I've been studying conformal field theory (CFT) and got the following "apparent" inconsistency. Let's take dilation ($D$) and translation ($P_\mu$, 4-momentum) generators that according to CFT are such that
$$
[D, P_\mu] = iP_\mu,\quad [D, \Phi(0)] = i\Delta\Phi(0),\quad \Delta \in \mathbb{R} \tag1
$$
Now we are looking for $[P_\mu, \Phi(x)]$ and $[D, \Phi(x)]$ for an arbitrary field $\Phi(x)$ (it may be scalar, spinor, vector, etc). Let's start with the first one and use the Heisenberg picture to make explicit the $x$-dependence of $\Phi$:
$$
\partial_\mu\Phi(x) = \partial_\mu e^{iPx}\Phi(0)e^{-iPx} = i[P_\mu, \Phi(x)] \implies [P_\mu, \Phi(x)] = -i\partial_\mu\Phi(x) \tag2
$$
Let's go now to the 2nd commutator:
$$
[D, \Phi(x)] = De^{iPx}\Phi(0)e^{-iPx} – e^{iPx}\Phi(0)e^{-iPx}D = e^{iPx}([\hat{D}, \Phi(0)])e^{-iPx} \tag3
$$
with
$$
\hat{D} = e^{-iPx}De^{iPx} = D – ix^\mu[P_\mu, D]\ \mbox{(obtained by Taylor expansion of exp.)}\tag4
$$
Now, using (1) and (4) into (3), the final result is
$$
[D, \Phi(x)] = i(\Delta + (x\partial))\Phi(x) \tag5
$$
But this contradicts first commutator in (1). Because of (5), $D = i(\Delta + (x\partial))$ whose commutator with $P_\mu = -i\partial_\mu$ is $[D, P_\mu] = [i(\Delta + (x\partial)), -i\partial_\mu] = -i(-i\partial_\mu) = -iP_\mu$
Isn't this a contradiction?
Best Answer
Let $Q_{\xi}$ be a charge that generates some symmetry $\delta_\xi$. Then, $$ [ Q_\xi , \Phi ] = \delta_\xi \Phi . $$ Then, \begin{align} [ [ Q_\xi , Q_{\xi'} ] , \Phi ] &= [ Q_\xi , [ Q_{\xi'} , \Phi ] ] + [ [ Q_\xi , \Phi ] , Q_{\xi'} ] \\ &= [ Q_\xi , \delta_{\xi'} \Phi ] + [ \delta_\xi \Phi , Q_{\xi'} ] \\ &= \delta_{\xi'} [ Q_\xi , \Phi ] - \delta_\xi [Q_{\xi'} , \Phi ] \\ &= \delta_{\xi'} \delta_\xi \Phi - \delta_\xi \delta_{\xi'} \Phi \\ &= - [ \delta_\xi , \delta_{\xi'} ] \Phi \end{align} Note that the commutator of charges generates the commutator of transformations with an extra minus sign! This resolves your contradiction.
Let me add in a bit of information regarding symmetries and transformations. The finite version of a symmetry transformation is $$ U \Phi U^{-1} = R(U^{-1}) \Phi $$ where $R(U)$ is the representation of $U$ under which $\Phi$ transforms. Note that the argument of $U$ on the RHS is $U^{-1}$ and not $U$. To understand this, we consider the transformation $$ (U_1U_2) \Phi (U_1 U_2)^{-1} = R( (U_1 U_2 )^{-1} ) \Phi $$ Alternatively, \begin{align} (U_1U_2) \Phi (U_1 U_2)^{-1} &= U_1U_2 \Phi U_2^{-1} U_1^{-1} \\ &= U_1 R(U_2^{-1}) \Phi U_1^{-1} \\ &= R(U_2^{-1}) U_1 \Phi U_1^{-1} \\ &= R(U_2^{-1}) R(U_1^{-1}) \Phi . \end{align} It follows that, $$ R(U_2^{-1} U_1^{-1} ) = R(U_2^{-1}) R(U_1^{-1}) $$ or equivalently, $R(U_1U_2)=R(U_1)R(U_2)$. In other words, $R$ satisfies the same group multiplication law as $U$ so it is indeed a representation.
We can now relate this to infinitesimal transformations. $$ U = \exp [ Q_\xi ] , \qquad R(U) = \exp [ R (Q_\xi ) ] = \exp [ - \delta_\xi ] . $$ Note that $- \delta_\xi$ is the representation of $Q_\xi$ (not $+\delta_\xi$). Substituting this and expanding to first order in $\xi$, we find $$ [ Q_\xi , \Phi ] = + \delta_\xi \Phi . $$ The extra minus sign between $Q_\xi$ and $\delta_\xi$ also implies that the algebra of the charges $Q_\xi$ has an extra sign compared to the algebra of $\delta_\xi$, \begin{align} [Q_{\xi_1} , Q_{\xi_2} ] &= Q_{\xi_3} \\ \implies R ( [Q_{\xi_1} , Q_{\xi_2} ] ) &= R ( Q_{\xi_3} ) \\ \implies [ R ( Q_{\xi_1} ) , R ( Q_{\xi_2} )] &= R ( Q_{\xi_3} ) \\ \implies [ -\delta_{\xi_1}, -\delta_{\xi_2} ] &= -\delta_{\xi_3} \\ \implies [ \delta_{\xi_1}, \delta_{\xi_2} ] &= -\delta_{\xi_3} \\ \end{align}