- #1
shinobi20
- 277
- 20
- TL;DR Summary
- I'm quite unsure about how the scaling dimension of a field is devised. I need clarifications on small details in order to make sure that my concepts are clear.
I'm studying CFT, and I find the lecture notes and books really confusing and devoid of explanations (more details).
In a scale transformation ##x' = \lambda x##, the field ##\phi(x)## should also be affected by the scale transformation, i.e., ##\phi'(x') = \phi'(\lambda x) = \lambda^{-\Delta} \phi(x)##. I think this should mean that we assume that the field should scale but we do not know by how much, and ##\Delta## quantifies this unknown, which is called the scaling dimension. We put a minus sign because when we plug it in the action, this will avoid a negative dimension (see below)?
If the scale transformation is a symmetry of the theory, then the action must be invariant under this. Particularly, in a free theory,
\begin{align*}
S' & = \int d^d x' \partial'_\mu \phi'(x') \partial'^\mu \phi'(x')\\
& = \int d^d x' g^{\mu \nu} \partial'_\mu \phi'(x') \partial'_\nu \phi'(x')\\
& = \int d^d x \Bigg| \frac{\partial x'}{\partial x} \Bigg| \lambda^{-2} g^{\mu \nu} \partial_\mu \phi'(x') \partial_\nu \phi'(x')\\
& = \int d^d x \lambda^d \lambda^{-2} \partial_\mu \phi'(x') \partial^\mu \phi'(x')\\
\end{align*}
Comparing this with ##S = \int d^d x \partial_\mu \phi(x) \partial^\mu \phi(x)##,
\begin{equation*}
\int d^d x \lambda^d \lambda^{-2} \partial_\mu \phi'(x') \partial^\mu \phi'(x') = \int d^d x \partial_\mu \phi(x) \partial^\mu \phi(x)
\end{equation*}
We can infer that,
\begin{equation*}
\lambda^{\frac{d-2}{2}} \phi'(x') = \phi(x), \quad \text{OR} \quad \phi'(x') = \lambda^{\frac{-(d-2)}{2}} \phi(x)
\end{equation*}
So the scaling dimension is ##\Delta = \frac{(d-2)}{2}##. If we were to not put a minus sign from the start, i.e., ##\phi'(\lambda x) = \lambda^{\Delta} \phi(x)##, then ##\Delta = \frac{-(d-2)}{2}##.
Questions:
1. Can anyone verify if what I'm saying (my statements) above are correct?
2. Can anyone explain more about the minus sign?
3. I'm not sure if what I wrote in the second line of the action is correct, i.e., ##\partial'_\mu \phi'(x') \partial'^\mu \phi'(x') = g^{\mu \nu} \partial'_\mu \phi'(x') \partial'_\nu \phi'(x')##. Is this correct or should it be ##\partial'_\mu \phi'(x') \partial'^\mu \phi'(x') = g'^{\mu \nu} \partial'_\mu \phi'(x') \partial'_\nu \phi'(x')##? If this is the case then ##g'^{\mu \nu}## should also transform, but then I would not get the correct scaling dimension.
In a scale transformation ##x' = \lambda x##, the field ##\phi(x)## should also be affected by the scale transformation, i.e., ##\phi'(x') = \phi'(\lambda x) = \lambda^{-\Delta} \phi(x)##. I think this should mean that we assume that the field should scale but we do not know by how much, and ##\Delta## quantifies this unknown, which is called the scaling dimension. We put a minus sign because when we plug it in the action, this will avoid a negative dimension (see below)?
If the scale transformation is a symmetry of the theory, then the action must be invariant under this. Particularly, in a free theory,
\begin{align*}
S' & = \int d^d x' \partial'_\mu \phi'(x') \partial'^\mu \phi'(x')\\
& = \int d^d x' g^{\mu \nu} \partial'_\mu \phi'(x') \partial'_\nu \phi'(x')\\
& = \int d^d x \Bigg| \frac{\partial x'}{\partial x} \Bigg| \lambda^{-2} g^{\mu \nu} \partial_\mu \phi'(x') \partial_\nu \phi'(x')\\
& = \int d^d x \lambda^d \lambda^{-2} \partial_\mu \phi'(x') \partial^\mu \phi'(x')\\
\end{align*}
Comparing this with ##S = \int d^d x \partial_\mu \phi(x) \partial^\mu \phi(x)##,
\begin{equation*}
\int d^d x \lambda^d \lambda^{-2} \partial_\mu \phi'(x') \partial^\mu \phi'(x') = \int d^d x \partial_\mu \phi(x) \partial^\mu \phi(x)
\end{equation*}
We can infer that,
\begin{equation*}
\lambda^{\frac{d-2}{2}} \phi'(x') = \phi(x), \quad \text{OR} \quad \phi'(x') = \lambda^{\frac{-(d-2)}{2}} \phi(x)
\end{equation*}
So the scaling dimension is ##\Delta = \frac{(d-2)}{2}##. If we were to not put a minus sign from the start, i.e., ##\phi'(\lambda x) = \lambda^{\Delta} \phi(x)##, then ##\Delta = \frac{-(d-2)}{2}##.
Questions:
1. Can anyone verify if what I'm saying (my statements) above are correct?
2. Can anyone explain more about the minus sign?
3. I'm not sure if what I wrote in the second line of the action is correct, i.e., ##\partial'_\mu \phi'(x') \partial'^\mu \phi'(x') = g^{\mu \nu} \partial'_\mu \phi'(x') \partial'_\nu \phi'(x')##. Is this correct or should it be ##\partial'_\mu \phi'(x') \partial'^\mu \phi'(x') = g'^{\mu \nu} \partial'_\mu \phi'(x') \partial'_\nu \phi'(x')##? If this is the case then ##g'^{\mu \nu}## should also transform, but then I would not get the correct scaling dimension.
