# About the order of indices in basis set expansion

In many books, e.g., Szabo and Ostlund, Helgaker Jorgensen and Olsen, etc etc, the expansion of Hartree-Fock orbital by basis set is written as a kind of $$\phi_i = \sum_{\mu} C_{\mu i} \chi_{\mu}$$

My question is, why the expansion coefficient $$C_{\mu i}$$ is not written in $$C_{i \mu}$$? I thought $$C$$ is coefficient, $$\chi$$ is a basis, if one wrote as $$\sum_{\mu} C_{ i \mu} \chi_{\mu}$$ the contracted indices come closer. But in reality, it is the opposite. Is that because people more commonly coded in Fortran, which is column-major, thus $$C_{\mu i}$$ looks better (the first index changes first)?

Let's start by writing the MO expansion in matrix form: $$\Phi=C^\dagger\Psi$$ Here, $$\Phi$$ and $$\Psi$$ are column vectors of MO and AO basis functions respectively. So we have to use the adjoint/transpose of the MO coefficients to transform between the two, but it seems like we could just as easily have defined the MO coefficients as $$B=C^\dagger$$ so that $$\Phi=B\Psi$$ and we wouldn't have to take a transpose. Let's continue on with this alternate definition and write out the Hartree-Fock Roothaan-Hall equations: $$FB^\dagger=SB^\dagger\epsilon$$ Now we have a transpose in these equations, so it seems like we are stuck with one either way and its just a matter of where we want to put it. Why might we prefer to have the transpose for the AO to MO transformation?
For one, we generally don't have to explicitly use this equation in a HF calculation, since we can solve the HF equations in AO basis and generally only need to transform to MO basis if we are proceeding on to some correlated post-HF method. Another reason is $$C$$ can be nicely thought of as a collection of column vectors that represent the MOs in AO basis. Notice the Roothaan-Hall equation doesn't directly use the MO functions, but instead passes around their coefficient representation.