# Why are there three $p$-orbitals?

This question is specifically about Schrödinger quantum mechanics, but if an answer in some other mode would illuminate it could be acceptable, as demonstrating a physical or mathematical reason for added axioms.

In short - since the p-orbital has rotational symmetry about only one axis, but the potential of a point charge has spherical symmetry, a specific solution corresponding to a p-orbital should also be a solution when arbitrarily rotated. That means there is an infinite number of p-orbital solutions in this context. However, the dimension of the solution space for the given energy, that is, the eigenspace for the given eigenvalue is presumably exactly three. One can use three axial p-orbitals to span the whole eigenspace.

Thus the exclusion principle for fermions seems to be that there are at most the dimension of the eigenspace number of particles in an eigenspace rather than 1 particle in an orbital, if an orbital is taken as a solution to the Schrödinger equation.

Can anyone confirm or deny this line of reasoning? And provide a reference to explicit statement in the literature?

## 3 Answers

Thus the exclusion principle for fermions seems to be that there are at most the dimension of the eigenspace number of particles in an eigenspace rather than 1 particle in an orbital, if an orbital is taken as a solution to the Schrödinger equation.

Your reasoning is correct. We don't normally frame things in this way because it is clunker than y the strictly-equivalent language of one particle per orbital in a linearly-independent set, but your description is somewhat more accurate.

The true underpinnings of this structure is the fact that multi-electron states must be antisymmetric with particle exchange; if you want to produce such a state given a set of orbitals, then you apply a procedure called antisymmetrization, and you end up with a state called a Slater determinant. If you start with more electrons than the dimension of the space spanned by your orbitals, then the Slater determinant will vanish.

Furthermore, as you correctly note, what really matters in a multi-electron state is strictly the subspace spanned by the constituent orbitals, and not the specific choice of orbitals as a basis for that subspace. For more on that, see Are orbitals observable physical quantities in a many-electron setting?.

More generally, the passage from the naive Pauli exclusion principle to the fully-grown version in terms of antisymmetrized Slater determinants is treated at length in any atomic physics textbook; if you want something specific I'll recommend Haken and Wolf's The physics of atoms and quanta, but any textbook should do.

• I thought about Slater Determinants and chased up a copy of Haken and Wolf. I have only spent a couple of hours on the book - but it plays out as very well grounded in experiment as well as the theory. Slater determinants is clearly the core of the answer. I will just need a bit of time to digest the point. Thanks. – Ponder Stibbons Jul 27 '18 at 22:39
• Ah, I believe I understand. The algebraic condition for the antisymmetrization of a sum of products of orbitals is identical to the structure of a determinant. So, if you use products of one too many orbitals there is nothing left but the zero element that is fully anti symmetric. – Ponder Stibbons Jul 27 '18 at 23:38
• Yes, that is precisely it. – Emilio Pisanty Jul 28 '18 at 5:27

… if an answer in some other mode would illuminate it could be acceptable, as demonstrating a physical or mathematical reason for added axioms.

There is a reason from chemistry which I want to discuss. Perhaps you know, that the observation of Methane and other chemical compounds led Linus Pauling to the concept of $sp^x$-hybridisations. The Carbon in Methane has 2 electrons in the s-subshell (to talk about orbitals is common but I think not precise because nothing is rotating nor moving) and 2 electrons in the p-subshells. But in the compound with the electrons from 4 Hydrogen atoms it was observed a tetrahedral structure:

So these observations show that in compounds the electrons from s- and p-subshells behave in the same manner, they are indistinguishable.

The more interesting point is the fact that the subshells are calculated from spherical harmonics and for Euclidean coordinates. What was calculated is something like this:

But would it be possible to calculate some spherical harmonics with a tetrahedral structure? The answer is clearly yes:

(From the visualization of spherical harmonics)

There are 8 equally distributed directions with positive and negative signs. Please pay attention to the fact that around every positive area are exactly 3 negative areas and on the corners are exactly 3 positive areas. The same in analogy holds of course for the all negative areas. If interprete the signs as the direction of magnetic dipoles one get 8 electrons in perfect equilibration and this is an amazing model of the Neon atom.

