Fourier transform of creation operator on the Schwartz-space Let $f\in S(\mathbb{R})$, where $S(\mathbb{R})$ denotes the Schwartz-space,
and $A=\frac{1}{\sqrt{2}}(x+\frac{\text{d}}{\text{d}x})$ be the creation operator and $A^\dagger=\frac{1}{\sqrt{2}}(x-\frac{\text{d}}{\text{d}x})$ the annihilation operator.
Now I have to show the following:
(a) $\mathcal{F}[Af]=iA\mathcal{F}[f]$
(b) $\mathcal{F}[A^\dagger f]=-iA^\dagger\mathcal{F}[f]$
and (c) $\mathcal{F}[Hf]=H\mathcal{F}[f]$ with $H=A^\dagger A$.
Here $\mathcal{F}$ represents the Fourier transform.
I plugged in the defintions and tried integration by parts but I think I am still missing something. So Help is very much appreciated!
 A: Recall that for $u\in\mathcal{S},$
$$D_{\xi}^\alpha (\mathcal{F} u)=\mathcal{F}((-x)^\alpha u)$$
and $$\mathcal{F}(D_x^\alpha  u)=\xi^\alpha \mathcal{F} u,$$ where $D^\alpha=\frac{1}{i^{|\alpha|}}\partial^\alpha$ and $\alpha$ is a multi-index. You can directly derive these facts (the former uses differentiation under the integral sign, and the latter uses integration by parts), and they can be applied easily to get your answers.
For example, for part (a), you can check that $$\mathcal{F}(xu)=i\frac{d}{d\xi} \mathcal{F}(u),$$ and $$\mathcal{F}\left(\frac{d}{dx}u\right)=i\xi\mathcal{F}(u).$$ Using linearity allows you to conclude your first identity.
In case you feel uneasy about using the identities that I wrote, I will explicitly demonstrate part (a) for you (this also shows you how the general identities are obtained):
We see that \begin{align*}
\mathcal{F}(xu)(\xi)&=(2\pi)^{-1/2}\int\limits_{-\infty}^\infty e^{-ix\xi} xu(x)\, dx\\
&=(2\pi)^{-1/2}\int\limits_{-\infty}^\infty \frac{d}{d\xi}\left(\frac{1}{-i}e^{-ix\xi}\right) u(x)\, dx\\
&=i\frac{d}{d\xi}\mathcal{F}u(\xi),
\end{align*} and \begin{align*}
\mathcal{F}\left(\frac{d}{dx}u\right)(\xi)&=(2\pi)^{-1/2}\int\limits_{-\infty}^\infty e^{-ix\xi}\frac{d}{dx}u(x)\, dx\\
&=-(2\pi)^{-1/2}\int\limits_{-\infty}^\infty \frac{d}{dx}(e^{-ix\xi})u(x)\, dx\\
&=i\xi\mathcal{F}u(\xi),
\end{align*}
where we have integrated by parts for the second part.
