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Let's have a definition of massive particle as an irreucible representation of the Poincare group. Then, let's have a spinor field $\psi_{\alpha \alpha_{1}...\alpha_{n - 1}\dot {\beta} \dot {\beta}_{1}...\dot {\beta}_{m - 1}}$, which is equal to $\left( \frac{m}{2}, \frac{n}{2}\right)$ representation of the Lorentz group. There is the hard provable theorem:

$\psi_{\alpha \alpha_{1}...\alpha_{n - 1}\dot {\beta} \dot {\beta}_{1}...\dot {\beta}_{m - 1}}$ realizes irreducible representation of the Poincare group, if $$ (\partial^{2} - m^{2})\psi_{\alpha \alpha_{1}...\alpha_{n - 1}\dot {\beta} \dot {\beta}_{1}...\dot {\beta}_{m - 1}} = 0, $$ $$ \partial^{\alpha \dot {\beta}}\psi_{\alpha \alpha_{1}...\alpha_{n - 1}\dot {\beta} \dot {\beta}_{1}...\dot {\beta}_{m - 1}} = 0. $$ Can this theorem be interpreted as connection between fields and particles?

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The definition is that a particle in Minkowski space is a unitary irreducible representation of the Poincare group. So one needs to see how various P.D.E.s are related to the classification of unitary irreducible representations of $iso(3,1)$ or $iso(d-1,1)$ in the case of $d$-dimensions instead of $4$.

Note that these are all the Poincare-invariant constraints that can be imposed on the given field without trivializing the solution space (one could imposed $\partial \psi=0$ (gradient), which is Poincare-invariant but too strong as the field must be a constant).

The theorem is not hard to prove. One has to know how to construct irreducible representations of the Poincare group, see chapter 2 of the Weinberg's QFT textbook. Then one solves the equations by standard Fourier transform and shows that the solution space indeed equivalent to what is called a spin-$m$ particle in Minkowski space.

There is nothing special about $4d$ in defining spin-$m$ field, so it is simpler to look at arbitrary dimension, where, say for bosons the above equations are equivalent to

$(\square-m^2)\phi_{\mu_1...\mu_m}=0$

$\partial_\nu \phi^{\nu \mu_2...\mu_m}=0 $

$ \eta_{\nu\rho} \phi^{\nu\rho \mu_3...\mu_m}=0$

$\phi^{\mu_1...\mu_s}$ is totally symmetric in all indices.

In $4d$ one can use $so(3,1)\sim sl(2,C)$ and the last algebraic constraint then trivializes - an irreducible spin-tensor is equivalent to an irreducible $so(3,1)$-tensor

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  • $\begingroup$ "...The theorem is not hard to prove...", - I didn't read this part of Weinberg book and only used spinor representation of the Lorentz group (without using some equations). After completing the proof I can construct field equation from this equivalence for cases of arbitrary spin. $\endgroup$ – user8817 Aug 26 '13 at 19:16
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    $\begingroup$ Just one comment, the correspondence between reasonable P.D.E.'s and particles is not one-to-one. One and the same particle can be desribed in many different ways. For example, a spin-one, photon, can be described with the help of gauge potential $A_\mu$ or field strength $F_{\mu\nu}$ $\endgroup$ – John Aug 27 '13 at 7:34
  • $\begingroup$ Because there are three representations of the spin 1: $\left( 1, 0 \right), \left( 0, 1\right), \left( \frac{1}{2}, \frac{1}{2}\right)$. $\endgroup$ – user8817 Aug 27 '13 at 8:02
  • $\begingroup$ it is even worse, there is infinitely many ways to describe a given particle, it can sit as a subrepresentation. I did not understand if I answered your question above or not? $\endgroup$ – John Aug 27 '13 at 18:00

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