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Electromagnetic Potential as an Observable

  1. Aug 23, 2011 #1
    Hi,

    I suppose I'm a little late to start here, but I just got hung up on the following: The field quanta in E&M is the photon and it comes from the gauge potential in QED [itex]A(x)[/itex]
    [tex]
    A(x)=\int \frac{d^3 p}{(2\pi)^3 \sqrt{2\omega_p}}\sum_{\lambda=1,2}\left[ \epsilon(p,\lambda)a_{p,\lambda}e^{-ipx}+\epsilon^{\ast}(p,\lambda)a^{\dagger}_{p. \lambda }e^{ipx} \right]
    [/tex]
    which is an operator on the fock space that creates a particle with helicity 1, momentum k, energy [itex]|k|[/itex], and no mass, the photon. But classically the potential [itex]A[/itex] is not an observable, so how come the photon is an observable?

    Thanks,
     
  2. jcsd
  3. Aug 23, 2011 #2

    Ben Niehoff

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    Science Advisor
    Gold Member

    But A is observable classically! Specifically, the combinations

    [tex]-\nabla \phi + \partial_t \vec A, \qquad \nabla \times \vec A[/tex]

    are observable. A has extra gauge degrees of freedom. But the true degrees of freedom, after removing the gauge redundancy, are observables. You'll notice that after removing the gauge redundancy, A (a four-vector) has two remaining degrees of freedom; these are exactly the two observable polarizations of light.
     
  4. Aug 23, 2011 #3
    Thanks Ben,

    That's what I thought, that A is observable after preforming those operations on it. I specifically recall that the potential field cannot be directly observed, only the field strength tensor, which is what you have above.

    Thanks,
     
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