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Does any one know the physical meaning of the following gauge invariant gauge coupling to the spinors? $$\bar \psi F_{\mu \nu} [\gamma^\mu, \gamma^\nu] \psi$$ This coupling is not minimal, as $$\bar \psi(\partial_\mu + ig A_\mu) \psi$$

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  • $\begingroup$ That's not a renormalizable coupling, in 4d. You'll see such terms in effective field theories, but generally only when you're fairly close to the scale where they cease to be effective.. $\endgroup$ – user1504 Jul 22 '16 at 17:43
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First, note that Your term is implicitly contained in an expression for the on-shell process $e \gamma\to e$ at tree level (with momenta $p, q \equiv p - k,k$ correspondingly), where $\gamma$ formally corresponds to external field $A_{\mu}$: $$ \tag 1 M_{e\gamma\to e} \sim \bar{u}(p)\gamma_{\mu}u(k)A^{\mu}(q) \equiv \bar{u}(p)\left(\frac{(p+k)_{\mu}}{2m_{e}} - \frac{i\sigma_{\mu\nu}}{2m}(p-k)^{\nu}\right)u(k)A^{\mu}(q) $$ Here $u$ is called spinor polarization vector, and $\sigma_{\mu\nu} \equiv \frac{i}{2}[\gamma_{\mu},\gamma_{\nu}]$. We see, that Your term is the second summand of $(1)$.

Second, note that loop corrections to process $e\gamma \to e$ generate form-factors $F_{1}((p-k)^2)$ and $F_{2}((p-k)^{2})$ near the first and the second summand in $(1)$. The second one comes explicitly, when we introduce effective interaction $~\sim\bar{\psi}[\gamma_{\mu},\gamma_{\nu}]\psi$ in the lagrangian, which is precisely Your term. Thus Your term brings no new fundamental effects, just corrections to known tree-level effects.

So, let's assume the physical sense of Your coupling by treating the second summand in $(1)$. Below I'll use the representation of gamma-matrices, for which $$ \gamma_{\mu} \equiv \begin{pmatrix} 0 & \sigma_{\mu} \\ \tilde{\sigma}_{\mu} & 0\end{pmatrix}, \quad \sigma_{\mu} = (1,\sigma), \quad \tilde{\sigma}_{\mu} = (1,-\sigma) $$ First, it is very attractive to study non-relativistic approximation, where $s$-th polarization $u_{s}$ (with $\epsilon_{s}$ being spin up down eigenvectors) is $$ \tag 2 u_{s}(p)\approx \begin{pmatrix}\sqrt{\sigma \cdot p}\epsilon_{s} \\ \sqrt{\tilde{\sigma} \cdot p}\epsilon_{s}\end{pmatrix} \approx \begin{pmatrix}\sqrt{m_{e}}\left(1-\frac{\mathbf q\cdot \mathbf \sigma}{4m_{e}}\right)\epsilon_{s} \\ \sqrt{m_{e}}\left(1+\frac{\mathbf q\cdot \mathbf \sigma}{4m_{e}}\right)\epsilon_{s} \end{pmatrix}, $$ $$ \tag 3 u_{s}(k)\approx \begin{pmatrix}\sqrt{\sigma \cdot k}\epsilon_{s} \\ \sqrt{\tilde{\sigma} \cdot\mathbf p}\epsilon_{s}\end{pmatrix} \approx \begin{pmatrix}\sqrt{m_{e}}\left(1+\frac{\mathbf q\cdot \mathbf \sigma}{4m_{e}}\right)\epsilon_{s} \\ \sqrt{m_{e}}\left(1-\frac{\mathbf q\cdot \mathbf \sigma}{4m_{e}}\right)\epsilon_{s} \end{pmatrix} $$ Second, it's convenient to divide $\sigma_{\mu\nu}(p-k)^{\nu}A^{\mu}$ term on two parts: $$ \tag 4 i\sigma_{\mu\nu}q^{\nu}A^{\mu}(q) = \alpha_{i}E_{i}+\Sigma_{i}B_{i}, $$ where $\alpha_{i}=\gamma_{0}\gamma_{i}$ is called velocity, while $\Sigma_{i} \equiv \frac{1}{2}[\gamma_{i},\gamma_{j}]$ is the spin operator.

It is not hard to obtain by using $(2)-(4)$, that $$ \bar{u}(p)i\sigma_{\mu\nu}F^{\mu\nu}(q)u(k)\simeq \text{const}_{1}(\bar{S}\cdot \mathbf B) + \text{const}_{2}([\bar{S}\times \mathbf q] \cdot \mathbf E (\mathbf q)), $$ where $$ \bar{S} \equiv \epsilon^{\dagger}\frac{\sigma}{2} \epsilon $$ The first summand corresponds to spin magnetic moment interaction with magnetic field, while the second summand is called spin-orbit interaction.

In conclusion, Your summand generates radiative corrections to spin-orbit coupling and magnetic moment coupling.

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