# Quantum Mechanical Meaning of Atomic Orbitals

According to quantum mechanics, for multi-electron atoms, a single electron around the nuclei can be in the state of linear combination of different eigen energy states. In that case, even the energy of the electron is not one of the eigen energies.

In chemistry, we often say that an electron is in an orbital (such as 1s, 2s), another electron is in another orbital, etc. How does it make sense if an electron is in a linear combination state?

As John Rennie points out, the picture of an electron in an orbital breaks down in multielectron atoms because of electron correlations. However, you can still assign a "fraction" of an electron to an orbital by diagonalizing the reduced one-particle density matrix (1RDM) of the system. This matrix is Hermitian, positive semi-definite, has eigenvalues $n_i \leq 1$, and describes the total electronic density of the system as \begin{equation} \rho (r) = \sum \limits_i n_i \psi_i^*(r) \psi_i(r) \end{equation} where $\psi_i (r)$ are the orbitals. Thus, $n_i$ is the "occupation number" of orbital $\psi_i$ and describes how much on average an electron is in that orbital.
Now, with respect to your question: "In chemistry, we often say that an electron is in an orbital (such as 1s, 2s), another electron is in another orbital, etc. How does it make sense if an electron is in a linear combination state?" Well, it makes sense because often we find that there are $N_e$ (number of electrons) eigenvalues $n_i$ which are very close to one, while the rest of the eigenvalues are very small. The reason for this is that the exact wavefunction is an infinite sum of Slater determinants (i.e., an antisymetric function which describes one electron configuration) considering all possible configurations. However, because of the variational principle, configurations which are very high in energy contribute very little to the wavefunction and hence have small weights $n_i$.