which means that the auxiliary fermion $f_{\sigma}$ now effectively acquires electric charge due to the well-developed Kondo hybridization in heavy fermion liquid state. As a matter of fact, this statement is more general and beyond the present mean-field level.[@Coleman] (In Appendix C, we give a brief discussion on this issue by path integral formalism.)

Therefore, when considering the superfluid response to external electromagnetic field, we should also include the contribution of auxiliary fermion and this will be analyzed in next subsection.

Superfluid response including auxiliary fermion
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In this section, we assume that beside the conduction electron, the auxiliary fermion also electromagnetically couples to $\vec{A}$, thus the total paramagnetic current is $$j_{i}^{p}(q,\tau)=j_{i}^{c}(q,\tau)+j_{i}^{f}(q,\tau)\nonumber$$ $$j_{i}^{c}(q,\tau)=e\sum_{k\sigma}v_{c}^{i}c_{k+q\sigma}^{\dag}(\tau)c_{k\sigma}(\tau)\nonumber$$ $$j_{i}^{f}(q,\tau)=e\sum_{k\sigma}v_{f}^{i}f_{k+q\sigma}^{\dag}(\tau)f_{k\sigma}(\tau).\nonumber$$ Here, $v_{c}^{i}=\partial\varepsilon_{k}/\partial k_{i}$ and $v_{f}^{i}=\partial\chi_{k}/\partial k_{i}$ are the effective velocity of conduction electron and auxiliary fermion, respectively.

Therefore, the paramagnetic response function is $$\begin{aligned}
\Pi_{ij}(q,\tau)&&=-\langle T_{\tau}j_{i}^{p}(q,\tau)j_{i}^{p}(-q,0)\rangle\nonumber\\
&&=\Pi_{ij}^{cc}(q,\tau)+\Pi_{ij}^{cf}(q,\tau)+\Pi_{ij}^{fc}(q,\tau)+\Pi_{ij}^{ff}(q,\tau)\nonumber.\end{aligned}$$ Note that there exist four terms, among which $\Pi_{ij}^{cc}$ and $\Pi_{ij}^{ff}$ denote the contribution of conduction electron and auxiliary fermion, respectively. The left terms $\Pi_{ij}^{cc}$ and $\Pi_{ij}^{ff}$ are the cross-term, which implies the quantum interference effect between different currents. In last subsection, we have calculated $\Pi_{ij}^{cc}$, $$\begin{aligned}
Re\Pi(q=0)_{ii}^{cc}=2e^{2}\sum_{k}(v_{c}^{i})^{2}[\alpha_{k}^{4}\frac{\partial f_ {F}(E_{k}^{A})}{\partial E_{k}^{A}}+\beta_{k}^{4}\frac{\partial f_ {F}(E_{k}^{B})}{\partial E_{k}^{B}}].\nonumber\end{aligned}$$

After a long but straightforward 