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Typo in definition of Fourier transform

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Ian Jauslin 2016-09-11 23:42:37 +00:00
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Giuliani_Jauslin_2015.tex
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@ -574,8 +574,8 @@ where $\mathcal H_0$ is the {\it free Hamiltonian} and $\mathcal H_I$ is the {\i
\end{array}\label{hamx}\end{equation} \end{array}\label{hamx}\end{equation}
Equation~(\ref{hamx}) can be rewritten in Fourier space as follows. We define the Fourier transform of the annihilation operators as Equation~(\ref{hamx}) can be rewritten in Fourier space as follows. We define the Fourier transform of the annihilation operators as
\begin{equation} \hat a_{k}:=\sum_{x\in\Lambda}e^{ikx}a_{x}\;,\quad \begin{equation} \hat a_{k}:=\sum_{x\in\Lambda}e^{ikx}a_{x}\;,\quad
\hat{\tilde b}_{k}:=\sum_{x\in\Lambda}e^{ikx}\hat{\tilde b}_{x+\delta_1}\;,\quad \hat{\tilde b}_{k}:=\sum_{x\in\Lambda}e^{ikx}\tilde b_{x+\delta_1}\;,\quad
\hat{\tilde a}_{k}:=\sum_{x\in\Lambda}e^{ikx}\hat{\tilde a}_{x-\delta_1}\;,\quad \hat{\tilde a}_{k}:=\sum_{x\in\Lambda}e^{ikx}\tilde a_{x-\delta_1}\;,\quad
\hat b_{k}:=\sum_{x\in\Lambda}e^{ikx}b_{x+\delta_1}\;\end{equation} \hat b_{k}:=\sum_{x\in\Lambda}e^{ikx}b_{x+\delta_1}\;\end{equation}
in terms of which in terms of which
\begin{equation} \begin{equation}