Are some of you guys able to instruct a curious solid-state physicist new to superconductivity about how the Higgs mechanism is related to the Meissner effect or more generally to gauge symmetry breaking in a superconducting phase? Links to some not too advanced articles or textbooks would also be welcome.

Thanks in advance

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    $\begingroup$ Well, Weinberg 1986 does the reverse (Superconductivity for Particular Theorists) quite well, so a bridge can be traversed in either direction... $\endgroup$ Jan 25 '20 at 20:38

In my view, the simplest way to see SSB (spontandeuos symmetry breaking) in superconductor is to start from path integral for superconductor. We can approximate attractive interaction between electrons (which comes from phonon exchange) to contact constant interaction with coupling $g$. Obtained theory consists 4-fermion interaction. This interaction can be decoupled by introducing complex boson field which describes cooper pairs. Symbolically, it means $$\bar{\psi}_{\uparrow}\bar{\psi}_{\downarrow}\psi_{\downarrow}\psi_{\uparrow}\rightarrow \bar{\Delta}\psi_{\uparrow}\psi_{\downarrow}+\Delta\bar{\psi}_{\uparrow}\bar{\psi}_{\downarrow}.\quad\quad(*)$$ You can easily understand the first expression in ($*$): two electrons with different spins go to point $(x,t)$ and scatter on each other (in fact, this process contain exchange photon but we use contact approximation!) and come out from point $(x,t)$. Introducing field $\Delta$, we explicitly allow fermion to form Cooper pairs and $\Delta$ correspond to Cooper pair. Terms with $\Delta$ describe processes of Cooper pair formation (first term, two electrons create Cooper pair $\bar{\Delta}$) and destruction (second term, Cooper pairt $\Delta$ decomposes into to fermions)

Gauge transformation for fermions is just rotation, $$\bar{\psi}\rightarrow e^{i\theta}\bar{\psi},$$ where $\theta$ is phase. How does field $\Delta$ transform? We explicitly assume that $\Delta\sim \psi\psi$, so $\Delta\rightarrow\Delta e^{2i\theta}$. Phase $\theta$ generally is the function which depends on $(x,t)$. This phase $\theta$ corresponds to U(1) gauge symmetry of theory. In simplified picture, we can set $\Delta=\Delta_0=\text{const}$ and after gauge transformation $\Delta\rightarrow\Delta_0e^{2i\theta}$. Then we consider small deviations near $\Delta_0$. Resulting action should be gauge invariant and has the following form: $$S[\theta,{\bf A}]=\int d\tau\int d^3r\left[c_1(\partial_{\tau}\theta)^2+c_2(\nabla\theta-{\bf A})^2\right],$$ where $c_1$ & $c_2$ are unknown (but it is possible to calculate them, $c_1=\nu_F$, $c_2=n_s/(2m)$) coefficients. Neglecting time-dependence and adding EM field (we suppose that EM field is constant and there is no electric field) action, $$S=\frac{1}{2}\int d^3r\left[\frac{n_s}{m}(\nabla\theta-{\bf A})^2+(\nabla\times{\bf A})^2\right].$$ Integrating out $\theta$-field, one can find effective action for field ${\bf A}$ and explicitly find that now motion equation for EM field is $$\frac{1}{2}\left(\partial^2+\frac{n_s}{m}\right){\bf A}=0.$$ Photon becomes massive, so U(1) symmetry in superconductor is broken. Therefore, in superconducting phase with non-zero Cooper pair field $\Delta_0$ gauge symmetry is broken.

Comprehensive and detailed discussion of this phenomenon you can see in Altland & Simons book but you should have basic knowledge in QFT.


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