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This is a follow-up to the question In what sense is the word quantum fluctuation used here?

In arXiv:0710.3787 it is stated on page 7 that there are "active fluctuations in the intrinsic degrees of freedom of gravity" and on page 13 that "Quantum stress tensor fluctuations lead to passive fluctuations of the gravitational field, which are in addition to the active fluctuations coming from the quantization of the gravitational field itself."

I will once again give my definition of quantum fluctuations:

"Quantum fluctuation" is an informal name for the fact that: In any quantum mechanical system with states in the Hilbert space $\mathcal H$, if you have a self-adjoint linear (densely defined, maybe?) operator $$\hat{\mathcal{O}}:\mathcal H\supset D(\hat{\mathcal O})\to\mathcal H$$ corresponding to some observable $\mathcal O$ and a state $\vert\psi\rangle\in\mathcal H$ which is not an eigenvector of $\hat{\mathcal O}$, then performing the same measurement on a physical system in the state $\vert\psi\rangle$ will lead to different results. For further discussion of my understanding see [1], [2], [3], [4]. In particular, the word "fluctuation" is misleading since the above fact need not have anything to do with a change in time (or space).

Now, what do the "active fluctuations" in the quoted text blocks above refer to? Are they fluctuations in my sense? Are they something else?

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The "fluctuations" language is addressed in my answer to the previous question, so in this answer I'll only address the "active" and "passive" language.

In classical general relativity, the metric is still dynamic even when/where the stress-energy tensor is zero, which is why gravitational waves are possible. "Active quantum fluctuations" refers to quantum fluctuations in those gravitational-wave degrees of freedom. "Passive" refers to fluctuations in the rest of the metric degrees of freedom, those that are tied to fluctuations in the stress-energy tensor as shown in equation (1) of arXiv:1703.05331.

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    $\begingroup$ For completeness, I will state the general-relativity-property that you mention here (please correct me if I made a mistake or understood you incorrectly): The Einstein field equations (EFE) with $\Lambda=0$ and $T=0$ are $$Rc-\frac12Sg=0.$$ Multiplying the $i,j$-th equation with $g^{j,i}$ and summing over $i,j$ yields $$\sum_{i,j=1}^4Rc_{i,j}g^{j,i}-\frac12Sg_{i,j}g^{j,i}=0,$$ which (exercise for the reader), is the same as $$S-2S =0.$$ Thus, $S=0$. Putting this into EFE gives $Rc=0$. But $Rc=0$ doesn't imply Riemannian curvature$=0$, so there are still non-trivial EFE solutions with $T=0$. $\endgroup$
    – Maximilian Janisch
    Commented Sep 26, 2021 at 17:13
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    $\begingroup$ @MaximilianJanisch That's all correct. $\endgroup$
    – Chiral Anomaly
    Commented Sep 26, 2021 at 17:24

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