Constructing a field theory action for the point particle in curved space The point particle action in the Hamiltonian formalism is
$$
S = \int d\tau \Big( -p_\mu \dot{x}^\mu - \frac{e}{2}(g^{\mu\nu} p_\mu p_\nu - m^2) \Big) \ ,\tag{1}
$$
where I explicitly displayed the metric, which depends on $x$. The sign convention is $(+,-,\ldots,-)$.  The Poisson brackets for the fields are $$\{x^\mu,p_\nu\}=\delta_\nu^\mu,\tag{2}$$ which can be replaced with commutation relations. We need some normal ordering prescription so I chose to put all $x$ to the left of $p$ (I hope this is consistent?).
I would like to construct a field theory action for this particle a la Siegel in his textbook, unnumbered equation on page 381. The first step is to find the BRST operator, which in this case is given by the $c$ ghost times the (first class) constraint:
$$
Q = c \frac{1}{2}(p^2-m^2) \ .\tag{3}
$$
Then the action should take the form
$$
S = \int d^d x dc\  \Phi(x,c)Q\Phi(x,c) \ .\tag{4}
$$
Since the metric appears on the left of the momenta, it is straight forward to take commutators with the fields $$\Phi(x,c) = \phi(x)+c\psi(x)\tag{5}$$ and to integrate over $c$, obtaining simply
$$
S = \frac{1}{2}\int d^d x \  \phi(x)(g^{\mu\nu}\partial_\mu\partial_\nu+m^2)\phi(x) \ .\tag{6}
$$
This is not the covariant action I was hoping to see, I thought I would find the covariant derivative in the kinetic term, $D^2$.
Does the commutator of $p_\mu$ with a tensor turn in to a covariant derivative, eg $$[p_\mu,A^\nu] = D_\mu A^\nu~?\tag{7}$$ If so, then why?
I think the resulting field-theory action should be generally covariant, since one can integrate out the fields $p$ and $e$ in the particle action and obtain the standard $\sqrt{\dot{x}^2}$ action which gives rise to the usual geodesic equation.
EDIT 1: It seems like being able to compute the successive Poisson brackets $\{p_\mu,\{p_\nu,\phi(x)\}\}$ is essentially the problem. The first bracket gives a partial derivative, while the second one needs to know how to act on a vector.
EDIT 2: This article basically says they do what I was after, the author says they will make the change $p_\mu\to\partial_\mu$, but then makes use of $$\Box = g^{-1/2}\partial_\mu g^{1/2}g^{\mu\nu}\partial_\nu,\tag{8}$$ which just gives the right answer for the scalar action. Well if we play a game of getting the right answer at any cost I would rather replace $p_\mu\to\nabla_\mu$. But I would like to know whats the reason for making replacements.
 A: *

*First of all, one should not forget that quantization is not a unique procedure. Nevertheless, some choices are more natural than other. Here is one line of reasoning.


*The volume form in configuration space (=spacetime) $M$ is
$$ \mu ~=~ \sqrt{|g|}~ \mathrm{d}x^0 \wedge \mathrm{d}x^1 \wedge \ldots \wedge \mathrm{d}x^{d-1}. \tag{A}$$


*OP's actions (4) & (6) should include a $\sqrt{|g|}$ factor to be invariant under general coordinate transformations $x\to x^{\prime}$.


*The Hilbert space is ${\cal H}=L^2(M,\mu)$. The Schrödinger representation of the momentum operator that (i) is self-adjoint wrt. the measure $\mu$ and (ii) satisfies the CCR reads
$$ \hat{p}_{\mu}~=~ \frac{\hbar}{i\sqrt[4]{|g|}} \frac{\partial}{\partial x^{\mu}} \sqrt[4]{|g|}, \tag{B}$$
cf. my Phys.SE answer here.


*Similarly, we should find a self-adjoint Schrödinger representation for the operator of the mass-shell condition$^1$
$$ p_{\mu}g^{\mu\nu}p_{\nu}+m^2~\approx~0.\tag{C}$$
The d'Alembertian that is self-adjoint wrt. the measure $\mu$ reads
$$ \Box~=~ \frac{1}{\sqrt{|g|}}\frac{\partial}{\partial x^{\mu}}\sqrt{|g|}~ g^{\mu\nu} \frac{\partial}{\partial x^{\nu}}. \tag{D}$$


*Finally, for the canonical momentum in the presence of a gauge potential, see e.g. this Phys.SE post.
References:

*

*Warren Siegel, Fields, arXiv:hep-th/9912205; p. 381.


*Baofa Huang, BRST quantization of a scalar particle in a curved background, Int. J. Theor. Phys. 30 (1991) 783.
--
$^1$ This answer, Ref. 1 p. 55, and Ref. 2 use $(-,+,\ldots,+)$ sign convention. Note that Ref. 1 p. 171 defines the Lagrangian $L=U-T$ and hence the action $S=\int dt~L$  oppositely of the standard convention. This explains the minus sign in the $p\dot{x}$ term of OP's eq. (1).
