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I am trying to understand some derivations involving the photon propagator, and I am having a lot of trouble with expressions in different gauges and also with terminology in general. Here is what I found in the literature, accompanied with my thoughts about it:

  1. In one paper, the authors only need to deal with the $D_{00}$ component of the photon propagator. Then they say that "it is most conveniently done in the Coulomb gauge" and they plug in $\dfrac{1}{|\boldsymbol{r_1}-\boldsymbol{r_2}|}$ for $D_{00}$ . (By the way, they also call it "the longitudinal part of the propagator". Is this terminology common/what is the reason behind it?)
  1. But then I read another paper on the same subject, and I see the following: $$\begin{align} & \mathrm{Feynman \; gauge:} \; D_{\mu \nu}(\omega,\boldsymbol{q}) = - \dfrac{g_{\mu \nu}}{q^2+i\epsilon} \; (g_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1), \; q^2=\omega^2-\boldsymbol{q}^2); \\ & \mathrm{Coulomb \; gauge:} \; D_{00}(\omega,\boldsymbol{q}) = \dfrac{1}{q^2+i\epsilon} , \; D_{ij}(\omega,\boldsymbol{q}) = \dfrac{1}{q^2+i\epsilon} \left( \delta_{ij} - \dfrac{q_i q_j}{|\boldsymbol{q}|^2} \right). \end{align}$$ It seems to me that the $D_{00}$ component is the same in both cases apart from the sign. Does this sign have any physical significance? Probably yes, otherwise it is not clear what exactly is "most convenient" about the Coulomb gauge when it comes to $D_{00}$ only.

  2. Then I find in another paper the following integration for the photon propagator in the Feynman gauge (with non-zero photon mass $\mu$): $$D_{\mu \nu}(\omega, \boldsymbol{x}-\boldsymbol{y}) = -g_{\mu \nu} \int \dfrac{d\boldsymbol{q}}{(2\pi)^3} \dfrac{\mathrm{exp}(i\boldsymbol{q}\cdot(\boldsymbol{x}-\boldsymbol{y}))}{\omega^2-\boldsymbol{q}^2-\mu^2+i0} = g_{\mu \nu} \dfrac{\mathrm{exp}(i\sqrt{\omega^2-\mu^2+i0}|\boldsymbol{x}-\boldsymbol{y}|)}{4\pi |\boldsymbol{x}-\boldsymbol{y}|}.$$ Setting $\mu=0$, I assume we would simply have something like: $$D_{\mu \nu}(\omega, \boldsymbol{x}-\boldsymbol{y})=g_{\mu \nu} \dfrac{\mathrm{exp}(i|\omega||\boldsymbol{x}-\boldsymbol{y}|)}{4\pi |\boldsymbol{x}-\boldsymbol{y}|},$$ but in exactly the same paper they also give for the Coulomb gauge: $$D_{00}(\omega, \boldsymbol{x}-\boldsymbol{y})=\dfrac{1}{4\pi |\boldsymbol{x}-\boldsymbol{y}|}, \; D_{ij}(\omega, \boldsymbol{x}-\boldsymbol{y}) =\int \dfrac{d\boldsymbol{q}}{(2\pi)^3} \dfrac{\mathrm{exp}(i\boldsymbol{q}\cdot(\boldsymbol{x}-\boldsymbol{y}))}{\omega^2-\boldsymbol{q}^2+i0} \left( \delta_{ij} - \dfrac{q_i q_j}{|\boldsymbol{q}|^2} \right),$$ where $D_{00}$ agrees with what I saw in [1] (up to a factor of $4\pi$), but then I do not understand why in the coordinate representation we have the factor $\mathrm{exp}(i|\omega||\boldsymbol{x}-\boldsymbol{y}|)$ only for the Feynman gauge, if in the momentum representation $D_{00}$ seems almost the same in both gauges from [2].

  3. Finally, if I go to Peskin and Schroeder, I find this general formula (using a different convention with the factor of $i$): $$D_{\mu \nu}(q) = \dfrac{-i}{q^2+i\epsilon} \left( g_{\mu \nu} - (1-\xi) \dfrac{q_{\mu} q_{\nu}}{q^2} \right).$$ First of all, another terminology question. They say that $\xi=0$ and $\xi=1$ correspond to the Landau and Feynman gauges, respectively. What value of $\xi$ corresponds to the Coulomb gauge? Is "Landau gauge"$=$"Coulomb gauge"? I am not sure because for $\xi=0$ we have: $$D_{00} (q)= \dfrac{-i}{q^2+i\epsilon} \left( 1 - \dfrac{q_{0} q_{0}}{q^2} \right),$$ which does not look like what I saw in [2] for the $D_{00}$ component in the Coulomb gauge.

