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SU(2) adjoint repr

  1. Jan 18, 2014 #1


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    Well I am trying to understand the adjoint representation of the su(2) algebra.
    We know that the algebra is given:
    [itex] [X_{i}, X_{j}]= ε_{ij}^{k} X_{k} [/itex]
    (maybe I forgot an [itex]i[/itex] but I am not sure).

    The adjoint representation is then ( in the matrix representation) defined by the [itex]ε_{ijk}[/itex] structure constants, via the identification [itex]X_{i}= [ε_{i}]_{j}^{k}[/itex]. Correct? Because by that we have:
    [itex] (adX_{i})^{k}_{j}= ad X_{i} X_{j}|_{X_{k}} = [X_{i}, X_{j}]|_{X_{k}}=[ε_{i}]_{j}^{k} [/itex]

    Now begins my question/problem. The matrices of [itex][ε_{i}]_{j}^{k}[/itex] are of dimension [itex]j_{max}\times k_{max}[/itex] so equal to the number of generators [itex]X_{i}[/itex].

    The su(2) algebra has [itex]n^{2}-1=4-1=3[/itex] generators, so the adjoint representation can be seen as [itex]3\times3[/itex] matrices, so it naturally acts on 3 component vectors. That means that the adjoint representation is 3 dimensional representation and so it is the spin J=1. (?)
    Is there another way to prove the last sentence? Eg using the existing isomorphism between su(2)-so(3) algebras? And one more question, what happens with the spin-1/2? The spin 1/2 representation (if my logic was correct) could only exist if the number of generators are 2!
    Can someone help? :confused::blushing:
  2. jcsd
  3. Jan 18, 2014 #2


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    Yes. The adjoint rep. is isomorphic to the standard weight =1 representation.

    You don't need it. Just write down the 3 generators of each representation and compare them. The basis of the rep. space is (1,0,0), (0,1,0) and (0,0,1).

    No. The dimension (weight) of the representation has to do with the dimension of the vector space on which the representation is built. Actually spin/weight 1/2 rep. is isomorphic to the fundamental representation. Both are built on C2.
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