3
$\begingroup$

I want to get a better understanding on particle being an irreducible representation. Does that mean one particular type of particles (say particle $A$) is a subspace of the "total" Hilbert space $H$ (which contains all the types?), and the restriction (to the vector space $V_A$ corresponds to particle $A$) of the representation of the Lorentz transformation group (which govern all the transformation of all types of particles) which has no proper subrepresentation.

Or is the space always stay as $H$ no matter what type of the particle we are talking about, then it is just a matter of choosing different representation on $H$, and different representations mean different particles? In another word, I am confused what vector space are we talking about if we say particle $A$ being an irreducible representation. is it the entire $H$? or some subspace $V_A$? Because we say spin $\frac{1}{2}$ corresponds to the two dimensional representation, I am unsure what is two dimensional.

$\endgroup$
3
  • $\begingroup$ Whether the irreducible representation is "the entire $H$" depends on your definition of "entire"! What is the context here? Are we doing non-relativistic QM, Fock spaces in QFT, something else? $\endgroup$ – ACuriousMind Nov 1 '20 at 12:03
  • $\begingroup$ QFT. I guess the question is "are we sticking with one vector space (which is the so-called "the entire $H$") and choosing different representation (on the same vector space) or we implicitly change the vector space when talking about another type of particles". $\endgroup$ – Tea_de Nov 1 '20 at 12:12
  • 1
    $\begingroup$ What is the difference between "changing the representation" and "changing the vector space", given that all the infinite-dimensional Hilbert spaces featuring in QFT as unitary representations of the Lorentz group are isomorphic as vector spaces anyway? This recent question seems relevant. $\endgroup$ – ACuriousMind Nov 1 '20 at 12:22
3
$\begingroup$

I think it's easiest to understand this from the bottom up. Suppose you have a vector in some Hilbert space which describes a state where there is just one particle in a pure momentum state. This particle is described by a momentum vector and has some internal spin degrees of freedom. If you act on this vector with elements of the Poincare group, you'll get new vectors in the Hilbert space, describing states where the original particle has been rotated, boosted, and translated. If you do this for every element of the Poincare group, the vectors you get will span a subspace of the Hilbert space. This subspace is going to be an irreducible representation. It has everything you need to fill out a representation, but -- if the particle has no substructure -- it doesn't have anything else. You can't change the invariant mass by boosting it, and you can't alter the spin.

So that corresponds more or less to your option 1. The Poincare irrep for a given particle is a subspace of the Hilbert space of all states.

Wigner built on this intuition and classified all the unitary, positive energy, discrete mass representations of the Poincare group. For massive particles of spin $j$ in 4d, you can build these representations by taking the direct sum of a whole bunch of copies of the spin $j$ irrep of $SU(2)$ (thought of as the double cover of the rotation group). The direct sum is over all the momentum states with a given invariant mass. For spin $1/2$, the $SU(2)$ irrep is $2$-dimensional, but the Poincare irrep is a sum of infinitely many copies of this, one for each momentum vector you can reach by Poincare actions. So the Poincare irrep is infinite-dimensional.

$\endgroup$
1
  • $\begingroup$ Thank you very much. $\endgroup$ – Tea_de Nov 2 '20 at 9:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.