\rightarrow \langle \hat\psi_b \rangle\approx
\psi^b_0(r,t),\;\; \hat \rho_b \rightarrow \langle \hat \rho_b
\rangle \approx N_b |\psi^b_0(r,t)|^2, \nonumber\\
\hat \rho_f \rightarrow \langle \hat \rho_f \rangle \approx
\sum_{n=1}^{N_f} |\psi^f_n(r,t)|^2,\nonumber\end{aligned}$$ we have that Eqs. (\[heiseqn2\]) reduce to $$\begin{aligned}
 && i \hbar \frac{\partial \psi^b_0}{\partial t}=\left[\hat
H_b^0+g_{b} N_b |\psi^b_0|^2 + g_{bf}\left( \sum_{n=1}^{N_f}
|\psi^f_n|^2 \right)\right] \psi^b_0, \nonumber
\\&& \sum_n \left[ i \hbar\frac{\partial \psi^f_n}{\partial t} -\hat
H_f^0 \psi^f_n- g_{bf} N_b |\psi^b_0|^2 \psi^f_n\right] \hat
c_n=0.\label{eq2}\end{aligned}$$ The completeness of the fermionic Fock space implies that the second equation in (\[eq2\]) can be satisfied only if the terms in the square bracket vanishes for all $n$ , this giving the following $N_f+1$ coupled equations $$\begin{aligned}
 && i \hbar \psi^b_t=
\left[-\frac{\hbar^2}{2 m_b}\nabla^2+V_{ext}^b+
g_{b}N_b|\psi^b|^2+g_{bf} \rho_{f} \right]\psi^b, \nonumber
\\&& i \hbar \psi^f_{n_t}=
\left[ -\frac{\hbar^2}{2 m_f}\nabla^2+V_{ext}^f + g_{bf} N_b
|\psi^b|^2 \right]\psi^f_n,\label{bose-fermi-eqns}\end{aligned}$$ with $\rho_f=\sum_{i=1}^{N_f} |\psi_i^f|^2$ the fermionic density and $\psi^b_0\equiv \psi^b$. From the above derivation it is clear that while bosons are treated in a classical mean field approximation the spin polarized fermions are in true quantum regime due to the lacking of interaction among them. Also note that for $g_{bf}=0$ the above equations separate into the Gross-Pitaevskii equation for the BEC part and $N_f$ identical (due to the lack of interaction among fermions) Schrödinger equations 