$\phi^4$ theory in higher dimensions

Question

For a scalar, the Lagrangian $$L = \partial_{\mu}\phi\partial^{\mu}\phi - m^2\phi^2 + \lambda \phi^4$$ seems to be particularly suited for 4 dimensional space-time. In four dimensions, the coupling constant, $\lambda$, ends up being dimensionless, and the theory is renormalizable. How does the interaction Lagrangian $\lambda \phi^4$ get modified for higher dimensions? Does it become $\lambda\phi^n$ in n-dimensions? Is the $\phi^4$ theory or $\phi^n$ theory renormalizable in higher dimensions? Also, any literature reference would also be helpful.

Answer 1

Let $D$ be the spacetime dimension. The action is dimensionless for any $D$ (since the action has units of $\hbar$, and we set $\hbar=1$)., Since $S = \int {\rm d}^D x \mathcal{L}$, and the volume element ${\rm d}^D x$ has mass dimension $-D$, this means the Lagrangian $\mathcal{L}$ has dimension $D$

Assuming we have a weakly coupled scalar field theory, the scaling dimension of the field $\phi$ will be determined by the kinetic term, $\mathcal{L} \sim (\partial \phi)^2$ (if you like, in the free theory only the kinetic term and maybe mass term are there, so in the free theory these determine the scaling of the field, and then perturbative quantum corrections will only lead to small changes to the free theory mass dimension). Since derivatives have mass dimension $1$ in any dimension, and the Lagrangian has mass dimension $D$, in order for things to work, the field must have dimensions $(D-2)/2$. You can check that in $D=4$, this works out to say that the field should have dimension $1$, which is the case.

Then we can consider a general operator (term in the Lagrangian) of the form \begin{equation} \mathcal{L} \sim \lambda \partial^{n_d} \phi^{n_\phi} \end{equation} where $\lambda$ is a (possibly dimensionful) coupling constant; $n_d$ is the number of derivatives; and $n_\phi$ is the number of powers of $\phi$. Then the dimension of $\lambda$ is \begin{equation} \Delta_\lambda = D - n_d - n_\phi \frac{D-2}{2} = D + \left(1-\frac{D}{2}\right)n_\phi - n_d \end{equation}

For $D=4, n_\phi=4, n_d=0$, this yields $\Delta_\lambda=0$, as you expect.

For an arbitrary $D$, with $n_\phi=4, n_d=0$, we have \begin{equation} \Delta_\lambda = 4 - D \end{equation} which is negative for all $D>4$; in other words, $\phi^4$ theory is power-counting non-renormalizable for all $D>4$.

Since we set up the formalism, we might as well look at a general interaction. Let's set $n_d=0$. Then the expression is \begin{equation} \Delta_\lambda = D + \left(1-\frac{D}{2}\right)n_\phi \end{equation} Then...

• For $D=1$, this is always positive.
• For $D=2$, the mass dimension is always $2$ for any operator. (2 dimensions is special in many ways).
• For $D=3$, this simplifies to $\Delta_\lambda = 3 - \frac{n_\phi}{2}$, which is only nonegative if $n_\phi \leq 6$. Therefore, $\phi^6$ theory is dimensionless in three dimensions. Higher powers of $\phi$ are nonrenormalizable.
• For $D=4$, this becomes $\Delta_\lambda = 4 - n_\phi$, so only $\phi^3$ and $\phi^4$ theories are renormalizable.
• For $D=5$, this becomes $\Delta_\lambda = 5 - \frac{3 n_\phi}{2}$, so only $\phi^3$ is renormalizable.
• For $D=6$, we have $\Delta_\lambda = 6 - 2 n_\phi$, so $\phi^3$ is dimensionless (and therefore analogous to $\phi^4$ theory in $4$ dimensions). This is the theory that Srednicki bases the first third of his textbook on.
• For $D \geq 7$, all terms with $n_\phi \geq 3$ have negative mass dimension, and so are not renormalizable.

Since $n_d$ contributes negatively to $\Delta_\lambda$, any interactions with derivatives can only possibly be renormalizable if the derivative-less version is.