# Schrodinger semigroup and conservation laws

Consider a sufficiently fast decaying and smooth function $$f\in L^1(\mathbb{R})\cap L^2(\mathbb{R})$$ such that $$\int_{\mathbb{R}} f(x)dx=0.$$ Now consider the Schrodinger semigroup $$e^{it\partial_x^2}$$, that is, the operator such that $$\mathcal{F}\big[e^{it\partial_x^2}f\big](\xi)=e^{-it\xi^2}\mathcal{F}[f](\xi).$$ Is it true that $$\int_{\mathbb{R}} e^{it\partial_x^2}\big(f(x-t)\big)dx=0?$$ I do believe that, if we put $$f(x)$$ instead of $$f(x-t)$$ in the last line it should be true since (sufficiently fast decaying) solutions to the equation $$i\partial_t u+\partial_x^2u=0$$ conserve the average $$\int_{\mathbb{R}}u(t,x)dx=\int_{\mathbb{R}}u(0,x)dx.$$ So in that sense, I would like to do something like $$\int_{\mathbb{R}} e^{it\partial_x^2}\big(f(x-t)\big)dx = \int_{\mathbb{R}} f(x-t)dx = \int_{\mathbb{R}} f(x)dx=0,$$ but I'm not 100% sure if that is allowed given the time dependence in $$f(x-t)$$. Does anybody know how to prove or disprove the above claim?

Denote Fourier transform by $$\hat{}$$. $$\hat{f}(0)=\int_{-\infty}^{\infty} f(x)dx=0$$ Let $$g_t(x)=f(x-t)$$ and $$h_t(x)=\left(e^{it\partial_x^2}g_t\right)(x)$$. You're looking for the value of $$\hat{h}(0)=\int_{-\infty}^{\infty} h(x)dx=0$$ Now \begin{align} \hat{g_t}(\xi)&=\int_{-\infty}^{\infty} g_t(x)e^{-ix\xi}dx\\ &=\int_{-\infty}^{\infty} f(x-t)e^{-ix\xi}dx\\ &=\int_{-\infty}^{\infty} f(x')e^{-i(x'+t)\xi}dx'\\ &=e^{-it\xi}\int_{-\infty}^{\infty} f(x')e^{-ix'\xi}dx'=e^{-it\xi}\hat{f}(\xi) \end{align} and $$\hat{h}(\xi)=e^{-it\xi^2}\hat{g_t}(\xi)=e^{-it\xi^2}e^{-it\xi}\hat{f}(\xi)=e^{-it(\xi^2+\xi)}\hat{f}(\xi)$$ so $$\hat{h}(0)=e^{-it(0^2+0)}\hat{f}(0)=1\cdot 0=0$$
Yes, that's ok. For each fixed $$t\in\mathbb R$$, the operator $$e^{it\partial_x^2}$$ commutes with translations, as you can see via the PDE or via the Fourier transform, as you prefer. Therefore, $$e^{it\partial_x^2}(f(\cdot - t))(x)=(e^{it\partial_x^2}f)(x -t).$$ Integrating in $$x$$, changing variable to $$y=x-t$$ (no problem here, since $$t$$ is fixed), you are back to the case you know.