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

+ 4 like - 0 dislike
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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 prevents 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.

This post imported from StackExchange Physics at 2014-11-27 10:37 (UTC), posted by SE-user glance
asked Nov 26, 2014 in Theoretical Physics by glance (65 points) [ no revision ]
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.

This post imported from StackExchange Physics at 2014-11-27 10:37 (UTC), posted by SE-user bechira
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

This post imported from StackExchange Physics at 2014-11-27 10:37 (UTC), posted by SE-user bechira
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.

This post imported from StackExchange Physics at 2014-11-27 10:37 (UTC), posted by SE-user jabirali
For the actual Lagrangians describing a vector field, google the Maxwell Lagrangian (massless spin-1 field) and Proca Lagrangian (massive spin-1 field).

This post imported from StackExchange Physics at 2014-11-27 10:37 (UTC), posted by SE-user jabirali
@glance i see i misundersyood the question a bit the first time. the author is saying that the lagrangian constructed is not in the desired 3+1 rep, as jabirali pointed out here

This post imported from StackExchange Physics at 2014-11-27 10:37 (UTC), posted by SE-user bechira
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 $A_0$ 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.

This post imported from StackExchange Physics at 2014-11-27 10:37 (UTC), posted by SE-user Qmechanic
Yes I understand that. However I'm trying to understand if there are also reasons/consistency arguments from the group-theoretical point of view. For example: why can't I say (or can I?) that $A^\mu$ is a spin-1 field for that choice of the Lagrangian? (despite the fact of it being unphysical for independent reasons). This is also addressed on that same page (p.115), and my question actually arises from that argumentation which I'm not sure I get.

This post imported from StackExchange Physics at 2014-11-27 10:38 (UTC), posted by SE-user glance

1 Answer

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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.

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

This post imported from StackExchange Physics at 2014-11-27 10:38 (UTC), posted by SE-user Andrew McAddams
answered Nov 26, 2014 by Andrew McAddams (340 points) [ no revision ]

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