Identifying gluon in Feynman diagram

[CAF123]

Consider the process in the picture below where an $r \bar r$ state goes to an $r \bar r$ state through mediation of a gluon. The gluon may carry the colour anticolour combination $r \bar r$. I'm just wondering...
1) Can we have a gluon with the colour assignments just $r \bar r$? If not, why not and if so, which of the eight combinations usually given in textbooks does it correspond to?

My thoughts are: when we speak of colour singlet states, what we really mean is the state has colour configuration $1/\sqrt{3}(r \bar r + g \bar g + b \bar b)$. So should I think of the incoming r rbar pair on the lhs of my diagram to be simultaneously including the case where we have b bar and g gbar so that we identify the intermediate gluon as the one synonomous with $(r \bar r - b\bar b -2g \bar g)$?

 

[CAF123]

Ok, I'll rephrase my question - I've seen in the literature a vertex of the form qqG where the q's carry the colour quantum number $r$ and the gluon G carries $(r \bar r - g \bar g)$. This made sense at first because there is no net colour into a vertex (red in, red out from the quarks and the gluon carries red antired e.g) but
1) Gluons are always in the colour octet representation and the vertex here shows that its coupling to the quarks did not change its colours. So the gluon being in the octet rep seems to be contradictory here, no?
2) In this vertex, can the gluon be chosen to be in the state $r \bar r - g \bar g - 2 b \bar b$ which is another of the eight allowed gluons with net colour out zero?

 

[mfb]

There is no $r \bar r$ gluon, as it would contain the singlet state. Just picking one of the two without net color charge looks arbitrary, but I'm not a theorist. One of those users will probably know more: @vanhees71, @Orodruin, @Vanadium 50, @Charles Link, @ohwilleke, @nrqed, @samalkhaiat

 

[Vanadium 50]

I'm not a theorist, but SU(3) color is a symmetry. You cannot even tell the colors of the initial state nor the final state, much less the gluon. There's no way to distinguish them, so you need to consider all allowed states in any physical process - what you wrote is unphysical, which might be why it is confusing. The two "colorless" gluons are chosen to make the SU(3) algebra work out OK. That means they have to be orthogonal, and the two states you wrote down are not. You need to flip one of the two signs for green.

 

[mfb]

Individual colors are meaningless, but "the two quarks have opposite colors" (or, better: the state is color-neutral) has a physical meaning. I guess the problem is the initial state? If we make some superposition of colors for the quarks then it could work out for the gluon.

 

[CAF123]

Individual colors are meaningless, but "the two quarks have opposite colors" (or, better: the state is color-neutral) has a physical meaning. I guess the problem is the initial state? If we make some superposition of colors for the quarks then it could work out for the gluon.

Yup I think this relates to what I was saying in my first post, when we write down r rbar for a meson for example, what we really mean is the colour state is not r rbar but a hybrid of rrbar+ggbar+bbbar which is the only colour singlet. But in a feynman diagram we pick a choice and work through the colour flows with a particular combination chosen. So, in this sense, I guess feynman diagrams can be slightly confusing but at the end of they day they are just the perturbative calculational tools.

There is no $r \bar r$ gluon, as it would contain the singlet state.

I think the rrbar gluon does exist - it's not a SU(3) colour singlet so is part of the colour octet. Indeed, it shows up in the two 'colourless' (but importantly not SU(3) singlets) combinations $r \bar r - g \bar g$ and $r \bar r + g \bar g - 2 b \bar b$. I think the reason we always have a rrbar gluon together with a ggbar or bbbar is for the same reasons above (maybe?)

I'm not a theorist, but SU(3) color is a symmetry. You cannot even tell the colors of the initial state nor the final state, much less the gluon. There's no way to distinguish them, so you need to consider all allowed states in any physical process - what you wrote is unphysical, which might be why it is confusing. The two "colorless" gluons are chosen to make the SU(3) algebra work out OK. That means they have to be orthogonal, and the two states you wrote down are not. You need to flip one of the two signs for green.

I see, yea I think what you also write is what I alluded to in my first post/ The minus sign before the ggbar was a typo, I corrected it above - I guess this means the choice of how we write down our gluons in the octet is somewhat arbitrary, like with suitable transformations of the other permissible gluons I could write rrbar-ggbar as rrbar-bbbar for example such that it too is orthogonal with all the other states.

