Does the lagrangian contain all the information about the representations of the fields in QFT?

Question

Given the Lagrangian density of a theory, are the representations on which the various fields transform uniquely determined?

For example, given the Lagrangian for a real scalar field $$ \mathscr{L} = \frac{1}{2} \partial_\mu \varphi \partial^\mu \varphi - \frac{1}{2} m^2 \varphi^2 \tag{1}$$ with $(+,-,-,-)$ Minkowski sign convention, is $\varphi$ somehow constrained to be a scalar, by the sole fact that it appears in this particular form in the Lagrangian?

As another example: consider the Lagrangian $$ \mathscr{L}_{1} = -\frac{1}{2} \partial_\nu A_\mu \partial^\nu A^\mu + \frac{1}{2} m^2 A_\mu A^\mu,\tag{2}$$ which can also be cast in the form $$ \mathscr{L}_{1} = \left( \frac{1}{2} \partial_\mu A^i \partial^\mu A^i - \frac{1}{2} m^2 A^i A^i \right) - \left( \frac{1}{2} \partial_\mu A^0 \partial^\mu A^0 - \frac{1}{2} m^2 A^0 A^0 \right). \tag{3}$$ I've heard$^{[1]}$ that this is the Lagrangian for four massive scalar fields and not that for a massive spin-1 field. Why is that? I understand that it produces a Klein-Gordon equation for each component of the field: $$ ( \square + m^2 ) A^\mu = 0, \tag{4}$$ but why does this prevent me from considering $A^\mu$ a spin-1 massive field?

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[1]: From Matthew D. Schwartz's Quantum Field Theory and the Standard Model, p.114:

A natural guess for the Lagrangian for a massive spin-1 field is $$ \mathcal{L} = - \frac{1}{2} \partial_\nu A_\mu \partial_\nu A_\mu + \frac{1}{2} m^2 A_\mu^2,\tag{8.17}$$ where $A_\mu^2 = A_\mu A^\mu$. Then the equations of motion are $$ ( \square + m^2) A_\mu = 0,\tag{8.18}$$ which has four propagating modes. In fact, this Lagrangian is not the Lagrangian for a amassive spin-1 field, but the Lagrangian for four massive scalar fields, $A_0, A_1, A_2$ and $A_3$. That is, we have reduced $4 = 1 \oplus 1 \oplus 1 \oplus 1$, which is not what we wanted.

Answer 1

What Qmechanic said in comments is pretty solid, "Lagrangian (2) is not bounded from below because the kinetic term of $A_0$ field has the wrong sign, and hence the theory is not physical in the first place", but I think your Question needs a change of emphasis. Your Lagrangian allows us to construct four equations of motion for four non-interacting fields. That the kinetic energy is not bounded below doesn't matter for a classical field theory if there are no interactions. We can say that $A_\mu$ transforms as a Minkowski 4-vector. But we can't do any significant classical physics with it because any Lorentz invariant interacting system would be unstable, and there have to be interactions of some sort (for us to be able to discuss measurement of the field by using its effects on other fields, for example, rather than just to talk about it as a theoretical object).

In the QFT context, however, we have to construct a positive semi-definite inner product (over the test function space, essentially the creation/annihilation operator commutator has to be positive-semidefinite in the 4-momentum coordinate space) for us to be able to construct a Fock space, which we cannot do for Lagrangian (2) even if there are no interactions. That is, QFT imposes an additional requirement even for a free quantum field, which forces us to the Proca equation or to the Maxwell equation for spin 1, because as well as a dynamics we also need a probability interpretation for observables (which is what the Hilbert space structure gives us).

EDIT: For local symmetries, if we use an n-dimensional vector space at each point, then the structure is defined by the ways in which we use the metric or other multilinear forms to construct a positive semi-definite inner product over the vector space at a point. If we use the metric to construct terms that can be written using the Lorentz metric, such as $g^{\mu\nu}A_\mu A_\nu$, then the field at a point can be taken to be a vector representation of the Lorentz group; if we use the metric to construct multinomials in $A_{\mu\nu}$, then we have a field that has a tensor structure, etc. If instead we introduce a different constant bilinear tensor $h^{\alpha\beta}$ and use the form $h^{\alpha\beta}A_\alpha A_\beta$, then the field is a vector representation of whatever symmetries $h$ has; if we use some higher degree multinomial in components $A_a$, then the field is a representation space of whatever the symmetries are of that higher degree multinomial.

Remembering, however, that we have to have a positive semi-definite energy for Physics, which requires work when we start from an indefinite metric.

There is also a global aspect of the Hilbert space, in the quantum case, that is not determined by the Lagrangian, at least insofar as one also has to specify the vacuum state, a thermal state, or some other state as a lowest energy state.

Answer 2

Field $\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}}$ with a given spin and mass (i.e. field which transforms under irrep of the Poincare group) must satisfy some determined conditions called irreducibility conditions: $$ \tag 1 \hat{W}^{2}\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}} = -m^{2}\frac{n + m}{2}\left(\frac{n + m}{2} + 1\right)\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}}, $$ $$ \tag 2 \hat{P}^{2}\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}} = m^{2}\psi_{a_{1}...a_{n}\dot{b}_{1}...\dot{b}_{m}}. $$ Here $\hat{W}$ is Pauli-Lubanski operator and $\hat{P}$ is translation operator. Representation with equal quantity $\frac{n + m}{2}$ are equivalent.

As for details, look here for the information about indices and Lorentz representations and here for fields as representations of Poincare group.

If you construct lagrangian which leads to $(1), (2)$, you will uniquely determine transformation properties of field with a given mass and spin.