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Color factors of color -- octet scalars

  1. Jun 25, 2015 #1
    Hi all,

    Has any one an idea how can we derive the color factor ## C_2## , eq. 27 in
    http://authors.library.caltech.edu/8947/1/GREprd07.pdf ?

    ## C_2 ## belongs to the second Feynman diagram in fig. 3 which includes two gluons and 4 octet scalars .

    Best.
     
  2. jcsd
  3. Jun 25, 2015 #2

    fzero

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    I don't recall any specific identities (other than symmetries) that would help. You can just look up the structure constants https://en.wikipedia.org/wiki/Structure_constants#SU.283.29 and compute the sums explicitly.

    Edit: I think you meant derive the color factor, rather than evaluate it. Have you tried deriving the Feynman rules for the theory described in the paper?
     
    Last edited: Jun 25, 2015
  4. Jun 25, 2015 #3
    Let me ask the question in other words, returning to the three gluon vertex, that why the color factor accompanied it is ## f^{abc} \simeq Tr ([T^a,T^b] T^c )## and not
    ## d^{abc} \simeq Tr (\{T^a,T^b\} T^c )## ? I mean why ## T^a## and ## T^b## has been subtracted and not added to each other..

    I just try to revise the QCD Feynman rules first to understand this new theory.
     
    Last edited: Jun 25, 2015
  5. Jun 25, 2015 #4

    fzero

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    OK, I will outline it, neglecting factors of ##i## and such that you should fill in depending on the conventions that you're using. The YM field strength is

    $$F_{\mu\nu} = \partial_{[\mu} A_{\nu]} + [ A_\mu, A_\nu],$$

    so the 3-gluon vertex comes from the cross-terms ## \text{Tr}(\partial_\mu A_\nu [ A^\mu, A^\nu]) \sim \text{Tr}(T_A [T_B,T_C])##. We could also have worked this out another way, in which we write

    $$F_{\mu\nu}^A T_A = \partial_{[\mu} A_{\nu]}^A T_A + A_\mu^B A_\nu^C [ T_B, T_C].$$

    Since ##[T_B,T_C] = f_{BCA} T^A##, we find that

    $$F_{\mu\nu}^A = \partial_{[\mu} A_{\nu]}^A + A_{B\mu} A_{C\nu} f^{ABC},$$

    which also leads to the vertex being proportional to ##f_{ABC}##.

    Finally, it might be useful for your computations with the octet to note that since the octet is the adjoint representation of ##SU(3)##, the generators themselves can be identified with the structure constants through ## (T_A)_{BC} = f_{ABC}##.

    I have a rough idea how that color factor comes out, but there is a slightly nontrivial calculation of the ##S-S-S## vertex that I would like you to try to do to verify that it is proportional to ##d_{ABC}##.
     
  6. Jun 25, 2015 #5
    Oh, thanx, I'll work out it again..
     
  7. Jun 26, 2015 #6
    Also you can see in this paper that not only S-S-S vertex can be proportional to ## d_{ABC}##, but also the amplitude of gluon-gluon decay into a single octet scalar via fermion loop as in eq. 27, ## C_1##, the first Feynman digram in fig. 3 ..

    I think here the reason for that not the Feynman rules because we have no for example g-g-S vertex, but the amplitude is proportional to something like : ## T^a T^b T^c+ T^b T^a T^c ##, as the answer in this thread:

    https://www.physicsforums.com/threads/color-factor.818519/

    do you agree?
     
  8. Jun 26, 2015 #7

    fzero

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    Yes, but the reason for the factors of ##T_A##s is because they appear in the Feynman rules for the gqq and Sqq vertices. For the gSS vertex we get a factor of ##f_{ABC}##.

    From the potential (10), we get, for example, a term ##\text{Tr} (\bar{S}_0 S_0 S_0)## which is definitely proportional to ##d_{ABC}##. I think the other terms also give this factor, so the SSS vertex probably gives a factor of ##d_{ABC}## as well. Then the middle diagram has an amplitude proportional to ##d_{CFG} f_{AEF}f_{BGE}##.

