# 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?

• Doesn't answer your question regarding how exactly the transform relates to an ordering, but here is its name: en.wikipedia.org/wiki/Weyl_transform Mar 19, 2013 at 8:59

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 $\hat{f}(\hat{Z})=g\left(\sum_{I=1}^n k_I\hat{Z}^I\right)$.

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.

• Can you give me a reference of Weyl ordering and related stuffs? Mar 19, 2013 at 17:05
• $\int {dx\over2\pi}{dk\over2\pi} e^{ixP + ikQ} \int dpdqe^{-ixp-ikq} (sq+tp)^n =(sQ+tP)^n$ @David Bar Moshe: In this equation what are the x and k? In fact, in my question there are also x and p. What are they in that context? For integration wrt x an k what are the upper and lower limits? Can you please write it explicitly? Mar 19, 2013 at 18:07
• @Ome The variables x and k are just dummy integration variables. The integration variables are between minus and plus infinity (This is just a Fourier transform). Mar 20, 2013 at 7:56
• @Ome Please see the following concise review: docs.google.com/…. Mar 20, 2013 at 8:01
• @Ome cont. Please see also the following essay on the subject by Terence Tao: terrytao.wordpress.com/2012/10/07/… Mar 20, 2013 at 8:04

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.