1
$\begingroup$

I am reading David Tong’s lecture notes on Application of Quantum Mechanics. My confusion is about the following paragraph:

Consider a function $f(\vec{x}) $, suppose $f(\vec{x})=f(\vec{x}+\vec{r})$ for all $\vec{r} \in \Lambda$. The Fourier transform is $$\tilde{f}(\vec{k})=\int d^3 x e^{-i\vec{k}\cdot \vec{x}} f(\vec{x})=\sum_{\vec{r} \in \Lambda}\int_\Gamma d^3 x e^{-i\vec{k}\cdot (\vec{x}+\vec{r})} f(\vec{x}+\vec{r})=\sum _{\vec{r} \in \Lambda}e^{-i\vec{k}\cdot \vec{r}} \int_\Gamma d^3 x e^{-i\vec{k}\cdot \vec{x}} f(\vec{x})$$

Then with some calculation, the author concludes that

$$\sum _{\vec{r} \in \Lambda}e^{-i\vec{k}\cdot \vec{r}} = V^*\sum_{\vec{q} \in \Lambda^*} \delta(\vec{k}-\vec{q})$$

I can work out and understand the left hand side is the sum of delta functions, but cannot work out the coefficient $V^*$, the volume of a reciprocal lattice cell. Can someone help me? Thanks.

The $\Lambda,\Lambda^*$ mean lattice points and reciprocal lattice points, and $\Gamma, \Gamma^*$ mean the corresponding cell.

In later discussion, the coefficient of inverse Fourier transform is $\frac{1}{(2\pi)^3}$, I think that maybe the problem arises from wrong coefficients.

$\endgroup$
0
$\begingroup$

The left hand side is: $$\sum_{\mathbf r \in \Lambda} e^{-i\mathbf k \cdot \mathbf r}=\sum_{n_1=-\infty}^{\infty}\sum_{n_2=-\infty}^{\infty}\sum_{n_3=-\infty}^{\infty}e^{-ik_1a_1n_1}e^{-ik_2a_2n_2}e^{-ik_3a_3n_3}$$ $$=\prod_{i=1}^3 \sum_{n_i=-\infty}^{\infty}e^{-ik_ia_in_i}=\prod_{i=1}^3 \sum_{n_i=-\infty}^{\infty}e^{-i2\pi m_in_i}$$ where I've used the fact that $k_i \equiv \frac{2\pi}{a_i}m_i$ for some $m_i\in \mathbb Z$. Now the inside sum is of the form $f(x)\equiv\sum_{n=-\infty}^\infty e^{-i2\pi nx}$. I claim that in fact: $$f(x)=\sum_{l=-\infty}^\infty \delta(x-l) \qquad (*)$$ How do we show this? Well, first of all, $f(x)$ is periodic with period $1$, because: $$f(x+1)=\sum_{n=-\infty}^\infty e^{-i2\pi n(x+1)}=\sum_{n=-\infty}^\infty 1\times e^{-i2\pi nx}=f(x)$$ Now, let's restrict ourselves to one period, for example $x\in(-\frac12,\frac12]$. The integral of $f$ in this region is: $$\int_{-1/2}^{1/2}dx f(x) = \sum_{n=-\infty}^\infty\int_{-1/2}^{1/2}dx e^{-i2\pi nx}=\sum_{n=-\infty}^\infty \frac{\sin(\pi n)}{\pi n}=1$$ The last sum is one since $\sin(n\pi)\equiv 0$, so the only term contributing is the $n=0$ term which looks like a $\frac{\sin x}{x} \to 1$ kind of limit. Also note that in this period, $f(x)=0$ for any nonzero $x$. This is because corresponding to each term $e^{-i2\pi nx}$ in the sum, there is also a term which cancels it out (evaluate the geometric sum explicitly if this doesn't convince you!).

Putting all this together, our function $f$ satisfies all of the defining properties of the Dirac distribution in each period (the integral is one, and it's zero everywhere other than the origin). All we now have to do is to put infinitely many of these Dirac deltas together (with a spacing of the period 1), to ensure periodicity; meaning that we have derived equation $(*)$.

Using this result, the desired sum is: $$\sum_{\mathbf r \in \Lambda} e^{-i\mathbf k \cdot \mathbf r}=\prod_{i=1}^3 \sum_{l_i=-\infty}^{\infty}\delta(m_i-l_i)$$ Finally, we want to express the result in terms of the momentum $k_i = \frac{2\pi}{a_i}m_i$. Using the fact that $\delta(ax)=\delta(x)/|a|$, we have: $$\sum_{\mathbf r \in \Lambda} e^{-i\mathbf k \cdot \mathbf r}=\prod_{i=1}^3 \sum_{l_i=-\infty}^{\infty}\delta(\frac{a_i}{2\pi}k_i-l_i)$$ $$=\prod_{i=1}^3 \sum_{l_i=-\infty}^{\infty}(\frac{2\pi}{a_i})\delta(k_i-\frac{2\pi}{a_i}l_i)$$ $$=\frac{(2\pi)^3}{a_1a_2a_3} \sum_{l_1=-\infty}^{\infty}\sum_{l_2=-\infty}^{\infty}\sum_{l_3=-\infty}^{\infty}\delta(k_1-\frac{2\pi}{a_1}l_1)\delta(k_2-\frac{2\pi}{a_2}l_2)\delta(k_3-\frac{2\pi}{a_3}l_3)$$ Finally, realizing that $\frac{(2\pi)^3}{a_1a_2a_3} \equiv V^*$, and that the $\frac{2\pi}{a_i}l_i$ terms are simply components of momenta situated on the reciprocal lattice, we get the final result: $$\sum_{\mathbf r \in \Lambda} e^{-i\mathbf k \cdot \mathbf r}=V^* \sum_{\mathbf q \in \Lambda^*}\delta(\mathbf k - \mathbf q)$$ Note that the $V^*$ is essential in making the units consistent!

$\endgroup$
  • $\begingroup$ Now the problem on calculation has been solved, thanks. Therefore, here $\mathbf{k}=\sum k_i\mathbf{\hat{b_i}}$, the basis are unit vectors instead of $\mathbf{b_i}$, is this notation universal? $\endgroup$ – user571299 Feb 4 at 7:48
  • $\begingroup$ Yes, I've used a normalized basis $\mathbf k = \sum_i k_i \hat{\mathbf{b}}_i$ with $k_i=\frac{2\pi}{a_i}m_i$ instead of maybe the more conventional $\mathbf k = \sum_i m_i \mathbf{b}_i$ with $\mathbf{b}_i = \frac{2\pi}{a_i} \hat{\mathbf{b}}_i$. I don't think it's a universal notation now that you mention it. $\endgroup$ – Sahand Tabatabaei Feb 4 at 16:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.