# Origin of exchage interactions

Can someone explain to me the origin of the exchange interaction between two electrically charged spin 1/2 fermions? Quantitative or qualitative accepted.

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possible duplicate of Simple description of exchange interaction? –  Manishearth Dec 7 '12 at 11:19

The wave function is antisymmetric under exchange of (all) the coordinates of each electron (we'll just call them electrons since that's shorter than "two electrically charged spin 1/2 fermions" and equivalent). We'll write the wave function as: \begin{align} \Psi(1,2) &= \psi_1(\mathbf{r}_1)\psi_2(\mathbf{r}_2)|s_1s_2\rangle -\psi_1(\mathbf{r}_2)\psi_2(\mathbf{r}_1)|s_2s_1\rangle. \end{align} (Check that this is antisymmetric under $\mathbf{r}_1 \leftrightarrow \mathbf{r}_2$ & $s_1 \leftrightarrow s_2$.)
Now let's calculate the force between these due to some two-body operator $V(\mathbf{r}_1,\mathbf{r}_2)$ (that might depend on spin). It's proportional to (I'm ignoring normalization): \begin{align} &\int d^3r_1 d^3r_2 \Psi^\dagger(1,2)V(\mathbf{r}_1,\mathbf{r}_2)\Psi(1,2)\\ &= 2\int d^3r_1 d^3r_2 \Big\{ |\psi(\mathbf{r}_1)|^2|\psi(\mathbf{r}_2)|^2 \langle s_1 s_2|V(\mathbf{r}_1,\mathbf{r}_2)|s_1 s_2\rangle\\ &- \mbox{Re}[\psi_1^*(\mathbf{r}_1)\psi_2(\mathbf{r}_1)\psi_2^*(\mathbf{r}_2)\psi_1(\mathbf{r}_2) \langle s_1 s_2|V(\mathbf{r}_1,\mathbf{r}_2)|s_2 s_1\rangle]\Big\} \end{align} The second term is the exchange term. Note that the square of the wave functions in the first term are proportional to particle densities at $\mathbf{r}_1$ and $\mathbf{r}_2$. While the second term has different wave functions, $\psi^*_1(\mathbf{r}_1)\psi_2(\mathbf{r}_1)$ at the point $\mathbf{r}_1$.