So we have a Maxwell field coupled to a complex $\Phi(x)$ (charged) scalar field with mass and self-interaction terms:

$$L=-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+D_{\mu}\Phi^*D^{\mu}\Phi-m^2\Phi^*\Phi-\frac{\lambda}{6}(\Phi^*\Phi)^2 $$

Determine the exact form of the covariant derivative $D_{\mu}$ so that the Lagrangian is invariant under the gauge transformation $\Phi \xrightarrow{} \Phi^{'}=e^{i \alpha(x)}\Phi$.

Since $D_{\mu}=\partial_{\mu}+ieA_{\mu}(x)$, isn't already determined?


And how does the mysterious field $A^{\mu}$ transform under the gauge transformation? This is what you want to find. Given that the scalar field transforms under the fundamental representation of the gauge group, i.e. $\Phi \rightarrow \Phi^{\prime} = e^{i\alpha(x)}\Phi$, then, demanding that the action of the covariant derivative on the scalar field also transforms in the fundamental representation, i.e. $D_{\mu}\Phi \rightarrow D_{\mu}^{\prime}\Phi^{\prime} = e^{i\alpha(x)}D_{\mu}\Phi$ (definition of the gauge covariant derivative), results to a very specific transformation rule for the mysterious field $A^{\mu}$, \begin{equation} A^{\mu} \rightarrow A^{\prime\mu} = A^{\mu} - \frac{1}{e}\partial^{\mu} \alpha \end{equation}

This transformation rule is actually a transformation according to the adjoint representation of the gauge group which is a spin-1 representation, so this is a hint of a vector boson $A^{\mu}$.

Proof of transformation law

Since $D_{\mu}=\partial_{\mu}+ieA_{\mu} \rightarrow D_{\mu}^{\prime}=\partial_{\mu}+ieA_{\mu}^{\prime}$ and $\Phi \rightarrow \Phi^{\prime} = e^{i\alpha}\Phi$, then, \begin{equation}\begin{aligned} D_{\mu}\Phi \rightarrow D_{\mu}^{\prime}\Phi^{\prime} &= (\partial_{\mu} + ieA_{\mu}^{\prime})e^{i\alpha(x)}\Phi(x) \\ &=i\partial_{\mu}\alpha e^{i\alpha}\Phi + e^{i\alpha}\partial_{\mu}\Phi + ieA_{\mu}^{\prime}e^{i\alpha}\Phi \\ &= e^{i\alpha}\left((i\partial_{\mu}\alpha + ieA_{\mu}^{\prime})\Phi + \partial_{\mu}\Phi\right) \end{aligned}\end{equation}

Now, the requirement is, \begin{equation}\begin{aligned} D_{\mu}\Phi \rightarrow D_{\mu}^{\prime}\Phi^{\prime} &= e^{i\alpha}D_{\mu}\Phi \\ &= e^{i\alpha}\left( ieA_{\mu}\Phi + \partial_{\mu}\Phi \right) \end{aligned}\end{equation}

Comparing the two relations above, tells us that, \begin{equation}\begin{aligned} &i\partial_{\mu}\alpha + ieA^{\prime}_{\mu} = ieA_{\mu} \\ &\Rightarrow A^{\prime}_{\mu} = A_{\mu} - \frac{1}{e} \partial_{\mu}\alpha \end{aligned}\end{equation}

  • $\begingroup$ How do you determine that $A_{\mu}\xrightarrow{} A^{' \mu}=A^{\mu}+\partial^{\mu}\alpha$ is the correct gauge transformation? By an ansatz? $\endgroup$ – Carl Apr 11 '18 at 10:52

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