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QFT: Higgs mechanisms covariant derivative under gauge transformation

Physics Asked on May 12, 2021

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

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

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?

One Answer

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. $Phi rightarrow Phi^{prime} = e^{ialpha(x)}Phi$, then, demanding that the action of the covariant derivative on the scalar field also transforms in the fundamental representation, i.e. $D_{mu}Phi rightarrow D_{mu}^{prime}Phi^{prime} = e^{ialpha(x)}D_{mu}Phi$ (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^{primemu} = 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 $Phi rightarrow Phi^{prime} = e^{ialpha}Phi$, then, begin{equation}begin{aligned} D_{mu}Phi rightarrow D_{mu}^{prime}Phi^{prime} &= (partial_{mu} + ieA_{mu}^{prime})e^{ialpha(x)}Phi(x) &=ipartial_{mu}alpha e^{ialpha}Phi + e^{ialpha}partial_{mu}Phi + ieA_{mu}^{prime}e^{ialpha}Phi &= e^{ialpha}left((ipartial_{mu}alpha + ieA_{mu}^{prime})Phi + partial_{mu}Phiright) end{aligned}end{equation}

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

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

Answered by Panos C. on May 12, 2021

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