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?


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.

  • $\begingroup$ Can you give me a reference of Weyl ordering and related stuffs? $\endgroup$ – rainman Mar 19 '13 at 17:05
  • $\begingroup$ $\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? $\endgroup$ – rainman Mar 19 '13 at 18:07
  • 1
    $\begingroup$ @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). $\endgroup$ – David Bar Moshe Mar 20 '13 at 7:56
  • 1
    $\begingroup$ @Ome Please see the following concise review: docs.google.com/…. $\endgroup$ – David Bar Moshe Mar 20 '13 at 8:01
  • 1
    $\begingroup$ @Ome cont. Please see also the following essay on the subject by Terence Tao: terrytao.wordpress.com/2012/10/07/… $\endgroup$ – David Bar Moshe Mar 20 '13 at 8:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.