calculation, we obtain $$\begin{aligned}
Re\Pi(q=0)_{ii}^{ff}=2e^{2}\sum_{k}(v_{f}^{i})^{2}[\beta_{k}^{4}\frac{\partial f_ {F}(E_{k}^{A})}{\partial E_{k}^{A}}+\alpha_{k}^{4}\frac{\partial f_ {F}(E_{k}^{B})}{\partial E_{k}^{B}}].\nonumber\end{aligned}$$

Since $Re\Pi(q=0)_{ii}^{cf}=Re\Pi(q=0)_{ii}^{fc}$, the remaining work is to calculate $\Pi(q=0)_{ii}^{cf}$, whose expression is $$\begin{aligned}
\Pi(q)_{ii}^{cf}&&=e^{2}\sum_{k}v_{c}^{i}v_{f}^{i}
[G_{13}(k)G_{31}(k+q)+G_{14}(k)G_{41}(k+q)\nonumber\\
&&+(1\leftrightarrow2)].\nonumber\end{aligned}$$ After inserting the explicit form of $G_{13}$ and $G_{14}$, the final result reads $$\begin{aligned}
Re\Pi(q=0)_{ii}^{cf}=2e^{2}\sum_{k}v_{c}^{i}v_{f}^{i}[\alpha_{k}^{2}\beta_{k}^{2}\frac{\partial f_ {F}(E_{k}^{A})}{\partial E_{k}^{A}}+\alpha_{k}^{2}\beta_{k}^{2}\frac{\partial f_ {F}(E_{k}^{B})}{\partial E_{k}^{B}}].\nonumber\end{aligned}$$

So, the paramagnetic current response is found to be $$\begin{aligned}
Re\Pi(q=0)_{ii}^{p}&&=2e^{2}\sum_{k}[(\alpha_{k}^{4}(v_{c}^{i})^{2}+\beta_{k}^{4}(v_{f}^{i})^{2}+2\alpha_{k}^{2}\beta_{k}^{2}v_{c}v_{f})\nonumber\\
&&\times\frac{\partial f_ {F}(E_{k}^{A})}{\partial E_{k}^{A}}
+(\beta_{k}^{4}(v_{c}^{i})^{2}+\alpha_{k}^{4}(v_{f}^{i})^{2}+2\alpha_{k}^{2}\beta_{k}^{2}v_{c}v_{f})\nonumber\\
&&\times\frac{\partial f_ {F}(E_{k}^{B})}{\partial E_{k}^{B}}]\nonumber\end{aligned}$$ and the diamagnetic current response is easy to obtain as $$\begin{aligned}
Re\Pi(q=0)_{ii}^{d}&&=-e^{2}\sum_{k}
[\alpha_{k}^{2}(\frac{1}{m_{c}})_{ii}+\beta_{k}^{2}(\frac{1}{m_{f}})_{ii} ]\frac{E_{k}^{+}}{E_{k}^{A}}\tanh\frac{E_{k}^{A}}{2T}\nonumber\\
&&+[\beta_{k}^{2}(\frac{1}{m_{c}})_{ii}+\alpha_{k}^{2}(\frac{1}{m_{f}})_{ii} ]\frac{E_{k}^{-}}{E_{k}^{B}}\tanh\frac{E_{k}^{B}}{2T}. \nonumber\end{aligned}$$

Therefore, our wanted superfluid density formula is $$\frac{\rho_{s}(T)}{m}=\frac{1}{e^{2}}(Re\Pi(q=0)_{ii}^{d}+Re\Pi(q=0)_{ii}^{p}).\label{eq10}$$

In Fig. \[fig:SFf2\], we show the calculation data of Eq. \[eq10\], which again exhibits incorrect result. (The parameters are the same to the ones in last subsection.) The reason for this may be due to the mismatch of temperature-dependence between bare particles ($v_{c}$, $v_{f}$, $m_{c}$, $m_{f}$) and the renormalized heavy fermion quasi-particle ($E_{k}^{A}$, $E_{k}^{B}$). In other words, the bare and renormalized particles have rather different temperature dependence, thus the paramagnetic part cannot cancel out the diamagnetic term when $T>T_{c}$. More generally, when calculating physical quantities at $T=0$, since no temperature effect is involved, the underlying breakdown of the corresponding formalism is hidden but at finite temperature, one 