Can I Weyl-order the following Hamiltonian?

Question

I am trying to perform a path integral but I am having trouble with the Weyl ordering of my Hamiltonian.

The Lagrangian of the system in question is

$$L~=~\frac{1}{2}f(q)\dot{q}^2,$$

where $f(q)$ is any function of the coordinate $q$. From this Lagrangian I obtain the Hamiltonian which is

$$H~=~\frac{p^2}{2f(q)},$$

where $p=f(q)\dot{q}$ is the canonical momenta.

Now, I want to perform a Path integral with this Hamiltonian. This is why I want that after quantization this Hamiltonian be Weyl-ordered.

My question is: Can I Weyl-order this Hamiltonian without knowing the explicit form of $f(q)$?

Since you have no explicit from of $f$, is your question whether you can claim that a Weyl-ordered symbol of $H$ exists without explicitly performing the Weyl order? — ACuriousMind, May 04 '15 at 18:17
You should take a look at the Weyl quantization from a mathematical standpoint. I would say that if the function $f(q)$ is sufficiently regular, you should not have problems. — yuggib, May 04 '15 at 18:22
@ACuriousMind i don't want to prove it exists. I want to know if I can somehow write it without knowing the exact form of $f(q)$ — Yossarian, May 04 '15 at 18:49
...I don't see what you mean by "writing it" if you don't know $f$. — ACuriousMind, May 04 '15 at 19:04
@ACuriousMind I just hoped that there might be a way, even without knowing the explicit form of $f(q)$, even though I guess it is not possible. — Yossarian, May 04 '15 at 19:08
Given a symbol $a(x,\xi)$, the Weyl quantization is the operator that acts as $a^W(x,D_x)\psi(x)=\int\int e^{2i\pi(x-y)\cdot\xi}a(\tfrac{x+y}{2},\xi)\psi(y)dyd\xi$. Take $a(x,\xi)=\tfrac{\xi^2}{2f(q)}$ and compute the integral. If it makes sense as an object of $L^2$ for $\psi$ in a dense subspace (e.g. of rapid decrease, smooth with compact support...), you get your Weyl quantization. — yuggib, May 04 '15 at 19:43

score 5 · Accepted Answer · edited Apr 13 '17 at 12:40

The answer is Yes. Define function $g(q):= \frac{1}{f(q)}$ for later convenience. Then the classical Hamiltonian reads $$2h~=~g(q)p^2.$$ One may show that the Weyl-ordered Hamiltonian reads $$2H_W~=~ (g(q)p^2)_W ~=~ \frac{1}{4}P^2 g(Q)+\frac{1}{2} Pg(Q)P+\frac{1}{4} g(Q)P^2$$ $$~=~ Pg(Q)P - \frac{1}{4}\hbar^2g^{\prime\prime}(Q),$$ see e.g. Ref. 1 and this Phys.SE post. Here $Q$ and $P$ denote the corresponding operators for the classical variables $q$ and $p$, respectively. $$ [Q,P]~=~i\hbar{\bf 1}, \qquad \{q,p\}_{PB}~=~1. $$
There exists another quantization method. If one chooses the Schrödinger representation for the momentum operator to be $$ Q~=~q, \qquad P~=~ \frac{\hbar}{i\sqrt[4]{f(q)}} \frac{\partial}{\partial q} \sqrt[4]{f(q)}, $$ it will become selfadjoint wrt. the measure $$\mu~=~\sqrt{f(q)}\mathrm{d}q.$$ The Hamiltonian in the Schrödinger representation is (up to a multiplicative constant) the Laplace-Beltrami operator $$ 2H~=~-\frac{\hbar^2}{2}\Delta~=~ -\frac{\hbar^2}{\sqrt{f(q)}}\frac{\partial}{\partial q}\frac{1}{\sqrt{f(q)}} \frac{\partial}{\partial q}, $$ which is selfadjoint. Therefore the quantum Hamiltonian becomes $$2H~=~ \frac{1}{\sqrt[4]{f(Q)}} P\frac{1}{\sqrt{f(Q)}}~P\frac{1}{\sqrt[4]{f(Q)}},$$ see e.g. Ref. 1 and my Phys.SE answer here.

References:

J. de Boer, B. Peeters, K. Skenderis and P. van Nieuwenhuizen, arXiv:hep-th/9511141; Section 2.

The two quantum Hamiltonians agree up to first loop-order in $\hbar$, but differ at second loop-order. — Qmechanic, May 05 '15 at 14:33
The Weyl ordering formula for arbitrary powers of p and q, and hence an arbitrary function g(q) was given 63 years before the de Boer et al ref, in McCoy, Neal H. "On the function in quantum mechanics which corresponds to a given function in classical mechanics." Proceedings of the National Academy of Sciences of the United States of America (1932), 674-676. — Cosmas Zachos, Sep 11 '15 at 23:48

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