Is $\Psi^0(\mathbb{R})$ (pseudodifferential operators with symbols obeying $ |\partial^\alpha_x \partial^\beta_\xi a(x,\xi)| \leq C_{\alpha,\beta} (1+|\xi|)^{-|\beta|} $ ) a $C^*$-algebra?

In other words, is $\Psi^0(\mathbb{R})$ is closed in $\mathcal{L}(L^2(\mathbb{R}))$ in the operator norm topology?

If not, then is there any nice characterization by the $C^*$-algebra generated by $\Psi^0$? Alternatively, what is the strongest (or just a reasonable) topology on $\mathcal{L}(L^2(\mathbb{R}))$ such that $\Psi^0$ is a closed subspace?

** Edit: **Per Yemon Choi's comments below, the above question seems somewhat hopeless. As described here, $\Psi^0(\mathbb{R})$ is a Fréchet $*$-algebra with a topology stronger than the operator topology. I assume that this is the topology given by the seminorms on symbols:
$$
\Vert a \Vert_{\alpha,\beta} = \sup_{x,\xi \in \mathbb{R}} (1+|\xi|)^{|\beta|} |\partial^\alpha_x \partial^\beta_\xi a(x,\xi)|.
$$

So, in addition to the above question, I am adding the following question, to make it so that there might be an answer:

Is there a reasonable description of the smallest $C^*$-algebra containing $\Psi^0$?

extremelyhazy recollection is that $\Psi^0$ is not norm-closed in ${\mathcal L}(L^2)$, and that there is no reasonable topology to put on the latter to make $\Psi^0$ closed therein. Very roughly, think of Schwarz functions inside $L^2(R)$ - the former are naturally a Frechet space, the latter is naturally a Banach space, and as far as I know no good has come of trying to change the topology on either to make it fit better with the other. $\endgroup$