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Photon propagator in an arbitrary gauge

  1. Jul 18, 2005 #1
    My aim is to derive the photon propagator in an arbitrary gauge. I follow Itzykson-Zuber Quantum Field Theory and start from the Lagrangian with gauge-fixing term:
    {\cal L}(x) = -\,\frac{1}{4}\,F_{\mu\nu}(x)F^{\mu\nu}(x) -
    I get the following equation of motion:
    \left(\Box g_{\mu\nu} - \Big(1-\frac{1}{\xi}\Big) \partial_\mu \partial_\nu \right) A^\nu (x) = 0

    We can write the solution as a superposition of plane waves:
    A^{\mu}(x) = \sum_{\lambda}\int \frac{d^3k}{(2\pi)^3 2\omega(\vec{k}\,)}\,
    \Big[c(\vec{k},\lambda)\,e^{\mu}( \vec{k},\lambda)\,e^{-ik\cdot x}
    + c^{\dagger}(\vec{k},\lambda)\,e^{*\mu}( \vec{k},\lambda)\,
    e^{+ik\cdot x}\Big],
    where [tex] e^{\mu} [/tex] is a polarization vector.

    Then I use the following commutation relations for the creation and annihilation operators to quantize the field:
    = (2 \pi)^3\, 2 \omega (\vec{k})\, \delta^{(3)} (\vec{k}-\vec{k}\,')\,\delta_{\lambda \lambda\,'}
    = [c^{\dagger}(\vec{k},\lambda),\,c^{\dagger}(\vec{k}\,',\lambda\,')] = 0

    I want to calculate the Green function as a vacuum expectation value of the time-ordered product:
    -\,i\,D^{\mu\nu}(x - y) = \langle 0|{\rm T}(A^{\mu}(x)A^{\nu}(y))|0\rangle

    To do this I need to derive the expression:
    \sum_\lambda e^{\mu}(\vec k, \lambda) e^{*\nu} (\vec k, \lambda) =
    \left(g^{\mu\nu} - (1-\xi) \frac{k^\mu k^\nu}{k^2}\right)

    I am novice at QFT. Please, could you advise on the methodology, how I could derive the expression for the sum of the products of polarisation vector components.
    Last edited: Jul 18, 2005
  2. jcsd
  3. Jul 18, 2005 #2
    \left(\Box g_{\mu\nu} - \Big(1-\frac{1}{\xi}\Big) \partial_\mu \partial_\nu \right) A^\nu (x) = 0

    You simply need to invert the quadratic differential operator above in momentum space (aka Fourier space).

    In momentum space the photon kinetic term looks like

    \int \frac{d^{4}k}{(2 \pi)^{4}}
    A^{\mu} (k) \left(k^{\beta} k_{\beta} g_{\mu\nu} - \Big(1-\frac{1}{\xi}\Big) k_\mu k_\nu \right) A^\nu (-k)

    The quadratic differential operator becomes a (0 2) tensor [itex]M_{\mu \nu} \equiv \left(k^{\beta} k_{\beta} g_{\mu\nu} - \Big(1-\frac{1}{\xi}\Big) k_\mu k_\nu \right)[/itex], which you can invert by linear algebra methods.
  4. Feb 28, 2010 #3
    I know this is an old post, but could anyone comment on how to invert the [tex] M_{\mu\nu} [/tex] matrix above by linear algebra methods?
  5. Feb 28, 2010 #4
    The inverse matrix has to be a linear combination of [tex]g^{\mu \nu}[/tex] and [tex]k^{\mu}k^{\nu}[/tex], so Ansatz [tex]Ag^{\mu \nu}+Bk^{\mu}k^{\nu}[/tex] should work.
  6. Feb 28, 2010 #5
    Thank you, that did work, how did you know it had to be a linear combination of those things?
  7. Mar 1, 2010 #6
    Well, what else could it be? :) To construct a Lorentz tensor with two indices, there really is no other choice. I cannot justify it any better than by noticing that it works.
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