Thanks for replies, would be nice to hear from some of the theorists too :)

 

[Orodruin]

Typically you would not write down the colour flow at all. Thinking in terms of rgb is essentially just a case of writing down simplifications that work well on the human mind. What you would do in practice would be to let the in- and out-states carry SU(3) indices and take care of the group structure with the help of the Feynman rules. As V50 said, you can never say that the in-state was $r\bar r$ or anything like that (also because it is arbitrary which direction in the SU(3) fundamental representation you call "r"). When putting everything into maths, you would not look at rgb, but at how the representations of the different states are related.

 

[Vanadium 50]

If your initial state (and final state) is supposed to be a color singlet, there can't be single gluon exchange, because the gluon is an octet. If you do this with the real color indices (and not shorthands like r, b and g), this will pop out of the calculation.

 

[CAF123]

Thanks for the responses! @Orodruin @Vanadium 50 I'd like to understand the colour constraints more at the level of the group structure that you mentioned in your posts - I put some diagrams in an attachment. On the left I show a single gluon exchange diagram and on the right a double one. I want to see through the feynman rules that indeed the left one does not permit colour singlet state in both initial and final and that the right one does.

On the left I show what I think are the relevant factors attached to the vertices - we have $t^a_{ij} t^b_{kl} \delta^{ab} = t^a_{ij} t^a_{kl}$.

On the right, basically I am thinking of a process where by a proton comes in, spits out a gluon that enters some hard interaction then a gluon reenters to ensure no colour break up of the proton, and the proton is therefore still in a colour singlet state. (Otherwise it breaks up, leading to DIS type events which I don't want to consider here). The net vertex factors are $t^a_{ij} t^b_{jk} = (t^a t^b)_{ik}$.

How to show that in the right case, I can arrange for the proton to be in a colour singlet state but in the left the initial and final states may not be? For the left diagram, a colour neutral state would be the hybrid of rrbar+bbbar+ggbar so I could demand, say that $i=j$ and $k=l$ and get a net factor of $(\text{Tr}(t^a))^2$ which is indeed zero but my problem is, $i,j,k,l$ are all fundamental representation indices so setting i=j, k=l doesn't seem to account for the fact that I want the initial state to be a antired, red say. It only implies red,red, no?

Thanks!

 

[CAF123]

The interaction vertex is $\bar \psi_j t^a_{jk} \psi_k$ , the $\psi$ accommodates an incoming quark/outgoing antiquark while the $\bar \psi$ accommodates for a outgoing quark or incoming antiquark (in the spinor decomposition over all momentum modes). If I have a quark/antiquark/gluon vertex then one of the indices on the t^a's is a fundamental index while the other is a conjugate fundamental index, while for quark/quark/gluon vertex both indices on the t^a are fundamental ones - is this the right way to think about it?

 

[vanhees71]

I'm not a theorist, but SU(3) color is a symmetry. You cannot even tell the colors of the initial state nor the final state, much less the gluon. There's no way to distinguish them, so you need to consider all allowed states in any physical process - what you wrote is unphysical, which might be why it is confusing. The two "colorless" gluons are chosen to make the SU(3) algebra work out OK. That means they have to be orthogonal, and the two states you wrote down are not. You need to flip one of the two signs for green.

QCD is the most mind-boggling part of the Standard Model. Weinberg once said: "The strong interaction is too complicated for the human mind", but this was indeed before the discovery of asymptotic freedom of (many) non-Abelian gauge theories, which makes it plausible on the basis of perturbation theory, why this kind of QFT may describe confinement, but one should keep in mind that confinement is a non-perturbative phenomenon and thus not really understood yet. However, lattice QCD indicates that QCD is indeed the correct theory describing the strong interaction, leading to the observed mass spectrum of the hadrons.

To understand, what's behind this question of color, one has to remember what's observable in QFTs: Observable in the sense of particles are asymptotic free states, i.e., strictly speaking stable particles that move freely in space and have a definite mass. The particles are further specified by intrinsic charge-like quantum numbers, describing the couplings mediated by gauge fields in the Standard Model, like electromagnetic charge and color. Then there are also other charge-like quantum numbers from "accidental symmetries" like the lepton number and the baryon number (the latter being approximately conserved, i.e., conserved under the strong interaction).

