The orbitals of an atom can be thought of as being formed from the probability of finding electrons in those orbitals. If the orbital is 1s (n = 1, l = 0), then it has a certain "volume" for which the electron can be at any position, and in some radial distances from the nucleus, it will be at those distances with more probability than in other distances. Imagine then a 2s orbital (n = 2, l = 0), it will have a bigger volume than the 1s orbital, which represents the probabilities of finding the electron at certain points in that volume.

Now, since the volume of 2s is greater than 1s, that means that if there are electrons in the 2s orbital, they will have a certain probability of being at a place where the 1s orbital is. Now, if this is a hydrogen or helium neutral atom with its electron in an excited state, then there would be no problem with this, I guess. However, if we talk about Lithium, which has 3 electrons, the aufbau's principle and Pauli's exclusion principle would imply that 2 electrons are at the 1s orbital, and 1 electron at the 2s orbital. Now, as I said in the first paragraph, that 1 electron at 2s will have a probability of being at 1s! Then, Pauli Exclusion Principle would be violated because only 2 electrons can occupy the 1s orbital, so this could not happen, if Pauli Exclusion Principle does hold. However, then, how can one explain this?

Why is there a probability of finding an electron of a 2s orbital at a same radius as one that holds for a 1s orbital? The only reasonable explanation that I can come up with is: If the electron in the 2s orbital shifts to the 1s orbital (since it would be possible since it has probability to be there), then one electron in the 1s orbital immediately shifts to the 2s orbital, and Pauli Exclusion Principle would hold. Is this reasoning correct? If not, why is there a probability of being at 1s orbital for an electron at the 2s orbital?

As an aside: I am struggling to think intuitively of why is energy quantized. If an electron is at 1s orbital, it will have a certain energy. But why is this energy fixed for that orbital? If the electron can be at points closer to the nucleus, then the potential energy of the system would be lower... Are the energy levels an average? I think the correct question would be: "Why is energy quantized and not continuous?"

If I said something that is wrong, do not hesitate to correct me please. As I understand that these topics deal with Quantum Mechanics, I can be completely wrong at some stuff because my knowledge of this area is little and it is just from what is usually learned in Chemistry. Thanks a lot in advance!


4 Answers 4


The Pauli Exclusion Principle (PEP) talks about two fermions not occupying the same state. For an electron bound to an atom, such state can be fully characterized by is position -- usually in spherical coordinates $(r,\theta,\phi)$ --, energy level $n$ and angular momentum $(l,m)$. This already answers your first question: even though the 1s and 2s electrons may be at the same place, they are not in the same energy level, hence this does not violate the PEP.

Regarding the quantization of energy, there are many ways to answer. The most “correct” explanation is that the time-independent Schrödinger equation for the electron-orbiting-nucleus system is a bounded equation, so it has discrete eigenvalues. That is, if you solve $H\psi=E\psi$, you find that $E$ can only have certain values that can be indexed with an integer $n$.

As for why this happens, I'm afraid that no intuition will quench your thirst for understanding this very crucial point, you have to learn the math behind it to get a grasp of what's going on.

  • 3
    $\begingroup$ "the 1s and 2s electrons may be at the same place" — they can't be in at the same place simultaneously, unless their spins are different. The fact that orbitals overlap doesn't imply that joint probability is nonzero. $\endgroup$
    – Ruslan
    Aug 12, 2021 at 11:21
  • $\begingroup$ A better way to analyze the "distunguishability" of two states would be to consider the wave function overlap $\left\langle\psi'\middle|\psi\right\rangle$ between those wave functions. $\endgroup$
    – schade96
    Aug 12, 2021 at 11:29
  • 3
    $\begingroup$ Your statement about what fully characterizes an electron bound state is incorrect - energy level $n$, angular momentum $l,m$ and spin $m_s$ fully characterize an energy eigenstate of an atom, position doesn't enter into it. A state localized at a certain position would not have definite energy or momentum. $\endgroup$
    – ACuriousMind
    Aug 12, 2021 at 14:06
  • $\begingroup$ "the 1s and 2s electrons may be at the same place" is flatly incorrect. This answer needs to be corrected. $\endgroup$ Aug 12, 2021 at 16:31
  • $\begingroup$ What do you mean by "such state can be fully characterized by is [sic] position?" Are you referring to the position-space representation of the state, as in $\langle x\vert n,l,m\rangle=\psi_{nlm}(x)$? $\endgroup$
    – Sandejo
    Aug 12, 2021 at 16:53

Now, since the volume of 2s is greater than 1s, that means that if there are electrons in the 2s orbital, they will have a certain probability of being at a place where the 1s orbital is.

Since neither of these distributions are bounded this probability is 100%. With probability 100% if you measure an electron in a 2s orbital it is in a location where you could also measure an electron in a 1s orbital if you had performed a position measurement on that.

Then, Pauli Exclusion Principle would be violated because only 2 electrons can occupy the 1s orbital, so this could not happen, if Pauli Exclusion Principle does hold. However, then, how can one explain this?

I think you might think that these are classical orbits at a fixed radius? They are not. Therefore observing an electron at one particular point covered by the 1s orbital would not put it into the 1s state, if you catch my drift.

