'Extra' terms in Abelian Higgs model

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SUMMARY

The discussion focuses on the Abelian Higgs model's Lagrangian, specifically the terms arising from the covariant derivative and field strength tensor. The Lagrangian is expressed as \(\mathcal{L} = |D_\mu\Phi|^2 - \frac{1}{4}F_{\mu\nu}F^{\mu\nu} - V(\Phi)\), with the potential defined as \(V(\Phi) = \lambda|\bar{\Phi}\Phi|^2 - \mu^2\bar{\Phi}\Phi\). The conversation highlights the cancellation of certain 'extra' terms during the expansion after symmetry breaking, confirming that these terms do indeed cancel each other out. The participants validate the calculations and clarify the behavior of the terms in question.

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bomanfishwow
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I'm taking 5 mins (hours) during some down-time to remind myself of some theory. Taking a simple Abelian Higgs model, where the Lagrangian is given by:

\mathcal{L} = |D_\mu\Phi|^2 - \frac{1}{4}F_{\mu\nu}F^{\mu\nu} - V(\Phi)

With the covariant derivative, field strength tensor and potential given by:

D_\mu = \delta_\mu - ig_\mu,
F_{\mu\nu} = \delta_\nu A_\mu - \delta_\mu A_\nu,
V(\Phi) = \lambda|\bar{\Phi}\Phi|^2 - \mu^2\bar{\Phi}\Phi.

I'm working in the unitary gauge, such that \Phi is given by:

\Phi = \frac{1}{\sqrt{2}}\left(v + H\right).

Taking the expanded potential after symmetry breaking, and plugging into |D_\mu\Phi|^2 like:

|D_\mu\Phi|^2 = D_\mu\Phi^*D^\mu\Phi = \frac{1}{2}\left[\left(\delta_\mu +igA_\mu\right)\left(v+H\right)\left(\delta^\mu - igA^\mu\right)\left(v+H\right)\right]

yields the expected interaction and mass terms. Some of the 'extra' terms trivially cancel as they contain derivatives of constants such as \delta_\mu v. However, there are some extra terms which I don't see mentioned in the standard texts:

-ig\left[\delta_\mu H\right] HA^\mu
-igv\delta_\mu HA^\mu
igHA_\mu\delta^\mu H
igvA_\mu\delta^\mu H.

Can anyone suggest a) if I've done something wrong b) if these terms also disappear c) Something else...

Thanks!
 
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You did it right, and those 4 terms cancel one another.

|D_{\mu}\phi|^2 = \frac{1}{2} [(\partial_{\mu} - ieA_{\mu})(v+H)] [(\partial^{\mu} + ieA^{\mu})(v+H)] = \frac{1}{2}[\partial_{\mu}H\partial^{\mu}H + e^2A_{\mu}A^{\mu}(v+H)^2 + ieA^{\mu}(v+H)\partial_{\mu}H - ieA_{\mu}(v+H)\partial^{\mu}H]

Because of the summation over \mu, we have for the third term ieA^{\mu}(v+H)\partial_{\mu}H = ieA_{\mu}(v+H)\partial^{\mu}H, cancelling the last term.

Cheers,
Sebas
 
Ah yes of course, thank you.
 

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