $L$ of $ (\mathbb R^n,+)$ which is discrete.

The word "discrete" means that every point $l\in L$ has a neighbourood which doesn't contain any other point of $L$. It is a theorem (see, e.g., [1]) that every lattice has generators, that is, there exist $k$ linearly independent vectors $\lbrace \mathbf e_i,i=1,2,\dots k\rbrace $ with $1\leq k \leq n$ such that the linear combinations $$\sum _{i=1} ^k m_i \boldsymbol e _i,$$with entire coefficients $m_i$ exhaust the lattice, so that the definition you are using is actually equivalent to the above one.

In any case for the three dimensional case, you can explicitly verify the "generators definition". Let $\mathbf a_1 ,\mathbf a_2,\mathbf a _3$ be generators of the direct lattice. Then: $$\mathbf b_3=2\pi\frac{\mathbf a _1 \times \mathbf a _2}{(\mathbf a _1 \times \mathbf a _2) \cdot \mathbf a _3} \ \, \text{and cyclic permutations}$$ are generators for the direct lattice. To see this, note that this three guys belong to the reciprocal lattice, are linearly independent, and any non-entire linear combination of them doesn't belong to the reciprocal lattice.

For a nice and simple introduction to solid state physics, I suggest you [2].

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[1] Arnold V.I., "Mathematical Methods of Classical Mechanics", §49, D. [2] Ashcroft N.W., Mermin D.N, "Solid State Physics"