At this point I am totally lost and confused. Is the $D_{00}$ component of the photon propagator the same in the Feynman and Coulomb gauges or not? I would really appreciate it, if someone more experienced could tie all this together!

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    $\begingroup$ Can you please provide a link/title/doi of the mentioned paper? $\endgroup$ Nov 19, 2021 at 16:46
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    $\begingroup$ "Landau gauge" = "Lorentz gauge" whic is $\partial_{\mu}A^{\mu}=0$. Coulomb gauge is $\nabla\cdot{\bf A}$, so $A^0$ is not fixed. $\endgroup$ Nov 19, 2021 at 16:53
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    $\begingroup$ physics.stackexchange.com/q/550570 this question can shed light on relations between different gauges $\endgroup$ Nov 19, 2021 at 16:58
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    $\begingroup$ From my point of view, Coulomb gauge is quite useful if one considers non-relativistic limit. So, for me it is required to introduce $\chi$-term. Finally, in order to check everything, it seems enough to derive propagators in two gauge. It is the best way because all the mentioned papers are quite large and several details about sign choices/etc/ can be missed $\endgroup$ Nov 19, 2021 at 19:05
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    $\begingroup$ For instance, see pages.uoregon.edu/soper/QFT/QED.pdf $\endgroup$ Nov 19, 2021 at 19:16

1 Answer 1

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Coulomb gauge is not Lorentz-covariant, so it cannot have a relation with the Feynman gauge which is Lorentz-covariant. In [1] the photon propagator is given as ($i,l=1,\ldots,3$):

$$\tilde{D}_{00}(\omega,\mathbf{q}) = - \frac{4\pi}{\mathbf{q}^2}\quad\text{and}\quad \tilde{D}_{il}(\omega,\mathbf{q}) = - \frac{4\pi}{\omega^2 -\mathbf{q}^2 }\left(\delta_{il} -\frac{q_i q_l}{\omega^2}\right)$$

The factor $4\pi$ comes from the normalization of the 4-potential operator which can vary according to the convention.

Compared to the Feynman gauge which can be indeed found from the formula of Peskin & Schroeder [2] ($q$ here is the 4-vector):

$$\tilde{D}^{\mu\nu}(q) = \frac{-i}{q^2 + i\epsilon}\left( g^{\mu\nu} - (1-\xi)\frac{q^\mu q^\nu}{q^2}\right)\quad\text{with}\quad \xi=1$$

it is different (apart from normalization factors and possible signs and $i$) that in the denominator in Coulomb gauge there is the square of the 3-vector $\mathbf{q}$ whereas in Feynman gauge it is the square of the 4-vector $q = (\omega,\mathbf{q})$.

Actually, the formulas given in your post are not "complete". Computing the propagator component $D_{00}$ in position space one has to take 4-times the fourier transform of it[3]($x$ and $y$ are 4-vectors):

$$D_{00}(x-y) \equiv D_{00}(x^0-y^0,\mathbf{x-y}) = - 4\pi \int^\infty_{-\infty} \frac{d^4 q}{(2\pi)^4} e^{-ik^0(x^0-y^0)} \frac{e^{i\mathbf{k\cdot (x-y)}}}{\mathbf{q}^2} = -\frac{\delta(x^0-y^0)}{|\mathbf{x-y}|}$$

Here the metric $g_{\mu\nu} =diag(-1,1,1,1)$ was used as in [3].

Used references:

[1] L.D. Landau, E.M.Lifshitz: Relativistic Quantum Theory (Vol.4)

[2] M.E. Peskin, D.V.Schroeder, Introduction to Quantum Field Theory

[3] Mark Srednicki, Quantum Field Theory

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  • $\begingroup$ I simply love how your answer clarifies everything, and in such a concise way! Only one suggestion to slightly improve it: you could explicitly mention that there is a mistake in the formula for $D_{00}$ in the Coulomb gauge that I cite in 2. (where I have $q^2$ instead of $\boldsymbol{q}^2$). Because this is exactly what caused all the confusion in the first place. $\endgroup$ Nov 21, 2021 at 22:10

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