Show that $\Phi^p$ transforms as a scalar primary field

Question

I am trying to solve the following exercise from an old worksheet but I don't even know where to start from:

A scalar primary field $\Phi(x)$ of scaling dimension $\Delta$ transforms under special conformal transformations as

$$\delta_{K_\nu} \Phi=2\Delta x_\nu \Phi - K_\nu \Phi \tag{1}$$

where $K_\nu= x^2 \partial_\nu -2x_\nu x^\mu \partial_\mu$

Show that $\Phi^p$ transforms as a scalar primary field for any positive integer $p$.

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My thoughts is that I must prove it transforms the way a scalar primary field transforms under special conformal transformations, as follows:

$$\Phi(x) = \frac{\Phi(x')}{(1+2b^\mu x_\mu +b^2 x^2)^\Delta} \tag{2}$$

where $\Delta$ represents the weight, but I don't know how to do this (that is , if it even is the right way of answering the question)

Answer 1

You can use that $\delta_{K_\nu}O(0) = 0$ if and only if the operator $O(x)$ is a primary (note that the operator is evaluated at the origin). One implication in this statement follows immediately from the formula you quoted (if a primary $O$ is evaluated at the origin, then it is annihilated by $K_\mu$) the reverse implication (i.e. if an operator evaluated at the origin is annihilated by $K_\mu$, then it is a primary) is a little less trivial but it is basically implied by the commutator between $K_\mu$ and the translation generator $P_\mu$ that can be used to move the operator from the origin to a generic point. Note that actually the definition that is typically used for "primary operator" is precisely that when evaluated at the origin it is annihilated by $K_\mu$, but in what I was saying above I was using your definition instead.

Once you convince yourself that this criterion works, you can apply it to $\Phi^p$, and use that $\Phi$ itself is a primary. So it only becomes a matter of knowing how to distribute the commutator with $K_\mu$ once you have this product of free fields at the same point (which is basically just the Leibniz rule).