What is a simple bare-bones description of exchange interaction between two electrons?

For instance, it seems to me that the only necessary ingredients are the Coulomb interaction and the requirement that the total wavefunction be antisymmetric.

  • 1
    $\begingroup$ Your intuition is correct. A mathematical description of how those two ingredients conspire to create exchange interactions can be found in Ashcroft & Mermin (chapter 32) [this is a pretty standard calculation and I'm sure it appears in a lot of other places too] $\endgroup$
    – wsc
    Commented Dec 17, 2011 at 20:34
  • 2
    $\begingroup$ Its in Griffiths intro quantum textbook as well. Somewhere. $\endgroup$ Commented Dec 17, 2011 at 20:59
  • 1
    $\begingroup$ It has nothing to do with the Coulomb force, there would be an exchange interaction between two uncharged but indistinguishable bosons as well. $\endgroup$ Commented Dec 18, 2011 at 5:41

1 Answer 1


Exchange interaction is an addition to other interactions between identical particles caused by permutation symmetry.

This addition is a result of specific form of multi-particle wave function. It gives no contribution to Hamiltonian unlike "usual" interactions but appears as an additional term in equations for single-particle wave functions (e.g. Hartree-Fock equation).

Interaction usually associated with energy and forces. We could find the exchange correction as a force added to Coulomb forces, but we should understand first what is force in quantum system.

Let's consider two fermions with single-particle coordinate wave functions $\psi_a(x)$ and $\psi_b(x)$ and spin wave fucntions $\phi_a(s)$ and $\phi_b(s)$. The possible two-particle wave fucntions are singlet with symmetric coordinate part $$ \Psi_S(x_1, x_2) = \frac{1}{\sqrt{2}}\left[ \psi_a(x_1)\psi_b(x_2) + \psi_a(x_2)\psi_b(x_1) \right] $$ and triplet with antisymmetric coordinate part $$ \Psi_A(x_1, x_2) = \frac{1}{\sqrt{2}}\left[ \psi_a(x_1)\psi_b(x_2) - \psi_a(x_2)\psi_b(x_1) \right] $$

Let the two-particle Hamiltonian do not depend on spins: $$ \hat{H} = \frac{\hat{\mathbf{p}}_1 + \hat{\mathbf{p}}_2}{2m} + V(x_1, x_2) $$ then the average energy of the interaction will be: $$ U_S = \left<\Psi_S\right|V\left|\Psi_S\right> = U + U_\text{ex} $$ $$ = \left<\psi_a(x_1)\psi_b(x_2)\right|V\left|\psi_a(x_1)\psi_b(x_2)\right> + \left<\psi_a(x_1)\psi_b(x_2)\right|V\left|\psi_a(x_2)\psi_b(x_1)\right> $$ $$ U_A = \left<\Psi_A\right|V\left|\Psi_A\right> = U - U_\text{ex} $$ $$ = \left<\psi_a(x_1)\psi_b(x_2)\right|V\left|\psi_a(x_1)\psi_b(x_2)\right> - \left<\psi_a(x_1)\psi_b(x_2)\right|V\left|\psi_a(x_2)\psi_b(x_1)\right> $$

The term $U_\text{ex}$ is not zero only if the particles are close enough to each other and their wave functions overlap (see picture below). In classical limit when distance $L$ is big the overlapping is zero and $U_S=U_A=U$

enter image description here

Let's assume that $\psi_a$ and $\psi_b$ are non-negative everywhere anv $V$ acts as Coulomb interaction (i.e. positive and decreases when the distance increases). Then $U$ and $U_\text{ex}$ are positive and energy of symmetric coordinate state (opposite spines) is higher than energy of antisymmetric coordinate state (similar spines). If the average positions of the particles are fixed the exchange interaction will put the spins same direction.

The force of interaction between the particles can be defined as the generalized force corresponding to the parameter L: $$ F = -\frac{\partial U}{\partial L} $$ Within our assumptions concerning $\psi_a$, $\psi_b$ and $V$ the derivative of both $U$ and $U_\text{ex}$ are negative. Hence the "usual" force is positive (repulsion) and the exchange force is positive for symmetric coordinate state and negative for antisymmetric coordinate state (attraction).

enter image description here

So the exchange interaction for the case of two particles can be considered as additional force depending on spin configuration. For multiple particles this is more complicated.

  • 3
    $\begingroup$ Hi, how to understand the effective force of exchange interaction for Fermion is attractive? Very counterintuitive. $\endgroup$
    – maplemaple
    Commented Aug 12, 2018 at 5:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.