# Does exchange energy affect completely filled orbitals?

In my textbook, it's written that exchange energy stabilises half filled and completely filled orbitals. But my teacher said electrons can't be exchanged in full filled orbitals because of pairing up of electrons. So which explanation is correct? My textbook has an explanation for half filled electron orbitals, by it had no explanation given for completely filled orbitals.

It seems like you are just being introduced to quantum mechanics and is confused with the concept of exchange energy. The concept can be confusing at first, since it neither involves exchanging electrons as classical particles nor has a classical analogue. I will try explaining the problem with a simple two electron system.

Before we discuss, here is a short introduction. In quantum mechanics, one solves the Schrodinger's equation and tries to estimate the wave function of the system. We are talking about time-independent case here, so we solve time independent Schrodinger's equation to obtain the wavefunction.

$$\hat{H}\psi=E\psi$$

Here $$\hat{H}$$ is the Hamiltonian (energy operator), $$\psi$$ is the wavefunction and $$E$$ is the energy. This equation is an eigenvalue equation, which essentially means any valid wavefunction should have a fixed energy irrespective of coordinates. In the equation, the $$\hat{H}$$ is the only known quantity (operator), $$\psi$$ and $$E$$ has to be estimated based on boundary conditions.

Let us consider a system with two electrons, with wavefunction $$\psi_1(r)$$ and $$\psi_2 (r)$$, where $$r$$ denotes the four dimensional space spin coordinates. Now, we know that electrons are indistinguishable and two electrons (fermions) cannot have same space-spin coordinates (Pauli's exclusion principle). Two electrons not having same space-spin coordinates is same as saying that the electrons should not have all same quantum numbers.

So, in order to consider the indistinguishability of the electrons, we can have two cases of the total wave function: (1) $$\psi_1(r_1)\psi_2(r_2)$$ or (2) $$\psi_1(r_2)\psi_2(r_1)$$. On a side note, these are known as Hartree products and are valid for bosons. However, in case we wish to include the Pauli's exclusion principle, we write the system as a Slater determinant, i.e.

$$\psi(r_1,r_2)=\frac{1}{\sqrt{2}}\begin{vmatrix} \psi_1(r_1) & \psi_2(r_1)\\\psi_1(r_2) & \psi_2(r_2)\end{vmatrix}=\frac{1}{\sqrt{2}}\left[\psi_1(r_1)\psi_2(r_2)-\psi_1(r_2)\psi_2(r_1)\right]$$

Now let us talk about the Hamiltonian (energy operator). The Hamiltonian for an atom/molecule can be divided into three parts:

• Constants: This include any constant terms, for example, nuclear-nuclear repulsion for poly-atomic molecules.
• One electron operator: This part contains terms dependent on single electrons, for example, kinetic energy and nuclear-electron attraction.
• Two electron operator: This part contains terms dependent on two electrons, and usually contains only electron-electron repulsion terms. This part gives rise to the exchange energy. We shall expand this part.

Before we go into details, if one is provided with a operator $$\hat{A}$$ and a wavefunction $$\psi$$, the expectation (average) value of the observable corresponding to the operator is given as $$\langle\psi|\hat{A}|\psi\rangle$$, where $$\langle\cdot\rangle$$ implies integral over all coordinates.

In atomic coordinates, the electron-electron repulsion is given as $$1/r_{12}$$, where $$r_{12}$$ is the distance between the electronic coordinates. Now, if we use the wavefunction in the Slater determinant, we get the total two electron interactions as:

$$\left\langle\psi(r_1,r_2)\left|\frac{1}{r_{12}}\right|\psi(r_1,r_2)\right\rangle \\= \frac{1}{2} \left[ \left\langle\psi_1(r_1)\psi_2(r_2)\left|\frac{1}{r_{12}}\right|\psi_1(r_1)\psi_2(r_2)\right\rangle + \left\langle\psi_2(r_1)\psi_1(r_2)\left|\frac{1}{r_{12}}\right|\psi_2(r_1)\psi_1(r_2)\right\rangle - \left\langle\psi_1(r_1)\psi_2(r_2)\left|\frac{1}{r_{12}}\right|\psi_2(r_1)\psi_1(r_2)\right\rangle - \left\langle\psi_2(r_1)\psi_1(r_2)\left|\frac{1}{r_{12}}\right|\psi_1(r_1)\psi_2(r_2)\right\rangle \right]$$

Now, in this cases the first two terms are called Coulomb terms and are easy to imagine classically, i.e. what is the repulsion of electron 1 if present in coordinate $$r_1$$ with electron 2 if present in coordinate $$r_2$$ for the first term and vice-versa for the second term. These are repulsive in nature.

The third and fourth terms are attractive terms, and are known as exchange terms. There is no classical analogue of these, and appears purely from the indistinguishability of electrons and Pauli's exclusion principle.

Now remember we discussed that $$r_1$$ and $$r_2$$ are space-spin operators, and $$r_{12}$$ depends only on space? This essentially means that we can separate the space and spin parts while calculating the terms. So the exchange terms vanish while the two electrons have different spins, and remains only when electrons have same spins. This is the origin of exchange energy and IT DOES NOT INVOLVE PHYSICALLY EXCHANGING THE ELECTRONS.

I hope I could explain you in brief the origin of exchange energy. In reality, it is very difficult to calculate the exact energy of a system with N-electrons and say half filled/full filled systems are more stable due to "exchange interactions" simply because:

• We do not exactly know how the wavefunctions look. We assuming hydrogenic wave functions but that may not work out always.
• Optimising the wavefunctions based on Hamiltonian is carried out self consistently and is costly.
• Let us say one resolves the first two issues somehow, still total energy has other terms like kinetic energy, electron-nuclear attraction, Coulomb terms, and even nuclear-nuclear repulsion for polyatomic molecules. So it is very difficult to assert only "exchange energy" is responsible for the extra stability.

Whatever I have tried to explain in short takes up at least two quantum chemistry courses in Bachelor's degree and Master's degree. In case you have some doubts, please feel free to ask in comments. In case you wish to study more, I will suggest the following books (in the order given):

1. D. J. Griffiths, Introduction to Quantum Mechanics
2. A. Szabo and N. Ostlund, Modern Quantum Mechanics
3. I. Levine, Quantum Chemistry
• Thanks a lot for the answer, actually I'm new to quantum mechanics, so I didn't get some points. But thanks a lot for the book refrences. Sep 5 '20 at 11:39