Solving a \phi^4 CFT Question: Conformal Group & Algebra

[tuu-tikki]

Homework Statement

Consider action S = $$\int d^4 x (\frac{1}{2} \partial^{\mu} \phi \partial_{\mu} \phi - \lambda \phi^4)$$. Consider dilatations $$\phi(x) \Rightarrow \alpha \phi(\alpha x), \alpha$$ is a transformation parameter; and special conformal transformations $$\phi(x) \Rightarrow \phi(x')$$, $$x'^{\mu} = \frac{x^{\mu}+ x^2 \alpha^{\mu}}{1+ \alpha^2 x^2 + 2 x^{\rho} \alpha_{\rho}}, \alpha^{\mu}$$ are 4 transformation parameters.

1) Show that such theory is invariant under dilatations and special conformal transformations;
2) Show that together with Poincare group of transformations these two transformations form another group of transformations called conformal group.
3) Find out commutation relations between generators of conformal algebra.
4) Show that conformal algebra is isomorphic to algebra of $$SO(2,4)$$ group.

Homework Equations

The Attempt at a Solution

1) It is trivial to show that theory is invariant under dilatations. Now - about special conf.transformations. The equations of motion are $$\square \phi + 4 \lambda \phi^3 = 0. \phi$$ doesn't transform under such transformations, only its argument does, so for equations to be invariant, d'Alembert operator must be invariant. Here we go: $$\square \phi(x') = \frac{\partial}{\partial x_{\mu}}\frac{\partial}{\partial x^{\mu}} \phi(x') = \frac{\partial x'^{\sigma}}{\partial x_{\mu}} \frac{\partial}{\partial x'_{\sigma}} \frac{\partial x'^{\tau}}{\partial x^{\mu}} \frac{\partial}{\partial x'_{\tau}} \phi (x') = (\frac{\partial x'^{\sigma}}{\partial x_{\mu}}) (\frac{\partial x'^{\tau}}{\partial x^{\mu}}) \frac{\partial^2 \phi(x')}{\partial x'^{\sigma} \partial x'^{\tau}}$$. So, for d'Alembert operator to be invariant, we need $$(\frac{\partial x'^{\sigma}}{\partial x_{\mu}}) (\frac{\partial x'^{\tau}}{\partial x^{\mu}})$$ to be equal to $$g^{\sigma \tau}.$$ But one can easily obtain that it isn't. Where's a mistake?

2) For it all to make a conformal group, we need at least that two consecutive transformations shouldn't move us out of a group - they should also be a transformation from the group with another parameters. For $$\phi(x^{\mu}) \Rightarrow \phi(\Lambda^{\mu}_{\nu} x^{\nu} + a^{\mu})$$ and $$\phi(x^{\mu}) \Rightarrow \alpha \phi(\alpha x^{\mu})$$ transformations it can be easily shown. Naturally, consider the transformation $$\phi(x^{\mu}) \Rightarrow \alpha \phi(\alpha (\Lambda^{\mu}_{\nu} x^{\nu} + a^{\mu}))$$ - such transformation is more general than two in the previous sentence (let's call it (1)). Two such consecutive transformations with different parameters lead us to $$\alpha \beta \phi(\beta(\Omega^{\mu}_{\rho} \Big(\alpha(\Lambda^{\rho}_{\nu} x^{\nu} + a^{\rho})\Big)+b^{\mu})) = \alpha \beta \phi(\alpha \beta(\Omega^{\mu}_{\rho} \Lambda^{\rho}_{\nu} x^{\nu} + \Omega^{\mu}_{\rho} a^{\rho}+\frac{b^{\mu}}{\alpha})) = \alpha' \phi(\alpha' (\Lambda'^{\mu}_{\nu} x^{\nu} + a'^{\mu})); \alpha' = \alpha \beta, \Lambda'^{\mu}_{\nu} = \Omega^{\mu}_{\rho} \Lambda^{\rho}_{\nu}, a'^{\mu} = \Omega^{\mu}_{\rho} a^{\rho} + \frac{b^{\mu}}{\alpha}$$. So we obtained that two consecutive transformations with different parameters are equal to one transformation of the similar type with some parameters. So it seems like a transformation group (but we still didn't check some other group properties). But we have three transformations in the group, not the first two: $$\phi(x^{\mu}) \Rightarrow \phi(\Lambda^{\mu}_{\nu} x^{\nu} + a^{\mu})$$ and $$\phi(x^{\mu}) \Rightarrow \alpha \phi(\alpha x^{\mu})$$ and $$\phi(x^{\mu}) \Rightarrow \phi(\frac{x^{\mu}+ x^2 \alpha^{\mu}}{1+\alpha^2 x^2 + 2 x^{\rho} \alpha_{\rho}})$$. I didn't manage to prove that the composition of two transformations from the group is the transformation from the group.
3) no question
4) I obtained all the 15 generators, but I can't understand why should it be ASO(2,4). Only from the dimensional thoughts. Any more ideas?

 

[mark726]

Thank you for your interesting post. it is my duty to help clarify any misunderstandings and provide guidance in your research.

1) Your attempt at showing the invariance under special conformal transformations is on the right track. However, you are missing a crucial step in your calculation. In order for the d'Alembert operator to be invariant, the transformation of the coordinates must also be taken into account. This means that the derivative terms in the d'Alembert operator should also be transformed. Once you take this into account, you will find that the d'Alembert operator is indeed invariant under special conformal transformations.

2) Your approach to showing that the dilatations and special conformal transformations form a group is correct. However, you are right in pointing out that we need to show that the composition of two transformations from the group is also a transformation from the group. This can be done by explicitly showing the composition of two transformations and then using the fact that the d'Alembert operator is invariant under these transformations.

3) I am glad to hear that you have no questions about this part.

4) Your approach to obtaining the 15 generators is correct. However, the reason why the conformal algebra is isomorphic to the algebra of SO(2,4) is because the conformal group is the group of transformations that preserve the Minkowski metric, which is the metric of the space-time in special relativity. This metric can be written in terms of the generators of the conformal algebra, which is why the two are isomorphic.

I hope this helps clarify your understanding of the subject. Keep up the good work in your research.