# Proof of momentum selection rule for solid-state transitions

I am reading Kittel's book, Quantum theory of solids, and it has a well-known theorem on its first chapter (page 3 of the second edition) which proof I do not understand fully.

The theorem is as follows,

THEOREM: If $f(\mathbf{x})$ has the periodicity of the lattice,

$$\int d^3 x f(\mathbf{x}) \exp(i\mathbf{K}\cdot\mathbf{x}) = 0,$$ unless $\mathbf{K}$ is a vector in the reciprocal lattice. The domain of integration is understood to be the crystal sample volume containing an integer number of unit cells.

PROOF: According to Kittel's book, this theorem is a direct consequence of the fact that $f(\mathbf{x})$ can be written as a Fourier series in the reciprocal lattice vectors $\mathbf{G} = m\mathbf{a}^* + n\mathbf{b}^* + p\mathbf{c}^*$, where $m,n,p$ are integers and $\mathbf{a}^*,\mathbf{b}^*,\mathbf{c}^*$ are the basis vectors of the reciprocal lattice, i.e.

$$\mathbf{a}^* \equiv 2\pi\frac{\mathbf{b} \times \mathbf{c}}{\mathbf{a}\cdot\mathbf{b}\times \mathbf{c}}, \\ \mathbf{b}^* \equiv 2\pi\frac{\mathbf{c} \times \mathbf{a}}{\mathbf{a}\cdot\mathbf{b}\times \mathbf{c}}, \\ \mathbf{c}^* \equiv 2\pi\frac{\mathbf{a} \times \mathbf{b}}{\mathbf{a}\cdot\mathbf{b}\times \mathbf{c}}.$$ $\mathbf{a},\mathbf{b},\mathbf{c}$ are the basis vectors of the direct lattice. The book introduces the $2\pi$ factor in the definition of the reciprocal basis for simplified notation.

According to the book, by writing $f(\mathbf{x})$ as its Fourier series, i.e.

$$f(\mathbf{x}) = \sum_{\mathbf{G}} a_{\mathbf{G}} \exp(i\mathbf{G}\cdot \mathbf{x}),$$ it follows that

$$\int d^3 x f(\mathbf{x}) \exp(i\mathbf{K}\cdot\mathbf{x}) = \sum_{\mathbf{G}} a_{\mathbf{G}}\int d^3 x \exp[i(\mathbf{K + G})\cdot\mathbf{x}] = \Omega \sum_{\mathbf{G}} a_{\mathbf{G}} \delta_{\mathbf{K},-\mathbf{G}},$$

where $\delta$ is the Kronecker delta and $\Omega$ is the crystal sample volume.

I agree that if $\mathbf{K} + \mathbf{G} = 0$ for some recriprocal lattice vector $\mathbf{G}$, then the integral of interest is non-zero and equal to $a_{\mathbf{G}}\Omega$. This is simply a consequence of the integrand in the middle termof last equation becoming unity for $\mathbf{G}$ satisfying $\mathbf{K} + \mathbf{G} = 0$ and integrating to zero for all other values of $\mathbf{G}$. However, I do not see how it has been shown that it is zero for $\mathbf{K} \neq \mathbf{G}$ for all reciprocal vectors $\mathbf{G}$.

Suggestions?

First of all, Kittel means that one should integrate over a macroscopic volume/specimen of a large number of primitive cells. Here we will only consider the ideal infinite lattice.

Let us for simplicity just discuss the 1D case with period $$2\pi$$. (The construction generalizes readily to higher dimensions.)

Note that a periodic integrand is only Lebesgue integrable over the entire real line $$\mathbb{R}$$ in trivial cases. In Fourier series, we only integrate over a period/a primitive cell:

$$f(\theta) ~\sim~ \sum_{n\in\mathbb{Z}} c_n e^{in\theta}, \qquad c_n ~:=~\int_{[0,2\pi]}\!\frac{\mathrm{d}\theta}{2\pi} e^{-in\theta}f(\theta). \tag{1}$$

Anyway, if we try to integrate over the whole real line $$\mathbb{R}$$ by breaking it into infinitely many primitive cells, we would get the Dirac comb/Shah distribution

$$\int_{\mathbb{R}}\!\frac{\mathrm{d}\theta}{2\pi} e^{-in\theta}f(\theta) ~=~c_n \sum_{m\in\mathbb{Z}}e^{-2\pi i nm}~=~c_n \delta(n-\mathbb{Z})~=~c_n\sum_{m\in\mathbb{Z}} \delta(n-m), \qquad n~\in~ \mathbb{R},\tag{2}$$

which is clearly only non-zero for $$n\in \mathbb{Z}$$ in the integer lattice. This is essentially the answer to OP's question.

• Thank you very much for the answer. It makes sense. I think I overlooked the fact that, if $f(x)$ is to be truly periodic, I should be considering an infinite lattice rather than a finite one. However, the question now turns to how to obtain that the integral is $a_{-\mathbf{G}} \Omega$ for $\mathbf{K} + \mathbf{G} = 0$. Should one now switch to the finite-lattice picture to obtain this result? Thanks again. Feb 28, 2018 at 22:19