A QED propagator in Coulomb gauge

1. Jul 21, 2017

lalo_u

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. Jul 25, 2017

vanhees71

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 $\xi=0$.

3. Jul 25, 2017

ChrisVer

Would it make sense to assume that the inverse matrix would be:
$K_{\mu \nu} = a \eta_{\mu \nu} + b k_{\mu} k_\nu$
and try to determine $a,b$?

4. Jul 25, 2017

vanhees71

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.

$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu}-\frac{1}{2 \xi} (\partial_{\mu} A^{\mu})^2.$$
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.

5. Jul 26, 2017

lalo_u

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
But, when i put the Green function in momentum space there's a leftover term, 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 in Coulomb Gauge. Are you agree with that?

On the other hand, there's a tensor i don't know where it comes from. Any idea?

6. Jul 27, 2017

vanhees71

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