In the paper I am reading, the following identity appears:
$$ e^{-it \Delta} f(x) e^{it \Delta} = f(x- 2it \nabla) $$
where $f \in \mathcal{D}(\mathbb{R}^{d})$ and $f(x)$ on the left hand side denotes the multplication operator, i.e., multiplying by $f(x)$.
I did not use this identity when I computed the following term:
$$ \operatorname{Tr} \left[\int_{0}^{t} e^{i(t-t')\Delta }\left[ w \ast \varphi(t'), f(-\Delta) \right] e^{i(t'-t)\Delta} dt'\right]$$
I just used the definition of $e^{it\Delta}$ as a Fourier multiplier operator, which is $$ e^{it\Delta} = \mathcal{F}^{-1} e^{-i t |\xi|^2} \mathcal{F}. $$
But I could not derive the first identity using the definition. Also, I could not find any references or textbooks regarding such identities. It will be appreciated if you suggest any references regarding this kind of calculus.
For a Schwartz function $g$, the following identity holds: $$\begin{aligned} x \mathrm{e}^{\mathrm{i} t \Delta} g(x) &= x \frac{1}{(2\pi)^d} \int_{\mathbb{R}^d} \mathrm{e}^{\mathrm{i} \langle x,\xi\rangle} \mathrm{e}^{-\mathrm{i} t |\xi|^2} \hat{g}(\xi) \mathrm{d}\xi \\ &= \frac{-\mathrm{i}}{(2\pi)^d} \int_{\mathbb{R}^d} \partial_\xi (\mathrm{e}^{\mathrm{i} \langle x,\xi\rangle}) \mathrm{e}^{-\mathrm{i} t |\xi|^2} \hat{g}(\xi) \mathrm{d}\xi \\ &= \frac{\mathrm{i}}{(2\pi)^d} \int_{\mathbb{R}^d} \mathrm{e}^{\mathrm{i} \langle x,\xi\rangle} \mathrm{e}^{-\mathrm{i} t |\xi|^2} (-2 \mathrm{i} t \xi \hat{g}(\xi) + \partial_\xi \hat{g}(\xi)) \mathrm{d}\xi \\ &= \mathrm{e}^{\mathrm{i} t \Delta} (x-2\mathrm{i} t \nabla)g(x). \end{aligned}$$ Thus, $\mathrm{e}^{-\mathrm{i} t \Delta} x \mathrm{e}^{\mathrm{i} t \Delta} = x-2\mathrm{i} t \nabla$. This motivates to define $$ \mathrm{e}^{-\mathrm{i} t \Delta} f(x) \mathrm{e}^{\mathrm{i} t \Delta} = f(x-2\mathrm{i} t \nabla) $$ for suitable functions $f$.