In QM, when treating a system with identical particles, the exchange term raised because the system must be the same after particle exchanged.

I am reluctant to use the term “exchange interaction” or “exchange force”, “exchange potential”, because there is no interaction, no force carrier, no potential at all.

Suppose the Hamiltonian for single fermion is

$H\psi = \epsilon \psi$

The 2-particle system, the Schrodinger equation is

$H_T \Psi = H_1 + H_2 + H'= E \Psi$

where $H'$ is the interaction between the 2 particles.

Now, guessing the total wave function to be

$\Psi = a \psi_1(r_1) \psi_2(r_2) + b \psi_1(r_2)\psi_2(r_1) = a \Psi_n + b\Psi_e$

The first term is “normal”, the second term is “exchanged”. substitute to the equation

$a H_T \Psi_n + b H_T \Psi_e = a E \Psi_n + b E \Psi_e$

multiply both side with $\Psi_1$ or $\Psi_2$, and integrate $r_1, r_2$, we have

$\begin{pmatrix} J & K \\ K & J \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = E \begin{pmatrix} a \\ b \end{pmatrix}$

where

$J = \int \psi_1^*(r_1) \psi_2^*(r_2) H_T \psi_1(r_1) \psi_2(r_2) dr_1 dr_2$

$K = \int \psi_1^*(r_2) \psi_2^*(r_1) H_T \psi_1(r_1) \psi_2(r_2) dr_1 dr_2$

Here, we assumed $H'$ is symmetric against $r_1, r_2$. This is a fair assumption as $H'$ is the mutual interaction, and $H_1, H_2$ are independent.

The $J$ is the energy from the interaction of the particles.

The $K$ is the energy due to EXCHANGE of the particles.

As we can see, the energy $J = \epsilon_1 + \epsilon_2 + J'$.

The $H_T$ in $K$ can be reduced to $H'$ when the eigen wave function $\psi_1 , \psi_2$ are in difference orbits. When $\psi_1 = \psi_2$, $K = \epsilon_1 + \epsilon_2 + K'$.

Thus the individual energy can be subtracted, yield,

$\begin{pmatrix} J' & K' \\ K' & J' \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \Delta E \begin{pmatrix} a \\ b \end{pmatrix}$

The solution is

$\Delta E = J' + K', (a,b) = (1,1)$

This is the symmetric state. For fermion, the spin-part is anti-symmetric. This is the spin-singlet state.

$\Delta E = J' - K', (a,b) = (1,-1)$

This is the anti-symmetric state. For fermion, the spin-part is symmetric, or spin-triplet state.

It is worth to notice that, in the symmetric state, the energy is higher because the two particles can be as close as possible, that create large energy. In the opposite, in the anti-symmetric state, the two particle never contact each other, thus, the interaction energy reduced.

We can summarized.

The energy $J'$ is the shift of energy due to the mutual interaction.

But the system of identical particle subjects to the exchange symmetry. The exchanged wave function is also a state, that the total wave function has to include the exchanged wave function. This exchanged wave function creates the exchange term or exchange energy $K'$.

In the system of Fermion, the exchange energy is related to Pauli exclusion principle.