Hausdorff formula for Dilatation operator

Question

Using the Hausdorff formula I can find the dilation operator at all position: $$ e^{i x^\rho P_\rho} D e^{-i x^\rho P_\rho}=D+\left[D,-i x^\rho P_\rho\right]+\frac{1}{2}\left[\left[D,-i x^\rho P_\rho\right],-ix^\sigma P_\sigma\right]+\ldots.\tag{1} $$

But isn't the second term zero?

By definition $$D=-ix^\mu \partial_\mu, \qquad P_\mu=-i\partial_\mu\tag{2}$$ then:

$$\begin{align} \left[D,-i x^\rho P_\rho\right] &= [-ix^\mu\partial_\mu, -ix^\rho(-i\partial_\rho)] \\&= -ix^\mu\partial_\mu(-ix^\rho(-i\partial_\rho)) - -ix^\rho(-i\partial_\rho)(-ix^\mu\partial_\mu) \\ &= (-i)^3 x^\mu\partial_\mu(x^\rho(\partial_\rho)) - (-i)^3 x^\rho\partial_\rho(x^\mu\partial_\mu)\\ &= i x^\mu\partial_\mu(x^\rho\partial_\rho) - i x^\rho\partial_\rho(x^\mu\partial_\mu)\\ &=0 \end{align}\tag{3} $$

With renaming of indices, this is exactly the same. Very confused because this shouldn't be the case.

Answer 1

As @Qmechanic points out, you seem to be interested in the subalgebra of the conformal group $$ [D,P_\mu]=iP_\mu, $$ and wonder what a translation of the dilation operator D would be, the translation parameter being $x^\rho$, so the Campbell identity truncates, $$ e^{i x^\rho P_\rho} D e^{-i x^\sigma P_\sigma}=D+\left[D,-i x^\rho P_\rho\right]+\frac{1}{2}\left[\left[D,-i x^\rho P_\rho\right],-ix^\sigma P_\sigma\right]+\ldots. \\ =D- x^\rho P_\rho.\tag{1} $$

Restricting yourself to the coordinate representation (2) can only confuse you, but, indeed, the operator outcome vanishes.