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Why does Lagrangian in QFT only include first order derivative of field?

  1. Nov 20, 2011 #1
    Please teach me this:
    Why the Lagrangian in QFT does not include high order derivative of field?Is it correct the reason being all fields obey the only Dirac and Klein-Gordon equations?
    Thank you very much for your kind helping.
     
  2. jcsd
  3. Nov 20, 2011 #2

    Fredrik

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    One explanation I've heard but don't fully understand is that the contribution from higher-order terms would be negligible at low energies anyway, and can therefore be ignored. These terms would also make the theories not renormalizable.
     
  4. Nov 21, 2011 #3

    Avodyne

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  5. Nov 28, 2011 #4

    A. Neumaier

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    A renormalizable QFT cannot have higher derivatives. Effective (nonrenormalizable) theories can.
     
  6. Nov 28, 2011 #5

    Ben Niehoff

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    Also, classically speaking, higher derivatives in the Lagrangian will result in higher derivatives in the equations of motion. In particular, if an equation of motion has more than two time derivatives, it becomes difficult to make physical sense out of it (because it seems to require too much initial data to define a unique solution).
     
  7. Nov 28, 2011 #6

    A. Neumaier

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    Not really. A k-th order differential equation requires as initial conditions the field and the first k-1 derivatives
    at the initial time. The case k=2 is nothing special here.
     
  8. Nov 28, 2011 #7

    Ben Niehoff

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    Yes, I'm aware of that. However, our vast experience with classical mechanics indicates that k=2 is indeed special. For example, for a classical system to be a Hamiltonian flow on phase space requires k=2.

    The one case I know of in which k>2, the radiation reaction on a moving, charged source, has issues with causality and runaway solutions, precisely because of the third time derivative that appears in the equation of motion.

    These are classical situations, but I think deep down they have some relationship to the nonrenormalizability of higher-derivative actions. There is something fundamentally quirky about a physical system that requires information about higher time derivatives.
     
  9. Nov 28, 2011 #8

    A. Neumaier

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    It depends very much on the variables one is using. The Hamiltonian formulation has in fact k=1, and not always can a Hamiltonian system be rewritten in second order form.

    Of course, Hamiltonian flow is very special, but it has nothing to do with the number of derivatives.
     
  10. Nov 28, 2011 #9
    Dear Ben Niehoff,

    I am afraid I am somewhat skeptical about your arguments for the following reason. For example, the Dirac equation is generally equivalent to a 4th order PDE for just one component (furthermore, this component can be made real by a gauge transform). Source:
    http://akhmeteli.org/wp-content/uploads/2011/08/JMAPAQ528082303_1.pdf (an article in Journ. Math. Phys.) Or another example: the equations of scalar electrodynamics (the Klein-Gordon-Maxwell electrodynamics) are generally equivalent (after algebraic elimination of the matter field) to higher order equations for electromagnetic field. Source: http://www.akhmeteli.org/akh-prepr-ws-ijqi2.pdf (an article in Int. Journ. Quantum Inf.). A somewhat cleaner proof can be found in my recent preprints.
     
  11. Nov 28, 2011 #10

    Ben Niehoff

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    Hmm, ok, maybe I'm wrong.
     
  12. Nov 28, 2011 #11

    atyy

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    Hollowood, http://arxiv.org/abs/0909.0859

    "... we should allow all possible operators consistent with spacetime symmetries. In the case of a scalar field, all powers of the field and its derivatives ... (p12)"

    "...the principle of universality that allows us to formulate our theories in terms of simple actions. All we need do is include the relevant couplings: all the irrelevant couplings can be taken to vanish. ... it is sufficient to write the simple action ... (p21)"
     
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