QED propagator in Coulomb gauge

[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.

 

[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 $\xi=0$.

 

[ChrisVer]

Taking the inverse of the matrix (I used Mathematica) leads to

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 $a,b$?

 

[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.

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.

 

[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 https://dl-mail.ymail.com/ws/download/mailboxes/@.id==VjJ-wEel3giyirs51tCc6stWMDCHQk66FApKzVSz8N35sfcyZLn8jbsULn3fO5PqCwxoGH4nkLovJSeSm65J69qU1w/messages/@.id==AIa-imIAIOcgWXjPXQDPCIhGVdc/content/parts/@.id==3/raw?appid=YahooMailNeo&ymreqid=6e23abce-7cd4-09ba-01d0-8800fa010000&token=zitEzqOML3j84e6ealFTT5U7-km5qEQF52lp7AcCuBZJX0cP3Smm0PLzoYxAZwyLAqSNYkOQUN8PUuanbftOsUJcnwk4dNUd4utEI3EJJsxRaGMi_a4dULZHRa_xD1WB
But, when i put the Green function in momentum space there's a leftover term, https://dl-mail.ymail.com/ws/download/mailboxes/@.id==VjJ-wEel3giyirs51tCc6stWMDCHQk66FApKzVSz8N35sfcyZLn8jbsULn3fO5PqCwxoGH4nkLovJSeSm65J69qU1w/messages/@.id==AIa-imIAIOcgWXjPXQDPCIhGVdc/content/parts/@.id==4/raw?appid=YahooMailNeo&ymreqid=6e23abce-7cd4-09ba-01d0-8800fa010000&token=zitEzqOML3j84e6ealFTT5U7-km5qEQF52lp7AcCuBZJX0cP3Smm0PLzoYxAZwyLAqSNYkOQUN8PUuanbftOsUJcnwk4dNUd4utEI3EJJsxRaGMi_a4dULZHRa_xD1WB compared with the result in the book.
In order to solve this i followed this reasoning: in configuration space this leftover term must be canceled because when the Green function is acting on the gauge field space, we have to evaluate https://dl-mail.ymail.com/ws/download/mailboxes/@.id==VjJ-wEel3giyirs51tCc6stWMDCHQk66FApKzVSz8N35sfcyZLn8jbsULn3fO5PqCwxoGH4nkLovJSeSm65J69qU1w/messages/@.id==AIa-imIAIOcgWXjPXQDPCIhGVdc/content/parts/@.id==5/raw?appid=YahooMailNeo&ymreqid=6e23abce-7cd4-09ba-01d0-8800fa010000&token=zitEzqOML3j84e6ealFTT5U7-km5qEQF52lp7AcCuBZJX0cP3Smm0PLzoYxAZwyLAqSNYkOQUN8PUuanbftOsUJcnwk4dNUd4utEI3EJJsxRaGMi_a4dULZHRa_xD1WB in Coulomb Gauge. Are you agree with that?

On the other hand, there's a https://dl-mail.ymail.com/ws/download/mailboxes/@.id==VjJ-wEel3giyirs51tCc6stWMDCHQk66FApKzVSz8N35sfcyZLn8jbsULn3fO5PqCwxoGH4nkLovJSeSm65J69qU1w/messages/@.id==AIa-imIAIOcgWXjPXQDPCIhGVdc/content/parts/@.id==6/raw?appid=YahooMailNeo&ymreqid=6e23abce-7cd4-09ba-01d0-8800fa010000&token=zitEzqOML3j84e6ealFTT5U7-km5qEQF52lp7AcCuBZJX0cP3Smm0PLzoYxAZwyLAqSNYkOQUN8PUuanbftOsUJcnwk4dNUd4utEI3EJJsxRaGMi_a4dULZHRa_xD1WB tensor i don't know where it comes from. Any idea?

 

[vanhees71]

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