Let us consider the classical action in $n$-dimensional (flat Minkowski) spacetime $$ S=\int dx^0\ L=\int d^nx\ \mathcal{L}\tag{1} $$ with Lagrange function $L$ and Lagrange density $\mathcal{L}$.
For a (real) scalar field $\phi$ we usually see a Lagrangian density of the form $$ \mathcal{L}=\frac{1}{2}(\partial\phi)^2-V(\phi)\tag{2} $$ where I use $(\partial\phi)^2=(\partial_\mu\phi)(\partial^\mu\phi)$ as a shorthand notation.
What constraints on $\mathcal{L}$ follow from fundamental physical considerations, i.e., why don't we see Lagrangians with, for instance, $\phi^6$?
For one, we need the dimensions to match but then we also want $V(\phi)$ to have a minimum.
Dimensional analysis
In natural units ($m_e=c=\hbar=\varepsilon_0=1$), the action is dimensionless, and it is common to express all other units in powers of energy. Time and length have energy dimension $-1$, thus, from $$ 0=[S]=[d^nx\ \mathcal{L}]=n[dx]+[\mathcal{L}]=n(-1)=-n+[\mathcal{L}]\tag{3} $$ we conclude that the Lagrange density is required to have energy dimension $n$. Assuming, the Lagrangian density of eq. (2) with $V(\phi)=0$ to be a valid field theory, we find $$ n=[\mathcal{L}]=[(\partial\phi)^2]=2\left([\partial]+[\phi]\right)=2(1+[\phi])\tag{4}. $$ Solving eq. (4) for $[\phi]$ yields $$ [\phi]=\frac{n}{2}-1\tag{5} $$ which tells us that the real scalar field has energy dimension $1$ in standard Minkowski $n=4$ spacetime as expected.
Series expansion of the potential
Assuming $\phi_0$ to be a local minimum configuration of the potential $V(\phi)$, we can perform a series expansion $$ V(\phi)=V(\phi_0)+V^\prime(\phi_0)\phi+\sum_{j=2}\frac{1}{j!}V^{(j)}(\phi_0)\phi^j\tag{6} $$ the linear term vanishing because $V^\prime(\phi_0)=0$ and $V^{\prime\prime}(\phi)>0$. We can drop the constant term $V(\phi_0)$ as it won't change the equations of motion, thus, we rewrite eq. (2) as $$ \mathcal{L}=\frac{1}{2}(\partial\phi)^2-\sum_{j=2}\frac{1}{j!}V^{(j)}(\phi_0)\phi^j\tag{7}. $$ Performing dimensional analysis on a term of the potential expansion, we find $$ n=[V^{(j)}(\phi_0)\phi^j]=[V^{(j)}(\phi_0)]+j[\phi]=[V^{(j)}(\phi_0)]+j\left(\frac{n}{2}-1\right)\tag{8} $$ solving eq. (9) for the dimension of the derivatives of the potential gives $$ [V^{(j)}(\phi_0)]=n-j\left(\frac{n}{2}-1\right)\tag{9} $$ For the quadratic (harmonic) term $j=2$, eq. (9) states that $V^{\prime\prime}(\phi_0)$ has energy dimension $2$, independent from the spacetime dimension $n$, and we commonly identify $V^{\prime\prime}(\phi_0)=m^2$ with the mass $m>0$ of the scalar field $\phi$. Adapting the mass, we write eq. (7) as $$ \mathcal{L}=\frac{1}{2}(\partial\phi)^2-m^2\phi^2-\sum_{j=3}\frac{1}{j!}V^{(j)}(\phi_0)\phi^j\tag{10}. $$
In many QFT textbooks, we consider the highest order to be $j=6$ in $n=4$ spacetime dimensions because then we have $\lambda=V^{(4)}(\phi_0)$ to be dimensionless allowing for renormalization. Does this have any relevance for classical relativistic field theories?
Are there any other fundamental physical requirements that constraint the form of our Lagrangian density? Why do we only "observe" a small set of possible Lagrangian densities in textbooks?
