The Lagrangian for pairs of fermions $\psi_+$ and $\psi_-$ is
$$
{\cal L}~=~i(\gamma^\mu\partial_\mu\bar\psi\gamma^\nu\partial_\nu\psi|_+~+~\gamma^\mu\partial_\mu\bar\psi\gamma^\nu\psi|_-)~-~m(\psi_+\psi_-~+~\bar\psi_+\bar\psi_-)
$$
This is a form of the Dirac equation. We now assume that this mass is due to the Higgs field. The mechanism of symmetry breaking and SUSY is too deep to enter into, but here we assume the symmetry of the gauge field and its fermions is broken, but SUSY remains. This means the part of the Lagrangian is that $m(\psi_+\psi_-~+~\bar\psi_+\bar\psi_-)~=~\frac{m}{2}\bar\psi\psi$ for the mass of the fermions determined by the Higgs field $H$ as $m~=~g\bar\psi H\psi$ which means this potential term is $\frac{g}{2}(\bar\psi\psi)^2$, which gives the Thirring fermion theory with the Lagrangian,
$$
{\cal L}~=~i\gamma^\mu\partial_\mu\bar\psi\gamma^\nu\partial_\nu\psi~-~\frac{g}{2}(\bar\psi\psi)^2
$$
This is the quartic Lagrangian of the Thirring theory of fermions.
The Thirring theory is a form of bosonization. Suppose that we have the fields $\psi_+(x)$ and $\psi_-(x')$ for two fermions, or usually electrons. Let us consider the product of states $\bar\psi_+\psi_-$ and $\bar\psi_-\psi_+$ $=~(\bar\psi_+\psi_-)^\dagger$ with $\psi_i~=~\psi_i^\dagger\gamma^0$, and the positions implicit. The anticommutator of products $\{\bar\psi_+\psi_-,~\bar\psi_-\psi_+\}$ can be computed by shifting the position of the fields with a minus sign so that
$$
\{\bar\psi_+\psi_-,~\bar\psi_-\psi_+\}~=~\bar\psi_+\psi_-\bar\psi_-\psi_+~+~\bar\psi_-\psi_+\bar\psi_+\psi_-
$$
$$
=~2\bar\psi_+\psi_-\bar\psi_-\psi_+,
$$
where several commutation operations were performed to get to the last step. The commutator is
$$
[\bar\psi_+\psi_-,~\bar\psi_-\psi_+]~=~0.
$$
This suggests this product of fields has bosonic quantum commutator properties
We then write the products as
$$
\bar\psi_+\psi_-~=~:e^{i\phi}:,~\bar\psi_-\psi_+~=~:e^{-i\phi^\dagger}:,
$$
where the colons mean normal ordering. Now evaluate the commutator
$$
[\bar\psi_+\psi_-,~\bar\psi_-\psi_+]~=~:[e^{i\phi},~e^{-i\phi^\dagger}]:~=~:[\phi,~\phi^\dagger]:
$$
where the normal ordering means this commutator is zero. This means the fermions have been paired into bosons. This is the basic mechanism behind superconductivity.
The fermionic term is then equivalent to a bosonic field with
$$
i\gamma^\mu\partial_\mu\bar\psi\gamma^\nu\partial_\nu\psi~=~\frac{1}{2}(\partial_\mu\phi)^2
$$
and we take the real part of the $\bar\psi_+\psi_-~=~:e^{i\phi}:$ so that
$$
\psi\psi~=~cos(\phi).
$$
Use the Fierz transformation $\psi\psi~=~2(\psi\gamma^\mu\psi)^2~+~c$ for $c$ a complex number, and $\psi\gamma^\mu\psi~=~\pi^{1/2}\epsilon^{\mu\nu}\partial_\mu\phi$. This results in the Lagrangian
$$
{\cal L}~=~\frac{2}{\pi}(\partial_\mu\phi)^2~+~cos(\phi)~+~c
$$
This is the Lagrangian for the sine-Gordon equation.
