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Tensor techniques in $3\otimes\bar 3$ representation of su(3)

  1. Sep 4, 2013 #1
    Hi everyone!

    I would like to ask you a very basic question on the decomposition [tex]3\otimes\bar 3=1\oplus 8[/tex] of su(3) representation.

    Suppose I have a tensor that transforms under the 8 representation (the adjoint rep), of the form [tex]O^{y}_{x}[/tex]
    where upper index belongs to the $\bar 3$ rep and the lower ones to the 3 rep (x,y=1,2,3).
    Now, I know that under $T_a$ (an element of the Lie algebra) this tensor transforms as
    [tex](T_a O)^y_x\equiv (T_a)_k^y O_x^k-(T_a)_x^k O_k^y[/tex]

    And that's by definition.

    But it is also true that, for any s-dimensional representation of a Lie algrebra I should have
    [tex][T_a,\mathcal O^{s}_i]=(T_a^{s})_{ij}\mathcal O_j[/tex]
    where -s≤ i ≤s, as an operatorial equation.
    This means that if we take the an operator that transforms under the adjoint rep (s=8) we should get
    [tex][T_a,\mathcal O^{adj}_i]=(T_a^{\text{adj}})_{ij}\mathcal O_j[/tex]
    [tex](T_a^{\text{adj}})_{ij}\equiv -if^{aij}[/tex]
    are the matrix element of the adjoint representation of the su(3) Lie algebra, and the f^{abc}'s are the structure constant.

    My problem is that I can't explicitly constuct the correnspondence between the tensorial approach and the operatorial one, that is
    [tex](T_a)_k^y O_x^k-(T_a)_x^k O_k^y\leftrightarrow (T_a^{\text{adj}})_{ij}\mathcal O_j=-if^{aij}\mathcal O_j [/tex]

    In other words, I cannot find an identification
    [tex] O^x_y\leftrightarrow \mathcal O_i[/tex]
    where again (x,y=1,2,3), (i=1,8) that could make the algebra strcture constants appear into
    [tex](T_a)_k^y O_x^k-(T_a)_x^k O_k^y[/tex]
    to check that the tensor actually transforms in the adjoint representation.

    I know that this may seem a rather cumbersome question, but it is in fact a ground question in group theory.
    Please tell me if I have not been clear enough!!

    Thanks for your time!
  2. jcsd
  3. Sep 4, 2013 #2


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    What you want is a set of eight 3 x 3 matrices such that Oi = (λi)xy Oxy. I believe you will find these are the so-called Gell-Mann matrices.
  4. Sep 4, 2013 #3


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    When dealing with su(2) the decomposition of a tensor representation into a direct sum of irreducible representations involves the Clebsch-Gordan coefficients. If you'll recall how one works that out, it involves identifying what's called the highest (or lowest) weight states and acting with the lowering (or raising) operators on both sides of the equivalence of representations. See this older post of mine where I explain in more words.

    For other Lie algebras, there is an analogous procedure, but the mechanics is a bit more complicated, since a general algebra has more structure than su(2). Some relevant keywords are Cartan-Weyl basis, roots, and weights. The details are discussed in Georgi or any more math-oriented text on representations of Lie algebras.

    The basic idea here is that one identifies the maximal subalgebra of mutually commuting elements, called the Cartan subalgebra. These matrices can be simultaneously diagonalized and are labeled ##H_i## in the Cartan-Weyl basis, where ##i=1,\ldots r##, and ##r## is called the rank of the algebra. The Cartan subalgebra is isomorphic to ##u(1)^{\oplus r}##.

    For su(2), the Cartan subalgebra is usually chosen to be generated by ##J_z##. In su(2), this corresponds to a u(1) subalgebra and there is an associated charge, which in the case of angular momentum corresponds to the "magnetic" quantum number ##m_j##.

    The remaining generators of the algebra are labeled ##E_\alpha## and can be chosen to satisfy

    $$[H_i, E_\alpha] = \alpha_i E_\alpha.~~~~~(*)$$

    The basis of generators ##\{ H_i, E_\alpha\}## is called the Cartan-Weyl basis. The collection of eigenvalues ##\alpha_i## can be considered vectors, called the root vectors or just roots, and define a lattice. A point in the lattice corresponds to a particular assignment of charges.

