I am going through Dr Frederic Schuller's course on Quantum Theory. The operators that follow are defined on Schwartz space $S(\mathbb{R}^d)$. $H_{free}$ is a self-adjoint operator $H=-\Delta$, $\mathfrak{F}$ and $\mathfrak{F}^{-1}$ are Fourier and inverse Fourier operators on the same space. $P^2$ is a multiplication operator such that: $$\mathfrak{F}(H_{free})(p)=|p|^{2}(\mathfrak{F}\psi)(p)=:P^{2}\mathfrak{F}(\psi)(p)=(P^{2}\hat{\psi})(p).$$ Also, $|p|^2=p_{1}^{2}+p_{2}^{2}+p_{3}^{2}$.
I know how to get to: $$H_{free}=\mathfrak{F}^{-1}P^{2}\mathfrak{F}$$
so
$$e^{-itH_{free}}=e^{-it\mathfrak{F}^{-1}P^{2}\mathfrak{F}}$$
I tried various things to get to the following but failed:
$$e^{-it\mathfrak{F}^{-1}P^{2}\mathfrak{F}}=\mathfrak{F}^{-1}e^{-itP^{2}}\mathfrak{F}.$$
I tried writing the exponential as a Lebesgue integral of the projection-valued measure from the Spectral Theorem and I tried using the definitions of the Fourier operator and the inverse Fourier operator, but I can's see a way forward.
Any help will be greatly appreciated. Thanks in advance.
Let $A = -itP^{2}$, then what you are asking is $e^{\mathfrak{F}^{-1}A\mathfrak{F}}=\mathfrak{F}^{-1}e^{A}\mathfrak{F}$. Or I may have misunderstood the question (?). If my assumption is correct:
Note that it is easy to show that $(\mathfrak{F}^{-1}A\mathfrak{F})^n=\mathfrak{F}^{-1}A^{n}\mathfrak{F}$ (expand the term and note that $\mathfrak{F} \mathfrak{F}^{-1}=1$ (1 being the the identity operator)). And by expanding the exponential:
$e^{\mathfrak{F}^{-1}A\mathfrak{F}}=1+\mathfrak{F}^{-1}A\mathfrak{F}+\frac{1}{2!}(\mathfrak{F}^{-1}A\mathfrak{F})^2+...=1+\mathfrak{F}^{-1}A\mathfrak{F}+\mathfrak{F}^{-1}\frac{1}{2!}A^2\mathfrak{F}+...=\mathfrak{F}^{-1}(1+A+\frac{1}{2!}A^2+...)\mathfrak{F}$,
which is $\mathfrak{F}^{-1}e^{A}\mathfrak{F}$.