Why $\psi(r_e, r_n) = \sum \psi_e^i(r_e|r_n) \chi^i(r_n)$ in Born-Huang expansion?
After reading this lecture note I have a better understanding.

The reason I'm so confused is that the form 
$$\psi(r_e, r_n) = \sum_{i = 1}^{\infty} \psi_e^i(r_e|r_n) \chi^i(r_N)$$
seems to be a linear combination with coefficients that are all $1$. And this is very importand for the derivation of the set of entangled nucleus time-independent SEs since there is no arbitrary coefficients in the set of equations. 

After I read the lecture note, here is my understanding:

For a fixed atomic configuration $R^{\alpha}$, the electronic state can be represented by a linear combination of the eigenstates of $\hat H_e(R^{\alpha})$
$$\psi(r_e| R^{\alpha}) = \sum_{i = 1}^{\infty} c_i(R^{\alpha})\psi^{i}_e(r_e|R^{\alpha})$$
although $\psi(r_e| R^{\alpha})$ is an electronic state, this is essentially a solution to $\hat H \psi = E \psi$ at a fixed atomic configuration. 

Since by varying $R^{\alpha}$, the set of linear coefficients {$c_i$} will change. So the general $\psi(r_e, r_n)$ can be represented as
$$\psi(r_e, r_n) = \sum_{i = 1}^{\infty} \chi^i(r_N) \psi_e^i(r_e|r_n)$$
in which the set of {$\chi^i(r_n)$} is just the set {$c_i({R^{\alpha}})$}.

Essentially, the $\psi(r_e, r_n)$ that we plug back into $\hat H \psi = E \psi$ is still an electron wavefunction.