    If we look at the ggSS term in the Lagrangian, we have ##\text{Tr}( [A_\mu,S_i]^2)## which gives a factor of ##f_{AFE}f_{BEG}##, so the third diagram is also proportional to ##d_{CFG} f_{AEF}f_{BGE}##. The total amplitude for the sum of all diagrams will take the form

    $$ \mathcal{A} = M_1 d_{ABC} + M_2 d_{CFG} f_{AEF}f_{BGE}.$$

    Then when we square the amplitude, we get ##C_1## from the square of the 1st term, ##C_2## from the cross term, and ##C_3## from the square of the last term. The only thing that puzzles me is why the complex conjugate of ##\lambda_{4,5}## does not appear in their expression.
     
  9. Jun 26, 2015 #8
    Hi fzero,

    I wonder why Tr (SSS) gives ## d^{abc}## ? it's proportional to Tr (T^a T^b T^c) . I think here it used the commutation of the scalar fields: ## S^a S^b = S^b S^a ##.

    Other conventions may consider :

    $$ T^a T^b = \frac{1}{2N}\delta^{ab}+\frac{1}{2}d^{abc}T^c +\frac{1}{2}if^{abc}T^c, $$

    and then ## Tr (T^a T^b T^c) = \frac{1}{4} (d^{abc}+i f^{abc}) ##. So now which convention is correct, because I think the Feynman rule of S-S-S vertex should be the same ..
     
  10. Jun 26, 2015 #9

    fzero

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    The term from the Lagrangian that I wrote is ##\text{Tr}(T_A T_B T_C) \bar{S}_0^A S^B_0 S^C_0##, so this picks the terms from the trace that are symmetric in ##BC##. This is precisely the ##d_{ABC}## term in the product. There is also a term ##\text{Tr}(T_A T_B T_C) \bar{S}_+^A S^B_- S^C_0## that it seems evaluates the same way if we consider the real and imaginary parts of ##S_\pm##.
     
  11. Jun 26, 2015 #10
    What I understand that in this term ## \text{Tr} ( T^a T^b T^c) S^a S^b S^c ## , the three indices a,b and c are symmetric, because the scalar fields commute..

    I think you mean some thing like ## \text{Tr} ( T^a T^b T^c) = T^a_{ij} T^b_{jk} T^c_ {ki} + T^c_{ik} T^b_{kj} T^a_{ij} =\text{Tr} ( \{T^b, T^c \} T^a)##, if we take the color loop in two directions.

    But when ## \text{Tr} ( T^a T^b T^c) \sim d^{abc} ## and when the last identity I listed in the previous reply should be used?

    Thanx :)
     
  12. Jun 26, 2015 #11

    fzero

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    Your identity ## Tr (T^a T^b T^c) = \frac{1}{4} (d^{abc}+i f^{abc}) ## is fine. Let me show in detail what I said in words before. Apply your identity to

    $$\text{Tr} ( T^a T^b T^c) S^a S^b S^c = \frac{1}{4} (d^{abc}+i f^{abc}) S^a S^b S^c.$$

    The second term has an antisymmetric tensor times a symmetric product so it has to vanish. I'm not saying that the trace gives ##d_{abc}## but that the sum with the fields only gives the ##d_{abc}S^a S^b S^c## term. The actual calculation is slightly more complicated since we have an ##SU(2)## index to track, but I suggested why it continues to work as in this example.
     
  13. Jun 26, 2015 #12
    Ok that is great. But what did you mean by:
    I think this term is symmetric in all color indices, also ## d^{abc } ## is that..
     
  14. Jun 26, 2015 #13

    fzero

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    If you break into real and imaginary parts you get terms like ##S^A_{0,R}S^B_{0,R}S^C_{0,I}## that are only symmetric in pairs of color indices. But that is enough for the argument.
     
  15. Jun 27, 2015 #14
    Hi fzero,

    Ok, thanks again, hope not to be so long question for a simple argument..

    But I were think that the scalar fields in general commute, so I'll be grateful if you have a reference notes that the real and imaginary parts of a scalar do not commute with each other or with the charged components ..
    I think this point is crucial because it explains why in new physics we get the symmetric structure constant .

