Time derivative of operator

Question

I have to compute, at least formally, the following derivative $$\partial_t \exp(it\Delta)f(x-ct)$$ where $\Delta$ is the Laplacian and $c$ is a constant. I know that $e^{it\Delta}$ is the Schrodinger propagator, but I have some problems with the fact that $f$ depends on $t$ too.

Answer 1

Since the laplacian is a linear differential operator it can be differentiated like any other linear operator. The way the differential works is the same way as for the product of functions : Let $D(t)$ be a linear differential operator then $$D(t+\epsilon)f(x,t+\epsilon)-D(t)f(x,t)=$$$$D(t+\epsilon)f(x,t+\epsilon)-D(t)f(x,t+\epsilon)+D(t)f(x,t+\epsilon)-D(t)f(x,t)$$ Dividing by $ \epsilon$ and taking the limit $\lim_{t \rightarrow 0}$ then yields: $$\frac{\mathrm{d}}{\mathrm{d}t}(D(t)f(x,t))=\left(\frac{\mathrm{d}}{\mathrm{d}t}D(t)\right)f(x,t)+D(t)\left(\frac{\mathrm{d}}{\mathrm{d}t}f(x,t)\right)$$

Answer 2

Although the problem can be treated in an elementary way I think the procedure below is rather elegant. Observe that \begin{equation*} f(x-ct)=\exp [-c\partial _{x}t]f(x) \end{equation*} as can be checked by expanding the exponential. Observe further that $\Delta =\partial _{x}^{2}$ and $\partial _{x}$ commute. Now \begin{equation*} \exp [it\Delta ]\exp [-c\partial _{x}t]=\exp [it\{\partial _{x}^{2}+ic\partial _{x}\}] \end{equation*} and \begin{eqnarray*} \partial _{t}\exp [it\Delta ]f(x-ct) &=&\partial _{t}\exp [it\{\partial _{x}^{2}+ic\partial _{x}\}]f(x) \\ &=&i\{\partial _{x}^{2}+ic\partial _{x}\}\exp [it\Delta ]f(x-ct) \end{eqnarray*}