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Tensor operators in QM

  1. Jul 30, 2012 #1
    Hello! I´m trying to read Georgi's book on Lie algebras in particle physics but am confused about the start of chapter 4.

    Georgi writes that "A tensor operator is a set of operators that transforms under commutation with the generators of some Lie algebra like an irreducible representation of the algebra. [...] A tensor operator transforming under the spin-s representation of SU(2) consists of a set of operators, O^s_l, for l=1 to l = 2s+1(or -s to s), such that:


    [tex] [J_a, O^s_l] = O^s_m[J^s_a]_{ml}"[/tex]

    I thought I understood 90-95% sofar in the book but I really dont see what he tries to define here.. Could somebody maybe help me and introduce the concept in Georgis way but with some more words and an example i will recognize from QM? I dont recognize what he is trying to construct from my QM courses.
     
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  3. Jul 30, 2012 #2

    Bill_K

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    In QM an example of a vector operator is the position operator x, a set of three operators x, y, z. Together they form a vector because under rotations they transform into each other. This fact can be expressed by writing out their commutators with the operators Ji that generate infinitesimal rotations:

    [Ji, xj] = i εijk xk

    Another example of a tensor operator is the electric quadrupole moment, which is a set of five operators Qm, m = -2, ... +2. Given a charge distribution ρ(x,y,z), you define its electric quadrupole moment by

    Qm = ∫ρ r2 Y2m dV

    where Y2m is a spherical harmonic. Under rotations the five operators Qm go into each other, and one can express this by writing out all the commutators of Ji with Qm.
     
    Last edited: Jul 30, 2012
  4. Jul 30, 2012 #3
    Of course! Didnt think of the three dimensional position operator as a set of operators. :)
    The equation of transformation however doesnt seem to intuitive.. how should I think about the indices in the equation I qouted above?

    For x_j, j is obviously the dimension of our space so for O^s_l, l specifies a operator in our set? What about s, is it just refering to which representation? :/
     
  5. Aug 2, 2012 #4

    king vitamin

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    I've been studying this book too so someone can (and should) correct any misunderstandings/misstatements below.

    The index s labels the representation. For SU(2) as Georgi defines his indices, s=0,1/2,1,... labels the representation. Since the dimension of these representations is 2s+1, then the lower indices on the operators run from 1 to 2s+1.

    In the example of the position operator, it clearly transforms under the s=1 representation, or the adjoint representation. Notice that i ε_ijk is exactly how he defines the adjoint representation, in terms of the structure constants of SU(2) in chapter 2.

    For a more intuitive (imo) introduction to tensor operators, check out chapter 3.10 in Sakurai. It shows how transforming under the commutator as Georgi defines it is equivalent to V -> (J^-1)VJ.
     
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