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 = 12f(q)˙q2,
where f(q) is any function of the coordinate q. From this Lagrangian I obtain the Hamiltonian which is
H = p22f(q),
where p=f(q)˙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):=1f(q) for later convenience. Then the classical Hamiltonian reads 2h = g(q)p2.
One may show that the Weyl-ordered Hamiltonian reads 2HW = (g(q)p2)W = 14P2g(Q)+12Pg(Q)P+14g(Q)P2= Pg(Q)P−14ℏ2g′′(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] = iℏ1,{q,p}PB = 1.There exists another quantization method. If one chooses the Schrödinger representation for the momentum operator to be Q = q,P = ℏi4√f(q)∂∂q4√f(q),
it will become selfadjoint wrt. the measure μ = √f(q)dq.The Hamiltonian in the Schrödinger representation is (up to a multiplicative constant) the Laplace-Beltrami operator 2H = −ℏ22Δ = −ℏ2√f(q)∂∂q1√f(q)∂∂q,which is selfadjoint. Therefore the quantum Hamiltonian becomes 2H = 14√f(Q)P1√f(Q) P14√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.
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