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)} $ ?
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)} $
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fourier-transform
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0What definition of Fourier transform do you use – 2017-01-14
1 Answers
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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.
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0I cannot make this work – 2017-01-14
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0@iheartalgebraa I transformed my hint into a full answer. – 2017-01-14