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I ##SU(2)## generators in ##1##, ##2## and ##3## dimensions

  1. Mar 16, 2017 #1
    The ##{\bf su}(2)## Lie algebra in a representation ##\bf R## is defined by

    ##[T^{a}_{\bf R},T^{b}_{\bf R}]=i\epsilon^{abc}T^{c}_{\bf R},##

    where ##T^{a}_{\bf R}## are the ##3## generators of the algebra.

    In ##2## dimensions, these generators are the Pauli matrices

    ##T^{1}_{\bf 1} = \frac{1}{2}\begin{pmatrix}0 & 1\\ 1 & 0 \end{pmatrix}, \qquad T^{2}_{\bf 1} = \frac{1}{2}\begin{pmatrix}0 & -i\\ i & 0 \end{pmatrix}, \qquad
    T^{3}_{\bf 1} = \frac{1}{2}\begin{pmatrix}1 & 0\\ 0 & -1 \end{pmatrix}.##

    In ##3## dimensions, these generators are

    ##T^{1}_{\bf 2} = \frac{1}{\sqrt{2}}\begin{pmatrix}0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix}, \qquad T^{2}_{\bf 2} = \frac{1}{\sqrt{2}}\begin{pmatrix}0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end{pmatrix}, \qquad
    T^{3}_{\bf 2} = \begin{pmatrix}1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}.##

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    1. How can you derive the generators in ##2## and ##3## dimensions?

    2. What are the generators in ##1## dimension?
     
  2. jcsd
  3. Mar 17, 2017 #2

    Orodruin

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    1. The two-dimensional representation is the fundamental representation. You can find out what the matrices J in its Lie algebra are by writing ##e^{-itJ}## and requiring that this matrix is part of SU(2). This will give you a set of matrices of which you can pick a basis.

    For the three-dimensional irrep, it is the symmetric part of the tensor product representation. You can construct the generators explicitly by constructing a basis for the tensor product space and checking how the representations act on this basis.

    2. I will give you a hint: All 1x1 matrices commute so [A,B]=0 for any A and B.
     
  4. Mar 24, 2017 #3
    So, are the matrices all equal to (0) for the one-dimensional representation?
     
  5. Mar 24, 2017 #4

    Orodruin

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    Yes. It is the trivial representation.
     
  6. Mar 24, 2017 #5
    It is also called non-faithful. Is this because every element of ##SU(2)## is being mapped to the same element, that is, ##1##?
     
  7. Mar 24, 2017 #6

    Orodruin

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    The three-dimensional representation is not faithful either - but it is not trivial. It maps ##A## and ##-A## to the same matrix.

    The trivial representation by definition maps all group elements to the identity.

    Also, you have a mistake it the third three-dimensional generator above - it should not be the identity.
     
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