Physics Asked by Taro on April 30, 2021
Others have asked how many quantum fields there are according to the Standard model (How many quantum fields are there?). In a comment on that post, it was claimed that in string theory there is only one “master” quantum field.
Is it true that there is only one master quantum field in string theory?
I suppose that when you write "master field" you mean the dynamical string field $Phi$ that appear in a given action of a particular string field theory. As in Witten's cubic string field theory the following linearized version of the action.
$$S(Phi) = -frac{1}{g^2}left[frac{1}{2}langle Phi,QPhi rangle + frac{1}{3}langle Phi, Phi * Phirangleright]$$
Then I guess the answer to your question is no.
For a simple reason. Over generic and non-Heterotic string vacua is expected that both open and unavoidable closed string sectors contribute to physical processes, an example is the famous tachyon condensation on the brane/anti-brane system. In such cases the formalism require the introduction of two different string fields, namely $Phi_{open}$ and $Phi_{closed}$. There is no (known) way to repackage them into a single string field because both obey different classical and quantum restrictions, different boundary conditions (such as the level-matching one), the moduli spaces of Riemann surfaces over they can be integrated are very different and the respective moduli spaces of picture changing and vertex operators are very different. It is very difficult to believe that they were two different components of a single string field.
I'm not sure about the heterotic cases. Even tough the entire Neveu-Schwarz sector has been constructed since many years ago I'm ignorant about the status of the construction of the Ramond-Ramond sector. I'm inconclusive here.
Some probably interesting references:
Maybe this is related to your question because those fermionic solitons are propagating fields not included explicitly in the type 0 action
Correct answer by Ramiro Hum-Sah on April 30, 2021
"Number of fields" is not a well-defined concept1. Here are some reasons that come to mind:
One can always introduce extra "auxiliary" fields, which can be integrated in/out, changing the number of fields.
Given two fields $A,B$, one can always define a tuple $vec C=(A,B)$, which now counts as "one field", or does it? Conversely, does a vector field $A_mu$ count as one field, or four? It seems reasonable to count fields as irreducible representations, but representation of what group? Only Lorentz? or also flavour symmetries? One could even introduce a "master" group under which all the fields transform together, as the components of a single field in the fundamental representation2.
Some systems have different descriptions (also known as dualities), where each description has different fields (and perhaps even different worldsheet dimension, as in AdS/CFT-type situations). So even the "field content" of a theory is not an intrinsic concept: it depends on the frame of reference, so to speak.
Etc. The claim "In String Theory there is only one master field" is meaningless. It is neither true nor false.
That being said, the standard presentation of String Theory contains 26 worldsheet scalars in the bosonic string, and 10 scalars plus 10 fermions in the supersymmetric case. Plus ghosts, if you want to count those. If you do not want to count ghosts, then it seems reasonable to gather the 26 scalars into a single vector (which is irreducible with respect to spacetime symmetry, namely the Lorentz group), and the 10 boson/fermion pairs into a single Wess-Zumino multiplet (which is also irreducible with respect to spacetime supersymmetry). So it is not unreasonable to claim that there is a single field, but again: this is so if you ignore ghosts, and think of the target manifold as that with the relevant symmetry group. With respect to the wordsheet, the fields are independent, and it is perhaps more natural to count them separately. An in the infrared the natural degrees of freedom are those of supergravity, which has a completely different field content.
1: This is why c-type functions, à la Zamolodchikov, are so useful: they give you an unambiguous definition of "number of degrees of freedom". In String Theory the fields are, in a sense, free bosons and free fermions in two dimensions. Roughly speaking, for free fields one has "central charge = number of fields", and so the latter is better defined than in typical QFTs. But this is a gauge theory, so it is still somewhat subtle. Anyway.
2: Needless to say, this "master" group is not a symmetry, but being a symmetry is a subtle concept, e.g. some groups that may appear to be actual symmetries are violated by quantum effects, and vice-versa.
Answered by AccidentalFourierTransform on April 30, 2021
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