Connection between particles and fields and spinor representation of the Poincare group - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-10-22T21:21:44Z https://physics.stackexchange.com/feeds/question/75321 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/75321 4 Connection between particles and fields and spinor representation of the Poincare group user8817 https://physics.stackexchange.com/users/0 2013-08-26T13:05:22Z 2013-08-26T17:58:06Z <p>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: </p> <p>$\psi_{\alpha \alpha_{1}...\alpha_{n - 1}\dot {\beta} \dot {\beta}_{1}...\dot {\beta}_{m - 1}}$ 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?</p> https://physics.stackexchange.com/questions/75321/-/75353#75353 4 Answer by John for Connection between particles and fields and spinor representation of the Poincare group John https://physics.stackexchange.com/users/22770 2013-08-26T17:58:06Z 2013-08-26T17:58:06Z <p>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 $4$.</p> <p>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). </p> <p>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.</p> <p>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 </p> <p>$(\square-m^2)\phi_{\mu_1...\mu_m}=0$</p> <p>$\partial_\nu \phi^{\nu \mu_2...\mu_m}=0$</p> <p>$\eta_{\nu\rho} \phi^{\nu\rho \mu_3...\mu_m}=0$</p> <p>$\phi^{\mu_1...\mu_s}$ is totally symmetric in all indices.</p> <p>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</p>