# Can I Weyl-order the following Hamiltonian?

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 4 '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 4 '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 4 '15 at 18:49
• ...I don't see what you mean by "writing it" if you don't know $f$. – ACuriousMind May 4 '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 4 '15 at 19:08

1. 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. $$[Q,P]~=~i\hbar{\bf 1}, \qquad \{q,p\}_{PB}~=~1.$$
2. 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.
• The two quantum Hamiltonians agree up to first loop-order in $\hbar$, but differ at second loop-order. – Qmechanic May 5 '15 at 14:33