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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

$$\hat{H}(\hat{P},\hat{Q}) \equiv \int {dx\over2\pi}\,{dk\over2\pi}\, e^{ix\hat{P} + ik\hat{Q}} \int dp\,dq\,e^{-ixp-ikq}\,H(p,q)\;$$

if we adopt the Weyl ordering.

How can I derive this equation?

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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 –  NikolajK Mar 19 '13 at 8:59
    
@NickKidman: Thanks a lot. –  Ome Mar 19 '13 at 9:11
    
Wow. I read the exactly, the same thing today, and was going to ask the same question. :) –  ramanujan_dirac Mar 20 '13 at 3:34
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2 Answers 2

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.

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Can you give me a reference of Weyl ordering and related stuffs? –  Ome Mar 19 '13 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? –  Ome Mar 19 '13 at 18:07
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@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). –  David Bar Moshe Mar 20 '13 at 7:56
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@Ome Please see the following concise review: docs.google.com/…. –  David Bar Moshe Mar 20 '13 at 8:01
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@Ome cont. Please see also the following essay on the subject by Terence Tao: terrytao.wordpress.com/2012/10/07/… –  David Bar Moshe Mar 20 '13 at 8:04
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Let the position and momentum operators in $n$ phase space dimensions be collectively denoted $\hat{Z}^I$, and let the corresponding symbol be denoted $z^{I}$, where $I\in\{1,\ldots,n\}$. The operator $\hat{f}(\hat{Z})$ corresponding to the Weyl-symbol $f(z)$ is

$$ \hat{f}(\hat{Z})~\stackrel{\begin{matrix}\text{symmetri-}\\ \text{zation}\end{matrix}}{=}~ \left.\sum_{m=0}^{\infty}\frac{1}{m!}\left[\hat{Z}^1\frac{\partial}{\partial z^1}+\ldots +\hat{Z}^n\frac{\partial}{\partial z^n} \right]^m f(z)\right|_{z=0} \qquad $$ $$~\stackrel{\begin{matrix}\text{Taylor}\\ \text{expan.}\end{matrix}}{=}~ \left.\exp\left[\hat{Z}^I\frac{\partial}{\partial z^I}\right] f(z)\right|_{z=0} \qquad $$ $$~=~\int_{\mathbb{R}^{n}} \! d^{n}z~\delta^{n}(z)~ \exp\left[\hat{Z}^I\frac{\partial}{\partial z^I}\right] f(z) $$ $$ ~\stackrel{\delta\text{-fct}}{=}~\int_{\mathbb{R}^{2n}} \! \frac{d^{n}z~d^{n}k}{(2\pi)^{n}} \exp\left[-ik_Jz^J\right] \exp\left[\hat{Z}^I\frac{\partial}{\partial z^I}\right] f(z)$$ $$~\stackrel{\text{int. by part}}{=}~\int_{\mathbb{R}^{2n}} \! \frac{d^{n}z~d^{n}k}{(2\pi)^{n}} f(z)~ \exp\left[-\hat{Z}^I\frac{\partial}{\partial z^I}\right] \exp\left[-ik_Jz^J\right] $$ $$~=~\int_{\mathbb{R}^{2n}} \! \frac{d^{n}z~d^{n}k}{(2\pi)^{n}} f(z)~ \exp\left[ik_I\hat{Z}^I\right] \exp\left[-ik_Jz^J\right] $$ $$~\stackrel{\text{BCH}}{=}~\int_{\mathbb{R}^{2n}} \! \frac{d^{n}z~d^{n}k}{(2\pi)^{n}} f(z)~ \exp\left[ik_I(\hat{Z}^I-z^I)\right].$$

The above manipulations make sense for a sufficiently well-behaved function $f(z)$.

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