 Solution

Use the Fourier transform $\tilde{f}(k)={\int }_{\mathbb{R}}f(x)e^{-ikx}dx$. For a rapidly decaying smooth function, integration by parts gives $\widetilde{f^{\prime }}=ik\tilde{f}$, and differentiation under the integral gives $\widetilde{xf}=id\tilde{f}/dk$. Both are justified here by Gaussian decay. Since $f^{\prime }=-2a^{2}xf$,The Gaussian integral supplied in the hint gives $\tilde{f}(0)=\sqrt{\pi }/a$. Solving this first-order equation yieldsThe calculation states the transform normalization and the two differentiation properties explicitly.