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Preliminary results
[D_tD_t e^{i\omega t}]^n = -\frac{4}{\Delta t^2}\sin^2\left(
\frac{\omega\Delta t}{2}\right)e^{i\omega n\Delta t}
By \omega\rightarrow k ,
t\rightarrow x , n\rightarrow q ) it follows that
[D_xD_x e^{ikx}]_q = -\frac{4}{\Delta x^2}\sin^2\left(
\frac{k\Delta x}{2}\right)e^{ikq\Delta x}
