# A Photon propagator in Coulomb gauge

1. Jul 20, 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,
1. Coulomb gauge: $\partial_{\mu}A^{\mu}-(n_{\mu}\partial^{\mu})(n_{\mu}A^{\mu})=0, \; n_{\mu}(1,0,0,0)$
2. 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.

Last edited: Jul 20, 2017
2. Jul 20, 2017

### dextercioby

First, it is Lorenz. Then why do you need to use the method in this book?

3. Jul 20, 2017

### lalo_u

If you read it again, the first component of de field has been removed, so it's Coulomb.

I'm trying to do this method, because seemed ellegant to me: it's done covariantly, even though we are dealing with Coulomb.

Last edited: Jul 20, 2017
4. Jul 20, 2017

### lalo_u

5. Jul 20, 2017

### George Jones

Staff Emeritus
@dextercioby 's point is that you have the named the gauge after the wrong person, i.e., "Lorenz gauge" (condition) is correct, and "Lorentz gauge" (condition) is incorrect.

6. Jul 20, 2017

### Avodyne

The role of $k^\mu$ in Lorenz gauge is played by $k^\mu-(n\cdot k)n^\mu$ in Coulomb gauge; make this replacement in $g^{\mu\nu}-k^\mu k^\nu/k^2$, including in the $k^2$ factor in the denominator. This will give you almost the desired expression, but you will have an extra term with $n^\mu n^\nu$. This is canceled by including the explicit charge-charge Coulomb interaction in Coulomb gauge.

7. Jul 22, 2017

### lalo_u

Thank you Avodyne, but there are two more things, a) we are calculating the photon free propagator, is it correct considering any interaction?, b) what about the extra $\delta_{\alpha\beta}$ at the beginning of the expression?

Last edited: Jul 22, 2017
8. Aug 4, 2017

### Avodyne

To compute the propagator in the interacting theory (order by order in perturbation theory), we must compute the corrections from Feynman diagrams with loops. The free propagator is the starting point.

The Kronecker delta presumably refers to the adjoint index of a non-abelian gauge field.