# Why doesn't the Standard Model have a whole Potential Zoo?

Is there any strong argument which stops us from including potential terms (may be polynomial or not) for other standard model fields besides the Higgs potential? In other words, why isn't there a potential for the other fields something like a fermionic potential? I would initially believe that one can construct functions that respect the symmetries of the theory for fields other than the Higgs doublet. Most likely one would avoid higher order derivative terms but besides that.

May be there is something obvious that I am missing here.

• While I see a bit of the general idea that this question is going after, ultimately, I am still unclear about what is being asked. – ohwilleke Feb 9 at 22:05

First of all, the crucial symmetry you need to satisfy is the Poincaré symmetry, so your Lagrangian should be a Lorentz scalar and translation invariant. Of course, there are gauge symmetries as you have already mentioned in the question.

Other than those, you need to have the appropriate dimensions for your Lagrangian, and the parameters (coupling constants) should be dimensionless, otherwise you can add higher and higher order operators of the fields. You would like to do such thing only if you are working in effective theories like chiral perturbation theory of mesons and hadrons.

So, if you are in 4 dimensions, the action functional of Standard Model Lagrangian would be in the following form: $$S = \int \mathrm{d}^4x \; \mathcal{L}_{\text{SM}}$$ where I assume the space-time is effectively flat, i.e. the metric Minkowski metric. The action has dimensions of $\hbar$.

Therefore, you need to have the dimensions of the Lagrangian density to be $\left[\mathrm{GeV}\right]^4$, if you set $\hbar=c=1$, because of the volume element which has dimensions of $\left[\mathrm{GeV}\right]^{-4}$. If you compare the kinetic terms of your fields, they would have the following dimensions: \begin{eqnarray} \tag{Higgs} \phi &\rightarrow & \left[\mathrm{GeV}\right]^1 \\ \tag{Fermions} \psi &\rightarrow & \left[\mathrm{GeV}\right]^{3/2} \\ \tag{Vector bosons} A_\mu^a &\rightarrow & \left[\mathrm{GeV}\right]^1 \\ \end{eqnarray} since the derivatives are $\left[\mathrm{GeV}\right]^1$ dimensions, and the coupling constants $g$ are $\left[\mathrm{GeV}\right]^0$, i.e., dimensionless.

So, in order to get $\left[\mathrm{GeV}\right]^4$ operators in the Lagrangian density, you could only write the following terms:

• second derivatives of bosonic fields (or first derivative with quadratic fields), $$\partial_\mu \phi^\dagger \partial^\mu \phi \\ \partial_\mu A_\nu \partial^\mu A^\nu \\ A^\mu A_ \nu \partial_\mu A_\nu$$
• first derivative of fermionic fields, $$\bar{\psi} \gamma^\mu \partial_\mu \psi$$
• fourth power of bosonic fields $$(\phi^\dagger \phi )^2 \\ (A^\mu A^\nu)^2$$
• Yukawa interactions of a scalar and fermions $$\phi \bar{\psi} \psi$$
• scalar-vector couplings $$\phi^\dagger \phi A_\mu A^\mu$$ where I omitted some indices and conjugates.

All these terms can have a dimensionless coupling constant. However, in Standard Model, we assume the kinetic terms are normalized and the interaction terms have their coupling constants as free parameters. On the other hand, in supersymmetric theories, the coupling constants become equal for super partners.

As you can see, there aren't so much terms you could add from the fields you have. Moreover, the Standard Model Lagrangian is all you can write for 3-family quarks and leptons, a Higgs doublet and $SU\left(3\right)\otimes SU\left(2\right) \otimes U\left(1\right)$ gauge bosons.

EDIT

The only two dimensionful constants in our understanding of nature are the $M_{\text{Pl}}$, Planck mass (or equivalently, the Newton's gravitational constant), and the vacuum expectation value of the Higgs field (or equivalently, the Higgs mass). If you add another dimensionful constant, it means you introduced another scale to the theory which you don't see in nature, yet.

• Thank you for your answer. So in the end the argument I was looking for is nothing more than renormalization and power counting I guess. Having dimensionfull constants is not an option. – ohneVal May 1 '18 at 13:09
• I have added a paragraph to my answer in order to make it a bit clearer regarding the last sentence of your comment. – Oktay Doğangün May 1 '18 at 14:23