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A Can classical EM be derived from QFT?

  1. Apr 23, 2017 #1
    In QFT, one can derive the equations for particles interacting electromagnetically by demanding phase invariance for the field when writing down the free field lagrangian for the klein-gordon or dirac equation.

    Question: Does classical EM follow from this method also? (At least theoretically, given that you know the fine structure constant). If not what comes in the way?
     
    Last edited: Apr 23, 2017
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  3. Apr 23, 2017 #2

    vanhees71

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    Almost :-). The most simple way to describe special relativistic point mechanics is to use a Lagrangian of the form
    $$L=\frac{m}{2} \dot{x}^{\mu} \dot{x}_{\mu} + L_{\text{int}}(x^{\mu},\dot{x}^{\mu}),$$
    where the dot denotes derivatives with respect to proper time and ##L_{\text{int}}## is homogeneous of degree 1 wrt. ##\dot{x}^{\mu}##, i.e.,
    $$\dot{x}^{\mu} \frac{\partial L_{\text{int}}}{\partial \dot{x}^{\mu}}=L_{\text{int}}.$$
    Then ##\tau## is an affine parameter, i.e., since ##L## is not dependent explicitly on ##\tau##, the quantity
    $$H=p_{\mu} \dot{x}^{\mu}-L=\frac{m}{2} \dot{x}^{\mu} \dot{x}_{\mu}=\text{const},$$
    and thus you can choose ##\tau## as the proper time, so that ##H=mc^2/2##.

    The most simple equation of motion, fulfilling this properties, is provided by
    $$L_{\int}=\frac{q}{c} A_{\mu} \dot{x}^{\mu},$$
    where ##A_{\mu}=A_{\mu}(x)## is a vector field and ##q## a parameter. Then the Euler-Lagrange equations give
    $$p_{\mu}=m \dot{q}_{\mu}+\frac{q}{c} A_{\mu},\\
    \dot{p}_{\mu} = m \ddot{q}_{\mu} +\frac{q}{c} \dot{u}^{\nu} \partial_{\nu} A_{\mu}=\frac{q}{c} \dot{q}^{\nu} \partial_{\mu} A_{\nu}$$
    or
    $$m \ddot{q}_{\mu} = \frac{q}{c} F_{\mu \nu} \dot{x}^{\nu}, \quad F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu},$$
    and that's the relativistic equation of motion for a particle in the electromagnetic field, represented by the four-potential ##A_{\mu}##.

    It is clear that only ##F_{\mu \nu}## is observable, and that indeed the equations of motion do not change under the gauge transformation
    $$A_{\mu}'=A_{\mu}+\partial_{\mu} \chi,$$
    where ##\chi## is an arbitrary scalar field. Indeed the change in the Lagrangian is a total derivative of a function of ##x## only:
    $$L_{\text{int}}'=L_{\text{int}} + \frac{q}{c} \dot{x}^{\mu} \partial_{\mu} \chi=L_{\text{int}} + \frac{q}{c}+\frac{\mathrm{d}}{\mathrm{d} \tau} \chi.$$
    The equation of motion is thus gauge invariant.
     
  4. Apr 24, 2017 #3
    Thanks vanshees71, that was interesting. I did, however mean the lagrangian density for the klein-gordon or dirac field.

    One gets the feynman diagram vertices for EM by demanding local gauge invariance in the lagrangian of the field(i read this in chap 11.3 in griffiths introduction to elementary particles, which there are pdfs of online).

    From Griffiths:
    "Thus the requirement of local gauge invariance, applied to the free Dirac Lagrangian, generates all of electrodynamics, and specifies the current produced by Dirac particles."

    Is this a simpler(in the sense of less parameters), yet way less practical, way of deriving classical electrodynamics(Maxwells eqs.)?
     
    Last edited: Apr 24, 2017
  5. Apr 24, 2017 #4

    vanhees71

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    Yes, sure. I obviously misunderstood the question. I thought you mean, how to derive the em. interaction in classical theory, and I took it to mean classical theory of point charges.

    Your case is, from an advanced point of view, simpler since in QFT you can start from the unitary irreducible representations of the Poincare group and formally derive how local microcausal relativistic QFTs with a stable ground state must look like. You are lead to massive and massless fields for the various spin 0, 1/2, 1,...

    It also turns out that a massless spin-1 field is necessarily described as a gauge field, if you demand that there are only a discrete (finite) number of intrinsic degrees of freedom as it should be for spin-like quantities. This immediately lead to the conclusion that any massless field of spin ##s \geq 1/2## has only two (and not ##2s+1##) polarizations (e.g., helicities ##\pm s##).

    Now this enforces you for an interacting theory of massless spin-1 field to obey U(1) gauge invariance and one very successful way is "minimal coupling". In this sense Griffiths is right when he says that from this the entire Maxwell theory of electromagnetism follows as the classical limit of QED.

    To get to the classical limit you have to do quantum many-body theory and some coarse-graining procedure (e.g., gradient expansion and Markov approximation of the Kadanoff-Baym equations). In linear-response approximation this leads to the usual constitutive relations for macroscopic electrodynamics.
     
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