The nature of the spectroscopic factor is discussed here (from the point of view of mean-field) and here (from an experimental point of view). Here are some updates and additional discussions on the topic.

Up to now, I still believe that the quenching of the SF simply reflects the mismatch between experiment and theory, due to experimental conditions that limited configurations can be observed (so that the sum of SF is less than 1), and also the reaction theory does not take into account the correlation (so that the SF of each state is quenched). If both limitations are removed, there is no quenching. Although the SF is quenched, careful normalization gives us a great deal about the nuclear structure.

Is the spectroscopic factor or the cross-section quenched?

It seems that since the spectroscopic factor and the cross-section are closely related by

\displaystyle \sigma_{exp} = C^2S \sigma_{th}

so, the question is meaningless, because when SF is quenched from unity, the cross-section is also quenched from the therotical cross-section. However, since cross-section is an observable while SF is not, people consider that it is more proper to say quenching of cross-section.

Whether SF is an observable or not, an article Spectroscopic Factors: Observability and Measurability by L. D. Blokhintsev stated that “Spectroscopic factors are non-invariant under the unitary transformations of nuclear forces“, and Non-observability of Spectroscopic Factors by B. K. Jennings said that “At the level of microscopic calculations, they can be varied by unitary transformations and we expect hard and soft nucleon-nucleon potentials to generate quite different values for them. At the phenomenological level, the spectroscopic factor can be varied by field redefinitions.“. Thus, the spectroscopic factor is a pure construction that is fixed on a specific basis and interaction. Another article Nonobservable nature of the nuclear shell structure: Meaning, illustrations, and consequences by T. Duguet, H. Hergert, J. D. Holt, and V. Somà is worth reading.

However, I think SF is an observable because there should be a natural basis and fixed interaction. when the basis is fixed, and the 1st order of the nuclear interaction is well-known, the SF is well-defined, just like the coefficient of fractional percentage. Another argument is that there are many articles on extracting the spectroscopic factors, and although the uncertainty can be as large as 50% due to various experimental conditions and analysis methods, the community has a consensus on the quenching. That is, the community reached a consensus on the basis and range of interaction. I think that means, we are reaching the natural basis and 1st order understanding of the interaction.

The situation is like if we have the theoretical framework but lack some fundamental parameters fixed. The theory can produce all kinds of phenomena. Say, neutrino mass, without that, we don’t know the neutrino mass hierarchy. But we cannot say the neutrino mass hierarchy should fall in either the normal or inverted hierarchy and therefore neutrino mass hierarchy is not well-defined. My point is once the nuclear interaction and the basis are fixed, the SF is uniquely determined. So, some people would say, SF is an quasi-observable and that it is not an observable yet, but it will. May be, we should call SF is an extractable/deductible, which is a extractable/deductible quantity with some will-be-fixed parameter sets.

The last comment is that, if SF is not an observable, and is a pure arbitrary construct, all the works on it become meaningless. But, I believe SF is not aether, it reveals the structure and nucleon configuration of a nucleus. Also, the effective single-particle energy (ESPE) is an SF weighted energy of an orbital. If SF is arbitrarily constructed and can be anything, so the ESPE. but the ESPE very much agrees with the energy levels from Woods-Saxon potential which is a 1st order approximation of the nuclear mean field. A similar thing to the asymptotic normalization coefficient (ANC), ANC is a ratio of the wavefunction to the Coulomb wave function outside the nuclear potential, and the wavefunction is not an observable, so ANC is also not. But the wavefunction is surely can be deduced/extracted from the experimental data by fixed some nuclear parameters.

So, if SF is an observable, it is meaningless to ask is SF quenched or cross-section quenched.

If, the theoretical cross-section is perfectly calculated, the SF would be unity and no quenching for either SF or cross-section, and the nucleon configuration is embedded into the coefficient of fractional percentage.