Now phenomenologically the strong interaction is confining, i.e., there are no asymptotic free states carrying color. So strictly speaking QCD diagrams with quarks and gluons as external legs are not representing physically observable S-matrix elements. What can be done in perturbative QCD is to describe the hard processes of scattering of quarks and gluons within hadrons which are colorless bound states of quark and antiquark (mesons) or three quarks (baryons). As any charge of a local gauge symmetry color must be strictly conserved, i.e., what's observable are only color-neutral states in the initial state and thus also only color-neutral states in the final state.

Concerning the question, why there are only 8 gluon states and not 9, it's mathematics. Looking at color symmetry as a global rather than a local symmetry, the mathematics of gauge fields imply that these fields have to transform according to the adjoint representation of this symmetry, and SU(3) has 8 generators, and it's a semisimple gauge group, i.e., there are no invariant Abelian subgroups. Thus the 8 gluons transform irreducible under color rotations, and thus there's no color-singlet gluon.

In contrast, looking at the color of states that are given as a Kronecker product of the fundamental representation $3$ and the anti-fundamental representation $\bar{3}$ (which are two inequivalent 3D representations of SU(3)). This refers to the color state of a system consisting of a quark and an antiquark. You can decompose the corresponding product representation $3 \otimes \bar{3}$ into irreducible representations: $3 \otimes \bar{3} = 1 \oplus 8$. The analysis shows that these are given by the color-antisymmetric singlet state, which is the color-neutral state as adequate for mesons and the totally symmetric state, which build the remaining 8 states and build the adjoint representation as for gluons, i.e., in this sense gluons have the same quantum numbers as quark-antiquark states in the color-octet representation.

 

[MathematicalPhysicist]

QCD is the most mind-boggling part of the Standard Model. Weinberg once said: "The strong interaction is too complicated for the human mind",

"The strong force is indeed strong". :-)

 

[CAF123]

@vanhees71
What is wrong with the feynman diagram that I have posted in an attachment in my OP? Basically I want to consider a colour neutral initial state and colour neutral final state. Group theory tells me that such a process cannot occur through mediation of a single gluon yet the diagram I have drawn seems to show it can happen through the gluon $r \bar r$, so can you (or anyone) point out the error in my conclusion?

 

[Orodruin]

@vanhees71
What is wrong with the feynman diagram that I have posted in an attachment in my OP? Basically I want to consider a colour neutral initial state and colour neutral final state. Group theory tells me that such a process cannot occur through mediation of a single gluon yet the diagram I have drawn seems to show it can happen through the gluon $r \bar r$, so can you (or anyone) point out the error in my conclusion?

There is no $r\bar r$ gluon. The gluon carries an 8 representation of SU(3) and color needs to be conserved at each vertex. If your in-state is color neutral - it transforms under the trivial representation (1) of SU(3) and therefore the color neutral state you started with cannot couple to a single gluon in the way you have depicted. The situation is generally more involved than just labelling each quark line with a color and every gluon line with a color and an anti-color.

 

[CAF123]

I see - so essentially the intuition behind why a single gluon exchange cannot mediate two colour neutral states is that we have an initial colour singlet state coupling to a colour octet. So there is no way to end up with a singlet in the final state too. OK?

Also, in general why is there no $r \bar r$ gluon? It's one of the 8 gluons is it not? (i.e it appears in the combinations $r \bar r - g \bar g$ and $r \bar r + g \bar g - 2b \bar b$ which are two of the eight orthogonal combinations?

 

[Orodruin]

You cannot make a pure $r\bar r$ from those combinations. The 8 representation is traceless.

 

[CAF123]

Let me see if I got this correct: If I want the initial state to be colour neutral then it must be in the state $\propto r \bar r + g \bar g + b \bar b$. If I choose one of these three combinations to represent the colour neutral state (i.e $r \bar r$ as in attachment or $g \bar g, b \bar b$) on the feynman diagram then I see there is no gluon that can couple to the vertex (it would have to be a gluon taking into account all possible colour flows $r \bar r, g \bar g, b \bar b$ into the vertex which is not a physical gluon). The only way to mediate between two colour neutral states in the tree like configuration shown is via a photon. Is it ok?

