Field transformation under Conformal transformation

[craigthone]

I am confused about the field transformation under conformal transformation. Consider the scale transformation of field $\phi$ (not necessarily scalar)

In CFT of Francesco et al, formula (2.121), the transformation is
$$ \vec{x}\rightarrow \vec{x}'=\lambda x,\,\,\,\phi(\vec{x}) \rightarrow \phi'(\lambda \vec{x}) =\lambda^{-\Delta}\phi(\vec{x})$$

In the AdS/CFT review of AGMOO https://arxiv.org/abs/hep-th/9905111, page 33, the transformation is
$$ x^{\mu}\rightarrow\lambda x^{\mu},\,\,\,\phi(x) \rightarrow \phi'(x) =\lambda^{\Delta}\phi(\lambda x)$$


Are these two kinds of transformation same and why?

 

[samalkhaiat]

Both are correct. However, the $\lambda$ in the first is $\lambda^{-1}$ in the second. To see this, start from the first transformation law $$\bar{\varphi} (\bar{x}) = \lambda^{- \Delta} \varphi (x) ,$$ substitute $x = \lambda^{-1} \bar{x}$ in the RHS and then re-name the coordinates labels in both sides as $x$: $$\bar{\varphi} (x) = \lambda^{- \Delta} \varphi ( \lambda^{-1} x) .$$ This gives you the second transformation law if you define $\alpha = \lambda^{-1}$. If you write $\lambda = 1 - \epsilon \$$\Rightarrow$ $\alpha = 1 + \epsilon$, then to first order in $\epsilon$ both transformation laws give you $$\delta \varphi (x) = i \epsilon [ D , \varphi (x)] = \epsilon \left( \Delta + x^{\mu} \partial_{\mu}\right) \varphi (x) ,$$ and more importantly $$[ D , \varphi (0) ] = - i \Delta \varphi (0) .$$

 

[craigthone]

Thanks for you explanation, samalkhaiat.
It still has some problem : under $x \rightarrow \bar{x}$, what does the field $\phi(x)$ change to? $\phi'(x)$ or $\phi'(\bar{x})$?
This should be certain.

 

[samalkhaiat]

When $x \to \bar{x} = \bar{x}(x)$, then $\varphi (x) \to \bar{\varphi} ( \bar{x})$. This does not mean that you cannot evaluate the new function $\bar{\varphi}$ at the coordinate value $\bar{x} = x$. So, $\bar{\varphi}(x) =\bar{\varphi}(\bar{x})|_{\bar{x} = x}$.

 

[craigthone]

Thanks samalkhaiat., I think I know your point. But if I consider a specific point, the transformation should be specific.
If $x_a \rightarrow \lambda x_a$, then $\phi(x_a) \rightarrow \phi'(\lambda x_a)$ rather than $\phi'(x_a)$.
So the second transformation has some problem.
Am I right?

 

[samalkhaiat]

If $x_a \rightarrow \lambda x_a$, then $\phi(x_a) \rightarrow \phi'(\lambda x_a)$ rather than $\phi'(x_a)$.
So the second transformation has some problem.
Am I right?

No, there is no problem because the symbol $\bar{\varphi}(x)$ stands for $\bar{\varphi}(x ; \lambda )$, i.e. it depends on the coordinate as well as the transformation parameter, with $\bar{\varphi}( x ; 1) = \varphi (x)$ representing the identity transformation. So, the statement $\varphi (x) \to \bar{\varphi}(\bar{x})$ is completely equivalent to the statement $\varphi (x) \to \bar{\varphi}(x)$, provided (of course) one knows the meaning of $\varphi (x) , \bar{\varphi} (\bar{x})$ and $\bar{\varphi} (x)$.

