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In flat space we have $$\hat{p}_\mu=-i\hbar \partial_\mu . $$ Does this still hold in a curved spacetime (particularly Schwarzschild space)?

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A short-ish answer to this : $\hat{p}_a = -i\hbar \partial_a$ is not appropriate for the momentum operator due to the fact that it is not self-adjoint. As a reminder, $\hat{p}$ is hermitian if

$$\langle \psi, \hat{p}_a \phi \rangle = \langle \hat{p}_a\psi, \phi \rangle$$

In a curved space, that will be

$$\langle \psi, \hat{p}_a \phi \rangle = \int \psi^* (-i\hbar \partial_a \phi) \sqrt{g}\ d^3 x$$

Integrating it by parts,

\begin{eqnarray} \langle \psi, \hat{p}_a \phi \rangle &=& -i\hbar [\psi^* \phi ] + i\hbar \int \partial_a (\psi^* \sqrt{g}) \phi d^3x\\ &=& -i\hbar [\psi^* \phi ] + i\hbar \int \left[ \sqrt{g} \partial_a (\psi^*) + \psi^* \partial_a (\sqrt{g}) \right] \phi d^3x\\ &=& -i\hbar [\psi^* \phi ] + i\hbar \int (-i\hbar (\partial_a + \frac{\partial_a \sqrt{g}}{\sqrt{g}}) \psi)^* \phi d^3x\\ \end{eqnarray}

It's not actually true that wavefunctions tend to 0 at infinity, but the boundary term can be neglected. We still have an extra term, though, which can be gotten rid of by using

$$\hat{p}_a = -i\hbar(\partial_a + \frac{1}{2} \frac{\partial_a \sqrt{g}}{\sqrt{g}})$$

From differential geometry, we also have that

$$\frac{\partial_a \sqrt{g}}{\sqrt{g}} = {\Gamma^b}_{ab} = \Gamma_a$$

so

$$\hat{p}_a = -i\hbar(\partial_a + \frac{1}{2} \Gamma_a)$$

This is both self-adjoint and obeys the canonical commutation relations.

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  • $\begingroup$ I ended up figuring this out independantly. But I sincerely appreciate this write up for future students! One thing to remember is the $\sqrt(g)$ is the determinant of the spacial metric, not the whole spacetime metric $\endgroup$ – Craig Aug 14 at 5:52
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In the paper [1], p.111, it is said that the operator of the square of momentum on a curved $n$-dimensional pseudo-Riemannian manifold is: $$ P^2 = -\hbar^2 \left( \square - \frac{n-2}{4(n-1)}R \right) $$ Here, $R$ is the scalar curvature. This $P^2$ is the analog of the classical one $g^{\alpha\beta}p_\alpha p_\beta = m^2 c^2$.

The inclusion of the scalar curvature $R$ is required by conformal invariance of zero mass scalar field. In the case where $n=4$, we have the analog of the Klein-Gordon equation on a curved space-time: $$ \left(\square - \frac16 R\right) \varphi + \left(\frac{mc}{\hbar}\right)^2\varphi = 0 $$ They say that this equation was considered by Penrose [3] in 1964. The "non-relativistic" limit of this should be Schrödinger equation where $R$ appears : $$ \hat{H}\psi = -\frac{\hbar^2}{2m} \left(\Delta - \frac16 R\right)\psi $$ This equation can also be found via geometric quantization of the hamiltonian responsible of the geodesic flow, see [2], eq. (7.114). Now, what is $\hat{p}_i$ on a curved space ? According to geometric quantization (see again [2], eq. (7.42) (7.82)) it is given, at least in the non-relativistic limit, by: $$ \hat{p}_i = -i\hbar \left(\partial_i + \frac12 \mathrm{Div}(\partial_i)\right) $$ Here, $\partial_i := \partial/\partial q^i$ and $\mathrm{Div}(\partial_i)=|\det[g]|^{-1/2}\partial_i(|\det[g]|^{1/2})=\Gamma^k_{ik}$ is the covariant divergence of the vector field $\partial_i$.

