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 $f(q)$?
Answer
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.
No comments:
Post a Comment