In short - since the p-orbital has rotational symmetry about only one axis, but the potential of a point charge has spherical symmetry, a specific solution corresponding to a p-orbital should also be a solution when arbitrarily rotated. That means there is an infinite number of p-orbital solutions in this context.

Science is a mix of learned knowledge and thinking of unthinkable. Your statement gives the thinking a new impulse but will be meet with resistance.

All images from Wikipedia.

• Thanks, this is interesting. I upvoted, but am not sure its the "one true" answer, as such, so I will cogitate. In particular, I did skim read Pauling's book a while ago, which was also interesting - with superposition shown to be required to explain chemical and thermodynamic states of gas - but, I want to follow up on the Methane connection. – Ponder Stibbons Jul 27 '18 at 7:00
• Like Terry Pratchett? – HolgerFiedler Jul 27 '18 at 7:03
• @Ponder Just for clarity, this beyond-wrong answer does not represent quantum mechanics as understood anywhere outside of this particular answerer's head. The f-shell spherical harmonic shown is only occupied in lanthanide elements, and it is no better as a model of neon than a play-doh figurine. – Emilio Pisanty Jul 27 '18 at 7:13
• I know that I’m not a scholar and my reputation is zero. But that doesn’t mean that it is not worth for a serious discussion about something different from the accepted state of art. My dry written point of view see here. – HolgerFiedler Jul 27 '18 at 7:29
• You continue to miss the point. Being "a scholar" or having "reputation" is irrelevant - what matters is providing a description of nature that fits into a larger, consistent framework with explicative power and which is able to account, with explicit and quantitative calculations, for the huge mountain of experimental evidence that atomic and molecular physics has accumulated over the past ninety years. Instead of linking to such calculations, though, you literally just chose to link to pictures of play-doh figurines. – Emilio Pisanty Jul 27 '18 at 10:37

but the potential of a point charge has spherical symmetry, a specific solution corresponding to a p-orbital should also be a solution when arbitrarily rotated. That means there is an infinite number of p-orbital solutions in this context.

When the Hamiltonian has a symmetry, the solution does not need to be invariant under the same symmetry. But it has to obey certain transformational properties under the group action. Mathematically speaking, a symmetry is a group action, and the solutions of a symmetric Hamiltonian are representations of the group. Any representation of a group can be written as a sum of irreducible representation of the group.

• Thanks. The solution not having the same symmetry as the Hamiltonian is precisely the matter at hand. When there is a symmetry in the equation, there is no need for any solution to have that symmetry - just that the symmetry operation applied to a solution is another solution. Hence, while the s-orbital has spherical symmetry, and so is only one solution under the symmetry group, the (group theory) orbit of any p-orbital has an infinite number of elements. That was the spur for my question. – Ponder Stibbons Jul 27 '18 at 7:04
• I think I understand your question better now. You already recognize that the three p orbitals are basis of an eigenspace, but you left the meaning of the eigenspace ambiguous. The eigenspace is the space of solutions with angular momentum L=1, so I guess I could rephrase your question as "why is the solution space with L=1 has dimension of three?" I'm not sure if there is an intuitive answer that does not involve representation theory of angular momentum... – wcc Jul 27 '18 at 7:33
• I have a strong understanding of the classical angular momentum space and have no problem transferring this to the hybrid discussion of p-orbitals. In the mode you use my question would be - why is the maximum number of fermions with orbitals in a given eigen space equal to the number of dimensions of that space rather than the number of solutions. The core of the answer to this has been given by Emilio Pisanty. I want to be able to derive the exclusion principle for higher dimensional eigen spaces from first quantum principles. – Ponder Stibbons Jul 27 '18 at 22:30