diamagnetic part, we obtain $$\begin{aligned}
Re\Pi(q=0)_{ij}&&=e^{2}\sum_{k}[\frac{-1}{m_{ij}}(\alpha_{k}^{2}\frac{E_{k}^{+}}{E_{k}^{A}}\tanh\frac{E_{k}^{A}}{2T}+\beta_{k}^{2}\frac{E_{k}^{-}}{E_{k}^{B}}\tanh\frac{E_{k}^{B}}{2T})\nonumber\\
&&+2v_{k}^{i}v_{k}^{j} (\alpha_{k}^{4}\frac{\partial f_ {F}(E_{k}^{A})}{\partial E_{k}^{A}}+\beta_{k}^{4}\frac{\partial f_ {F}(E_{k}^{B})}{\partial E_{k}^{B}})]. \nonumber\end{aligned}$$

Finally, by using of the definition of superfluid density $\frac{\rho_{ij}e^{2}}{m}=Re\Pi(q=0)_{ij}$, one gets the desirable superfluid density formula $\rho_{s}\equiv\rho_{ii}$, $$\begin{aligned}
\frac{\rho_{s}(T)}{m}&&=\sum_{k}[\frac{\partial^{2}\varepsilon_{k}}{\partial k_{x}^{2}}(-\alpha_{k}^{2}\frac{E_{k}^{+}}{E_{k}^{A}}\tanh\frac{E_{k}^{A}}{2T}-\beta_{k}^{2}\frac{E_{k}^{-}}{E_{k}^{B}}\tanh\frac{E_{k}^{B}}{2T})\nonumber\\
&&+2(\frac{\partial\varepsilon_{k}}{\partial k_{x}})^{2}(\alpha_{k}^{4}\frac{\partial f_ {F}(E_{k}^{A})}{\partial E_{k}^{A}}+\beta_{k}^{4}\frac{\partial f_ {F}(E_{k}^{B})}{\partial E_{k}^{B}})]. \label{eq9}\end{aligned}$$ Unfortunately, as what can be seen in Fig. \[fig:SFf1\], for both $d_{x^{2}-y^{2}}$ and extended-s wave pairing symmetry, the superfluid density does not vanish at superconducting transition temperature T$_{c}$, thus the formula Eq. \[eq9\] is undoubtedly not valid. (The parameters we used are $t=1$, $t'=0.3$, $J_{K}=2$, $J_{H}=0.6$ and $n_{c}=0.9$, which are typical ones in the mean-field calculation of Kondo-Heisenberg model.[@Liu2012]) So, we have to answer why such formula is not correct although the derivation is rather straightforward and standard.

![\[fig:SFf1\] Normalized superfluid density $\rho_{s}(T)/\rho_{s}(0)$ versus normalized temperature T/T$_{c}$ for $d_{x^{2}-y^{2}}$ (a) and extended-s (b) wave pairing symmetry by using Eq. \[eq9\].](SF22.eps){width="1.0\columnwidth"}

A careful reader may note that in the normal-state mean-field Hamiltonian Eq. \[eq1\], conduction electron is hybridized with auxiliary fermion via the formation of Kondo screening ($V\neq0$ and is real), thus repeating the argument of U(1) charge gauge transformation discussed in the beginning of this subsection, the auxiliary fermion has to transform like $$f_{i\sigma}\rightarrow f_{i\sigma}e^{i\theta_{i}},\nonumber$$ 