You cannot make a pure $r\bar r$ from those combinations. The 8 representation is traceless.

I see! So essentially from the interaction term $\bar \psi A \psi = \bar \psi A^a T^a \psi$, if I demand $\bar \psi = ( \bar r \, 0\, 0)$ and $\psi = ( r \, 0 \, 0)^T$ then I would need to find a T^a such that $A^a ( \bar r \, 0\, 0) T^a ( r \, 0 \, 0)^T \overset{!}{=} \bar r A^a r$. There does exist traceless T^a for this to happen (T^3 and T^8) but it doesn't represent a pure $r \bar r$ gluon.

Is that all fine?

 

[Orodruin]

If I choose one of these three combinations to represent the colour neutral state

You cannot do this. The color neutral state is $r\bar r + g\bar g + b\bar b$. It is like saying that you pick the x-component of a vector to represent it.

The only way to mediate between two colour neutral states in the tree like configuration shown is via a photon. Is it ok?

Or anything else that is color neutral - like a Z boson. Note that you could also mediate the interaction at loop level with two gluons. The intermediate state would be part of an $8\otimes 8$ representation that includes a color singlet when split into irreps.

I see! So essentially from the interaction term $\bar \psi A \psi = \bar \psi A^a T^a \psi$, if I demand $\bar \psi = ( \bar r \, 0\, 0)$ and $\psi = ( r \, 0 \, 0)^T$ then I would need to find a T^a such that $A^a ( \bar r \, 0\, 0) T^a ( r \, 0 \, 0)^T \overset{!}{=} \bar r A^a r$. There does exist traceless T^a for this to happen (T^3 and T^8) but it doesn't represent a pure $r \bar r$ gluon.

If you make that demand your in-state is not a color singlet. The color singlet is represented by the trivial representation and your $\psi \bar\psi$ is not.

 

[CAF123]

Thanks!

Ok so really what I wanted to do was to demand that my in state was the colour singlet represented by $\bar \psi = ( \bar r \,\, \bar b \,\, \bar g)$ and $\psi = (r \, b \, g)^T$ which implies the combination $\bar \psi \psi$ is now a colour singlet. Then I want to find a traceless T^a such that $A^a ( \bar r \,\, \bar b \,\, \bar g) T^a (r \, b \, g)^T$ gives an expression such that T^a may be interpreted as one of the eight gluons. But no such traceless T^a should exist. Is that ok now?

Also, I understand that the two gluon exchange allows the colour singlet-colour singlet interaction to happen but could you explain what the colour singlet $\mathbf 1$ in $\mathbf 8 \otimes \mathbf 8 = \mathbf 1 \oplus \dots$ physically means? Is it sort of like a mediation of a 8x8 state allows a transition from two singlet states which is what the 1 represents?

 

[vanhees71]

No! The expression $\bar{\psi} T^{a} \psi$ represents 8 components of a "color tensor" that transforms according to the octet representation of color SU(3) transformations. That's why you can contract it with the gluon field in the Lagrangian which consists of color-singlet (neutral) expressions as it must be to fulfill gauge symmetry.

 

[CAF123]

No! The expression $\bar{\psi} T^{a} \psi$ represents 8 components of a "color tensor" that transforms according to the octet representation of color SU(3) transformations. That's why you can contract it with the gluon field in the Lagrangian which consists of color-singlet (neutral) expressions as it must be to fulfill gauge symmetry.

Ok, thanks for mentioning this - but if I write the colour indices on the expression $\bar{\psi} T^{a} \psi$ I get $\bar{\psi}_i T^{a}_{ij} \psi_j$. In the matrix notation this is (1x3)(3x3)(3x1) type product in colour space which is just a number. So I would have thought based on that it is already a SU(3) singlet (since results in a number after contraction of colour indices).

(I know I'm wrong somewhere I just want to see where) . So given that the expression $\bar{\psi} T^{a} \psi$ transforms under octet SU(3), the questions above in response to Orodruin's message remain, where I remove the $A^a$ term now (I've written the questions below in quote):

Ok so really what I wanted to do was to demand that my in state was the colour singlet represented by $\bar \psi = ( \bar r \,\, \bar b \,\, \bar g)$ and $\psi = (r \, b \, g)^T$ which implies the combination $\bar \psi \psi$ is now a colour singlet. Then I want to find a traceless T^a such that $( \bar r \,\, \bar b \,\, \bar g) T^a (r \, b \, g)^T$ gives an expression such that T^a may be interpreted as one of the eight gluons. But no such traceless T^a should exist. Is that ok now?

