I have a 2D square crystal which has a continuous potential $U(\vec r)$. It stretches to infinity. I want to find the diffraction pattern for this crystal. The potential I use is $$U(\vec r )=U(x \hat x+ y \hat y)=2 U_0(\cos(q \; x)+\cos(q \; y)) = \\2 U_0(\cos(\vec q_x \cdot \vec r)+\cos(\vec q_y \cdot \vec r))$$

This potential can be written in a more useful form,

$$U(\vec r )= \sum_{\vec G} U_{\vec G} e^{i \vec G \cdot \vec r }\;\;\;\;\;\;\;\;\;\;\;\;\;(*)$$ The sum can be taken over $\vec G = \vec q_1, \vec q_2, \vec q_1 + \vec q_2$ and so on, i.e. $\vec G = \sum_{i=1,2,3,...} n_i \vec q_i$. In this case, $U_{\vec G} = U_{0} $ if $\vec G \in \{ \pm q \hat x,\pm q\hat y\} $ and $U_{\vec G} =0$ otherwise.

Using this, it is easy and well known how to find the diffraction pattern.

The amplitude that a photon with wavevector $\vec k$ scatters to $\vec k'$ due to the crystal is $$F=F_{\Delta \vec k=\vec k' - \vec k}=\langle \vec k'|\hat U| \vec k \rangle \propto \int_{\text{all space}} d^3\vec r \;e^{-i(\Delta \vec k)\cdot \vec r} U(\vec r)\\$$ From $(*)$, we see that the integral is equal to $$\sum_{\vec G}U_{\vec G} \int_{\text{all space}} d^3\vec r \;e^{-i(\Delta \vec k- \vec G)\cdot \vec r}\\ =\sum_{\vec G \in \{ \pm q \hat x,\pm q\hat y\}}U_{\vec G} \int_{\text{all space}} d^3\vec r \;e^{-i(\Delta \vec k- \vec G)\cdot \vec r} \;\;\;\;\;\;\;\;\; (**)$$ In the last equality I used the fact that all but 4 of the $U_{\vec G}$ are $0$, so we only have to sum over the 4 relevant $\vec G$.

From $(**)$ scattering only happens when F is nonzero, i.e. when the arguments in the exponential are 0. $$\Delta \vec k = \{ \pm q \hat x,\pm q\hat y\}$$

This result seems wrong. Considering a potential of just delta functions instead of cosines, we would have expected scattering equally at every reciprocal lattice site, unrestrained $$\Delta \vec k = \vec G$$

Clearly, the potential I am using should be similar. What is the correct diffraction pattern here, and what am I doing wrong?