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Suppose you have a system described by the following Lagrangian: $$L=(1-gq²)\dot{q}^2/2.$$

How would you quantize this theory? Do you need to symmetrize the Hamiltonian before promoting the coordinate and momentum to operators?

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  • $\begingroup$ What is usually understood as "canonical quantization" depends very much on whether the system is constrained. Can you check this? $\endgroup$
    – DanielC
    Dec 15, 2019 at 16:43

1 Answer 1

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  1. The 1D particle $L=\frac{1}{2}m(q)\dot{q}^2$ with a position-dependent mass $m(q)$ has classical momentum $p=m(q)\dot{q}$ and Hamiltonian $H=\frac{p^2}{2m(q)}$.

  2. The Hamiltonian operator should be a self-adjoint operator. However this does not make quantization unique.

  3. Following the same quantization strategy as my Phys.SE answer here where the mass $m(q)$ is viewed as the lone component $g_{qq}$ of a 1D metric, in the Schrödinger representation the momentum operator becomes $$ \hat{p}~=~ \frac{\hbar}{i} m(q)^{-1/4}\frac{\partial }{\partial q} m(q)^{1/4}, $$ and the Hamiltonian operator becomes $$ \hat{H}~=~ -\frac{\hbar^2}{2} m(q)^{-1/4}\frac{\partial }{\partial q} m(q)^{-1/2} \frac{\partial }{\partial q}m(q)^{-1/4}. $$

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