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I SU 2 and SU 3

  1. Jul 25, 2016 #1
    Hi, I'm recently reading some text on particle physics and there is a section on symmetries and group theory. It gave the definition of SU2 as the group of unitary 2*2 matrices and that SU3 is the group of unitary 3*3 matrices. However, it kind of confuses me when it mentioned representations of higher orders. What's the difference between a 3*3 representation of SU2 and SU3? Also, I don't really understand what it means when it mentioned something like "invariant under SU2 transformations", can anyone give an example of a vector that's invariant under SU2 transformations?

    Thanks!
     
  2. jcsd
  3. Jul 25, 2016 #2

    fresh_42

    Staff: Mentor

    A representation is a mapping from the given group into a group of (regular, invertible) transformations of a vector space. In mathematical terms: A representation ##(G,V,φ)## of a group ##G## is a vector space ##V## together with a group homomorphism ##φ: G \longrightarrow GL(V).##
    So the number you mentioned, "##3 \times 3## representation" refers to the dimension of the vector space (here ##3##), not to the group! Thus it has nothing to do with whether you consider ##SU(2)## or ##SU(3)##. An invariant vector ##v## under ##SU(2)## transformation means, that ##φ(X)(v) = v## for all ##X \in SU(2)##, here unitary ##2 \times 2##-matrices with determinant ##1##. The mapping ##φ## in this context is often omitted and the equation is noted ##X.v = v## or ##v^X = v##. Things become a bit messy if the vector space ##V## itself is a vector space of (not necessarily regular, since ##0 \in V##) matrices.

    An example for a representation of ##SU(n)## would be ##φ: SU(n) \longrightarrow GL(\mathfrak{su}(n))## where ##φ: u \longmapsto uAu^{-1}## for ##u \in SU(n) \, , \, A \in \mathfrak{su}(n).##
    It shouldn't be too difficult to find invariant vectors here or in a simplier representation ##V##.

    One last remark: A representation ##(G,V,φ)## is often simply called by "##G## operates on ##V##".
     
  4. Jul 25, 2016 #3
    Thanks a lot! So I guess I've mixed up the definitions.

    But for the invariant vector question, I think that's what I thought initially until I saw a line stating that (1, 1, 1) is invariant under SU3 transformations. Actually, it stated η = (u(ubar) + d(dbar) + s(sbar))/√3 is invariant under SU3. This really confuses me.
     
  5. Jul 25, 2016 #4

    fresh_42

    Staff: Mentor

    I'm not sure here, what ##u,d,s## are. Unipotent, diagonal, symmetric matrices? And I haven't generators of ##SU(3)## in mind to verify that ##(1,1,1)## is invariant under ##SU(3)## by its natural representation (matrix multiplication / application on ##\mathbb{C}^3##) ##u.(1,1,1) = (u_{1i},u_{2i},u_{3i})##, i.e. all the row sums of ##u## should be equal to ##1##. Seems wrong to me, so either it's another representation on ##\mathbb{C}^3## or the diagonal matrix ##\mathbb{1} = (1,1,1)## is meant, which is of course invariant under ##SU(3)##.
     
  6. Jul 25, 2016 #5
    Actually, since I'm reading a particle physics text, the u d s refers to quarks. Maybe I should raise this on the physics section.

    Thanks for the reply!
     
  7. Jul 25, 2016 #6
    Oh crap... Just found that I've got wrong idea about that issue...

    Thanks again, the explanation is very nice and clear!
     
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