2

Given a lattice $L=\{x = j_1a_1 + j_2a_2 : j_1,j_2 \in \mathbb{Z}\}$ with $a_1,a_2 \in\mathbb{R}^2$ beeing linearly independent.

We can define a torus by $\mathbb{T}=L\,/ (\mathbb{Z}C + \mathbb{Z}T)$, where $C=c_1a_1 + c_2a_2$ and $T=t_1a_1+t_2a_2$, $c_1,c_2,t_1,t_2 \in\mathbb{Z}$, are linearly independent as well.

Thus, we have for $x,y \in \mathbb{T}$: $x=y \Leftrightarrow \exists z_1,z_2 \in\mathbb{Z} $ with $x=y + z_1C+z_2T$.

Let $C^\perp$ and $T^\perp$ such that $$ 0 = \langle C^\perp, C\rangle = \langle T^\perp, T\rangle $$ and $$ 2\pi = \langle C^\perp, T\rangle = \langle T^\perp, C\rangle. $$

I would like to show that the functions $e_k(x) =\frac{1}{\sqrt{|\mathbb{T}|}} e^{i \langle k, x\rangle }$, $e_k:\mathbb{T}\rightarrow \mathbb{C}$, $k=\mathbb{Z}C^\perp + \mathbb{Z}T^\perp$ are orthogonal, i.e.,

$$\langle e_k, e_{\tilde{k}} \rangle = \sum_{x\in\mathbb{T}} e_k(x) \overline{e_{\tilde{k}(x)}} = \sum_{x\in\mathbb{T}} e^{i \langle k - \tilde{k}, x\rangle } = 0 \Leftrightarrow k\neq \tilde{k},$$ but I don't quite know how to do that.

Eventually I would like to show that these functions form an orthonormal basis of $\{f:\mathbb{T}\rightarrow\mathbb{C}\}$ and the orthogonality is the missing part.

Maybe the proof of the cardinality helps in some sense: The number of wavefunctions is equal to the number of lattice points:

Let $b_1,b_2$ the reciprocal lattice vectors, i.e. $\langle a_i,b_j \rangle = 2\pi \delta_{ij}$. Then we have $e_k(x)= e_{k+j_1b_1+j_2b_2}(x)$ for every $x \in\mathbb{T}, j_1,j_2\in\mathbb{Z}$.

Thus, the number $K$ of different wavefunctions $e_k, k=\mathbb{Z}C^\perp + \mathbb{Z}T^\perp$ is equal to $$ K= |\frac{\det\begin{pmatrix}b_1|b_2\end{pmatrix}}{\det{\begin{pmatrix}C^\perp|T^\perp\end{pmatrix}}}|.$$

The number of lattice points in $\mathbb{T}$ is given by $$ |\mathbb{T}| = |\frac{\det{\begin{pmatrix}C|T\end{pmatrix}}}{\det\begin{pmatrix}a_1|a_2\end{pmatrix}}|.$$

It is a straightforward calculation to see that $|\mathbb{T}|=K$ if you write $k$ in terms of the reciprocal lattice vectors $b_1,b_2$.

Nils
  • 133
  • 9
  • Your torus is a discrete parallelogram, think to it as a discrete rectangle, and you'll get $$\langle e_k, e_m \rangle = \frac{1}{\sqrt{N_1N_2}}\sum_{n_1=0}^{N_1-1} \sum_{n_2=0}^{N_2-1} e^{-2i \pi (n_1\frac{k_1-m_1}{N_1}+n_2\frac{k_2-m_2}{N_2})} = 1_{m=k}$$ – reuns Jun 08 '17 at 09:41
  • I don't quite understand how to get to this point. I know that the lattice points are located in a paralellogram spanned by $C$ and $T$, but this is not the direction of the basis vectors $a_1$ and $a_2$. So I don't know how to split the sum in a useful way. – Nils Jun 08 '17 at 12:06
  • No matter how are the lattice, the parallelogram will be essentially $\mathbb{Z}^2 / (N_1 \mathbb{Z}+N_2\mathbb{Z}) = (\mathbb{Z}/N_1\mathbb{Z})\times (\mathbb{Z}/N_2\mathbb{Z})$ and the basis of orthogonal functions is $e_{k_1,k_2}(n_1,n_2) = e^{-2i\pi (n_1k_1/N_1+n_2 k_2/N_2)}$ for $(k_1,k_2) \in (\mathbb{Z}/N_1\mathbb{Z})\times (\mathbb{Z}/N_2\mathbb{Z})$ – reuns Jun 08 '17 at 12:15
  • I'm not quite sure if this is the case and I don't know how to show the connection (if it's true). $C$ and $T$ are not perpendicular. – Nils Jun 08 '17 at 12:28

1 Answers1

2

In matrix/vector notation :

  • We have a lattice $L = M \mathbb{Z}^2$ for some inversible matrix $M \in \mathbb{R}^{2 \times 2}$

  • If $L_2= M_2 \mathbb{Z}^2$ is another lattice then $L_2 \supseteq L$ iff $MM_2^{-1} \in \mathbb{Z}^{2 \times 2}$.

