Conformal Field Theory: Questions & Answers

[Ace10]

Hi all,

my question is rather a simple one and regards conformal transformations. On "Applied CFT" by P.Ginsparg, http://arxiv.org/pdf/hep-th/9108028.pdf , on page 10, gives the transformation rule of a quasi primary field and relates the exponent of 1.12 to the one of 1.10. My first question is how can I obtain 1.10 and secondly, how the first equation of 1.11 is related to the one of 1.12..

I know that under dilatations: x'→λx , but how can I write this for a field? It has to do with the Jacobian 1.10? Is this somehow related to the volume element? (I see the determinant of the metric in the denominator and I think that is related to the volume element but I'm not sure..)
Thank you very much in advance for your help.

 

[samalkhaiat]

Hi all,

my question is rather a simple one and regards conformal transformations. On "Applied CFT" by P.Ginsparg, http://arxiv.org/pdf/hep-th/9108028.pdf , on page 10, gives the transformation rule of a quasi primary field and relates the exponent of 1.12 to the one of 1.10. My first question is how can I obtain 1.10 and secondly, how the first equation of 1.11 is related to the one of 1.12..

I know that under dilatations: x'→λx , but how can I write this for a field? It has to do with the Jacobian 1.10? Is this somehow related to the volume element? (I see the determinant of the metric in the denominator and I think that is related to the volume element but I'm not sure..)



Thank you very much in advance for your help.

