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

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 $$p$$.

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)

• Did you try to use the Leibniz rule satisfied by $\delta$ and $K_{\nu}$? – Blazej Apr 27 at 10:15
• I have never used the Leibniz rule. Just searched for it. Does that mean I should integrate $(1)$? I don't understand how that would relate to the power $p$, or how I would preform the integration in this case. I'm sorry for being this lost and thank you so much for your help. – user256673 Apr 27 at 10:45

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).