# Scale Invariant Classical Field Theory

## Homework Statement

A class of interesting theories are invariant under the scaling of all lengths by $x^{\mu} \rightarrow (x')^{\mu}=\lambda x^{\mu}$ and $\phi(x) \rightarrow \phi'(x) = \lambda^{-D}\phi(\lambda^{-1}x)$.

Here $D$ is called the scaling dimension of the field.

Consider the action for a real scalar field given by $S = \int d^{4}x\ \frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^{2}\phi^{2}-g\phi^{p}$.

Find the scaling dimension $D$ such that the derivative terms remain invariant.

For what values of $m$ and $p$ is the scaling $x^{\mu} \rightarrow (x')^{\mu}=\lambda x^{\mu}$ and $\phi(x) \rightarrow \phi'(x) = \lambda^{-D}\phi(\lambda^{-1}x)$ a symmetry of the theory? How do these conclusions change for a scalar field living in an $(n+1)$-dimensional spacetime instead of a $3+1$-dimensional spacetime?

In $3+1$ dimensions, use Noether's theorem to construct the conserved current $D^{\mu}$ associated to scaling invariance.

## The Attempt at a Solution

Under the scaling of all lengths by $x^{\mu} \rightarrow (x')^{\mu}=\lambda x^{\mu}$ and $\phi(x) \rightarrow \phi'(x) = \lambda^{-D}\phi(\lambda^{-1}x)$, where $D$ is called the scaling dimension of the field, the given action for the real scalar field transforms as follows:

$S = \int d^{4}x\ \frac{1}{2}\partial_{\mu}\phi(x)\partial^{\mu}\phi(x)-\frac{1}{2}m^{2}\phi^{2}(x)-g\phi^{p}(x)$

$\rightarrow \int d^{4}x\ \Big|\frac{1}{\lambda^{4}}\Big| \Big[ \frac{1}{2}({\Lambda^{\rho}}_{\mu}\lambda^{-D}\partial_{\rho}\phi)(\lambda^{-1}x)({\Lambda_{\sigma}}^{\mu}\lambda^{-D}\partial^{\sigma}\phi)(\lambda^{-1}x)-\frac{1}{2}m^{2}(\lambda^{-D}\phi)^{2}(\lambda^{-1}x)-g(\lambda^{-D}\phi)^{p}(\lambda^{-1}x)\Big]$, where the factor of $\Big|\frac{1}{\lambda^{4}}\Big|$ is the Jacobian for the coordinate transformation

Therefore, the scaling dimension $D$ such that the derivative terms remain invariant is found as follows:

$\Big|\frac{1}{\lambda^{4}}\Big| \Big[ \frac{1}{2}({\Lambda^{\rho}}_{\mu}\lambda^{-D}\partial_{\rho}\phi)(\lambda^{-1}x)({\Lambda_{\sigma}}^{\mu}\lambda^{-D}\partial^{\sigma}\phi)(\lambda^{-1}x)\Big]=\frac{1}{2}(\partial_{\nu}\phi)(\lambda^{-1}x)(\partial^{\nu}\phi)(\lambda^{-1}x)$

$\Big(\frac{1}{\lambda^{4}}\Big)(\lambda^{-D})(\lambda^{-D}) \Big[ \frac{1}{2}({\Lambda^{\rho}}_{\mu}{\Lambda_{\sigma}}^{\mu})(\partial_{\rho}\phi)(\lambda^{-1}x)(\partial^{\sigma}\phi)(\lambda^{-1}x) \Big]=\frac{1}{2}(\partial_{\nu}\phi)(\lambda^{-1}x)(\partial^{\nu}\phi)(\lambda^{-1}x)$

$\lambda^{-4-2D} \Big[ \frac{1}{2}{\Lambda^{\rho}}_{\mu}{(\Lambda^{-1})^{\mu}}_{\sigma}(\partial_{\rho}\phi)(\lambda^{-1}x)(\partial^{\sigma}\phi)(\lambda^{-1}x) \Big]=\frac{1}{2}(\partial_{\nu}\phi)(\lambda^{-1}x)(\partial^{\nu}\phi)(\lambda^{-1}x)$

$\lambda^{-4-2D} \Big[ \frac{1}{2}{\eta^{\rho}}_{\sigma}(\partial_{\rho}\phi)(\lambda^{-1}x)(\partial^{\sigma}\phi)(\lambda^{-1}x) \Big]=\frac{1}{2}(\partial_{\nu}\phi)(\lambda^{-1}x)(\partial^{\nu}\phi)(\lambda^{-1}x)$

$\lambda^{-4-2D} \Big[ \frac{1}{2}(\partial_{\rho}\phi)(\lambda^{-1}x)(\partial^{\rho}\phi)(\lambda^{-1}x) \Big]=\frac{1}{2}(\partial_{\nu}\phi)(\lambda^{-1}x)(\partial^{\nu}\phi)(\lambda^{-1}x)$

so that

$\lambda^{-4-2D}=\lambda^{0}$
$D=-2$

Am I correct so far?