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Gauge invariant Lagrangian: unique?

  1. Feb 19, 2012 #1
    Hi all! Long story short, my QFT class recently covered gauge equivalence in QED, and this discussion got me thinking about more general gauge theory. I spent last weak reading about nonabelian symmetries (in the context of electroweak theory), and I like to think I now have a grasp on the basics: gauge fields are introduced to maintain invariance of the Lagrangian when the parameters in front of the symmetry generators are allowed to vary over spacetime. (Though please correct me if this is a misunderstanding!) What's confusing me now is how to "derive" a Lagrangian from ground up. It's my (perhaps misled) understanding that L is determined solely by the symmetries we choose to impose (which are motivated by experiment, of course). Here's my line of thought:

    It seems that we should start with the free fermion (or similarly, scalar) Lagrangian. It is invariant under global rotations of the fermion field PSI in SU(N) space. But L is not generally invariant under local SU(N) transformations, i.e. if we choose a different phase convention (element of SU(N)) at each spacetime point, L does not transform nicely: there are leftover partials of the transformation parameters. To maintain invariance (that is, to compensate for these extra terms), we introduce gauge fields G. If the G transform in a "very nice way", they cancel the aforementioned extra terms, leaving the Lagrangian fixed. So when we say that a theory "has SU(N) gauge symmetry", we mean that its Lagrangian is unchanged when we (1) transform the fermion fields under SU(N), i.e. left-multiply each fermion n-plet by an element of the n-dimensional representation of SU(N) and (2) transform the gauge fields in this "very nice way". Am I on track so far?

    In particular the gauge fields are introduced as terms in a covariant derivative D, which we substitute for the partial derivative(s) in L. Why is this the standard way of introducing the fields? It seems that I could cancel the bad terms in many other ways. Is the answer related to getting D(PSI) to transform as PSI?

    Once the fields are included in the Lagrangian, it is possible to derive the particular form of the "very nice" transformation (invariance of L determines the transformation properties of the G). After a bit of math, we get an answer in terms of the structure constants of the symmetry algebra and partials of the PSI transformation parameters. Can I interpret the structure constant term as an infinitesimal rotation in SU(N) space? Is there a natural way to interpret the partial term other than that it is needed to cancel the other bad terms? I guess I'm really wondering whether there is a more natural or direct way of determining how the gauge fields should transform. Are they even transforming "under SU(N)"?

    Since, under Lorentz transformations, the covariant derivative should transform like the partial derivative, we require the gauge fields to be vectors.

    Possibly related to the above is the question of the kinetic energy term. I see how, written in terms of field tensors (e.g. the F terms in the QED Lagrangian), this term is consistent with all of the relevant symmetries. But why is this term required? Does the reason have anything to do with renormalizability?

    TL;DR: I think I see how gauge equivalence works, given a Lagrangian. But are this Lagrangian and its associated method of introducing gauge fields unique?

    Thank you for the clarification in advance. This is really cool stuff, and I can't wait to understand!
  2. jcsd
  3. Feb 19, 2012 #2

    Ben Niehoff

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    This is called "minimal coupling". In principle, one can write down non-minimal couplings as well, such as

    [tex]\bar \psi \psi F^2[/tex]
    Of course, these terms might not be renormalizable, but they can still turn up in effective theories.

    The most elegant way is to think of the gauge field as a connection on a principle G-bundle (where G is the gauge group). The field strength F is then the curvature of this connection. The structure constants turn up in a natural way.

    It's required if you want the gauge field to be dynamic. If instead you want to put the theory in a background gauge field that stays fixed, you don't need it.
  4. Feb 21, 2012 #3
    This all makes sense. Thank you very much for the help.
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