Now, what is $\hat{p}_\mu$ ? I guess it would be $-i\hbar \left(\partial_\mu + \frac12 \mathrm{Div}(\partial_\mu)\right)$, but I'm not sure.

So, to answer your question : Does $\hat{p}_\mu = -i\hbar\partial_\mu$ still hold in a curved spacetime? I would say no. There should be at least some divergence term $\mathrm{Div}(\partial_\mu)$ somewhere.


Remark : According to the Wikipedia sign convention : $$ \Gamma^k_{ij} := \frac12 g^{km}(\partial_i g_{jm} + \partial_j g_{im} - \partial_m g_{ij}) $$ $$ R^l_{kij} := \partial_i\Gamma^l_{jk} - \partial_j\Gamma^l_{ik} + \Gamma^l_{im}\Gamma^m_{jk} - \Gamma^l_{jm}\Gamma^m_{ik} $$ $$ R_{ij} := R^k_{ikj} = \partial_k \Gamma^k_{ij} - \partial_j \Gamma^k_{ik} + \Gamma^l_{ij}\Gamma^k_{kl} - \Gamma^l_{ik}\Gamma^k_{jl} $$ $$ R := g^{ij} R_{ij} $$

the "good" sign is $\square - R/6$ as in [2], p.180, and not $\square + R/6$ as in [1] and [3]. Also, even if [2], p.133, has a different definition of the Riemann curvature and scalar curvature, at the end of the day the scalar curvature according to [2] is the same as the one written here.


  • [1] : Quantum theory of scalar field in de Sitter space-time, N. A. Chernikov and E. A. Tagirov, 1968
  • [2] : Geometric quantization and quantum mechanics, J. Sniatycki, 1980.
  • [3] : Conformal treatment of infinity, R. Penrose, 1964.
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This operator exists in flat spacetime because one has translational symmetry $x^\mu \to x^\mu + a^\mu$. This is no longer true in generic curved spacetimes. So no, there is no momentum operator in curved spacetimes.

That being said, quantum mechanics/field theory is well-defined only in spacetimes which are globally hyperbolic, i.e. those that have a globally time-like Killing vector $\xi$ (with $\xi^2 < 0$). In such spacetimes, there is a time-translation symmetry so there exists a Hamiltonian operator ${\hat H} = - i \hbar \xi^\mu \partial_\mu$. In Schwarzschild spacetime, $\xi = \partial_t$.

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    $\begingroup$ I don't know why you say the momentum operator cannot apply in curved spacetime. You can write the full energy-momentum relation as in context with a Dirac Hamiltonian in curved spacetimes. Such things can exist. $\endgroup$ – Gareth Meredith Apr 3 at 22:09
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    $\begingroup$ Existence of a Hermitian operator such as ${\hat p}$ implies (via exponentiation) existence of a unitary operator ${\hat U} = \exp ( i a {\hat p} )$ and existence of such a unitary operator implies a symmetry (since we can transform all operators via the unitary transformations $O \to O' = U O U^{-1}$ and unitary transformations preserves correlation functions). In the case of the momentum operator the corresponding symmetry is space or time translations. In a generic curved background such symmetries simply do not exist! $\endgroup$ – Prahar Apr 4 at 0:34
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    $\begingroup$ @GarethMeredith - In any local quantum field theory on a curved background, there continues to exist a local energy-momentum tensor $T_{\mu\nu}$ which is covariantly conserved $\nabla^\mu T_{\mu\nu} = 0$. This does not however imply existence of a translation generator $p_\mu$. $\endgroup$ – Prahar Apr 4 at 0:37
  • $\begingroup$ So explain why we can construct the Dirac Hamiltonian on curved spacetimes? That involves an energy and momentum operator. $\endgroup$ – Gareth Meredith Apr 4 at 9:01
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    $\begingroup$ The existence of a unitary operator does not imply the existence of a symmetry. It is only a symmetry if it preserves the Hamiltonian. The absence of translational invariance does not mean that momentum doesn't exist, just that it isn't conserved. $\endgroup$ – Subhaneil Lahiri Apr 4 at 14:15

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