What is more interesting in quantum mechanics is when a transition exists between two non-orthogonal states. And the mathematics for determining that is very similar to what you're talking about. We take every point in space that the electron could be measured at, and the wave function assigns to that point an complex number that we call a transition amplitude, where the volume integral of the squared magnitude of the complex number gives a probability to be found in that volume, while the phase of the complex number gives a way for wavy interference effects to still happen.

Rule one for amplitudes is that the amplitude for a parallel process, A can go to B via either Z1 or Z2, is the sum of the amplitudes for each of the individual processes. Rule two for amplitudes is that the amplitude for a sequential process, A can go to B by going first to Z1 and then from there to B, is complex multiplication: you take these two complex numbers and you multiply their magnitudes and add their angles.

You may already see the punchline appearing here, but let me be explicit about it. When you try to do this, you get a volume integral over all space, of the amplitude for the 2s electron to go to some point (x,y,z) times the amplitude to go from that point into 1s. You do this integral and you find perfect destructive interference. The resulting amplitude is zero. No transition is possible.

And indeed you can prove that it must be zero for a different reason: because these are both eigenvectors with different eigenvalues for an operator, which operator happens to be the Hamiltonian. It does not have to be the Hamiltonian for that result to hold, but there is an added strength to the statement when it is the Hamiltonian: eigenstates of the Hamiltonian do not evolve in a measurable way over time. So it stops being just “there is no instantaneous way for this to happen” and it's strengthened to “even if you let it sit for a certain period of time this does not happen.”

Quantum mechanics has a couple of other tricks that are worth mentioning as well, one of them is that electrons are indistinguishable particles, so there would actually be no way to find out if an electron in the 1s state got there from the 2s state or not. So maybe you detect an electron in the 1s state, you cannot tell me with any certainty that you are sure it was the one that used to be in the 2s state. Maybe it was, maybe it wasn't. Another tricky aspect like this is that, if you are going to lock down one of these electrons’ positions very tightly, this is going to require firing a large amount of light at the atom, and that light is going to be enough to kick those electrons out of their orbits, so finding one of them not where it's supposed to be is actually going to be very normal given what you're trying to do to measure where that electron is.


"Why is energy quantized and not continuous?"

The great Feynman has answered this in a very didactic way: https://www.youtube.com/watch?v=36GT2zI8lVA

To rephrase his speech: asking "Why" as a layperson usually has no useful answer in physics, at least not if asking a very general question sort of "out of context", like this here.

To cite Feynman:

Now, when you explain a "Why", you have to be in some framework that you allow something to be true. (sic)

This means that looking at a mathematical proof, you can, at each step, ask "why" this step works. For an answer, you can then check the previous step, convince yourself that everything up to that one is correct; also check that whatever mechanism the current step is has also been proven or postulated as an axiom; and then you can with confidence say "that's why it is so". But this only holds for statements within this mathematical/abstract framework, it says nothing about reality.

The same approach is not possible for questions about the universe. To take an easier, everyday question:

"Why does a rock fall down if I let it go while standing on Earth?"

The answer is obviously "because gravity makes bodies move towards each other".

Unfortunately, my next question will be "OK, but why does gravity make bodies move towards each other?"

Answer: "because mass bends spacetime, and stuff moves along trajectories in spacetime."

My next question will be "Why does mass bend spacetime?"

And at this point you're out. No can defence. Maybe you're an actual physicist and can come up with an extra answer to that, but I guarantee you that I as a lay person will always be able to come up with another, more basic "Why" question which eventually (and usually very quickly) will lead you where there simply is no further answer. Heck, children do this as favourite sport at a very young age.

The problem is not that it is somehow hard for us to come up with the answers (which of course is also true), but that eventually, the universe does not contain the answers to "Why" questions if you drill deep enough. The axioms are not there for us to see.

So, while "Why" is the great driver for many scientific endeavours, as a question that actually looks for a concrete answer, it can be very frustrating.

Finally: the maths and our theories are most definitely not the proper answer (except if we rephrase the question within the context of the theory; which then has little to do with objective reality anymore). The universe does not care about our formulas.

To finish up with Feynman:

I'm not going to be able to give you an answer to why [...] except that I ensure you that they do, and to tell you that that's one of the elements of the world [...] but I really can't do a good job - any job - of explaining [...] in terms of something you're familiar with, because I don't understand it in terms of anything else that you're more familiar with. [big smile]


'why is there a probability of being at 1s orbital for an electron at the 2s orbital?'

Pauli exclusion states that only one fermions can occupy one orbital. Due to spin, this means that two electrons can occupy a single spatial orbital. In your example of Helium that is indeed the case. However, you are right that there is a probability of two or more electrons to be found at the same location. Note that although this gives infinite repulsion, the volume in which this occurs vanishes hence this still gives a finite contribution to the repulsion energy. Note that electrons not only have Pauli correlation but also Coulomb correlation, allowing for electrons to avoid each other somewhat better than the configuration $1s^2 2s$ suggests. The $1s^2 2s$ configuration is the Hartree-Fock-Slater, single determinant, approximation. A more accurate wave function will have admixtures of determinants like $1s 2s^2$ etc.


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