# Why don't electrons occupy infinite degenerate states with the same energy?

I have a question about the degeneracy of energy levels in atoms and the Pauli exclusion principle.

I understand that, according to the Pauli exclusion principle, each orbital can host a maximum of two electrons with opposite spins. However, I am wondering how this principle applies in the presence of degeneracy.

If there are multiple degenerate states (with the same energy), in theory, I could form linear combinations of these degenerate states and obtain infinite states with the same energy. So why don't electrons occupy infinite states with the same energy? Am I missing something in my understanding of degeneracy and the linear combinations of quantum states?

• I mean, you have already said what happens in the case of degeneracy due to spin. There is really no difference if there is degeneracy due to other factors Commented May 23 at 11:59
• forming a linear combination of states still only applies to one electron. the moment you consider two or more you get antisymmetric tensor products which will vanish if two states are the same, no matter for what reason. Commented May 23 at 12:18
• @BySymmetry indeed now I have the same question also for the spin Commented May 23 at 12:52
• @paulina can you explain it in an answer? I don't understand Commented May 23 at 12:55
• Why? Because you don't fall through your chair into the center of the Earth. Therefore, the theory must include some physics that prevents this. Commented May 24 at 12:49

Pauli exclusion comes from the fact that the wave function of the $$N$$-electron system has to be anti-symmetric under exchange of the particle labels; this is a consequence of the fact that electrons are indistinguishable (and that they are spin-1/2 particles).

Let's see what happens when you try to put three electrons into three states that all live in a two-dimensional space. Suppose that the space is spanned by the orthonormal basis $$\lvert v_1\rangle$$ and $$\vert v_2\rangle$$. Then, consider three different states in this space, \begin{align} \lvert\psi\rangle &= a_1\lvert v_1\rangle + a_2\lvert v_2\rangle\,,\\ \lvert\varphi\rangle &= b_1\lvert v_1\rangle + b_2\lvert v_2\rangle\,,\\ \lvert\chi\rangle &= c_1\lvert v_1\rangle + c_2\lvert v_2\rangle\,. \end{align} Putting three distinguishable particles in these three states would correspond to the state $$\lvert1:\psi\rangle\lvert2:\varphi\rangle\lvert3:\chi\rangle\,,$$ where the number is the particle label. Now, if these three particles are fermions, then we need to anti-symmetrize this state. This systematic way of doing that is to form the Slater determinant, $$\lvert\Psi\rangle = \begin{vmatrix} \lvert1:\psi\rangle&\lvert1:\varphi\rangle&\lvert1:\chi\rangle \\ \lvert2:\psi\rangle&\lvert2:\varphi\rangle&\lvert2:\chi\rangle \\ \lvert3:\psi\rangle&\lvert3:\varphi\rangle&\lvert3:\chi\rangle \end{vmatrix}\,.$$ We would then plug in the expressions for $$\lvert\psi\rangle$$, $$\lvert\varphi\rangle$$, and $$\lvert\chi\rangle$$, but upon doing so, we'd see that the columns (or rows) are not linearly independent, and so we would get zero for the determinant.

This is, fundamentally, what it means that fermions need to occupy different states. You have to put $$N$$ particles in $$N$$ linearly independent states; otherwise, it's impossible to anti-symmetrize the wave function.

This is why we can only put up to two particles in 1s shell, up to six particles in 2p shell, etc. It's not that we're "using up" hydrogen orbitals (for multi-electron atoms, the hydrogen orbitals are generally not great approximations); rather, it's that these spaces are 2 dimensional (1s), 6 dimensional (2p) and so on (as long as we're including spin), and we can only use linearly independent states to put the particles in.

In the case of orbital degeneracy only two electrons per orthonormal orbital can be accommodated.

For identical fermions the many-particle wave function must be antisymmetric under the exchange of particles. Such a wave function can be represented as a linear combination of determinants. For N fermions, each determinant must contain N orbitals. It is possible to choose overlapping orbitals, as long as an orthogonalisation gives N linearly independent orbitals. If this number is less than N, the determinant vanishes. Note that orbitals with opposite spin are orthogonal regardless of their spatial parts.

So, to reply to your comment, overlap is allowed but is only meaningful between orbitals of opposite spin.

There is a mathematical connection between half integer spin and antisymmetry under particle exchange.

• Ok, only two electrons can be accommodated per orthonormal orbital. But why can't another electron stays in a state that is a linear combination of them? Commented May 23 at 13:16
• In general, you can have one electron per orthonormal state. Spin eigenstates are orthonormal, energy eigenstates are orthonormalm, etc. If you make linear combinations of them, they will not be orthonormal, but have overlap with the orthonormal states. If you make a linear combination of spin states, that will have overlap with the spin up and spin down eigenstates. No overlap is allowed, ergo only two electrons allowed. Commented May 23 at 13:26
• why no overlap is allowed? Commented May 23 at 13:32

Indeed, if you had electrons with a continuous degree of freedom, you could obtain infinitely many states, all of the same energy, as long as they differ in that other continuous degree of freedom. For free particles, this leads to no contradictions. As soon as you consider electrons in bound states however, you only have a finite set of positions, momenta, spin orientations etc. available for each energy level, meaning that you can only have finitely many electrons in each energy eigenstate, differing in all those other quantum numbers. This is how you get the electron orbitals around the atomic nucleus.