    For su(2), the ##E_\alpha## are simply ##J_\pm## and the root lattice is one-dimensional.

    For a general representation of the algebra, a given state in the representation will satisfy

    $$ H_i |w\rangle = w_i |w\rangle,$$

    where we now the eigenvalues ##w_i## again form a vector which we now call a weight vector, or just weight. For the adjoint representation, we can see from the relation (*) that the weights will correspond to the roots. If we apply (*), we also see that

    $$ H_i E_\alpha |w\rangle = (w_i + \alpha_i) E_\alpha |w\rangle,$$

    so the ##E_\alpha## can still be thought of as raising and lowering operators for a general algebra.

    So the decomposition of the tensor product will follow in an analogous way to the familiar procedure in su(2). Identify the ##H_i## and ##E_\alpha##. For an explicit description, it will be clear which of the ##E_\alpha## are raising and which are lowering operators. We must identify the highest and lowest weight states in each description of the representation (these states are annihilated by some ##E_\alpha##. We identify these states up to normalization on both sides. Then we can work out the rest of the states by acting with the ##E_\alpha##. We will find certain linear combinations of states map from the tensor product to the direct sum representation. By exhausting the list of states, we can work out the 1-1 correspondence.

    Again, I'm pretty sure Georgi does most of this explicitly in at least one example. If not, it would be a great exercise for you to work through.
  5. Sep 5, 2013 #4
    Thank you both for your quick and kind replies!
    Anyway I am still a bit confused..

    So the correspondence is simply [itex]O_i=\lambda_{xy}O_{xy}=Tr(\lambda_i^T O)[/itex] as you suggest? I still cannot verify the correspondences explicitly starting from

    Moreover the [itex]T_a[/itex]s above are already the su(3) fundamental representation i.e. the Gell-Mann matrices! Where am I mistaking?

    Thanks again for your kindness!
  6. Sep 5, 2013 #5


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    Looks to me like you've got it all right, but you just have to decide how your quantity O is going to transform, and not be surprised that there's more than one choice.

    a) If it's a matrix Oxy, then like you said, it's (Ta)yk Okx − (Ta)kx Oyk where (Ta)kx are the generators in the 3 representation, namely Ta = λa/2.

    b) If, on the other hand, Oa is a vector in the 8 representation, then its transform is (like you said) −ifabc Oc.

    c) On the other hand, if O is going to be regarded as both, Oaxy, then you'll get transform terms from all three indices. But in fact, these terms will cancel out, because in this case Oaxy is a fixed linear combination of the (λa)xy's, and the λ's are invariant (once the basis is chosen), them being just a set of Clebsch-Gordan coefficients. The cancellation can also be proved using the identity [λi, λj] = i fijk λk.

    The fact that the transforms in (c) cancel shows that (a) and (b) are equivalent!
    Last edited: Sep 5, 2013
  7. Sep 8, 2013 #6
    Thanks a lot Bill_K, now I see that the two ways of transformation are not mutually exclusive!
    I have taken some time to work to the bottom of this matter, but I cannot still set my mind at ease..

    Can you be a little more specific on how the total transformation of [itex]O^a_{xy}[/itex] looks like?

    Thanks again for your time, and please excuse me if I keep on insisting on this topic!
  8. Sep 8, 2013 #7


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    Ok, the quantity O can either be regarded as a 3x3 matrix Oxy or a vector quantity in 8 dimensions Oi (but not both!). Apply an infinitesimal group transformation:

    a) As an 8-vector, Oi → i fijk Ok.
    b) As a 3 x 3 matrix, Oxy → ½ λjxz Ozy - ½ λjyz Oxz.

    The relationship between the two forms is Oi = λixy Oxy, so from (a) we have λixy Oxy → i fijk λkxy Oxy, and from (b) we have λixy Oxy → ½ λixy λjxz Ozy - ½ λixy λjyz Oxz.

    Now use the identity i fijk = [λi, λj] λk and the orthonormality of the λ's and see that these two things are the same. This shows the two forms for O are equivalent under the group transformations.
  9. Sep 8, 2013 #8
    Thanks a lot pal, you have been very kind and helpful!

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