# 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.

• Chapter IX.B.2 of the linked textbook suggests using $\pi_a:=e_a^\mu\partial_\mu$ for the momenta. It still does not generate the right action, but it can introduce some parts of the spin connection in to the mix. Sep 30, 2022 at 9:39

1. 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.

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

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

4. 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.

5. 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 $$\Box~=~ \frac{1}{\sqrt{|g|}}\frac{\partial}{\partial x^{\mu}}\sqrt{|g|}~ g^{\mu\nu} \frac{\partial}{\partial x^{\nu}}. \tag{D}$$

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

References:

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

2. Baofa Huang, BRST quantization of a scalar particle in a curved background, Int. J. Theor. Phys. 30 (1991) 783.

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$$^1$$ 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).