The defining relation of the conformal group $C ( 1 , n - 1 )$ is given by $$\bar{g}_{a b} ( x ) = \frac{\partial \bar{x}^{c}}{\partial x^{a}} \frac{\partial \bar{x}^{d}}{\partial x^{b}} \ \eta_{c d} = S( x ) \ \eta_{a b} . \ \ \ (1)$$ Taking the determinants and assuming even-dimensional space-time with signature $(1 , n - 1)$, we find $$( - \bar{g} ) = | \frac{\partial \bar{x}}{\partial x} |^{2} = S^{n} ,$$ or $$\frac{1}{\sqrt{- g}} = S^{- \frac{n}{2}} = | \frac{\partial x}{\partial \bar{x}} | .$$ From this, we obtain $$\frac{1}{\sqrt{S( x )}} = | \frac{\partial x}{\partial \bar{x}} |^{\frac{1}{n}} . \ \ \ \ \ \ \ \ (2)$$ In order to understand how the fields transform, you really need to study the representation theory of the conformal algebra. You can find more details in
www.physicsforums.com/showthread.php?t=172461
However, we can do it loosely in here. Let us rewrite (1) in the form $$\left( \frac{1}{\sqrt{S}} \frac{\partial \bar{x}^{c}}{\partial x^{a}} \right) \left( \frac{1}{\sqrt{S}} \frac{\partial \bar{x}^{d}}{\partial x^{b}} \right) \eta_{c d} = \eta_{a b} .$$ Therefore, it is clear that the matrix $$\Lambda ( x ) \equiv \frac{1}{\sqrt{S}} \frac{\partial \bar{x}}{\partial x} ,$$ is an element of the Lorentz group $SO(1,n-1)$. Moreover, this $\Lambda (x)$ forms a linear representation of the conformal group. This is because both $\frac{ \partial \bar{x}^{a}}{\partial x^{b}}$ and $\sqrt{S}=| \frac{\partial \bar{x}}{\partial x}|^{\frac{1}{n}}$ are themselves linear representations. This allows us to extend any linear representation of the Poincare group to the full conformal group. Therefore, given the finite-dimensional (matrix) representation $\Lambda \rightarrow D(\Lambda) , \ \forall \Lambda \in SO(1,n-1)$, the conformal transformation $x \rightarrow \bar{x}$ can be represented by $$\mathcal{C}(\frac{\partial \bar{x}}{\partial x})= \left( \sqrt{S(x)} \right)^{ - \Delta} \times D( \Lambda(x)) = \left( \sqrt{S(x)} \right)^{ - \Delta} \times D( \frac{1}{\sqrt{S(x)}}\ \frac{\partial \bar{x}}{\partial x} ) ,$$ where $\Delta$ is a real number (the scaling dimension) if $D(\Lambda)$ is irreducible (Schur's lemma), otherwise a matrix satisfying $[\Delta,D(\Lambda)]=0$. Indeed, all finite-dimensional representations of $C(1,3)$ are completely specified by the finite-dimentional irreducible representations $(j_{1},j_{2})$ of the Lorentz group $SO(1,3)$ and those of the non-compact group of pure dilatations $SO(1,1)$ labelled by the scaling dimension $\mathcal{R}(\sqrt{S(x)}) = \left(\sqrt{S} \right)^{- \Delta}$. For example, if $V^{a}(x)$ is a field transforming in the vector representation of the Lorentz group: $\tilde{V}^{a}(\tilde{x}) = \Lambda^{a}{}_{b} \ V^{c}(x) ; \ x^{a}\rightarrow \tilde{x}^{a} = \Lambda^{a}{}_{c} \ x^{c}$, then, under a conformal transformation $x \rightarrow \bar{x}$, we have $$\bar{V}^{a} ( \bar{x} ) = \mathcal{C}^{a}{}_{c} ( \frac{\partial \bar{x}}{\partial c} ) \ V^{c} ( x ) , \ \ \ \ (3)$$ where $$\mathcal{C}^{a}_{c} ( \partial \bar{x} / \partial x ) = \left( \sqrt{S} \right)^{- \Delta} \ D^{a}{}_{c}( \frac{1}{\sqrt{S(x)}} \ \frac{\partial\bar{x}}{\partial x} ) = \left( \sqrt{S} \right)^{- \Delta - 1} \frac{\partial \bar{x}^{a}}{\partial x^{c}} . \ \ \ (4)$$ So, under pure dilatations; $\bar{x}^{a} = e^{-\alpha}\ x^{a}$, we have $$\frac{\partial \bar{x}^{a}}{\partial x^{c}} = e^{- \alpha} \delta^{a}_{c} , \ \ \Rightarrow \ | \frac{\partial \bar{x}}{\partial x} | = e^{- n \alpha} , \ \ \ (5)$$ and, therefore $$\left( \sqrt{S} \right)^{- \Delta - 1} = | \frac{\partial \bar{x}}{\partial x} |^{\frac{- \Delta - 1}{n}} = e^{\alpha \Delta + \alpha} . \ \ \ (6)$$ Putting eq’s (4), (5) and (6) in equation (3), we find $$\bar{V}^{a}( \bar{x} ) = e^{\alpha \Delta} \ V^{a} ( x ) \equiv | \frac{\partial x}{\partial \bar{x}} |^{\frac{\Delta}{n}} V^{a} ( x ) .$$

Sam

 

[Ace10]

Wow Sam, thank you very much for your detailed reply! I realized I was treating the representations the wrong way! Thank you very much!

By the way, just to ask, if we have an odd dimensional spacetime, say:
g=diag{...-1... , ...+1...} (n-times -1 with n odd),
then we use absolute value at the deteminant under the squareroot, right? So one can say that in general case we follow the notation with the absolute value?

 

[samalkhaiat]

For $(p,q)$ signature in $n = |p| + |q|$ dimensions, you have $$\det (g_{ab}) = (-1)^{|q|} |g| = (-1)^{n} (-1)^{- |p|} |g| .$$ So
$$\det(g_{ab}) = (-1)^{|p|} |g| , \ \ \mbox{for} \ \ n = 2k ,$$ $$\det(g_{ab}) = (-1)^{|p| + 1} |g| , \ \ \mbox{for} \ \ n = 2k + 1 .$$

 

[Ace10]

Right, thank you very much!