Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Electromagnetic Potential as an Observable

  1. Aug 23, 2011 #1

    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]
    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]
    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?

  2. jcsd
  3. Aug 23, 2011 #2

    Ben Niehoff

    User Avatar
    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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook