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I am reading Solid State Properties, From Bulk to Nano by Dresselhaus in Springer's Graduate texts in Physics Series. There I got stuck in the derivation of Weak Binding or Nearly Free Electron Approximation.

  1. How is the equation (3.11) derived from the eq. (3.9)?
  2. Please correct me if I am wrong. I think they have made an assumption that each unit cell is primitive hence each unit cell has only one atom and therefore when they came at an eq. (3.8) the summation is run at the no. of unit cells and $\mathbf R_n$ follows the same.

Can someone derive the eq. (3.11) from eq. (3.9)?

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1 Answer 1

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To get $(3.11)$ use the expressions for $\mathbf R_n$ and $\mathbf q$ given in $(3.10)$ and the relations that the unit vectors $\mathbf a_j$ and $\mathbf b_j$ in real and reciprocal space satisfy (see here in Wikipedia)

$$\mathbf a_j\cdot \mathbf b_k=2\pi\delta_{jk}\ ,$$

so that

\begin{align}\mathbf q\cdot\mathbf R_n&=\left(\sum_{j=1}^{3}n_j\mathbf a_j\right)\cdot\left(\sum_{k=1}^{3}\alpha_k\mathbf b_k\right)\\ &=\sum_{j,k=1}^3n_j\alpha_k\ \mathbf a_j\cdot \mathbf b_k\\&=2\pi(n_1\alpha_1+n_2\alpha_2+n_3\alpha_3).\end{align}

Then

$$e^{i\mathbf q\cdot\mathbf R_n}=e^{i2\pi n_1\alpha_1}e^{i2\pi n_2\alpha_2}e^{i2\pi n_3\alpha_3}$$

and summing over $n$ means summing over $n_1$, $n_2$ and $n_3$ (over all lattice vectors). If the lattice is finite, $n_1$, $n_2$ and $n_3$ go from $0$ to $N_1-1$, $N_2-1$ and $N_3-1$ respectively, where $N_j$ are the number of lattice points in each of the three directions

$$\sum_ne^{i\mathbf q\cdot\mathbf R_n}=\left(\sum_{n_1=0}^{N_1-1}e^{i2\pi n_1\alpha_1}\right)\left(\sum_{n_2=0}^{N_2-1}e^{i2\pi n_2\alpha_2}\right)\left(\sum_{n_3=0}^{N_3-1}e^{i2\pi n_3\alpha_3}\right).$$

Each one of these three sums is given by

$$\sum_{n=0}^{N-1}e^{inx}=\frac{1-e^{iNx}}{1-e^{ix}},$$

hence getting $(3.11)$.

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