Time derivative of operator 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.
 A: 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:
$$\frac{\mathrm{d}}{\mathrm{d}t}(D(t)f(x,t))=\left(\frac{\mathrm{d}}{\mathrm{d}t}D(t)\right)f(x,t)+D(t)\left(\frac{\mathrm{d}}{\mathrm{d}t}f(x,t)\right)$$
A: 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*}
