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  One question about Weinberg's derivation of arbitrary spin fields expressions

+ 3 like - 0 dislike

In his book "QFT" (vol. 1) Weinberg writes the expression for an arbitrary spin massive field: $$ \hat {\Psi}_{a}(x) = \sum_{\sigma = -s}^{s} \int \frac{d^{3}\mathbf p}{\sqrt{(2 \pi)^{3}2 \epsilon_{\mathbf p}}}\left( k_{1}F_{a}(\mathbf p) \hat {a}^{\sigma}(\mathbf p )e^{-ipx} + k_{2}G_{a}(\mathbf p) {\hat {b}^{\sigma}}^{\dagger}(\mathbf p )e^{ipx} \right).$$ He writes a  very long derivation (which consists of lorentz transformation law for fields) before he concludes that $$ F_{A}^{\sigma}(\mathbf p ) = (-1)^{s + \sigma}G_{A}^{-\sigma}(\mathbf p).$$ How to prove this faster than Weinberg does it? Let's assume that equations of motion and degrees of freedom decreasing relations for corresponding field are known.

This post imported from StackExchange Physics at 2014-04-13 11:25 (UCT), posted by SE-user Andrew McAddams

asked Apr 12, 2014 in Theoretical Physics by Andrew McAddams (340 points) [ revision history ]
edited Apr 18, 2014 by Arnold Neumaier

1 Answer

+ 1 like - 0 dislike

In (5.7.1)-(5.7.15), Weinberg derives the form of a massive covariant field operator of any spin in terms of Clebsch-Gordan coefficients, but the English version available to me uses quite different notation. The relation $F^\sigma=\pm G^{-\sigma}$ mentioned in the question is an intermediate result of that calculation, based on the Lorentz covariance properties form Section 5.1.

The relation possibly follows more directly from the causality requirement used by Weinberg later to deduce (5.7.27). Just assuming the first formula above and requiring the commutator at spacelike arguments to vanish gives relations between $F$ and $G$ that must be satisfied. They should be strong enough to force the above relation and the connection between spin and statistics, but I didn't do the calculations. Treating the equal time case might be enough.

But this causality argument is unlikely to give the explicit form (5.7/14/15) of $F$ and $G$, so the more detailed calculations of Weinberg don't become superfluous.

answered Apr 20, 2014 by Arnold Neumaier (15,787 points) [ revision history ]

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