Derivative of a Fock state Let us suppose we have the operators $a(\omega)$ and $a^\dagger(\omega)$ with $[a(\omega),a^\dagger (\omega')] = \delta(\omega-\omega')$. Does a derivative of the Fock state $a^\dagger(\omega) |0\rangle$ with respect to $\omega$ exist, that is $\partial_{\omega} a^\dagger(\omega) |0\rangle $. Or more generally, do these expressions make senes
\begin{equation}
\partial_{\omega} a^\dagger(\omega)\\
\partial_{\omega} a(\omega)
\end{equation}
Intuitively, I would guess that the derivative does not exist because of the orthogonality of Fock states for different $\omega$. But maybe the derivative can be expressed in terms of a delta function.
My idea to find an expression for the derivative is the following. First let us rewrite the annihilation operator as
\begin{equation}
a(\omega) = \int d\omega' \delta (\omega' - \omega) a(\omega').
\end{equation}
Differentiation of the above expression with $\partial_{\omega}\delta (\omega' - \omega) = -\delta' (\omega'-\omega) = \frac{1}{\omega'-\omega} \delta (\omega' - \omega)$ yields
\begin{equation}
\partial_{\omega}a(\omega) = \int d\omega' \partial_{\omega}\delta (\omega' - \omega) a(\omega') = \frac{1}{\omega-\omega} a(\omega) .
\end{equation}
So one could argue that the derivative does not exist or that is equals $\infty$, is that correct?
 A: In quantum field theory deriving operators with respect to parameters is routine. (Recall how the canonical momentum field is produced.)
So, if you are careful, of course you may define such objects.
For instance,
$[a(\omega),a^\dagger (\omega')] = \delta(\omega-\omega')$ can yield
$$[\partial_{\omega} a(\omega),a^\dagger (\omega')] = \partial_{\omega} \delta(\omega-\omega').$$
Given the identity
$$
(\partial_{\omega} + \partial_{\omega’})\delta(\omega-\omega')=0,
$$
you then see that. e.g.,
$$
[\partial_{\omega} a(\omega),\int\!\! d\omega'~ f(\omega')~a^\dagger (\omega')] = \int\!\! d\omega'~ f(\omega')~ \partial_{\omega} \delta(\omega-\omega')\\ = -\int\!\! d\omega'~ f(\omega')~ \partial_{\omega'} \delta(\omega-\omega') = \int\!\! d\omega'~\partial_{\omega'}  f(\omega')~ \delta(\omega-\omega') = f'(\omega).$$
NB Response to comment ancillary question  Yes, of course. Since
$$\partial_x a(\omega/x)= -{\omega \over x} \partial_\omega a(\omega/x),$$
and analogously for $a^\dagger (\omega'/x)$, you indeed have, obviously,
$$
[\partial_x a(\omega/x) , \partial_x a^\dagger (\omega'/x)]={\omega \omega' \over x^2}\partial_\omega \partial_{\omega '} [ a(\omega/x) ,   a^\dagger (\omega'/x)]= -{\omega \omega ' \over x}\partial^2_\omega \delta (\omega-\omega').
$$
