Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
  2. jcsd
  3. Jul 30, 2012 #2


    User Avatar
    Science Advisor

    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

    User Avatar
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook