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I Why can't the real scalar field and the EM be coupled?

  1. Jun 19, 2016 #1
    According to David Tong's notes the real scalar field can't be coupled to the electromagnetic field because it doesn't have any "suitable" conserved currents. What does "suitable" mean? The real field does have conserved currents, why aren't those suitable?
     
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  3. Jun 20, 2016 #2

    vanhees71

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    Can you explicitly give a conserved current for a real scalar field? What's the corresponding symmetry?
     
  4. Jun 20, 2016 #3
    The rotational and spatial symmetries both give currents, namely the components of the energy-momentum and angular momentum tensors. Is there something wrong with them?
     
  5. Jun 20, 2016 #4

    vanhees71

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    Well, they are the sources of gravity in GR but not of the electromagnetic field. You need a current of a gauge field.
     
  6. Jun 20, 2016 #5
    Unfortunately I haven't studied GR yet. Could you explain what do you mean by the currents being the sources of the fields?

    Buy why is that required? In Tong's notes http://www.damtp.cam.ac.uk/user/tong/qft/six.pdf a condition is derived for the interaction term of the coupling of the EM field to matter, and it just says we have to use a conserved current.
     
  7. Jun 20, 2016 #6

    vanhees71

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    What you are trying to do is to gauge Poincare symmetry, but this leads to GR and the gravitational field described by a massless spin-2 field ("graviton") but not electrodynamics. Have a look at P. Ramond, QFT a modern primer (2nd edition), for the treatment of GR as a gauge theory. Concerning electromagnetism you rather like to end up with a U(1) gauge theory, i.e., Maxwellian electrodynamics and then quantize it to get QED!
     
  8. Jun 20, 2016 #7

    samalkhaiat

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    Not any conserved current. The electromagnetic field couples to electrically charged fields. The real scalar field is electrically neutral, because the (real scalar field) Lagrangian is not invariant under the global phase transformation [itex]\exp (i \alpha)[/itex], i.e., there is no conserved Noether current, [itex]J^{\mu}[/itex], associated with the transformation [itex]\varphi (x) \to \exp (i \alpha) \varphi (x)[/itex]. So, in the case of real scalar field, there is no coupling of the form [itex]e A_{\mu} J^{\mu}[/itex].
     
  9. Jun 20, 2016 #8
    While that explains the physics of it what I am trying to understand now is a mathematical reason why we need the matter field to be invariant under a phase transformation. I guess the reason is this:

    but I don't understand most of it, so I'll wait until I have studied GR and come back here.

    Just a final question: from your answers I gather that the reason the real scalar field can't be coupled to the Dirac field is not that there are no conserved currents, as I thought, but that there is no phase symmetry. Is that right?
     
  10. Jun 20, 2016 #9

    samalkhaiat

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    The electromagnetic field, [itex]A_{\mu}[/itex], is a gauge field. Gauge field can only couple to matter fields whose free Lagrangians are invariant under the global [itex]U(1)[/itex] phase transformation. For, if the free matter field Lagrangian is not invariant under global [itex]U(1)[/itex] phase, then Noether theorem will not provide you with the conserved vector current [itex]J^{\mu}[/itex] which is necessary to form the interaction Lagrangian [itex]e A_{\mu}J^{\mu}[/itex]. The Lorentz index [itex]\mu[/itex] is crucial here.
    When you study gauge field theories, you will realise that the gauge fields and their coupling to matter fields actually arise naturally from enlarging the global “phase” symmetry of (the free matter fields) Lagrangian to local (gauge) symmetry.
    Real scalar field can (and actually does) couple to Dirac’s fields: [itex]g \ \bar{\psi} (x) \varphi (x) \psi (x)[/itex] is such coupling. The pion [itex]\pi^{0}[/itex], which is described by real scalar field, interacts strongly with the protons and neutrons which are Dirac fields: [itex]g \ \pi^{0} \ (\bar{p}p - \bar{n}n)[/itex]. In this case, coupling to a conserved symmetry current is not an issue because you are not dealing with a gauge theory.
     
    Last edited: Jun 20, 2016
  11. Jun 21, 2016 #10
    Oh, sorry, I meant the electromagnetic field, not the Dirac.

    Wait, so pions are described by the scalar field? I thought the Higgs boson was the only particle that was described by it. Do you know any place where I can read which particles are described by what fields?
     
  12. Jun 21, 2016 #11

    vanhees71

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    The pion is not an elementary particle but a complicated bound state of quarks and gluons. At low energies, however, we can ignore this compositeness and treat hadrons (i.e., mesons and baryons) as if they were elementary particles with some effective QFT model. The only link to QCD are the adequate symmetries. For the "light hadrons" the symmetry is the (approximate) chiral symmetry of the light quarks (to some extent also including strange quarks), and you can build effective field theory models based on this symmetry. The oldest is the linear sigma model involving protons neutrons, pions and sigma mesons. A marvelous book, explaning this approach nicely is

    Donoghue, J. F., Golowich, E., Holstein, B. R.: Dynamics of the Standard Model, Cambridge University press, 1992

    For a nice introduction to chiral perturbation theory, see

    https://arxiv.org/abs/nucl-th/9706075
     
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