# Color factors of color -- octet scalars

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.

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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?

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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.

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fzero
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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.
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}##.

Oh, thanx, I'll work out it again..

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}.
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:

do you agree?

fzero
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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:

do you agree?
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.

a term Tr(S¯0S0S0) which is definitely proportional to dABC
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 ..

fzero
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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 ..
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##.

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 :)

fzero
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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 :)
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.

Ok that is great. But what did you mean by:
## \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
I think this term is symmetric in all color indices, also ## d^{abc } ## is that..

fzero
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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..
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.

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.

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fzero
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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 ..
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##.

I think this point is crucial because it explains why in new physics we get the symmetric structure constant .
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.

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 ?
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.

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.
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] ## .

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.

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 ?

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fzero