    A last notice that, it seems the scalar and gluon fields don't commute, since gSS has ##f^{ABC}##, which can appear in the octet 's covarieant derivative, or from Tr( T T T) by using my identity .. I mean it's not right a scalar do commute ( like numbers) with any other field, as it commutes with fermion field . But how did we know that? did octet scalars treated with gluon field just as another gluon field with the same Lie Algebra ?


    Cheers.
     
    Last edited: Jun 27, 2015
  16. Jun 27, 2015 #15

    fzero

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    Let's look at a pair of terms in

    $$\begin{split} f_{ABC}S^B_{0,R}S^C_{0,I} &= f_{A12} S^1_{0,R}2^C_{0,I} + f_{A21} S^2_{0,R}1^C_{0,I} + \cdots \\
    &= f_{A12} ( S^1_{0,R}2^C_{0,I} - S^2_{0,R}1^C_{0,I}).\end{split} $$

    Because of the different ##R,I## labels, there is no cancellation. Suppose that we had both ##R## components, then we would have ##S^1_{0,R}2^R_{0,I} - S^2_{0,R}1^R_{0,I}=0##.

    As I said, we only need symmetry in a pair of indices for the ##f## term to drop, we don't need it in all three.

    The gSS vertex comes from the kinetic term for ##S## including the gauge covariant derivative, specifically the terms like ##\partial S^\dagger_i [A_\mu,S_i]##, so we explicitly get ##f_{ABC}## when we compute the color trace.
     
  17. Jun 27, 2015 #16
    Actually I tried that in two different ways. As the kinetic term is given by ## (D^\mu S)^\dagger (D_\mu S) ##, with ## D_\mu =\partial_\mu + ig A^a_\mu ##, so first:
    $$ \mathcal{L}_{gs} = (\partial^\mu - ig A^{\mu } ) S^\dagger (\partial_\mu + ig A_\mu) S, $$
    Now the gSS terms are: ## ig ~ ( \partial^\mu S^\dagger A_\mu S - A^{\mu } S^\dagger \partial_\mu S ) ##,
    which looks as two different terms, also what 'll be the right color indices to have the right color structure for gSS or for ggSS

    Second if we start by:
    $$ D_\mu S^a = \partial_\mu S^a + ig A^b_{\mu } S^c f^{abc}, ~~ \to (2)$$

    then the gSS terms come directly as : ig ## \partial^\mu S^{a \dagger} A^b_{\mu } S^c f^{abc} ##, and
    ggSS term: ## g^2 A^b_{\mu }~ S^c A^{\mu~ d} ~ S^e ~f^{abc}~ f^{ade},##

    but in (2) we started by ##A^b## and ## S^c ## as they are commutators from the beginning, I mean as ## [A_\mu,A_\nu] ## .
     
  18. Jun 28, 2015 #17

    fzero

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    You can derive (2) from your original expression by noting that S is in the adjoint representation, so that ##S = S^a T_a##, where ##(T_a)_{bc}=f_{abc}##. This was the reason I wrote the expression with the commutator explicitly; since it represents the adjoint action of the gauge group on S.
     
  19. Jun 28, 2015 #18
    I agree that ## (T_a)_{bc}=f_{abc}##, where a,b,c =1,...,8, but still some thing unclear, as you said ##S = S^a T_a##, not ##(T_a)_{bc}##, i mean from where the extra b,c indices have come ? is it a good question ! also as the gauge boson ## A_\mu = A_\mu^a T_a ## in the adjoint representation, why we didn't substitute by another ## f_{abc}## instead of the A's generator ..
    Have i missed some thing in the SU(3) Algebra ?
     
    Last edited: Jun 28, 2015
  20. Jun 28, 2015 #19

    fzero

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    Because of the color charge (and weak isospin), ##S## is a matrix. When we write ##S=S^aT_a##, we let the generators ##T_a## carry the color indices. In the fundamental representation of ##SU(3)## (like for quarks) the generators are 3x3 matrices, but for the adjoint (for the gauge bosons and these scalars), the generators are 8x8 matrices. It is typical to suppress the matrix indices whenever they can be inferred by context. For the gluons, we would choose ##(T_a)_{bc} =f_{abc}## if we had a calculation that required us to specify the generators explicitly. For instance, you could use the explicit form of the generators to derive the ggg vertex.
     
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