Schrödinger's equation yields us the solutions of 90% probability of presence of electrons in orbitals. Hence, the electrons in theory should be able to exist elsewhere in space as well.

In that case, they could exist in another atoms' vicinity, possibly in its orbital as well. So, what would be the quantum numbers in this case? Shouldn't they be affected by possible interference, hence changin quantum properties like spin, etc.

P.S. I have very limited knowledge on quantum mechanics, so please excuse me if any statement of mine is incorrect or sounds rudimentary.

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    $\begingroup$ This is incorrect from the very start: "Schrödinger's equation yields us the solutions of 90% probability of presence of electrons in orbitals." Schrodinger's Eq gives us wavefunctions, not just probabilities, and especially not "90% probabilities". Thought this the statement is unrelated to your question, which is about the quantum numbers if I understand well. $\endgroup$
    – Greg
    Jul 27, 2020 at 14:44
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    $\begingroup$ Orbitals aren't "90% probability regions". Orbitals are wavefunctions, which are functions that have a given value at every point in space, let's say $f = f(x, y, z)$, (although the conventional symbol is $\psi$ not $f$). The shapes that you see are our mere human attempts to visualise, or graphically represent these functions, and because the functions extend to infinity, we have to choose some cutoff for our visualisation. The 90% thing is related to this cutoff and is solely for visualisation purposes; it does not actually have much to do with the orbitals themselves. $\endgroup$ Jul 27, 2020 at 14:44
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    $\begingroup$ Heisenbergs hydrogen atom is a hypothetical entity, living in a universe that is otherwise completely empty. Its orbitals properties differ a lot from those of a single hydrogen molecule, which still differ a bit from those of a molecule in a drop of liquid hydrogen. $\endgroup$
    – Karl
    Jul 27, 2020 at 15:01
  • $\begingroup$ Ok, misconception cleared. I knew the fact that Ψ^2 represents the probability density and that orbital is a hypothetical region of space that we assume to be present for better understanding, however didn't seem to correlate the two facts together. $\endgroup$
    – sfsg
    Jul 28, 2020 at 4:37

1 Answer 1


The issue here is that Schrödinger's equation doesn't yield us a "solution of 90% probability". It doesn't yield a probability at all. It yields a wave function, $\psi(x,y,z,t)$, which has units $\frac{1}{\sqrt{m^3s}}$. To get a probability density, we square its norm: $|\psi|^2=\psi^*\psi$. Note: even this is not a probability. A probability is dimensionless, whereas this has units $\frac{1}{{m^3s}}$. We can represent the probability of the particle described by $\psi$ being in the range (x, y, z, t) and (x+dx, y+dy, z+dz, t+dt) by multiplying the norm squared by dxdydz*dt: $P(x,y,z,t) = |\psi(x,y,z,t)|^2dxdydzdt$. Of course, these increments are infinitesimal, so this alone would evaluate to zero. However, this approach allows you to integrate over a range of x,y,z,and t: $P(x+\Delta{x},y+\Delta{y},z+\Delta{z},t+\Delta{t}) = \int{dx\int{dy\int{dz\int{dt|\psi(x,y,z,t)|^2}}}}$.

A better understanding of this will help you understand the issues with your second question. However, we can also rely on the Pauli exclusion principle to assert that there cannot be two Fermions that share the same set of quantum numbers (note that some of the quantum numbers may be shared, but not all simultaneously.

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    $\begingroup$ Why do I think you're wasting time on this OP? $\endgroup$
    – Mithoron
    Jul 27, 2020 at 17:44
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    $\begingroup$ Ideally, the answer will help more than just the OP $\endgroup$
    – jezzo
    Jul 27, 2020 at 18:02
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    $\begingroup$ Yeah, it sure could, but under better question (or at least more popular) chances would be better. There was lots of q. "what are these orbitals" already. $\endgroup$
    – Mithoron
    Jul 27, 2020 at 19:08

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