Does graviton loops affect the seperately covariant conservation of energy momentum of two noninteracting sectors of matter

Question

Consider the action $$\int \sqrt{-g}\left[R[g]+\mathcal{L}_{m1}(g,\psi_1)+\mathcal{L}_{m2}(g,\psi_2)\right]$$ Classically we have $$\nabla^\mu T^1{}_{\mu\nu}=0,\,\,\,\,\nabla^\mu T^2{}_{\mu\nu}=0$$ seperately. Point is that at the level of classical action $\psi_1,\psi_2$ does NOT interact directly but they both interact with gravity, that is to say, there are no terms of the form $\mathcal{L}_{int}(g,\psi_1,\psi_2)$, which, if present, would enable the two sectors to covariantly exchange energy-momentum and reduce the relation above to the weaker form $$\nabla^\mu(T^1{}_{\mu\nu}+T^2{}_{\mu\nu})=0$$ My question is whether quantum corrections can have such effects as well? Because naively it seems quantum mechanically many possible direct interaction terms between $\psi_1$ and $\psi_2$, even if no such terms are present in the classical action. Thanks a lot!

An answer to a simpler problem might equally be of value to me: the reduced case of abelian gauge coupling, the action $S=\int -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\Psi_1 \gamma^\mu{D_\mu}\Psi_1+\Psi_2\gamma^\mu{D_\mu}\Psi_2$, whether $J_1$ and $J_2$ retains individual conservation.

Comments

What do you mean by direct interaction? There's no $\psi_1-\psi_2$ vertex in your lagrangian. However one can typically have loop diagrams with only fermion external lines, is that what you are thinking?

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Hi @xavier, the \slashed commands weren't working in your post, so I replaced them with \(\gamma^\mu D_\mu\), which is the same thing. Hope it's OK.

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yes exactly, because classically we'll have separate conservation laws for the two flows which is derivable from the fact that there's no $\Psi_1 - \Psi_2$ vertex in the lagrangian. I'm wondering if this is still true quantumally, or if there's still two separate conservation laws at all. The reason being what you said.

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and what is the order of the violation of separate conservation

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@xavier, My impression is that they will still be conserved separately in the quantized case, unless there are anomalies. I will write an answer shortly.

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In your simple example I can see the symmetries which guarantee the separate conservation (modulo anomalies). In fact it is more traditional to talk not about the separate "electric currents" of $\Psi_1$ and $\Psi_2$ but about the total electric current which comes with the global gauge transformations (and you must keep it anomaly free) and under which the fields have the same charge, and the flavor current under which the fields are charged oppositely. I cannot see however why the stress tensors would be conserved separately with dynamical gravity (and if you want graviton exchange, you need dynamical gravity) even classically. Am I missing something obvious here?

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Indeed the cases seem very different, the QED case has two obvious global symmetries guaranteeing the conservation. whereas in the gravity case there's no global symmetry, I find that confusing too. However, the conservation of each sector is a consequence of matter equation of motion and has nothing to do with whether gravity is dynamical (in my opinion). The conservation also follows from seperate coordinate invariance of the two matter parts of the action, we can apply the following procedure separately to each sector, because it is merely a (infinitesimal) change of coordinate. ($X$ is arbitrary vector field which vanishes on $\partial D$) \begin{align} 0&=\int_D L_X\mathcal{L}\nonumber\\ &=\int_D (E_\psi L_X\psi + T^{ab}L_X g_{ab})Vol\nonumber\\ &=\int_D (2T^{ab} \nabla_a X_b)Vol\nonumber\\ &=\int_{\partial D}boundaryterms-2\int_D (\nabla_a T^{ab})X_b\cdot Vol \end{align} Please keep me up to date with your opinion

Answer 1

There are two layers underlying the issue: The first layer is, formally we can show every classical conservation law has a quantum counterpart through, say, canonical quantization(see Weinberg's QFT chap7). The second layer is, if we go beyond the formal level, it is really that we would want the conservation law to hold and strive to make it hold in the quantized case too, unless there are obstructions such as quantum anomalies. 

On the surface of the issue, if we canonically quantize a classical Lagrangian, all conservation laws should be preserved, and this is fairly simple to see: the field operator equations have exactly the same forms of their classical versions, hence if any classical quantity has a vanishing 4-divergence, so will its quantized version.

However, reading between OP's lines, I suspect OP is having the misconception that a QFT can be defined by "a Lagrangian + a quantization recipe(canonical, path integral etc.)" This is not entirely true, a regularization and renormalization scheme must also be specified to make the QFT unambiguous. During this procedure, many symmetries are broken in the intermediate stages, and it is not a priori clear that the symmetries can all be restored after the procedure is finished.

The common practice is, through carefully controlling the regularize-renormalize procedure, we define a QFT preserving as many symmetries and conservations as possible. It is not always possible to preserve all the classical symmetries at hands, in such cases we will have the quantum anomalies.

Hence my earlier comment under your question:

My impression is that they will still be conserved separately in the quantized case, unless there are anomalies.

Comments

Thank you for the answer and the points you made, it's been beneficial~ In that case is there any quick way to see if a symmetry can be preserved in the quantum theory?

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In the abelian case the separate conservation seems to remain true, because all vertices respects the conservation, because photons don't carry charge. However the nonabelian case look more subtle.

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 In that case is there any quick way to see if a symmetry can be preserved in the quantum theory?

@xavier, This is beyond my expertise, I suspect there has been notable progress in detecting anomalies mathematically. Maybe @UrsSchreiber can shed some light on the question?

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Is the @ functioning?

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@xavier yes, and it works everywhere and has some additional useful features compared to SE for example..