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I've spent a lot of times trying to show that $$ \mathcal{F}[e^{-x^2/2} G(x,t)] = e^{-k^2/2} G(k, -it) $$ with $G(x,t)$ being the generating function of Hermite polynomial, $$ G(x,t) = e^{2tx - t^2} $$

My attempt

We want to show that \begin{align*} \mathcal{F}[e^{-x^2/2} G(x,t)] &= e^{-k^2/2} G(k, -it) \\ &= e^{-k^2/2} e^{-2itk+t^2} \\ &= e^{-\frac{k^2}{2} - 2itk + t^2} \end{align*}

Computing the Fourier transform: \begin{align*} \mathcal{F}[e^{-x^2/2} G(x,t)] &= \mathcal{F}[e^{-x^2/2} e^{2tx - t^2}] \\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-x^2/2} e^{2tx - t^2} e^{-ikx} dx \end{align*}

We know that \begin{align*} \sqrt{2\pi} &= \sqrt{\frac{\pi}{1/2}} \\ &= \int_{-\infty}^{+\infty} e^{-\frac{1}{2}x^2} dx \end{align*}

I've been trying to factor the exponential isolating something like $-\frac{1}{2}x^2$. In fact I should get something like $e^{-\frac{1}{2}x^2} e^{-\frac{k^2}{2} + 2itk + t^2}$ in the integrant so I can get rid of the $\frac{1}{\sqrt{2\pi}}$ and get the desired result but I can't do it.

Mathieu
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  • You should indicate the variables ($x \leftrightarrow k$) involved in your Fourier Transform (for fixed $t$). – Jean Marie Aug 27 '20 at 09:10
  • Your generating function $G$ is not integrable

    see related computations https://math.stackexchange.com/questions/2114701/computing-the-fourier-transform-of-h-kxe-x2-2-where-h-k-is-the-hermit

     – prolea Aug 27 '20 at 10:36
  • I do not really understand the computation, I cannot resolve this exercice using Fourier transforms ? – Mathieu Aug 27 '20 at 12:18
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    Well, I found a solution. Edited my post with the answer. – Mathieu Aug 27 '20 at 14:22

1 Answers1

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We can use the following relation, \begin{equation*} \int_{-\infty}^{\infty} e^{-ax^2 + 2bx} dx = \sqrt{\frac{\pi}{a}} e^{\frac{b^2}{a}} \end{equation*}

Therefore, \begin{align*} \mathcal{F}[e^{-x^2/2} G(x,t)] &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty}e^{-x^2/2} e^{2tx - t^2} e^{-ikx} dx \\ &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty}e^{-x^2/2} e^{2tx} e^{-ikx} e^{-t^2} dx \\ &= e^{-t^2} \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty}e^{-x^2/2} e^{2tx} e^{-ikx} dx \end{align*}

We can rewrite the exponential argument as a 2nd degree polynomial tor the exponential to fit the above relation, \begin{align*} -\frac{x^2}{2} + 2tx - ikx &= -\frac{1}{2}x^2 + (2t - ik)x \\ &= -\frac{1}{2}x^2 + 2(t - \frac{ik}{2})x \end{align*}

We have $a = \frac{1}{2}$ et $b = t-\frac{ik}{2}$.

Therefore, \begin{align*} \mathcal{F}[e^{-x^2/2} G(x,t)] &= e^{-t^2} \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty}e^{-x^2/2} e^{2tx} e^{-ikx} dx \\ &= e^{-t^2} \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} e^{-\frac{1}{2}x^2 + 2(t - \frac{ik}{2})x} dx \\ &= e^{-t^2} \frac{1}{\sqrt{2\pi}} \sqrt{2\pi} e^{2(t-\frac{ik}{2})^2} \\ &= e^{-t^2} e^{2(t^2 - 2\frac{ikt}{2} - \frac{k^2}{4})} \\ &= e^{-t^2} e^{2t^2} e^{-2ikt} e^{-\frac{k^2}{2}} \\ &= e^{-\frac{k^2}{2}} e^{-2ikt} e^{t^2} \\ &= e^{-\frac{k^2}{2}} G(k, -it) \end{align*}

Mathieu
  • 607