In section 2.B of Metlitski and Vishwanath's paper:

"Generally condensation of a dyon with charges $(q,m)$ gives rise to an analogue of a Meissner effect for the gauge field combination $q\vec{b}-2\pi m \vec{e}$, with $\vec{b}=\nabla \times \vec{a}$ - the magnetic field of $a_\mu$ (an emergent u(1) gauge field), and $\vec{e}=\partial_t \vec{a} - \nabla a_t$ - the electric field of $a_\mu$."

My question: how can we obtain the $q\vec{b}-2\pi m \vec{e}$?

I'm fine with the condensation of electric charges and ordinary Meissner effect, but I'm not familiar with the condensation of monopoles and the so called "dual Meissner effect". I found this is related to Nambu, Mandelstam and t' Hooft's idea back in 1970s, which states that confinement is a dual Meissner effect upon condensation of monopoles.

I believe there must be some simple explanation for the $q\vec{b}-2\pi m \vec{e}$ and hope to avoid digging into the QCD literature...

(References probably relevant:

  1. Dyon condensation in topological Mott insulators: arXiv:1203.4593.

  2. Constructing symmetric topological phases of bosons in three dimensions via fermionic projective construction and dyon condensation: arXiv:1303.3572.)

(Another question on StackExchange related to mine: 't Hooft duality)


Here is a very "quick and dirty" way to get the result, too long for a comment so I'll post it as an answer:

Although condensation is essentially a quantum phenomena, for many purposes it is sufficient to think at the classical level, e.g. Meissner effect. The Maxwell electrodynamics with both electric and magnetic charges famously has a $S$-duality symmetry, namely $(\mathbf{e}, \mathbf{b})\rightarrow (-\mathbf{b}, \mathbf{e})$ (and a similar transformation on the charges $(q_e, q_m)\rightarrow (-q_m, q_e)$). In fact, the classical equations of motions has a full $\mathrm{SO}(2)$ symmetry:

$ \begin{pmatrix} \mathbf{e}\\ \mathbf{b} \end{pmatrix}\rightarrow \begin{pmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{pmatrix}\begin{pmatrix} \mathbf{e}\\ \mathbf{b} \end{pmatrix} $

In principle one could start from a quantum theory of a dyons field interacting with the EM field (which is highly nontrivial thing to write down), and derive the Meissner effect. However, since we know that Meissner effect can be derived using equations of motion, we can exploit the symmetry of the classical equations of motion as a short-cut to the answer. We use the transformation to "rotate" the dyon $(q,m)$ to a pure electric charge $(\sqrt{q^2+m^2}, 0)$:

$ \begin{pmatrix} q\\ m \end{pmatrix}\rightarrow \dfrac{1}{\sqrt{q^2+m^2}}\begin{pmatrix} q & m\\ -m & q \end{pmatrix}\begin{pmatrix} q\\ m \end{pmatrix}= \begin{pmatrix} \sqrt{q^2+m^2}\\ 0 \end{pmatrix} $

Apply the same transformation to the field strength:

$ \begin{pmatrix} \mathbf{e}\\ \mathbf{b} \end{pmatrix}\rightarrow \dfrac{1}{\sqrt{q^2+m^2}} \begin{pmatrix} q\mathbf{e}+m\mathbf{b}\\ -m\mathbf{e}+q\mathbf{b} \end{pmatrix}=\begin{pmatrix} \mathbf{e}'\\ \mathbf{b}' \end{pmatrix} $

Now, we know that if charge $Q$ condenses then it causes $Q\mathbf{B}$ to have a Meissner effect, and in this case $\sqrt{q^2+m^2}\mathbf{b}'=-m\mathbf{e}+q\mathbf{b}$.

I miss some factors of $2\pi$ compared to Metlitski but I think that's not essential.


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