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Normalization of SU(N) group

  1. Mar 3, 2013 #1
    I am reading my textbook of QFT (Maggiore, Modern Introduction in QFT), and there is this statement:

    "If [itex] T^a_R [/itex] is a representation of the algebra and V a unitary matrix of the same dimension as [itex] T^a_R [/itex] , then [itex] V T^a_R V^\dagger [/itex] is still a solution o the Lie algebra and therefore provides an equivalent representation. We can fix V requiring that it diagonalizes the matrix [itex] D^{ab}(R) ≡ Tr (T^a_R T^b_R) [/itex], so that [itex] Tr (T^a_R T^b_R) = C(R) \delta^{ab} [/itex]."

    I can't understand the second sentence, matrix D has different dimensions than V, how can it be used to diagonalize? Putting V within the trace doesn't make any sense, it would give unit matrix.

  2. jcsd
  3. Mar 3, 2013 #2


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    No you're right, there's no connection between the first and second sentences. It can't be the same V.

    The correct statement is that Dab = Tr(TaTb) is a symmetric matrix, and so there exists a linear transformation on the group indices a which brings it to a multiple of the identity.
  4. Mar 3, 2013 #3
    Just a side question: The similarity transformation on particular representation generates all possible representations of the same dimensionality? If it is true then no way to tinker with representation of a given dimension to render D diagonal, right? But in that case how do you normalize it?
  5. Mar 3, 2013 #4


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    Georgi's book on group theory treats this normalization issues in great detail in, if I remember correctly, chapter 2.
  6. Mar 3, 2013 #5


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    The similarity transformation generates equivalent representations of a given set of generators Ta. But you can also reparametrize SU(N), leading to a different set of generators T'a with different structure constants (but they still generate the same group.) The relationship between Ta and T'a is a linear transformation, T'a = Cba Tb where Cba is a constant matrix. It's this change of basis, Cba which gives you the freedom to diagonalize Dab.
  7. Mar 3, 2013 #6
    Thank you, Bill, I understand now.

    I got Georgi's book, it seems to be a great book for a physicist. Thanks, haushofer.
  8. Mar 9, 2013 #7
    Just one more question that i could not find in Georgi's book (probably don't look close enough). How do we prove that for fundamental representation C(R) = 1/2 for SU(N)? what difference "fundamental representation" makes? Can't we just choose any factor with appropriate linear transformation?

    Georgi's 2nd chapter deals with normalization of the adjoint representation mostly.

    Thanks again.
  9. Mar 22, 2013 #8
  10. Mar 24, 2013 #9


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    I believe you can get them by playing around with traces. You need to use the fact that the fundamental representation is n x n, so the n2-1 generators along with I form a complete set, and any n x n matrix can be expressed as a linear combination of these. Applying this:

    TaTb ≡ αab I + βabc Tc

    Choosing the generators to obey the orthonormality condition Tr(TaTb) = ½ δab gives you αab = (1/2n) δab. Next,

    Tr(TaTbTc) = Tr((αab I + βabd Td)Tc) = ½ βabc.

    This gives you the fact that βabc is cyclically symmetric, which means it is of the form ifabc + dabc where f is totally antisymmetric and d is totally symmetric.

    Finally, look at Tr(TaTbTcTd). Reduce this two ways: (1) Tawith Tb, then Tc with Td, and (2) Tbwith Tc, then Tawith Td should give you the expression for dacedbce.

  11. Mar 24, 2013 #10
    thanks, Bill!

    One thing that really bothers me is how do we get the one half in the exponential of the representation: [itex] e^{\frac{i}{2} \alpha_c T^c} [/itex]? It is very important to clearly understand why special unitary condition yealds this 1/2.

    For SU(2) with straightforward algebraic derivation it is easy to see, but how about the general case?
  12. Mar 24, 2013 #11


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    It's a question of normalization of the generators, and what convention you use. For example one of the generators of SU(3) is λ3:
    for which Tr(λ3λ3) = 2. This is NOT the convention used in the Wikipedia page. Rather they use T3 = λ3/2, for which Tr(T3T3) = 1/2. So you can write the finite rotation as Wikipedia would: exp(iαT3), or as exp(iαλ3/2) with the 2 present.
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