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A QED propagator in Coulomb gauge

  1. Jul 21, 2017 #1

    lalo_u

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    My aim is to derive the photon propagator in an Coulomb gauge following Pokorski's book method.
    In this book the photon propagator in Lorenz gauge was obtained as follows:

    1. Lorenz gauge: ##\partial_{\mu}A^{\mu}=0##
    2. It's proved that ##\delta_{\mu}A^{\mu}_T=0##, where ##A^{\mu}_T=(g^{\mu\nu}-\frac{\partial^{\mu}\partial{\nu}}{\partial^2})A^{\mu}## is the transverse field.
    3. Then, ##\partial^2A^T_{\mu}=0\rightarrow (\partial^2-i\epsilon)D_{\mu\nu}(x-y)=-(g_{\mu\nu}-\frac{\partial_{\mu}\partial_{\nu}}{\partial^2})\delta(x-y)##, is the equation for the corresponding the Green's function in the transverse space.
    4. After a Fourien transformations this becomes ##(-k^2-i\epsilon)\tilde{D}_{\mu\nu}(k)=-(g_{\mu\nu}-\frac{k_{\mu}k_{\nu}}{k^2})##.

    Now, in Coulomb gauge,

    5. Coulomb gauge: ##\partial_{\mu}A^{\mu}-(n_{\mu}\partial^{\mu})(n_{\mu}A^{\mu})=0, \; n_{\mu}(1,0,0,0)##

    6. I've tried to do the same program as before but i'm stuck. It's supose the propagator we have to obtain is:

    $$\tilde{D}^{\alpha\beta}_{\mu\nu}=\frac{\delta^{\alpha\beta}}{k^2+i\epsilon}\left[g_{\mu\nu}-\frac{k\cdot n(k_{\mu}n_{\nu}+k_{\nu}n_{\mu})-k_{\nu}k_{\mu}}{(k\cdot n)^2-k^2}\right]$$.

    The reference,
    Gauge Field Theories, 2000. Stefan Pokorski. Pages: 129-132.

    I'll appreciate any help.
     
  2. jcsd
  3. Jul 25, 2017 #2

    vanhees71

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    I checked the book. I also don't understand it ;-). I'd derive it in a very straight-forward way. Just do the usual Faddeev-Popov quantization. Since we deal with an Abelian gauge symmetry and use a linear gauge (Coulomb gauge),
    $$\vec{\nabla} \cdot \vec{A}=u^{\mu} \partial_{\mu} u_{\nu} A^{\nu}-\partial_{\mu} A^{\mu}=0, \quad (u^{\mu})=(1,0,0,0)$$
    the Faddeev-Popov ghosts decouple, i.e., are free fields and can thus be omitted for the calculation of Green's functions.

    The upshot is that the final gauge fixed QED Lagrangian reads
    $$\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu} -\frac{1}{2 \xi} (\vec{\nabla} \cdot A)^2 + \mathcal{L}_{\text{mat}}.$$
    The propagator comes from evaluating the bilinear part for the photon fields. The inverse propagator after Fourier transformation to momentum space yields
    $$(D^{-1})^{\mu \nu}=-k^2 \eta^{\mu \nu} + k^{\mu} k^{\nu} -\frac{1}{\xi} k_{\perp}^{\mu} k_{\perp}^{\nu}.$$
    For convenience I have defined
    $$k_{\perp}^{\mu}=k^{\mu} - u^{\mu} (u \dot k)=(0,k^1,k^2,k^3).$$
    Taking the inverse of the matrix (I used Mathematica) leads to
    $$D_{\mu \nu} = \frac{1}{k^2+\mathrm{i} 0^+} \left [-\eta_{\mu \nu} + \frac{(k \cdot u)(k_{\mu} u_{\nu}+k_{\nu} u_{\mu})-k_{\mu} k_{\nu}}{\vec{k}^2} \right]-\xi \frac{k_{\mu} k_{\nu}}{\left (\vec{k}^2 \right)^2}.$$
    The Coulomb gauge in the sense of canonical quantization you get for ##\xi=0##.
     
  4. Jul 25, 2017 #3

    ChrisVer

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    Would it make sense to assume that the inverse matrix would be:
    [itex]K_{\mu \nu} = a \eta_{\mu \nu} + b k_{\mu} k_\nu[/itex]
    and try to determine [itex]a,b[/itex]?
     
  5. Jul 25, 2017 #4

    vanhees71

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    That's not sufficient, because in the Coulomb (and also various axial) gauges there's an extra constant four-vector. In the case of the Coulomb and time-like axial gauges it introduces a preferred reference frame.

    Your ansatz is valid for the covariant gauges leading to
    $$\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu}-\frac{1}{2 \xi} (\partial_{\mu} A^{\mu})^2.$$
    Then you get
    $$D_{\text{inv}}^{\mu \nu} = -k^2 \left (\eta^{\mu \nu} - \frac{k^{\mu} k^{\nu}}{k^2} \right)-\frac{1}{\xi} \frac{k^{\mu} k^{\nu}}{k^2}.$$
    The inverse is very easily found since the matrix structures on the right-hand side are mutual Minkowski-orthogonal projectors (transverse and longitudinal degrees of freedom):
    $$D_{\mu \nu} = -\frac{1}{k^2+\mathrm{i} 0^+} \left (\eta_{\mu \nu} - (1-\xi) \frac{k_{\mu} k_{\nu}}{k^2+\mathrm{i} 0^+} \right).$$
    The most common special cases are ##\xi=0## (Landau gauge), where the propgator is transverse and ##\xi=1## (Feynman gauge). It's also a good check to do calculations with arbitrary ##\xi## to check whether the S-matrix elements come out gauge invariant, i.e., independent of ##\xi##, as it should be.
     
  6. Jul 26, 2017 #5

    lalo_u

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    Thanks @vanhees71 i agree with your calculation.

    However I was insisting on the calculation of Pokorski, and redefine the projection operator in Coulomb gauge as follows [​IMG]
    But, when i put the Green function in momentum space there's a leftover term, [​IMG] compared with the result in the book.
    In order to solve this i followed this reasoning: in configuration space this leftover term must be cancelled because when the Green function is acting on the gauge field space, we have to evaluate [​IMG] in Coulomb Gauge. Are you agree with that?

    On the other hand, there's a [​IMG] tensor i don't know where it comes from. Any idea?
     
  7. Jul 27, 2017 #6

    vanhees71

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    I don't have the book here. So can't check his calculation right now.
     
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