#### [SOLVED] Screening of the electric field in the Higgs phase

Dvali states in his paper on Three-Form Gauging of axion Symmetries and Gravity that

“As usual, in the Higgs phase the electric field is completely screened in the vacuum,”

and in another occasion in the same paper that

“There is a well known way to get rid of any constant electric field in the vacuum (i.e. to screen it), which is by putting the gauge theory in the Higgs phase.”

It may sure be well-known among others, but I for one, am not aware of it. Can someone explain to me with an illustrative example (QED) how this works?

Thanks.

#### @ACuriousMind 2019-03-03 15:24:37

As Dan Yand says in a comment, the reason the electric field is "screened" in the Higgs phase is because the Higgs mechanism endows the photon(s) with mass, and massive bosons mediate forces whose classical potential is a Yukawa potential proportional to $$\mathrm{e}^{-mr}\frac{1}{r}$$, i.e. it is the potential for the massless boson damped by the mass of the boson as the damping (or "screening") factor.

That massive bosons mediate such exponentially damped forces is completely general. I explain the logic of obtaining the classical potential from the quantum field theory in this answer of mine, where the tree-level scattering between charged particles essentially tells us the potential is the Fourier transform of the propagator. The Fourier transform of $$\frac{e^2}{q^2 + m^2 - \mathrm{i}\epsilon}$$ in 3 dimensions and after $$\epsilon\to 0$$ is precisely the Yukawa potential:

Carrying out the angular integrations in the 3d integral, one arrives at (modulo factors of $$\pi$$) $$V(r) = \frac{e^2}{\mathrm{i}r}\int \left(\frac{q\mathrm{e}^{\mathrm{i}qr}}{(q+m +\sqrt{\mathrm{i}\epsilon})(q+m -\sqrt{\mathrm{i}\epsilon})}\right)\mathrm{d}q,$$ which is now interpreted as a complex contour integral in the upper half-plane with the contour being the $$x$$-axis and a half-circle "pushed out to infinity" (to do this properly, you should really take a finite contour and then take a limit, but whatever). The integrand vanishes on the half-circle at infinity, so only the real axis contributes and this is indeed still the same integral. Now we apply the residue theorem - there is a singularity enclosed by the contour at $$q_0 = m + \sqrt{\mathrm{i}\epsilon}$$ and it is a simple pole, i.e. has multiplicity one.

We apply a standard result about the residues of simple poles that says that $$f(z) = \frac{g(z)}{h(z)}$$ has a residuum of $$\mathrm{Res}(f,z_0) = \frac{g(z_0)}{h'(z_0)}$$ at a simple pole $$z_0$$, and obtain that $$V(r) = \frac{e^2}{\mathrm{i}r}2\pi\mathrm{i}\frac{m + \sqrt{i\epsilon}}{2(m + \sqrt{i\epsilon})}\mathrm{e}^{\mathrm{i}(m + \sqrt{\mathrm{i}\epsilon})r}$$ Taking the limit $$\epsilon\to 0$$ now yields the Yukawa potential as claimed.