Undergrad Determine Scaling Dimension of Field Theory

[ergospherical]

It is given that a theory is invariant under the length scaling:\begin{align*}
x &\rightarrow \lambda x \\
\phi(x) &\rightarrow \lambda^{-D} \phi(\lambda^{-1} x)
\end{align*}for some $D$ to be determined. The action of a real scalar field is here:\begin{align*}
S = \int d^4 x \dfrac{1}{2}\partial_{\mu} \phi \partial^{\mu} \phi - \dfrac{1}{2}m^2 \phi^2 -g\phi^p
\end{align*}Since $\partial_{\mu} = \frac{\partial x'^{\nu}}{\partial x^{\mu}} \partial'_{\nu} = {(\Lambda^{-1})^{\nu}}_{\mu} \partial'_{\nu}$ then would I be correct in thinking that the derivative of the field transforms as:\begin{align*}
\partial_{\mu} \phi(x) \rightarrow \partial_{\mu} \phi'(x) &= \lambda^{-D} \partial_{\mu} \phi(\lambda^{-1} x) \\
&= \lambda^{-D} {(\Lambda^{-1})^{\nu}}_{\mu} \partial'_{\nu} \phi(x')
\end{align*}so the derivative term in the action transforms as \begin{align*}
(\partial_{\mu} \phi)^2 &\rightarrow \lambda^{-2D} {(\Lambda^{-1})^{\nu}}_{\mu} {\Lambda^{\mu}}_{\rho} (\partial'_{\nu} \phi(x'))( \partial'^{\rho} \phi(x')) \\
&= \lambda^{-2D} (\partial'_{\mu} \phi(x'))^2
\end{align*}Meanwhile $d^4 x = \lambda^{-4} d^4 x'$, and this would imply scale invariance when $D=-2$? That feels wrong and I worry that I have transformed the wrong things.

 

[vanhees71]

It's much simpler. For the scale transformation to be a symmetry, it's sufficient that the action is invariant. Now start with the massless free field,
$$S_0=\int \mathrm{d}^4 x \frac{1}{2} (\partial_{\mu} \phi) (\partial^{\mu} \phi).$$
Since $\mathrm{d}^4 x \rightarrow \lambda^4 \mathrm{d}^4 x$ and $\partial_{\mu} \rightarrow \frac{1}{\lambda} \partial_{\mu}$, you must have $D=1$ to get $S_0$ invariant.

Then you see that the mass term is "forbidden" by the symmetry, because $\mathrm{d}^4 x m^2 \phi^2$ is not invariant. This is no surprise, because $m$ is a dimensionful parameter, which breaks scale invariance to begin with.

For the interaction term $\mathrm{d}^4 x \phi^p$ must be invariant, and thus $p=4$. Indeed, only for $p=4$ the coupling constant $g$ is dimensionless too.

So the only allowed theory of this kind is a massless field $\phi$ with a $\phi^4$ interaction.

Some further ideas to think about:

(a) What's the Noether current of the scale symmetry?

(b) If you quantize it, you have to renormalize this massless (!) field theory, and then what happens with scale invariance?

 

[ergospherical]

I was thinking about the two questions earlier, I wonder if you can guide me a little since this is new material. What I understand is that if the Lagrangian varies by a total derivative $\delta \mathcal{L} = \partial_{\mu} F^{\mu}$ under a variation $\delta \phi_a(x) = X_a (\phi)$ of the fields, then one obtains a Noether current $j^{\mu} = \frac{\partial \mathcal{L}}{\partial \phi_{a,\mu}} X_a(\phi) - F^{\mu}(\phi)$ satisfying $\nabla \cdot j = 0$.

For this problem, is it correct to proceed as follows? I will put $\lambda = 1+ \epsilon$ where $\epsilon$ is a small positive or negative number, then write\begin{align*}
\tilde{\phi}(x) = \lambda^{-1} \phi(\lambda^{-1}x) &= (1 - \epsilon + O(\epsilon^2))\phi((1- \epsilon) x + O(\epsilon^2 x^2)) \\
&= (1 - \epsilon + O(\epsilon^2)) (\phi(x) - \epsilon x^{\nu} \phi_{,\nu}(x) + O(\epsilon^2 x^2) ) \\
&= \phi(x) - \epsilon \phi(x) - \epsilon x^{\nu} \phi_{,\nu}(x)
\end{align*}Then $\delta \phi = - \epsilon (\phi + x^{\nu} \phi_{,\nu})$. For $\phi_{,\mu}$ I write\begin{align*}
\delta \phi_{,\mu} &= -\epsilon \partial_{\mu}(\phi + x^{\nu} \phi_{,\nu}) \\
&= - \epsilon \phi_{,\mu} - \epsilon \partial_{\mu}(x^{\nu} \phi_{,\nu}) \\
&= - \epsilon \phi_{,\mu} - \epsilon \phi_{,\mu} - x^{\nu} \phi_{,\nu \mu} \\
&= -\epsilon(2\phi_{,\mu} + x^{\nu} \phi_{,\nu \mu})
\end{align*}I write for the Lagrangian,\begin{align*}
\mathcal{L} = \dfrac{1}{2} \phi_{,\mu} \phi^{,\mu} -g \phi^4
\end{align*}from which follows $\frac{\partial \mathcal{L}}{\partial \phi_{,\mu}} = \phi^{,\mu}$ and $\frac{\partial \mathcal{L}}{\partial \phi} = -4g\phi^3$, so that\begin{align*}
\delta \mathcal{L} &= -\epsilon \phi^{,\mu}(2\phi_{,\mu} + x^{\nu} \phi_{,\nu \mu}) + 4 g \epsilon \phi^3 (\phi + x^{\nu} \phi_{,\nu}) \\
&= -4\epsilon( \frac{1}{2} \phi^{,\mu} \phi^{,\mu} - g \phi^4) - 4\epsilon x^{\nu} (\frac{1}{4}\phi^{,\mu} \phi_{,\nu \mu} - g \phi^3 \phi_{,\nu}) \\
&= -4 \epsilon \left[ \mathcal{L} + x^{\nu} (\frac{1}{4}\phi^{,\mu} \phi_{,\nu \mu} - g \phi^3 \phi_{,\nu}) \right] \\
&= -4 \epsilon \left[ \mathcal{L} + x^{\nu} \dfrac{\partial}{\partial x^{\nu}} (\frac{1}{2}\phi^{,\mu} \phi_{,\mu} - \dfrac{1}{4}g \phi^4) \right] \\
\end{align*}which is almost $- 4\epsilon \partial_{\nu}(x^{\nu} \mathcal{L})$ but fails because of the factor of $1/4$ in front of the $g\phi^4$ term...

 

[vanhees71]

Looks good (though I haven't checked the calculation in dateail). But now you also have to take into account that you have to express $\mathrm{d}^4 x'$ through $\mathrm{d}^4 x$ in the action integral, i.e., there's an additional contribution from the corresponding Jacobian, which you also get by expanding to first order in $\epsilon$.