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

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carllacan
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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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Can you explicitly give a conserved current for a real scalar field? What's the corresponding symmetry?
 
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vanhees71 said:
Can you explicitly give a conserved current for a real scalar field? What's the corresponding symmetry?

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?
 
Well, they are the sources of gravity in GR but not of the electromagnetic field. You need a current of a gauge field.
 
vanhees71 said:
Well, they are the sources of gravity in GR but not of the electromagnetic field. .

Unfortunately I haven't studied GR yet. Could you explain what do you mean by the currents being the sources of the fields?

vanhees71 said:
You need a current of a gauge field.

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.
 
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!
 
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carllacan said:
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?
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].
 
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samalkhaiat said:
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].

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:

vanhees71 said:
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!

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?
 
carllacan said:
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.

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 realize 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.
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?
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.
 
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samalkhaiat said:
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.

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?
 
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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