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

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  • $\begingroup$ what works for orbits works for orbitals, orbitals are just probability loci with the quantum numbers assigned to bohr type orbits. $\endgroup$ – anna v Mar 27 at 10:09
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    $\begingroup$ @annav That's just... wrong. Orbitals are substantially more than "just probability loci". If you only look at the probability, then you're missing information about the phase of the wavefunction, and that is essential to capture the full quantum state. $\endgroup$ – Emilio Pisanty Mar 27 at 10:17
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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.

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  • $\begingroup$ Yes,by non-stable state I mean state e.g. consist of 1s and 2s wave functions.Indeed I never learn the multi electron wave function chapter in detail because it is not easy to accept the concept,so I'm not very sure whether this is the right answer.But at least I know that the key is that you cannot talk about one electron alone in this case. $\endgroup$ – jw_ Mar 27 at 11:29
  • $\begingroup$ BTW can you up vote my quesstion?After asking some (may be stupid) questions,I have only 7 votes left ,maybe I will lose the right to ask later,I just want to learn something... $\endgroup$ – jw_ Mar 27 at 11:33
  • $\begingroup$ @jw_ You don't seem to be in immediate danger of a question ban (unless you have deleted posts which don't show up in your public profile). You've been around for two days - how about you take it easy, and you read more than you ask for a while? It will give you a better idea of the sort of things (say, starting with proper grammar and punctuation, but also how to properly provide the context and background of your core query) that help questions succeed here. $\endgroup$ – Emilio Pisanty Mar 27 at 12:00
  • $\begingroup$ "I never learn the multi electron wave function chapter in detail because it is not easy to accept the concept" $-$ so you're saying you've been explicitly presented with the material but you didn't feel like learning about it? If so, why should others spend time and effort trying to explain it to you? $\endgroup$ – Emilio Pisanty Mar 27 at 12:01
  • $\begingroup$ @ Emilio Pisanty Its not I don't feel like learn it,it's that after reading some of the answer I already know that I need to read more in the text book before some day I finally read the answer----I don't have the knowledge that is needed to fully understand the answer.Your answer is always valuable for others and maybe in some day for me and your time won't be wasted.Indeed I'm sure I have to learn basic QM thoroughly before I ask more question about QM.I'm not an native English speaker so I have many grammar error and there is no solution for this. $\endgroup$ – jw_ Mar 27 at 12:09

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