Also, I understand that the two gluon exchange allows the colour singlet-colour singlet interaction to happen but could you explain what the colour singlet $\mathbf 1$ in $\mathbf 8 \otimes \mathbf 8 = \mathbf 1 \oplus \dots$ physically means? Is it sort of like a mediation of a 8x8 state allows a transition from two singlet states which is what the 1 represents?

Thanks!

 

[Orodruin]

So I would have thought based on that it is already a SU(3) singlet (since results in a number after contraction of colour indices).

It is not an SU(3) singlet. The a is an index of the 8-irrep.

 

[CAF123]

Yup, I realize it cannot be a singlet but why is my argument about the colour indices not correct?

 

[Orodruin]

Because the color singlet is not a product of any two given triplet states. It is the linear combination of the colorless products.

 

[CAF123]

Ok! So why when the expression is contracted with $A_{a}$ it results in a colour singlet? Is it similar to why when terms like $a_{\mu} b^{\mu}$ involve a contraction in the mu index results in a lorentz scalar?

 

[Orodruin]

Ok! So why when the expression is contracted with $A_{a}$ it results in a colour singlet? Is it similar to why when terms like $a_{\mu} b^{\mu}$ involve a contraction in the mu index results in a lorentz scalar?

Yes.

 

[CAF123]

Thanks! So I guess my final questions would be the ones I quoted in #21, can you please provide your comments to them?

 

[Orodruin]

Thanks! So I guess my final questions would be the ones I quoted in #21, can you please provide your comments to them?

I already did.

Because the color singlet is not a product of any two given triplet states. It is the linear combination of the colorless products.

 

[CAF123]

Ah sorry, I meant these ones, within the quote in #21. It's the same as the ones in #19 but with $A^a$ removed from the first question, in light of what vanhees71 mentioned.

1) Ok so really what I wanted to do was to demand that my in state was the colour singlet represented by $\bar \psi = ( \bar r \,\, \bar b \,\, \bar g)$ and $\psi = (r \, b \, g)^T$ which implies the combination $\bar \psi \psi$ is now a colour singlet. Then I want to find a traceless T^a such that $( \bar r \,\, \bar b \,\, \bar g) T^a (r \, b \, g)^T$ gives an expression such that T^a may be interpreted as one of the eight gluons. But no such traceless T^a should exist. Is that ok now?

2) I understand that the two gluon exchange allows the colour singlet-colour singlet interaction to happen but could you explain what the colour singlet $\mathbf 1$ in $\mathbf 8 \otimes \mathbf 8 = \mathbf 1 \oplus \dots$ physically means? Is it sort of like a mediation of a 8x8 state allows a transition from two singlet states which is what the 1 represents?

 

[vanhees71]

I think you have to study groups and group representations more thoroughly. An expression like $\psi=(r,b,g)^T$ is not a color singlet but a color triplet. It has 3 components, so how can it be a color singlet? It transforms according to the fundamental representation of SU(3). Also $\psi^*=(\bar{r},\bar{b}, \bar{g})^T$ is not a color singlet but an antitriplet, transforming with the conjugate complex fundamental transformation (which is inequivalent to the fundamental representation). This means that if $\hat{U} \in \mathrm{SU}(3)$ then you have the transformations $\psi'=\hat{U} \psi$ and $\psi^{* \prime}=\hat{U}^{*} \psi^*$. Now you have
$$\psi^{\prime \dagger} \psi' =\psi^{\prime *} \psi'=\psi^{*T} \hat{U}^{*T} \hat{U} \psi =\psi^{\dagger} \hat{U}^{\dagger} \hat{U} \psi = \psi^{\dagger} \psi,$$
because $\hat{U}$ is unitary and thus $\hat{U}^{\dagger} \hat{U}=\hat{U} \hat{U}^{\dagger} = \mathbb{1}_3$. The expressen $\psi^{\dagger} \psi$ is thus transforming under the trivial representation, i.e., it doesn't change at all under the SU(3) transformation. One also says it transforms according to the singlet representation.