# Mean-field approach to quantum phase transitions in Fermi systems

I have a basic confusion concerning the mean-field theory of quantum phase transitions in Fermi systems. Consider as an example the BCS theory of superconductivity in a Dirac fermion system, considered by Sachdev in Sec. 17.1 of his QPT book. The variational ground state energy is given by

\begin{align} E_{BCS}=\frac{J_1}{2}\left(|\Delta_x|^2+|\Delta_y|^2\right)-\int\frac{d^2k}{(2\pi)^2}[E_k-\varepsilon_k], \end{align}

obtained by integrating out the fermions, where $\Delta_x$ and $\Delta_y$ are $d$-wave order parameters. The single-particle energy $E_k$ is

\begin{align} E_k=\sqrt{\varepsilon_k^2+|J_1(\Delta_x\cos k_x+\Delta_y\cos k_y)|^2} \end{align}

which is positive and always greater than $\varepsilon_k$. Suppose I would like to derive a Ginzburg-Landau energy by expanding the last term (fermion determinant) in powers of the order parameters. If I expand the integrand and perform the momentum integrals, I will obtain

\begin{align} E_{BCS}=r(|\Delta_x|^2+|\Delta_y|^2)+u(|\Delta_x|^4+|\Delta_y|^4)+\text{mixed terms}. \end{align}

Because the term coming from the fermion determinant comes with a minus sign, it is clear that there will be a negative contribution to $r$ that offsets the positive bare mass'' $J_1$, thus giving the possibility of a transition as $J_1$ is varied.

Now, imagine we would like the transition to be continuous. This requires $u>0$, and $u$ comes entirely from the fermion determinant. It is not clear at all to me how this is possible, because $-\int\frac{d^2k}{(2\pi)^2}[E_k-\varepsilon_k]$ is always negative. In particular, if the order parameter is very large the quartic terms will dominate over the quadratic terms, and it seems that we generically get $u<0$.