How to show that the Fourier Transform of $ f(x)=e^{-mx^2} $ satisfies $ d/d\psi $ $ \hat{f} {(\psi)} $ = $ -\psi/2m $ $\hat{f} {(\psi)} $

Question

How to show that the Fourier Transform of the Gaussian function $ f(x)=e^{-mx^2} $ satisfies $ d/d\psi $ $ \hat{f} {(\psi)} $ = $ -\psi/2m $ $\hat{f} {(\psi)} $ ?

Answer 1

The function $f$ satisfies the equation $$ f'(x) + 2mx f(x) = 0.$$ Take the fourier transform of both sides of the equation:

$$\mathcal{F}[f'(x) + 2mxf(x)](\xi) = \mathcal{F}[0](\xi) = 0$$

Using linearity, this simplifies to

$$\mathcal{F}[f'(x)](\xi) + 2m\mathcal{F}[xf(x)](\xi) = 0$$

Using that $\mathcal{F}[f'(x)](\xi) = i\xi \mathcal{F}[f(x)](\xi)$ and $\mathcal{F}[xf(x)](\xi) = i\frac{d}{d\xi} \left(\mathcal{F}[f(x)](\xi)\right)$ we obtain

$$i\xi \mathcal{F}[f](\xi) + i2m \frac{d}{d\xi}\left(\mathcal{F}[f](\xi)\right) = 0$$ Dividing by $i2m$ yields the desired equation.