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I've been looking for quite some time an expression for the Wightman functions for a massless vector field in the Coulomb gauge $\nabla\cdot\mathbf{A}=0$ (I think it is equivalent to the Feynman gauge since using Maxwell's equations and Coulomb gauge we can make $\phi=\text{const}$, and therefore have $\nabla_\mu A^\mu=0$) for the Minkowski vacuum, but I have not quite succeeded. Instead, I have found thermal Wightman functions for the Fulling vacuum here, and the same for an arbitrary gauge in Eq. (2.17) here (I believe that in this case it would suffice to do $\xi\to 1$ and $\beta\to 0$?).

The closest I have got is with the aid of Eqs. (6), (7) of this, in which a general form is given which depends on a given function and another (unspecified) dependent on the gauge.

Could anyone therefore point me out which is the expression for the Wightman functions of a massless, free vector field in the Minkowski vacuum and in the Coulomb gauge $\left<0\right|A_i(\mathbf{x},t)A_j(\mathbf{y},t')\left|0\right>$, or at least some guidance on how to obtain it from the results above? Thank you really much.

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  • $\begingroup$ I strongly believe that the appropriate expression is the one given in the third reference when doing $\Phi_\mu=0$, i.e, being just the scalar Wightman function times the metric. A similar result is found in Eq. (12) here, but could anyone confirm it? $\endgroup$
    – Alex
    Commented Nov 17, 2015 at 9:54

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So actually I just computed it by hand from the very beginning. Starting from the expression of the field $A_i(\mathbf{x},t)=\sum_s\int\frac{\text{d}^3\mathbf{k}}{\sqrt{(2\pi)^3 2|\mathbf{k}|}}\left[a_s(\mathbf{k})\epsilon_i(\mathbf{k},s)e^{i(|\mathbf{k}|t-\mathbf{k\cdot x})}+\text{H.c.}\right]$ where $\epsilon_i(\mathbf{k},s),\,s=1,2,\, i=1,2$ are orthonormal polarization vectors perpendicular to $\mathbf{k}$ and $[a_s(\mathbf{k}),a^\dagger_s(\mathbf{k}')]=\delta_{s,s'}\delta(\mathbf{k}-\mathbf{k}')$, the Wightman function takes the form

$W_{ij}(\mathbf x,\mathbf y,t,t')=\int\frac{\text{d}^3\mathbf k}{(2\pi)^32|\mathbf k|}\sum_{s}\epsilon_i(\mathbf k,s)\epsilon_j(\mathbf k,s)e^{-i |\mathbf{k}|(t-t')}e^{i \mathbf{k}\cdot (\mathbf{x}- \mathbf{y})}$

And now, using for example Eq. (1) in page 36 of "Photons and Atoms: An introduction to Quantum Electrodynamycs" by Cohen-Tannoudji we simply get

$W_{ij}(\mathbf x,\mathbf y,t,t')=\left(\delta_{ij}-\frac{\partial}{\partial x_i}\frac{\partial}{\partial x_j}\right)\int\frac{\text{d}^3\mathbf k}{(2\pi)^32|\mathbf k|}\frac{e^{-i |\mathbf{k}|(t-t')}e^{i \mathbf{k}\cdot (\mathbf{x}- \mathbf{y})}}{|\mathbf{k}|^2}$

The only drawback of this is that the integral diverges at the origin, but I guess that there is not much more one can do.

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