Let us consider a simple example (which you should consider doing whenever you get confused about something). We take our field to be a vector field $V$ on $\mathbb{R}^{2}$, i.e., for every point $p \in \mathbb{R}^{2}$, we have $$V(p) = \begin{pmatrix} V_{1}(p) \\ V_{2}(p) . \end{pmatrix}$$ Let $S$ be a system in which the point $p$ has coordinate value $x = (x_{1} , x_{2})$, and the field $V(p)$ has the following simple form $$V(x) \equiv \begin{pmatrix} V_{1}(x_{1} , x_{2}) \\ V_{2} (x_{1} , x_{2}) \end{pmatrix} = \begin{pmatrix} x_{2} \\ 0 \end{pmatrix} .$$ So, in the $S$-system, the components of the field at $p$ are given by $$V_{j}(x) = \delta_{j1} x_{2} . \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
Notice that our vector field has length given by $$\left(V_{i}(x)\right)^{2} = (x_{2})^{2}$$
Now, let $\bar{S}$ be another system in which the same point $p \in \mathbb{R}^{2}$ has coordinates $\bar{x} = (\bar{x}_{1} , \bar{x}_{2})$ and the vector field at the same $p$ is represented by $\bar{V}(p) = \bar{V}(\bar{x})$. Let us assume that the system $S$ is related to $\bar{S}$ by infinitesimal $SO(2)$ rotation. This means that the coordinates of the point $p$ in the two systems are related by $$\bar{x}_{i} = x_{i} + \theta \ \epsilon_{ij} x_{j} , \ \ \ \ \ \ \ \ (2a)$$ where $| \theta | \ll 1$ is the infinitesimal rotation angle (infinitesimal means you put $\theta^{2} = \theta^{3} = \cdots = 0$), and $\epsilon_{ij} = - \epsilon_{ji} \ , \epsilon_{12} = 1$ is the antisymmetric $SO(2)$-invariant tensor. The inverse transformation is obtained by letting $\theta \to - \theta$ $$x_{i} = ( \delta_{ij} - \theta \ \epsilon_{ij}) \bar{x}_{j} . \ \ \ \ \ \ \ (2b)$$
Being a vector, our field $V_{i}(x)$ must transform in the vector representation of $SO(2)$, i.e., it obeys the same transformation law of the coordinates: $$\bar{V}_{i}(\bar{x}) = V_{i}(x) + \theta \ \epsilon_{ij} V_{j}(x) .$$ Substituting eq(1), we get $$\bar{V}_{i}(\bar{x}) = \delta_{i1} x_{2} + \theta \ \epsilon_{i1} x_{2} . \ \ \ \ \ \ (3)$$ This equation looks very ugly because the RHS is written in terms of untransformed coordinate $x_{2}$ only. To give eq(3) a face-left, we use eq(2b) to write $x_{2} = \bar{x}_{2} + \theta \bar{x}_{1}$ and substitute it in the RHS of (3). Thus, to first order in $\theta$ we get $$\bar{V}_{i}(\bar{x}) = \delta_{i1} \left( \bar{x}_{2} + \theta \bar{x}_{1} \right) + \theta \ \epsilon_{i1} \bar{x}_{2} \ . \ \ \ \ (4)$$
Now, we check that eq(4) gives us the expected actions of rotations on a vector field, these are
A) No rotation corresponds to $\theta = 0$. Indeed putting $\theta = 0$ in eq(4) gives us $$\bar{V}_{i}(\bar{x}) = \delta_{i1} \bar{x}_{2} = \delta_{i1}x_{2} = V_{i}(x) .$$
B) Rotation mixes the components of the field. Indeed, while $V_{i}(x)|_{x \neq 0}$ has only one non-zero component (see eq(1)), the transformed field in eq(4) $\bar{V}_{i}(\bar{x})|_{\bar{x} \neq 0}$ has two non-zero components.
C) More importantly, rotation preserves the length of our vector field. Indeed, from eq(4) it follows $$\left( \bar{V}_{i}(\bar{x})\right)^{2} = \left( \bar{x}_{2} + \theta \ \bar{x}_{1}\right)^{2} = \left(x_{2}\right)^{2} = \left( V_{i}(x) \right)^{2} .$$
Now, I don’t like those barred coordinates in eq(4)! The question then: Do I loose any of the above 3 properties, if I evaluate function $\bar{V}_{i}(\bar{x})$ at $\bar{x} = x$? As you can check for yourself, you lose absolutely nothing if you write $$\bar{V}_{i}(x) = \delta_{i1} \left( x_{2} + \theta \ x_{1}\right) + \theta \ \epsilon_{i1} x_{2} \ . \ \ \ \ (5)$$
Not just you don’t loose any property, in fact you gain more insight when you consider the transformation law eq(5): Since $\bar{V}_{i}(x)$ and $V_{i}(x)$ are evaluated at the same coordinate value $x$, the difference $$\bar{V}_{i}(x) - V_{i}(x) = \theta \left( \delta_{i1} x_{1} + \epsilon_{i1} x_{2}\right) \equiv \delta V_{i}(x) \ ,$$ gives you the infinitesimal change in the functional form of the field.

Why did you post this in the Homework section? Can some one move this thread to the Relativity or QM forums?