Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lorentz invariant theory, irreducible representations

  1. Jun 20, 2012 #1
    "In a Lorentz invariant theory in d dimensions a state forms an irreducible representation under the subgroups of [itex]SO(1,d-1)[/itex] that leaves its momentum invariant."

    I want to understand that statement. I don't see how I should interpret a state as representation of a group. I have learned that representations of groups are maps which assign a linear transformation between vector spaces (or matrix) to each group element. But why should a state correspond to such a mapping? I am completely lost here.

    I would be happy if someone could clear this up for me.
  2. jcsd
  3. Jun 20, 2012 #2

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    Yet another example of physicists mixing up the language. States in a QFT are elements of the vector space that is acted upon by representations of some group. For example, scalars are acted upon by the trivial representation, spinors are acted upon by the spin-1/2 representation, etc.
  4. Jun 21, 2012 #3


    User Avatar
    Science Advisor

    physicus, it means that there is an infinite-dim., reducible rep. of the Poincare group defined by a family of unitary operators U[θa] (where θa are the group parameters) acting on the full Hilbert space H, which can be decomposed into irred. reps., corresponding to subspaces of H which are classified w.r.t. to their Casimir invariants PαPα and WαWα (with eigenvalues m² and m²s(s+1) in the massive case m²>0; refer to Wigner's classification for more details).

    In theoretical physics one often says that such a (finite dim.) subspace of H is a rep., whereas is mathematics one says that a rep. acts on a (finite dim.) vector space.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook