Weyl Ordering Rule While studying Path Integrals in Quantum Mechanics I have found that [Srednicki: Eqn. no. 6.6] the quantum Hamiltonian $\hat{H}(\hat{P},\hat{Q})$ can be given in terms of the classical Hamiltonian $H(p,q)$ by 
$$\hat{H}(\hat{P},\hat{Q}) \equiv \int {dx\over2\pi}\,{dk\over2\pi}\,
e^{ix\hat{P} + ik\hat{Q}} \int dp\,dq\,e^{-ixp-ikq}\,H(p,q)\; \tag{6.6}$$
if we adopt the Weyl ordering.
How can I derive this equation?
 A: The basic Weyl ordering property generating all the Weyl ordering identities for polynomial functions is:
$$
((sq+tp)^n)_W = (sQ+tP)^n
$$
$(q, p)$ are the commuting phase space variables, $(Q, P)$ are the corresponding noncommuting operators (satisfying $[Q,P] = i\hbar $).
For example for n = 2, the identity coming from the coefficient for the
$st$ term is the known basic Weyl ordering identity:
$$
(qp)_W = \frac{1}{2}(QP+PQ)
$$
By choosing the classical Hamiltonian as $h(p,q) = (sq+tp)^n$ and carefully performing the Fourier and inverse Fourier transforms, we obtain the Weyl identity:
$$
\int {dx\over2\pi}{dk\over2\pi} e^{ixP + ikQ} \int dpdqe^{-ixp-ikq} (sq+tp)^n =(sQ+tP)^n
$$
The Fourier integral can be solved after the change of variables:
$$
l = sq+tp, m = tq-sp
$$
and using the identity
$$
\int dl e^{-iul} l^n  =2 \pi \frac{\partial^n}{\partial v^n} \delta_D(v)|_{v=u}
$$
Where $\delta_D$ is the Dirac delta function.
A: Let the position and momentum operators in $n$ phase-space dimensions be collectively denoted $\hat{Z}^I$, and let the corresponding symbols be denoted $z^{I}$, where $I\in\{1,\ldots,n\}$. The operator $\hat{f}(\hat{Z})$ corresponding to the Weyl-symbol $f(z)$ is
$$ \hat{f}(\hat{Z})~\stackrel{\begin{matrix}\text{symmetri-}\\ \text{zation}\end{matrix}}{=}~ \left.\sum_{m=0}^{\infty}\frac{1}{m!}\left[\hat{Z}^1\frac{\partial}{\partial z^1}+\ldots +\hat{Z}^n\frac{\partial}{\partial z^n} \right]^m f(z)\right|_{z=0} \qquad $$
$$~\stackrel{\begin{matrix}\text{Taylor}\\ \text{expan.}\end{matrix}}{=}~ \left.\exp\left[\sum_{I=1}^n\hat{Z}^I\frac{\partial}{\partial z^I}\right] f(z)\right|_{z=0} \qquad $$
$$~=~\int_{\mathbb{R}^{n}} \! d^{n}z~\delta^{n}(z)~ \exp\left[\sum_{I=1}^n\hat{Z}^I\frac{\partial}{\partial z^I}\right] f(z)  $$
$$ ~\stackrel{\delta\text{-fct}}{=}~\int_{\mathbb{R}^{2n}} \! \frac{d^{n}z~d^{n}k}{(2\pi)^{n}} \exp\left[-i\sum_{J=1}^n k_Jz^J\right] \exp\left[\sum_{I=1}^n \hat{Z}^I\frac{\partial}{\partial z^I}\right] f(z)$$
$$~\stackrel{\text{int. by parts}}{=}~\int_{\mathbb{R}^{2n}} \! \frac{d^{n}z~d^{n}k}{(2\pi)^{n}} f(z)~ \exp\left[-\sum_{I=1}^n\hat{Z}^I\frac{\partial}{\partial z^I}\right] \exp\left[-i\sum_{J=1}^n k_Jz^J\right]  $$
$$~=~\int_{\mathbb{R}^{2n}} \! \frac{d^{n}z~d^{n}k}{(2\pi)^{n}} f(z)~ \exp\left[i\sum_{I=1}^n k_I\hat{Z}^I\right] \exp\left[-i\sum_{J=1}^n k_Jz^J\right]  $$
$$~\stackrel{\text{BCH}}{=}~\int_{\mathbb{R}^{2n}} \! \frac{d^{n}z~d^{n}k}{(2\pi)^{n}} f(z)~ \exp\left[i\sum_{I=1}^n k_I(\hat{Z}^I-z^I)\right].$$
The above manipulations make sense for a sufficiently well-behaved function $z\mapsto f(z)$.
Example: If the Weyl-symbol is of the form $f(z)=g\left(\sum_{I=1}^n k_I z^I\right)$ for some analytic function $g:\mathbb{C}\to \mathbb{C}$, then the corresponding operator is $\hat{f}(\hat{Z})=g\left(\sum_{I=1}^n k_I\hat{Z}^I\right)$.
A: Another way to look at this:
$e^{ix\hat{P}+ik\hat{Q}}$ is automatically Weyl-ordered. This is because each term in the Taylor expansion, $\frac{1}{n!}(ix\hat{P}+ik\hat{Q})^n$, is Weyl-ordered. You can see this just by multiplying out the terms. For example, $$(\hat{P}+\hat{Q})(\hat{P}+\hat{Q})=\hat{P}^2+\hat{P}\hat{Q}+\hat{Q}\hat{P}+\hat{Q}^2.$$
More generally, if $\hat{P}^m\hat{Q}^l\hat{P}^p\cdots$ is a term in $(\hat{P}+\hat{Q})^n$, then every unique permutation of those factors is also a term in $(\hat{P}+\hat{Q})^n$.
Therefore, if we take the Fourier transform of a classical Hamiltonian,
$$\int dp\,dq\,e^{-ixp-ikq}\,H(p,q),$$
and then take the inverse Fourier transform, only replacing the variables $p$ and $q$ with operators,
$$\int {dx\over2\pi}\,{dk\over2\pi}\,
e^{ix\hat{P} + ik\hat{Q}} \int dp\,dq\,e^{-ixp-ikq}\,H(p,q),$$
we end up with a Hamiltonian $\hat{H}(\hat{P},\hat{Q})$ that is Weyl-ordered and naturally associated with the classical Hamiltonian.
