\circ E: L^2(\mu) \to L^2(\lambda)$$ where $$\begin{array}{c}
E: L^2(\mu) \to L^2(\lambda)\\
f \mapsto \left(x \mapsto f(x) \exp(-\frac{\phi(x)}{2})\right)\\
\\
A: L^2(\lambda) \to L^2(\lambda)\\
f \mapsto \left(x \mapsto f(cx)\right)\\
\end{array}$$

\[defnUp\] Define the isometry $U_p: L^p(\mu) \to L^p(\lambda)$ by $$f \mapsto \left(x \mapsto f(x) \exp(-\frac{\phi(x)}{p})\right)$$

The generalised Weyl pairs we will consider in this context will be $$\begin{array}{c}
Q_i = \tilde{U}_2^{-1} \circ X_i \circ \tilde{U}_2\\
P_i = \tilde{U}_2^{-1} \circ D_i \circ \tilde{U}_2
\end{array}$$ with domains $\tilde{U}_2^{-1}D(X_i)$, $\tilde{U}_2^{-1}D(D_i)$ respectively. That $(Q,P)$ satisfy the requirements of Definition \[defnWeylPair\] on $L^2(\mu)$ follows from their unitary equivalence to the standard pair. It is a simple exercise to determine the action of $Q$ and $P$ on their domains. They act as $$\begin{array}{c}
(Q_i f)(x) = \frac{x_i}{c}f(x)\\
(P_i f)(x) = -ic\left(\frac{\partial}{\partial x_i} - \frac{1}{2}\frac{\partial \phi}{\partial x_i}\right)f(x).
\end{array}$$ The integral kernel of operators $a(Q,P)$ can be calculated explicitly via unitary equivalence to the standard Weyl Calculus (see, for example, [@HormanderWeylCalc]), leading to the formula:

\[thmKernel\] For $a \in \mathcal{S}\left(\mathbb{R}^{2d}\right)$, $Q, P$ as above and $f \in L^2(\mu)$, we have for all $y \in \mathbb{R}^d$ $$(a(Q,P)f)(y) = \frac{1}{(2\pi)^{d}c^d} \int_{\mathbb{R}^{2d}}a\left(\frac{x+y}{2c}, \xi\right) \exp\left(-i\xi \left(\frac{x-y}{c}\right)\right) \exp\left(\frac{1}{2}(\phi(y)+\phi(x))\right)f(x) d\xi d\mu(x).$$

The inclusion of $c$ is to satisfy physicists. What many physicists would consider as THE 