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How does one derive the Lagrangian densities used in QFT?

  1. May 30, 2015 #1
    I've been working through a qft book by Sadovskii (while I wait for my Peskin book to come in) and I've used some later chapters of Griffith's Into to Elementary Particles as an introduction to some qft. My issue with both of these is that, where in classical mechanics we have the Lagrangian defined as the difference in potential and kinetic energy, we have no explicit definition of the Lagrangian density in qft. Everything I've seen has been, "so here's the basic Lagrangian density we want for this situation" with no derivation or even justification beyond, "we use this form because it gives us what we want later". This hardly seems like a legitimate method to me, and they have the tendency to imply that the additional terms in the density used ensure invariance are just happened upon, as though someone just kind of noticed it worked out if they added it on. I was wondering if there's a way to derive the Lagrangian density for, say, the free Klein-Gordon equation and how the additional terms are methodically sought out?
     
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  3. May 30, 2015 #2

    Orodruin

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    I would see it more in the light of "we use this form because it gives a good description of what experiments tell us". This is not particular for QFT, it goes for all of science, including Newton's equations. Of course there may be some leads as to what kind of terms are allowed and provide consistent QFTs, but all in all, science is mainly about trying to guess how the Universe works and checking if it does.
     
  4. May 31, 2015 #3
    In classical physics we first learn Newtonian mechanics. Then we learn that the same physics can be represented in a different way, with Lagrangian mechanics. We prove that if you set L = T - V, Lagrangian mechanics gives the same results as Newtonian mechanics. But note that we could have gone the other way. We could take Lagrangian mechanics as fundamental, and then derive Newton's laws from the Lagrangian. This would be perfectly valid; after all the two formulations are totally equivalent so you can take either one as fundamental and then derive the other.

    In field theory it's convenient to take the Lagrangian formulation as the fundamental one. So our approach is basically, "let's write down all the possible Lagrangians and see what physics they produce." The reason it's convenient to start from a Lagrangian is that the symmetries of a theory are manifest in its Lagrangian. So for example if we want a Lorentz-invariant theory, we restrict ourselves to Lorentz-invariant Lagrangians. And the way we come up with the Klein-Gordon Lagrangian is that it's the simplest possible Lorentz-invariant Lagrangian for a spin-0 field.
     
  5. May 31, 2015 #4

    vanhees71

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    As any fundamental physical law we take the Lagrangians of elementary particle physics from experience. Anything fundamental in physics is a (often extremely condensed) description of a lot of observations in nature.

    There are of course some guide lines from mathematics. In this case it's the symmetries of space-time you start with (which are of course also just a description of empirical facts). In HEP you use Minkowski space-time with the Poincare symmetry. This already gives a lot of hints, which possible action functionals can be written down in terms of fields. Then there are more symmetries, all empirical facts deduced from observed conservation laws, which constrain the actions further.
     
  6. Jun 1, 2015 #5

    ChrisVer

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    The Lagrangian you write down, still remains the Legendre transformation of the Hamiltonian of your system...
    So even if you want to write down the Lagrangian of a photon field, you could as well take the Hamiltonian (the "photons' energy") and do the transformation to get the Lagrangian. It's not so unreasonable afterall...but more convenient...
     
  7. Jun 2, 2015 #6

    vanhees71

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    The approach via the Lagrangian is usually more convenient, because it's manifestly Lorentz invariant. Of course, the right way to quantize is via the Hamiltonian formalism. You come back to the Lagrangian formulation in almost all practically important cases by integrating out the canonical field momenta in the path-integral formula for the generating functional for Green's functions, but one must be careful, if there are time derivatives in couplings. Then you may not end up with the classical Lagrangian you expect!
     
  8. Jun 2, 2015 #7

    ChrisVer

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    Is there an example about that? I don't remember coming across any (it sounds like something new, or maybe it's the phrasing of something known that is changed)
     
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