# Why does covalent bonding not break down if observer effect can be applied to atomic electrons? [closed]

The observer effect in quantum mechanics states that when unobserved, quantum particles such as electrons can simultaneously occupy two different states. In an atom of any element, where there are electrons around the nucleus, this effect is hinted at in some publications:

Electrons can occupy several energy levels, or orbitals, simultaneously. (Scientific American.(2012, October 9). Bringing Schrödinger's Cat to Life. Retrieved from https://www.scientificamerican.com/article/bringing-schrodingers-quantum-cat-to-life/.)

This idea is so weird. It is equivalent of saying that when unobserved, all the electrons of an atom can both be in their normal configuration and be in a totally different configuration. Only the act of observation forces the electrons into their normal configuration. Take the example of neon atom. According to the observer effect, when unobserved, its electron configuration can both be 1s4 2s3 2p3 (one of many possible alternative configurations) and 1s2 2s2 2p6. Only when observed will the electron configuration collapse into its normal version of 1s2 2s2 2p6.

I don't know if this understanding of observer effect on atomic electrons is valid. If it is, there are at least two serious problems. First, the Pauli exclusion principle will break down, because the alternative electron configuration allows more than 2 electrons in the same orbital (specified by the principal quantum number, the angular momentum quantum number, and the magnetic quantum number). Second, the foundation of covalent bonding is that when two or more atoms pool together electrons in orbitals that overlap, the total energy of the system is reduced to the minimum. If the observer effect on atomic electrons is valid, the energy required to maintain the system may get even lower than that in covalent bonding, and then covalent bonding will break down. All that's needed is for each atom to achieve an alternative electron configuration.

Thanks, everybody. I now understand this question better. Yes, the observer effect does apply to atomic electrons by allowing each of them to occupy several orbitals, subshells, or even shells at the same time. However, if we look at all the electrons and summarize their states into an electron configuration scheme, the scheme will still be the same as that obtained through the aufbau principle.

• Why do you think that a superposition of different states must include invalid states? It can't. – Ruslan Aug 24 '20 at 12:49
• So, does superposition suggest two simultaneous states when both states are valid? In this way, I can see that in an atom, all the electrons can be in different orbitals at the same time, but when we give the electron configuration, both states conform to the ground-energy scheme. – nothingnessEMPTY Aug 24 '20 at 12:54
• All of the states that are simultaneously existing describe the same overall electron density and the same total energy. The only difference is that each electron does not have defined quantum numbers. – Andrew Aug 24 '20 at 13:22
• This is a quite confusing interpretation of quantum mechanics of atoms and solids. Perhaps one should consider that covalent bonding is, in fact, the alternative electron configuration that minimizes energy of the system. – Jon Custer Aug 24 '20 at 14:34
• Let me add that in a molecule, not only quantum particles are involved (electrons) but also very massive "things" (nuclei), with a slow motion compared to electrons and a non-quantistic behavior. Electrons might even go crazy for some fractions of femtosecond, but their lowest-energy state is due to the environment in which they are found (around nuclei) – The_Vinz Aug 24 '20 at 15:32

To start with a simple example, consider an atom of helium in its singlet ground state. Prior to measurement, neither electron has a defined spin. Both are a superposition of $$+1/2$$ and $$-1/2$$ states, but that does not mean that the helium atom can have $$+1$$ or $$-1$$ spin. The net spin is still 0. If we measure one electron's spin to be $$+1/2$$, the other will necessarily have $$-1/2$$, even though neither had defined spin prior to measurement.