- #1

Markus Kahn

- 112

- 14

- Homework Statement
- Consider the real scalar field with interaction

$$\mathcal{L}=-\frac{1}{2} \partial^{\mu} \phi \partial_{\mu} \phi-\frac{1}{2} m^{2} \phi^{2}-\frac{1}{6} \mu \phi^{3}-\frac{1}{24} \lambda \phi^{4}.$$

The coordinate scaling transformation ##x'=\alpha x## with some ##\alpha\in\mathbb{R}^+## can be extended to the scalar field by ##\phi'(x')=\alpha^{-\Delta}\phi(x)## for some ##\Delta \in\mathbb{R}##. Assume there exist values of the parameters ##\{m,\mu,\lambda,\Delta\}## for which the action scale is invariant.

Derive the scale current using Noether’s procedure for invariance under the scaling transformation. Show explicitly that it is conserved

- Relevant Equations
- None

This is my first time dealing with scaling symmetry, so I'm sorry if the following is fundamental wrong. My approach was the same as if I was trying to show the same for translation or Lorentz symmetry.

We have

$$\delta\phi(x)= \phi'(x')-\phi(x)= (a^{-\Delta}-1)\phi(x)\approx-\Delta\log(a)\phi(x)\equiv -\Delta \beta \phi(x).$$

We can now go on and show that we have

$$\delta\mathcal{L}= -\Delta\beta \partial_\sigma \left(\frac{\partial\mathcal{L}}{\partial\phi_{,\sigma}}\phi(x)\right).$$

Now the problems start here... When I derive the current for, let's say, translation symmetry (##x\mapsto x+a##), I would know equate this with ##\delta\mathcal{L}=a_\lambda\partial^\lambda\mathcal{L}## and therefore get

$$\partial^\lambda\mathcal{L}=\partial_\sigma \left(\frac{\partial\mathcal{L}}{\partial\phi_{,\sigma}}\phi(x)\right) \Longleftrightarrow \partial_\sigma \underbrace{\left(\frac{\partial\mathcal{L}}{\partial\phi_{,\sigma}}\phi(x)-\eta^{\sigma\lambda}\mathcal{L}\right)}_{\equiv T^{\sigma\lambda}}=0 \Longleftrightarrow \partial_\sigma T^{\sigma\lambda}=0.$$

When I try applying this to the scaling symmetry given in the problem I end up with

$$\mathcal{L}=\partial_\sigma \left(\frac{\partial\mathcal{L}}{\partial\phi_{,\sigma}}\phi(x)\right),$$

since ##\delta\mathcal{L}=-\Delta\beta \mathcal{L}##. The issue here is that it is impossible to rewrite this equation in the from ##\partial_\sigma (\dots)=0##, which -- as far as I undertand -- makes it impossible to determine the conserved current...

Where exactly does my misstep happen? Any help appreciated.

We have

$$\delta\phi(x)= \phi'(x')-\phi(x)= (a^{-\Delta}-1)\phi(x)\approx-\Delta\log(a)\phi(x)\equiv -\Delta \beta \phi(x).$$

We can now go on and show that we have

$$\delta\mathcal{L}= -\Delta\beta \partial_\sigma \left(\frac{\partial\mathcal{L}}{\partial\phi_{,\sigma}}\phi(x)\right).$$

Now the problems start here... When I derive the current for, let's say, translation symmetry (##x\mapsto x+a##), I would know equate this with ##\delta\mathcal{L}=a_\lambda\partial^\lambda\mathcal{L}## and therefore get

$$\partial^\lambda\mathcal{L}=\partial_\sigma \left(\frac{\partial\mathcal{L}}{\partial\phi_{,\sigma}}\phi(x)\right) \Longleftrightarrow \partial_\sigma \underbrace{\left(\frac{\partial\mathcal{L}}{\partial\phi_{,\sigma}}\phi(x)-\eta^{\sigma\lambda}\mathcal{L}\right)}_{\equiv T^{\sigma\lambda}}=0 \Longleftrightarrow \partial_\sigma T^{\sigma\lambda}=0.$$

When I try applying this to the scaling symmetry given in the problem I end up with

$$\mathcal{L}=\partial_\sigma \left(\frac{\partial\mathcal{L}}{\partial\phi_{,\sigma}}\phi(x)\right),$$

since ##\delta\mathcal{L}=-\Delta\beta \mathcal{L}##. The issue here is that it is impossible to rewrite this equation in the from ##\partial_\sigma (\dots)=0##, which -- as far as I undertand -- makes it impossible to determine the conserved current...

Where exactly does my misstep happen? Any help appreciated.