Pauli exclusion principle for non-stable states

Question

If an electron partially occupy $1s$ orbital, can other electron occupy $1s$ partially, too?

Answer 1

The "grown-up" version of the Pauli exclusion principle reads as follows:

Electrons are identical particles which obey Fermi-Dirac statistics, so that the wavefunction needs to change sign if two electrons are exchanged.

For the specific case of two electrons, the wavefunction $\Psi$ is a function of two positions $\mathbf r_1$ and $\mathbf r_2$, with one electron at $\mathbf r_1$ and the other at $\mathbf r_2$, and the Fermi-Dirac condition is that $$ \Psi(\mathbf r_2, \mathbf r_1) = - \Psi(\mathbf r_1, \mathbf r_2). \tag1 $$

When we say that "one electron is in state $a$ and the other electron is in state $b$", what we normally mean is that the global wavefunction is in the specific form of a Slater determinant, $$ \Psi(\mathbf r_1, \mathbf r_2) = \frac{1}{\sqrt{2}} \bigg[ \psi_a(\mathbf r_1)\psi_b(\mathbf r_2) - \psi_b(\mathbf r_1)\psi_a(\mathbf r_2) \bigg]. \tag 2 $$ It should be obvious that this satisfies the condition $(1)$. (However, it is important to remark from the outset that this is not the only possible way for this to happen $-$ there are antisymmetric wavefunctions which satisfy $(1)$ but which cannot be expressed as a single Slater determinant as in $(2)$; those are generally only required in post-Hartree-Fock theories.)

This is where the "pre-grown-up" version of the Pauli exclusion principle fits: trying to have "two electrons in the same state" would require you to have $\psi_b=\psi_a$, and if you try to put that into the Slater determinant in $(2)$ it will lead to a vanishing two-electron wavefunction, so this state is not possible for fermions (as opposed to bosons!).

---

It's unclear what you mean by

an electron partially occupy $1s$ orbital

and why you think this is related to "non-stable" states. (Unstable states in quantum mechanics are tricky beasts - see the paper cited at the end of this answer to see just how tricky.) If by that quote you simply mean, say, that you have one electron in an orbital that has only a partial overlap with the $1s$ orbital, such as e.g. $$ \psi_a(\mathbf r) = \frac{1}{\sqrt{2}} \bigg[ \psi_{1s}(\mathbf r) + \psi_{2s}(\mathbf r) \bigg], \tag 3 $$ say, then the overall answer is yes: it is indeed possible for this orbital to be part of a two-electron state in combination (through a Slater determinant) with a second orbital which includes a nonzero overlap with the $1s$ state; the simplest such state is just $$ \psi_b(\mathbf r) = \frac{1}{\sqrt{2}} \bigg[ \psi_{1s}(\mathbf r) - \psi_{2s}(\mathbf r) \bigg]. \tag 4 $$

However, it is important to note that this type of manipulation should be handled with extreme care, since individual orbitals typically do not have physical meaning in multi-electron states, and indeed if you substitute in $(3)$ and $(4)$ into the formula $(2)$ for the Slater determinant, you will find that $$ \frac{1}{\sqrt{2}} \bigg[ \psi_a(\mathbf r_1)\psi_b(\mathbf r_2) - \psi_b(\mathbf r_1)\psi_a(\mathbf r_2) \bigg] = \frac{1}{\sqrt{2}} \bigg[ \psi_{2s}(\mathbf r_1)\psi_{1s}(\mathbf r_2) - \psi_{1s}(\mathbf r_1)\psi_{2s}(\mathbf r_2) \bigg], \tag 5 $$ or, in other words, "rotating" in this way to "mixed" orbitals does not actually accomplish anything $-$ because of the antisymmetry of the state, and the linearity of quantum mechanics (a.k.a. "the wave nature of matter", as that property is called in the earliest introductory texts and in popular-science presentations), this does not affect the real, physical state in any way.