Why Must $w(Q_n) = -w(Q_m)$ for $Z_{nm} \neq 0$ in Weinberg III?

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SUMMARY

The discussion centers on the necessity of the condition $w(Q_n) = -w(Q_m)$ for the non-zero scalar $Z_{nm}$ in Weinberg III. The user derives this condition by analyzing the commutator $\{Q_n, Q_m\}$ using equation 32.1.5 and the Poincare algebra, ultimately leading to the equation $(-w(Q_n) + w(Q_m))Z_{nm} = 0$. The conclusion drawn is that for $Z_{nm} \neq 0$, the weights of the charges $Q_n$ and $Q_m$ must be opposites, which is critical for maintaining consistency in the definitions of $J_{d1}$ and the weights involved.

PREREQUISITES
  • Understanding of Poincare algebra
  • Familiarity with Weinberg III, particularly equations 32.1.1, 32.1.2, and 32.1.5
  • Knowledge of commutators in quantum field theory
  • Concept of weight in the context of quantum charges
NEXT STEPS
  • Study the implications of the Poincare algebra on quantum charges
  • Review the derivation of equation 32.1.5 in Weinberg III
  • Explore the concept of weight in representation theory
  • Investigate the role of scalar quantities in quantum field theory
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The discussion is beneficial for theoretical physicists, particularly those specializing in quantum field theory and representation theory, as well as graduate students seeking to deepen their understanding of Weinberg's work.

Si
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Hi, and thanks in advance for reading this. I've been bashing my head on my desk for days on this now.

My problem is the first sentence of the paragraph after equation 32.1.5 in Weinberg III:

"... note that for a given $Z_{nm}$ to be non-zero, since it is a scalar all of the $\sigma$s in Eq. (32.1.1) must be opposite for $Q_n$ and $Q_m$."

This is not obvious to me. To prove it I calculate the commutator of $\{Q_n,Q_m\}$ with e.g. $J_d1$ using Eq. (32.1.5) and the Poincare algebra, then use Eq. (32.1.5) to eliminate $\{Q_n,Q_m\}$. Then from the resulting equation I extract the weight $w=0$ terms (where $w$ is defined in Eq. (32.1.2)) to get $(-w(Q_n)+w(Q_m))[Z_{nm} +\sum_{i=2}^{d-1} \Gamma^i_{nm}P_i]=0$. The second term vanishes if I rotate the 1-direction to point in the spatial part of $P$. If this was OK, the rest is easy: $(-w(Q_n)+w(Q_m))Z_{nm}=0$ so for $Z_{nm}\neq 0$ to be possible we need $w(Q_n)=-w(Q_m)$. But this is not OK, because a rotation of the 1-direction changes the definition of $J_{d1}$, the $Q_n$ and the $w(Q_n)$.

So how do I prove that for $Z_{nm}\neq 0$ to be possible we need $w(Q_n)=-w(Q_m)$?

Thanks in advance for any help.
 
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Maybe I wrote too much... quite simply, can anyone explain the first sentence of the paragraph after equation 32.1.5 in Weinberg III?