# What is the “exchange energy of electrons”?

First, do not try to give me the perplexing explanations involving higher QM. I am new to QM, so please give a gentle explanation.

Now, my textbook see page 36 states that electrons of degenerate orbitals of same subshell and with same spin, tend to exchange their positions. So, do they physically exchange their position? If yes, then why and how does this lead to release of energy? If no, then what is really happening that causes this release of energy (please give a gentle explanation).

I find some explanations stating that this leads to lesser shielding. But how?

• Commented Jul 1, 2021 at 13:38
• This question is a good one in my opinion, though it could be worded much better and clearer. Commented Jul 1, 2021 at 13:43
• It's not the heck. LOL :) Commented Jul 1, 2021 at 14:12

In quantum mechanics, you don't know the exact position of either electron. So when computing the energy of an orbital, we really compute an average of the energy over possible positions of the electrons, weighted by the probability of each configuration.

The ultimate origin of the exchange energy for electrons is the Pauli exclusion principle, which states that two electrons cannot be in the same position at the same time. This in turn comes from a fact in quantum mechanics, that is difficult to explain in simple terms, that the quantum mechanical wavefunction describing the system must change sign when we interchange two electrons.

While the origin of the Pauli exclusion principle is admittedly mysterious, it has a clear effect on the average over electron positions. The electrons have a repulsive energy $$$$U = \frac{ke^2}{r}$$$$ where $$r$$ is the distance between the electrons. If we were to ignore the Pauli exclusion principle, the average would include configurations where the electrons were very close and had a large repulsive energy. The Pauli exclusion principle means that configurations where the electrons touch are not allowed, and since the wavefunction is smooth is also means the probability for the electrons to be "close" is reduced relative to the case where we ignore the exclusion principle. So the average energy is smaller when accounting for the exclusion principle, because less weight is given to configurations with a large repulsive energy between the electrons. (Note that a full calculation will involve other terms in the energy, and the antisymmetry of the wavefunction is crucial to understand what is going on in full detail, but here I am just focused on getting across the main point in a simple way).

Even though the Pauli exclusion principle is not really a contribution to the energy, the difference between the average energy you compute accounting for the Pauli exclusion principle, and the energy you compute without accounting for it, is called the exchange energy. It encodes the net effect of the Pauli exclusion principle as if it were a contribution to the energy, even though it is not, strictly speaking. The exchange energy is negative, because the actual energy is lower than what you get ignoring the exclusion principle.

Note that the exchange energy is negative for electrons because they are identical fermions. For non-identical particles, the exchange energy is zero. For identical bosons, the exchange energy is positive, because the symmetry (instead of antisymmetry) of the boson wavefunction assigns additional probability to configurations where the bosons are near each other, compared to the non-identical case.

• Isn't the exchange energy of electrons positive in ferromagnets? Commented Nov 27, 2021 at 3:10
• @JuanPerez No. The exchange term in the Hamiltonian is often written $H_{ex} = \sum_{jk} - J/2 \vec{S}_j \cdot \vec{S}_k$, where $j, k$ label lattice sites, $\vec{S}$ is electron spin, and $J$ is the exchange integral. It is true that $J$ is positive for ferromagnets, but this means that the exchange energy $H_{ex}$ is negative. This makes sense since to minimize the energy, you then want the spins to align on neighboring lattice sites, so that $\vec{S}_i \cdot \vec{S}_j$ is maximized. If the exchange energy were negative, you would want the spins antialigned and you'd get antiferromagnets. Commented Nov 27, 2021 at 3:19
• Some sources: southampton.ac.uk/~rpb/thesis/node18.html (see the table of exchange energies at the bottom of the page), and farside.ph.utexas.edu/teaching/sm1/Thermalhtml/node89.html Commented Nov 27, 2021 at 3:19

Talking naively as you are not familiar to QM.

In the atoms, the electron's spin can either align or antialigned depending on which is energetically favorable.

The Pauli's exclusion principle says that two electrons can't be in the same state. Now if the electrons pointing opposite that means they can come closer to each other without violating Pauli's exclusion principle. If the electrons pointing in the same direction that means they can't come closer to each other otherwise the state of the two electrons becomes equal.

Why aligned spins have lower energy?

When the two electrons have opposite spins, the effective charge of the nucleus seen by the fixed electron is reduced by the screening provided by the other electron.

However, when the spins are aligned, the two electrons can't come close to each other and the fixed electron sees more of the nucleus is stronger in the case where the two electrons are spin aligned, therefore it's a lower-energy configuration.

What's exchange energy?

The energy difference between having two spins aligned versus antialigned is called exchange energy.

Note: Quantum mechanically, The spin aligned state corresponds to triplet state while spin antialigned state corresponds to singlet : $$E_\text{exchange}=E_\text{singlet}-E_\text{triplet}$$

$$|\psi_1\rangle =|\psi_\text{symmetric}\rangle_\text{spartial}=|AB\rangle +|BA\rangle$$

$$|\psi_2\rangle =|\psi_\text{anti-symmetric}\rangle_\text{spartial}=|AB\rangle -|BA\rangle$$

$$E_\text{exchange}=\langle\psi_2|V|\psi_2\rangle-\langle \psi_1|V|\psi_1\rangle =-4\text{Re}\langle AB|V|BA\rangle$$ In the cross term, $$\langle AB|V|BA\rangle$$ the two electrons have exchanged place.

• The stability of half filled orbitals is often explained by the concept of exchange energy. In half filled orbitals, there are more pairs of electrons with same spin....hence more exchange energy is released....hence less total energy.....thus more stable. Where does that fit? Commented Jul 1, 2021 at 14:51
• Have you even read the explanation? Commented Jul 1, 2021 at 16:43
• This does not seem to be an explanation for a naive.
– Shub
Commented Jul 26, 2021 at 12:53