chiral case in subsection \[subsec:chiral\]. In this case the relevant free energy (for orthorhombic, tetragonal, and hexagonal materials) is $$\begin{aligned}
f &=& \sum_{s=\uparrow\!\uparrow, \downarrow\!\downarrow}
\left[-\alpha|\Delta_s|^2 + \beta_1 |\Delta_s|^4/2 + \left(\sum_{i=x,y,z} K_i|D_i\Delta_s|^2\right)\right]\nonumber\\
&-& \beta_2\!|\Delta_{\uparrow\!\uparrow}|^2 |\Delta_{\downarrow\!\downarrow}|^2
- \tilde{\kappa} h (|\Delta_{\uparrow\!\uparrow}|^2 - |\Delta_{\downarrow\!\downarrow}|^2)\nonumber\\
&+& \gamma[\Delta_{\uparrow\!\uparrow}^*\Delta_{\downarrow\!\downarrow}
(\textbf{D}\Delta_{\uparrow\!\uparrow})\!\cdot\!(\textbf{D}\Delta_{\uparrow\!\uparrow})^*+{\rm c.c}]
+ \frac{h^2}{8\pi} - \frac{H h}{4\pi}.\nonumber\\
\label{EQ:freeLLL1}\end{aligned}$$ As mentioned before, we assume that we have re-scaled lengths so that we can take $K_z=K_x=K_y=K$. First look at the upper critical field problem. The linearized GL equation is: $$\frac{\alpha l^2}{K} \Delta_{s}= \left(1+2\Pi_+\Pi_- -s\frac{Hl^2}{K}\tilde{\kappa}\right) \Delta_{s}.$$ The largest $H_{c2}$ occurs when the $\Delta_s$ are in the lowest Landau level. This leads to two possible values for the upper critical field, $$H_{c2}^{\pm}=\frac{\alpha \Phi_0}{2\pi(K\pm Hl^2\tilde{\kappa})}$$ We assume that $\tilde{\kappa}>0$ so that $\Delta_{\uparrow\!\uparrow}$ has the larger $H_{c2}$. We do not assume that the splitting between these two critical fields is large since it is given by the small Zeeman term $\tilde{\kappa}$.

Now consider the expectation value of $f$ in terms of the Landau level wave functions. Applying Eq. (\[EQ:genLinGL\]) to the first two terms of Eq. (\[EQ:freeLLL1\]) gives the lowest Landau level solutions $\Delta_{\uparrow\!\uparrow} = C_{\uparrow\!\uparrow} \phi_0$ and $\Delta_{\downarrow\!\downarrow} = C_{\downarrow\!\downarrow} e^{i 2\alpha} \tilde{\phi}_0$. Inserting this solution as determined at $H=H_{c2}$ gives $$\begin{aligned}
\langle 