  • The dual lattice is $L^\perp = M^{-1} \mathbb{Z}^2$.

We are looking at the vector space of $L$-periodic functions $L_2 \to \mathbb{C}$, ie. the finite dimensional vector space of functions $L_2/L \to \mathbb{C}$

We have a basis of complex exponentials $$e_k(x) = e^{-2i \pi \langle k, x \rangle}, \qquad k \in L^\perp/L_2^\perp,x \in L_2 $$

They are orthonormal for the inner product

$$\langle e_k,e_m \rangle = \frac{1}{|L_2/L|} \sum_{x \in L_2/L} e_k(x) \overline{e_m(x)} = \frac{1}{|L_2/L|}\sum_{x \in L_2/L} \exp(-2i \pi \langle k-m,x \rangle)\\ =\frac{1}{|L_2/L|}\sum_{x \in L_2/L} \exp(-2i \pi \langle (\underbrace{MM_2^{-1} }_{\in \mathbb{Z}^{2 \times 2}})^{-1}\underbrace{M(k-m)}_{\in \mathbb{Z}^2},\underbrace{M_2^{-1} x}_{\in \mathbb{Z}^2} \rangle)$$

And you can check this is $ = 0$ whenever $M_2(k-m) \not \in \mathbb{Z}^2$ ie. whenever $k \ne m$ in $L_2^\perp/L^\perp$.

reuns
  • 77,999
  • THanks for the answer. The second point is wrong though. I think you meant $L_2 \subset L$. Could you recommend literature on that topic? – Nils Jun 09 '17 at 08:39
  • @Nils $L_2$ is a lattice containing $L$ which means $L = A L_2$ for some integer matrix $A$. So that $M \mathbb{Z}^2 = A M_2 \mathbb{Z}^2$ and $A = M M_2^{-1} \in \mathbb{Z}^{2 \times 2}$. Since $L$ is a subgroup of $L_2$ it makes sense to look at $L_2/L$ (a discrete parallelogram with $\det(A)$ points) – reuns Jun 09 '17 at 08:46
  • Search for 'dual lattice' on google, you'll find many pdf using this vocabulary. – reuns Jun 09 '17 at 08:51
  • This confuses me. Let $M=\operatorname{id_2}=\begin{pmatrix}1 & 0 \ 0 & 1 \end{pmatrix}$, i.e. $L=\mathbb{Z}^2$ and $M_2 = 2M$, i.e. $L_2 = 2\mathbb{Z}^2$. This means for every $v \in L_2$ we have $v\in L$, i.e. $L_2\subset L$. Additionaly we have $M M_2^{-1} = \frac{1}{2} \operatorname{id_2} \notin \mathbb{Z}^{2\times2}$. – Nils Jun 09 '17 at 08:54
  • @Nils : Yes but I did the opposite. Let $L_2 = \mathbb{Z}^2$ and $L = 2 \mathbb{Z}^2 =\begin{pmatrix}2 & 0 \ 0 & 2 \end{pmatrix}L_2$. Then $L \subset L_2$ and the parallelogram is $L_2/ L = {(0,0), (1,0),(0,1),(1,1)}$ – reuns Jun 09 '17 at 08:58
  • Ah I finally got this point. Thanks for the clarification. – Nils Jun 09 '17 at 09:00
  • @Nils $L_2$ is a set of points of $\mathbb{R}^2$ and an abelian group, and with my definition we have $L \subset L_2$ ($L_2$ contains $L$) – reuns Jun 09 '17 at 09:01
  • Unfortunately I still don't get the last step, i.e. checking that the sum is 0. I still need to find a way to rewrite this sum as a geometric series $\sum_{n=0}^{N-1} r^{n} = \frac{1-r^N}{1-r}$ where $r=e^{2\pi i \frac{1}{N}}$, right? To that end I need an isomorphism between $\mathbb{Z} / (N_1\mathbb{Z} + N_2\mathbb{Z})$ and $L_2 / L$ but I am not able to find one. – Nils Jun 09 '17 at 14:48