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?


[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 $$ \mathcal{L} = - \frac{1}{2} \partial_\nu A_\mu \partial_\nu A_\mu + \frac{1}{2} m^2 A_\mu^2,$$ where $A_\mu^2 = A_\mu A^\mu$. Then the equations of motion are $$ ( \square + m^2) A_\mu = 0,$$ 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.

  • Why would writing out the Lagrangian for each component of a vector field prevent you from viewing the vector field as a vector field? I think whatever you heard about not being able to do so is wrong. – bianchira Nov 26 '14 at 17:21
  • Also re the title, at least within the scope of what you're asking, the Lagrangian specifies the representation by the virtue that it is written in terms of a field in some specific rep, e.g. a scalar field Lagrangian specifies dynamics of a scalar field not a vector one. But of course that says nothing about not being able to view components of a vector field as scalar fields – bianchira Nov 26 '14 at 17:28
  • If each component of A satisfies the Klein-Gordon equation, that doesn't necessarily mean that the components of A transform like a vector under Lorentz transformations. – jabirali Nov 26 '14 at 18:05
  • 3
    Comment to the question (v5): As M. Schwartz mentions on top of p. 115, the energy density for the Lagrangian (2) is not bounded from below because the kinetic term of the $A_0$ field has the wrong sign, and hence the theory is not physical in the first place. Therefore the discussion of possible representations and interpretations of (2) seems somewhat academic. On the other hand, if $A_0$ did not have the wrong sign, then $A_{\mu}$ could not be viewed as a 4-covector, but could only be interpreted as 4 scalars. – Qmechanic Nov 26 '14 at 20:21
  • 1
    @glance, "why can't I say (or can I?) that Aμ is a spin-1 field for that choice of the Lagrangian?" Yes you definitely can, then the reason to reject (2) becomes the energy bounded below condition, instead of the one Schwartz gave. Qmechanic's comment is right on. – Jia Yiyang Nov 27 '14 at 13:20

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.

  • thanks for the answer. However, you focused on a particular example I provided. Can you also say something about the main question (and the other example I gave)? – glS Nov 27 '14 at 20:24
  • @glance Right, and apologies. Read the Question carefully. I take it that there are two aspects to the question: the local and the global. See my Edit, to appear momentarily, I hope. – Peter Morgan Nov 28 '14 at 23:13

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.

  • could you elaborate a bit? What kind of field is $\psi$? Why did you used different kind of labels (some dotted and some not) for its indices? Can you also give some reference for your statements (or show their validity)? Finally, you didn't directly addressed the point of the question: how does the Lagrangian lead to the representation of the fields you are mentioning? – glS Nov 27 '14 at 21:51
  • @glance : as for the references, I've added some into the answer. As for the last question, the answer is following. Lagrangian is secondary quantity in QFT formalism, because equations of motion are obtained by using requirement of irreducibility of representation which is realizes by field. Equations of motion determine the representation uniquely. Lagrangians are constructed by the way that they must lead to the equations of motion, which determine the representation uniquely. – Andrew McAddams Nov 28 '14 at 8:31

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.