Maybe this does it:

$\begin{array}{rccl}
\textrm{Translation:}&P_\mu&=&-i\partial_\mu\\
\textrm{Rotation:}&M_{\mu\nu}&=&i(x_\mu\partial_\nu-x_\nu\partial_\mu)\\
\textrm{Dilation:}&D&=&ix^\mu\partial^\mu\\
\textrm{Special Conformal:}&C_\mu&=&-i(\vec{x}\cdot\vec{x}-2x_\mu\vec{x}\cdot\partial)
\end{array}$

Then the commutation relation gives:

$[D,C_\mu] = -iC_\mu$

so $C^\mu$ acts as raising and lowering operators for the eigenvectors of the dilation operator $D$. That is, suppose:

$D|d\rangle = d|d\rangle$

By the commutation relation:

$DC_\mu - C_\mu D = -iC_\mu$

so

$DC_\mu|d\rangle = (C_\mu D -iC_\mu)|d\rangle$

and

$D(C_\mu|d\rangle) = (d-i)(C_\mu|d\rangle)$

But given the dilational eigenvectors, it's possible to define the raising and lowering operators from them alone. And so that defines the $C_\mu$.

P.S. I cribbed this from:

http://web.mit.edu/~mcgreevy/www/fall08/handouts/lecture09.pdf

This post imported from StackExchange Physics at 2014-07-28 11:17 (UCT), posted by SE-user Carl Brannen