# What dictates the Higgs potential of the Standard model?

As far as I know, there is no convincing answer for why does the Higgs potential has a 'Mexican hat' shape. Apart from Lorentz invariance and gauge invariance, the Higgs potential of the Standard model has been chosen somewhat arbitrarily.

Do more advanced theories that go beyond the Standard model such as Grand Unified theories, supersymmetry or the theory of inflation have any deeper answer?

Working in natural units ($c=\hbar=1$), each physical quantity has a mass dimension. For the action $S=\int d^4 x \mathcal{L}$ in 4-dimensional spacetime, it's $0$; for $x^\mu$, it's $-1$; for $d^4 x$, it's $-4$; for $\mathcal{L}$, it's 4. A kinetic term $\partial_\mu\phi^\ast\partial^\mu\phi$ in this Lagrangian density implies $\partial_\mu\phi$ is of dimension $2$ while $\partial_\mu$ is of dimension $1$, so $\phi$ is of dimension $1$. Thus if $\lambda_n\phi^n$ is in the potential $\lambda_n$ has mass dimension $4-n$. Renormalisation typically requires this to be $\ge 0$, so we stop at $\phi^4$; this is called scalar $\phi^4$ theory.

Well, technically if we want a $U(1)$ theory for the Higgs, we're limited to $(\phi^\ast\phi)^k$ terms with $k\in\{0,\,1,\,2\}$. As long as neither of the $k>0$ terms has coefficient $\neq 0$, we're done. We can shift the $k=0$ term by an arbitrary constant, so feel free to set it to $0$ or make the potential a perfect-square function of $\phi^\ast\phi$.

You may wish to work out what happens in $d$-dimensional spacetime.

• Is it essential to impose renormalizability? @J.G.
– SRS
Mar 25, 2018 at 15:58
• @SRS A predictive theory needs a finite number of parameters, so it has to cut the potential at some finite power of $\phi$. Luckily, renormalizability provides a natural placement for this cap. If you want to learn more about it, look up relevant, marginal & irrelevant couplings.
– J.G.
Mar 25, 2018 at 16:24
• @J.G. That's not entirely true: you can have a predictive theory with an infinite number of parameters, as long as only a finite number of them are relevant. Moreover, you don't need to cut the potential at some finite power of $\phi$ to have a finite number of parameters: for example, the sine-Gordon model has a single parameter, and a potential with an infinite number of terms. Mar 25, 2018 at 16:38
• @AccidentalFourierTransform That's true, although "my Higgs has a sine-Gordon potential" is a bit of a "because I say so" hypothesis. Ideally we want to say, "I don't know what the power series looks like, but I do know which coefficients are allowed to be relevant, so I'll truncate there."
– J.G.
Mar 25, 2018 at 16:42
• @J.G. So are you saying there is no scope of changing the Higgs potential within Standard model if it is assumed to be renormalizable?
– SRS
Mar 28, 2018 at 9:33