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In solid state physics we have a periodic function over the lattice as $$f(\boldsymbol{r})=\sum_{\boldsymbol{G}}f_{\boldsymbol{G}}\mathrm{e}^{2\pi i\boldsymbol{G}\cdot\boldsymbol{r}}$$ where $$\boldsymbol{G}=v_1\boldsymbol{b}_1+v_2\boldsymbol{b}_2+v_3\boldsymbol{b}_3+v_4\boldsymbol{b}_4$$ in which $f(\boldsymbol{r})$ is invariant with respect to a translation

$$\boldsymbol{T}=u_1\boldsymbol{a}_1+u_2\boldsymbol{a}_2+u_3\boldsymbol{a}_3+u_4\boldsymbol{a}_4$$ and $$\boldsymbol{a}_i\cdot \boldsymbol{b}_j=\delta_{ij}.$$

In my example I am trying to find the coefficients $f_{\boldsymbol{G}}$ when $$\{\boldsymbol{a}_i\}_{i=1}^4=\{(-\ell_p,0,0,-\frac{\ell_q}{k}), (0,-\ell_p,-\frac{\ell_q}{k},0), (0,0,\ell_q,0), (0,0,0,\ell_q)\},$$ where $\ell_p$ and $\ell_q$ are constants and $k\in\mathbb{Z}\setminus\{0\}$. I have found the vectors $\boldsymbol{b}_i$, which have the form $$\boldsymbol{b}_1=-\frac{e_1}{\ell_p}\\ \boldsymbol{b}_2=-\frac{e_2}{\ell_p}\\ \boldsymbol{b}_3=\frac{e_3}{\ell_q}-\frac{e_2}{k\ell_p}\\ \boldsymbol{b}_4=\frac{e_4}{\ell_q}-\frac{e_1}{k\ell_p}$$ where $\{e_i\}_{i=1}^4$ is the canonical basis for $\mathbb{R}^4.$ With $\boldsymbol{r}=(P_1,P_2,Q_1,Q_2)=(\boldsymbol{p},\boldsymbol{q})$ we therefore have $$f(\boldsymbol{P},\boldsymbol{Q})=\sum_{\boldsymbol{v}\in\mathbb{Z}^4}f_{\boldsymbol{v}}\mathrm{e}^{2\pi i\left(\frac{v_3Q_1+v_4Q_2}{\ell_q}-\frac{v_3P_2}{k\ell_p}-\frac{v_4P_1}{k\ell_p}-\frac{(v_1P_1+v_2P_2)}{\ell_p}\right)}.$$ Now what is not clear to me is how to get $f_{\boldsymbol{G}}$. But usual Fourier analysis we have

$$f_{\boldsymbol{G}}=\frac{1}{\ell_p^2\ell_q^2}\int_{\text{cell}}d\boldsymbol{r}f(\boldsymbol{r})\mathrm{e}^{-2\pi i\left(\frac{v_3Q_1+v_4Q_2}{\ell_q}-\frac{v_3P_2}{k\ell_p}-\frac{v_4P_1}{k\ell_p}-\frac{(v_1P_1+v_2P_2)}{\ell_p}\right)},$$ where $V_{\text{cell}}=\ell_p^2\ell_q^2$ but my main issue is the $\int_{\text{cell}}d\boldsymbol{r}$ term? Some authors use a form which involves the length of $\boldsymbol{a}_i$, i.e., $|\boldsymbol{a}_i|$, but his doesn't give the correct volume of the cell. How does one define the integration limits over this cell?

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  • $\begingroup$ You expressed $a_i$, $b_i$ and $r$ in canonical basis. Would it be easier if you convert $r$ in the $a_i$ basis, and do the r-integral in the $a_i$ basis? Then the cell would be naturally bounded inside four $a_i$. $\endgroup$ – Jinchen Aug 22 '19 at 12:28
  • $\begingroup$ Thank you for your comment. Do you mean something like $\boldsymbol{r}=P_1\boldsymbol{a}_1+P_2\boldsymbol{a}_2+Q_1\boldsymbol{a}_3+Q_2\boldsymbol{a}_4$? then $\int_{\text{cell}}dV=\int_0^1dP_1\int_0^1dP_2\int_0^1dQ_1\int_0^1dQ_2$? $\endgroup$ – Lewis Proctor Aug 22 '19 at 13:23
  • $\begingroup$ Yep, that's what I would try, use the lattice basis. It also simplifies the $G \cdot r$ part since if r is expressed using $a_i$ and G is expressed using b_j and $a_i \cdot b_j = \delta_{ij}$. I also noticed your lattice basis is not an orthogonal one. If possible, chose the orthogonal basis, that would further simplify the calculation. Cheers. $\endgroup$ – Jinchen Aug 22 '19 at 15:45
  • $\begingroup$ cool thank you! The lattice basis I have chosen is do to the problem rather than a specific choice. Thank you. $\endgroup$ – Lewis Proctor Aug 23 '19 at 9:29
  • $\begingroup$ Doesn't this reduce the problem into the usual Fourier series though with the exponent $2\pi(P_1v_1+P_2v_2+Q_1v_3+Q_2v_2)$ and thus is on the unit square in 4d i.e., $2\pi((P_1+1)v_1+(P_2+1)v_2+(Q_1+1)v_3+(Q_2+1)v_2)=2\pi(P_1v_1+P_2v_2+Q_1v_3+Q_2v_2)$? $\endgroup$ – Lewis Proctor Aug 27 '19 at 20:36

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