3
$\begingroup$

I am currently trying to understand the representations of the conformal group. I am following the script by J. D. Qualls.

At page 29, the author finds the effect of $L_{\mu\nu}$ by "studying the Lorentz group–the subgroup of the Poincaré group that leaves the point $x = 0$ invariant", and postulates: $L_{\mu\nu}\Phi(0)=S_{\mu\nu}\Phi(0)$. He uses here the Hausdorff formula \begin{align} e^{ix^{\lambda}P_{\lambda}}L_{\mu\nu}e^{-ix^{\lambda}P_{\lambda}} &=L_{\mu\nu}-x_\mu P_\nu+x_\nu P_\mu \end{align} and then concludes that \begin{align} P_\mu\Phi(x)&=-i\partial_\mu \Phi(x) \\ L_{\mu\nu}\Phi(x)&=(i(x_\mu\partial_\nu - x_\nu \partial_\nu) +S_{\mu\nu})\Phi(x). \end{align} I am having trouble understanding the reasoning here. I tried to derive the $P_\mu$ by \begin{align} \Phi(x)&=e^{-ix^{\lambda}P_{\lambda}}\Phi(0) \\ \partial_\mu\Phi(x)&=-iP_{\mu}e^{-ix^{\lambda}P_{\lambda}}\Phi(0) \\ &\implies P_\mu=i\partial_\mu \end{align} Already here I don't find a minus in front of the derivative (for the expression of $P_{\mu}$). I tought,whether the equation $\Phi(x)=e^{-ix^{\lambda}P_{\lambda}}\Phi(0)$ had to be $\Phi(x)=e^{ix^{\lambda}P_{\lambda}}\Phi(0)$. If I do that way, I find the equation for $P_\mu$ right, but then I find $L_{\mu\nu}\Phi(x)=(-i(x_\mu\partial_\nu - x_\nu \partial_\nu) +S_{\mu\nu})\Phi(x)$ Is my calculation here wrong? If so, what is the right method?

My second question related to this topic would be the following: Some references like https://arxiv.org/abs/1602.07982 uses commutators to describe the same problem as (page 16, Poincare Representations) $$ [M_{\mu\nu},\mathcal{O}^a(0)]=(S_{\mu\nu})^a_b\mathcal{O}^b(0) $$ Why are here now commutators used? How is it related to the first description? Are they the same description, but is the Heisenberg picture hidden in the first description? I am quite new in the field theories in general and I would appreciate any help.

I am aware that there is a possible duplicate Derivation of the full generator of the Lorentz transformations, but I did not quite understand the derivations and reasoning there.

$\endgroup$
3
  • 1
    $\begingroup$ The coordinate dependence of a function always transforms under the inverse representation as the components, hence the minus sign. The commutator is basically just the Poisson bracket. The Noether charge associated with a sym. generates that sym. via the bracket. $\endgroup$ Jun 12, 2021 at 3:07
  • $\begingroup$ Your second question is the leading commutator in Hadamard's lemma. You've already been there, and your reference is crushingly explicit about it. $\endgroup$ Jun 12, 2021 at 14:38
  • $\begingroup$ Note that when operators transform as $M \mapsto U M U^{-1}$, the change of basis matrix affecting the states is $U$, not $U^{-1}$. $\endgroup$ Oct 7, 2021 at 16:32

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.