# Question about conformal transformation on fields

Textbook says for scale transformation: $$x^\mu\rightarrow x'^\mu=e^\sigma x^\mu$$ The field transforms like $$\phi^A(x)\rightarrow \phi'^A(x')=\left[e^{-\mathbf{D} \sigma} \right]^A_{\quad B} \phi^B(x)$$

where $\mathbf{D}=\frac{d-2}{2} \mathbb{I}$ for Bose field and $\mathbf{D}=\frac{d-1}{2} \mathbb{I}$ for Fermi field.

But why is there such matrix $\mathbf{D}$ for field? From the definition of scalar, vector or tensor field in differential geometry.

For scaler field $\phi(x)$, it's a map $\phi: \mathcal{M} \rightarrow \mathbb{R}$. It should be invariant under coordinate transformation. So under scale transformation $x^\mu\rightarrow x'^\mu=e^\sigma x^\mu$, $\phi(x)\rightarrow \phi'(x')=\phi(x)$.

For vector field $A^\mu(x)$, $A^\mu(x)\rightarrow A'^\mu(x')=\frac{\partial x'^\mu}{\partial x^\nu}A^\nu(x)=e^\sigma A^\mu(x)$

• Could you mention the textbook that you are referring to ? – M111 Jan 13 '18 at 18:16

Your book are not taking the $A^\mu(x)\rightarrow A'^\mu(x')=\frac{\partial x'^\mu}{\partial x^\nu}A^\nu(x)=e^\sigma A^\mu(x)$ transformation. Your book are defining the scaling of each field in order to preserve the kinetic term of the Lagrangian of those fields under the transformation $x^\mu\rightarrow x'^\mu=e^\sigma x^\mu$.
$$\int d^{d}x\mathcal{L}=\int d^{d}x\left(\frac{1}{2}\eta^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi\, +\,...\right)$$
and under $x^\mu\rightarrow x'^\mu=e^\sigma x^\mu$, we have $d^dx\rightarrow (e^\sigma)^dd^dx\,$ and $\,\partial_{\mu}\rightarrow (e^{\sigma})^{-1}\partial_{\mu}$. So in order to preserve the action/lagragian we need $\phi\rightarrow(e^\sigma)^{(2-d)/2}$. You can check that this match with the scaling of yours:
$$\phi^A(x)\rightarrow \phi'^A(x')=\left[e^{-\mathbf{D} \sigma} \right]^A_{\quad B} \phi^B(x)$$