Fourier transform of the square of a function using convolution theorem

I have the function $$\psi$$ defined with the Fourier(FT) and inverse Fourier transform(IFT) between the $$x$$-$$k$$ space such that $$\psi(x)=\mathcal{F}\{\psi_{k}\}=\frac{1}{\sqrt{L}}\int_{-\infty}^{\infty}dk\,e^{ikx}\psi_{k} \\ \psi_{k}=\mathcal{F}^{-1}\{\psi(x)\}=\frac{1}{\sqrt{L}}\int_{-\infty}^{\infty}dx\,e^{-ikx}\psi(x)$$ where $$L$$ is a normalization constant. How can I show, and is it true, that $$\psi_{k}^{2}$$ is given as $$\psi_{k}^{2}=\mathcal{F}^{-1}\{\psi^{2}(x)\}=\int_{-\infty}^{\infty}dx\,e^{2ikx}\psi^{2}(x)$$ ? Otherwise, what should be the correct approach in writing $$\psi^{2}_{k}$$ in $$x$$-space? I understand the convolution theorem can be used here, but I'm not sure how to proceed or relate that to solving the problem.

• @flyinginsectleopard Thanks for your comment. So how should one evaluate $\psi_{k}^{2}$ in the fourier basis? Aug 8 at 20:37