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Usually the Higgs potential is given as $$ \frac{1}{2}\mu^2\phi^2 - \frac{1}{4}\lambda^2\phi^4 $$ but I never quite understood if this just serves to give us an idea of how symmetry breaking works, or if it is actually the correct expression for the potential.

Does the potential above agree with experiments? Is that the Higgs potential?

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Yes, that is the Higgs potential of the Standard Model.

Note that a $\phi^3$ term is forbidden by symmetry (it would not be an $\mathrm{SU}(2)$ scalar), and $\mathcal{O}(\phi^5)$ terms would be non-renormalizable, so this is really the only potential we can write down that does not need other new physics.

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  • $\begingroup$ but the effective potential can look a bit different a la Coleman-Weinberg $\endgroup$
    – innisfree
    Commented May 5, 2015 at 16:37
  • $\begingroup$ Is there an easy/intuitive way of understanding why the $\mathcal{O}(\phi^5)$ term in non-renormalisable? $\endgroup$
    – SuperCiocia
    Commented May 5, 2015 at 22:14
  • $\begingroup$ @SuperCiocia: If you know the power-counting renormalizability argument, it's obvious from the mass dimension the coupling must have for terms of order higher than 4, if not, then not. $\endgroup$
    – ACuriousMind ♦
    Commented May 6, 2015 at 11:00
  • $\begingroup$ And why is the $\phi$ term at power one is forbidden ? $\endgroup$
    – Mathieu Krisztian
    Commented May 28, 2020 at 20:03
  • $\begingroup$ @MathieuKrisztian 1. It would also not be a scalar. 2. Linear terms can be transformed away anyway. $\endgroup$
    – ACuriousMind ♦
    Commented May 28, 2020 at 21:15

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