Let us take a many-body quantum system, whose phases in the configuration basis are labeled by $\mathbf {\hat q}=(q_1,\cdots, q_N)$ and momenta $\mathbf {\hat p}=\left(-i\frac{\partial}{\partial \hat q_1},\cdots, -i\frac{\partial}{\partial \hat q_N}\right)$. Let us then consider the operator \begin{equation*} f(\mathbf {\hat q}, \mathbf {\hat p})\equiv \hat q_1^{n_1}\cdots \hat q_N^{n_N}\left(-i\frac{\partial}{\partial \hat q_1}\right)^{m_1}\cdots \left(-i\frac{\partial}{\partial \hat q_N}\right)^{m_N} \end{equation*} of powers of configurations and positions, $n_i, m_i\in \mathbb N^0$.
Is it correct that the object \begin{equation*} \tilde{\mathrm{tr}}\left\{f(\mathbf {\hat q}, \mathbf {\hat p})\right\}\equiv\int_{\mathbb{R}^N} \mathrm d\mathbf {\hat q} \left\langle\mathbf q\middle|f\left(\mathbf {\hat q} ,-i\frac{\partial}{\partial {\mathbf {\hat q} }}\right) \middle|\mathbf q\right\rangle \end{equation*} is NOT defined (i.e. it is not a well posed trace)?
In particular, for infinite-dimensional Hilbert spaces $H$, an operator is trace class if it is bounded. In my case this is not supposed to be the case, as \begin{equation} \sup_{|\mathbf q\rangle\in\mathcal D(H), ||\mathbf r||\neq 0}\frac{||f\left(\mathbf q,-i\frac{\partial}{\partial {\mathbf q}}\right) |\mathbf q\rangle ||}{|| |\mathbf q\rangle ||}=+\infty \end{equation} where $\mathcal D(H)$ is the (unbounded) domain in the Hilbert space of definition of the operator; in particular, $\hat f$ is the product of powers of unbounded operators.
Instead, in case one includes a canonical weight and defines \begin{equation*} \mathrm{tr}\{e^{-\beta\hat H}f(\mathbf {\hat q} ,\mathbf {\hat p} )\}\equiv \int_{\mathbb{R}^N} \mathrm d\mathbf {\hat q} \left\langle{\mathbf{\hat q}} \middle|e^{-\beta\hat H}f\left(\mathbf {\hat q} ,-i\frac{\partial}{\partial {\mathbf {\hat q} }}\right) \middle|\mathbf {\hat q} \right\rangle \end{equation*} is the equation above a well defined trace?