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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
up vote 2 down vote accepted

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 $\mathbf{r}_1$.

The short answer is: "The origin of the exchange interaction is the property of definite parity under coordinate exchange of the wave function." (Often the answer is given that it's the Pauli principle. But this is incomplete. Systems of bosons with internal coordinates (flavor, spin, etc.) enjoy the exchange interaction too. Prove this.) Note too that this holds for any two-body operator.

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Thank you, I appreciate it! This is great. – Dylan Sabulsky Dec 7 '12 at 7:13

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