2
$\begingroup$

Gluons are bicolored objects. They are made out of one color and one anticolor. Therefore, there seems to be nine possible states $r\bar{r},r\bar{b},r\bar{g},b\bar{r},b\bar{b},b\bar{g},g\bar{r},g\bar{b}, g\bar{g}$.

$1.$ Why are there only 8 gauge bosons instead of nine? What is the problem with the counting of states above?

$2.$ These color-anticolor combinations are obtained by taking the product of the fundamental (to which the quark of a three colors belong) and conjugate (to which the antiquarks of a three anticolors belong) representations of $SU(3)$ as ${\bf 3}\otimes {\bf 3^*}={\bf 8}+{\bf 1}.$ But this suggests that gluons are made of quarks and antiquarks. But I think this is wrong. What is the correct interpretation of this then?

$3.$ If these ${\bf 8}$ states represensent gluons, what does the singlet state ${\bf 1}$ represent? I mean, which particle.

| cite | improve this question | | | | |
$\endgroup$
  • 3
    $\begingroup$ May be "Gluon - Color singlet states" answers your question. $\endgroup$ – Thomas Fritsch Sep 5 '19 at 14:20
  • $\begingroup$ @ThomasFritsch Does it say that the color singlet state ${\bf 1}$ does not exist in nature? I am more disturbed question 2 though. $\endgroup$ – mithusengupta123 Sep 5 '19 at 14:30
  • 1
    $\begingroup$ Short answer: gluons don't live in $3\otimes 3^*$, they live in the adjoint representation, which is 8-dimensional. The color-anticolor thing is a quick intuitive statement, but it doesn't correspond precisely to reality. $\endgroup$ – Javier Sep 5 '19 at 15:31
  • 1
    $\begingroup$ Yes and no. That 8 is certainly the adjoint, but being in the adjoint is the more fundamental definition. It's not a consequence of that identity. $\endgroup$ – Javier Sep 5 '19 at 15:36
  • 2
    $\begingroup$ Having color and anti-color is not the same as being “made out of” color and anti-color. Gluons, being elementary particles, should not be thought of as being made out of anything. $\endgroup$ – G. Smith Sep 5 '19 at 18:13
4
$\begingroup$

1:

The answer to any question along the lines of "why is this term not present in the Standard Model Lagrangian" is always the same: because it works experimentally.
You could make a theory with 9 gluons, that would be a Lagrangian with a $U(3)$ symmetry and hence $9$ gauge fields. Then you would go to the experiment and see that you cannot adequately describe what you see.

2:

The correct interpretation is as follows.

  • You choose an $SU(3)$ symmetry for your Lagrangian. I.e. if you Lagrangian has terms like $\partial \psi \cdot ( \partial \psi)^\dagger$, you say this is invariant under $\psi \rightarrow g\psi$ where $g\in SU(3) = \mathrm{e}^{\mathrm{i}\sum_j\phi_j \lambda_j}$. $\lambda_i$ are the generators of the group, in this case known as Gell-Mann matrices. For $SU(N)$, you have $N^2-1$ generators, so $SU(3)$ has $8$ Gell-Mann matrices.

  • You choose to make this phase symmetry local, i.e. the phases $\phi_j$ are not global but a function of space $\phi_j(x)$. If you plug $g\psi$ in the Lagrangian now, you have extra terms $\propto \partial \phi_j$.

  • You requre local phase invariance, for it has been phenomenologically deemed a law of nature (read: it has never been observed to break down experimentally). For this invariance, you need to "get rid' of the spurious $\propto \partial \phi_j$ terms that spoil the local invariance.
    How? You include gauge fields $A_j$ in the original Lagrangian, which you know can undergo gauge transformations $A_j\rightarrow A_j + \partial\Lambda_j$ without affecting the physics. You then match $\Lambda_j$ and $\phi_j$ to be equal and opposite, so as to cancel out.

  • How many gauge fields do you need? As many as required to cancel all $8$ independent phase terms for $g\in SU(3)$. So $8$ gauge fields. Which we call gluons.

This is why there are $8$ gluons.
You look at experimental data, and the results fit.

2a

Why did I choose $SU(3)$, and not $U(3)$?

Short answer: experiments agree with $SU(3)$.

Long answer:

$U(3) = SU(3)\times U(1)$, i.e. the same reasoning above would apply but we'd just need another Gell-Mann matrix to generate the extra $U(1)$ symmetry. The generator for $U(1)$ is the identity. So the extra gauge field you'd need to balance this local phase is independent on the specifics of $SU(3)$, i.e. it would be the same were $SU(3)$ symmetry not to exist at all. Mathematically, we say that it would be a singlet under $SU(3)$ (more about this in 2b).

Now, particles and fields participating in the strong force $SU(3)$ have a charge associated with it, called colour. The singlet would have no colour. This is analogous to neutral particles having no electrical charge and therefore not participating in electromagnetic interactions (and vice versa).

$SU(3)$ has a particular property (due to its non-linearities) called colour confinement. The strong force increases with distance, meaning it would take infinite energy to have two coloured objects at an infinite separation. Hence, you cannot have free coloured objects. They can only be bound and hence confined.

So if $U(3)$ were correct, you'd have a gluon that is a colour singlet and that could therefore be observed free-roaming around, not tightly confined to the nucleus.
No experiment so far has seen this, hence $SU(3)$ is taken instead.

2b

The decomposition thing that you quote is now a different story.

$SU(3)$ deals with $3\times 3$ matrices. The 'S' means that you restrict these matrices only to those which have determinant $=1$.

But let's consider all $3\times 3$ matrices, i.e. constructed by the product of two $3$-vectors (denoted by bold numbers $\mathbf{3}$).

Matrices and vectors are meaningless in physics unless you specify a basis. So let's choose bases.

  • Colour basis:

red : $r = (1,0,0)$, green: $g = (0,1,0)$ and blue: $b = (0,0,1)$.

Let's take the product between a generic colour and anti-colour vector:

$$ \mathbf{3} \otimes \bar{\mathbf{3}} = \mathbf{8} \oplus \mathbf{1}, $$

which means you can decompose the representation of the LHS into two independent bits. Leaving aside all the group theory related maths, physically you interpret this as $8$ gluons and the $1$ colour singlet discussed in 2a.
And you can draw these pretty "Eightfold way" diagrams (source):

enter image description here

  • Quark basis:

up $u = (1,0,0)$, down $d = (0,1,0)$ and strange: $s = (0,0,1)$.

Again, let's take the product between a generic quark and anti-quark vector:

$$ \mathbf{3} \otimes \bar{\mathbf{3}} = \mathbf{8} \oplus \mathbf{1}, $$

which we interpret as the $\boldsymbol{\eta}'$ meson singlet, and the $\;\boldsymbol{\lbrace}\boldsymbol{\pi}^{+},\boldsymbol{\pi}^{-},\boldsymbol{\pi}^{0},\mathbf{K}^{+},\mathbf{K}^{-},\mathbf{K}^{0},\overline{\mathbf{K}}^{0},\boldsymbol{\eta}\boldsymbol{\rbrace}\;$ octet (see here for maths).

enter image description here

3

Depending on which basis you choose, i.e. on what $\mathbf{3}$ corresponds to physically, it could be a meson, a yet-experimentally-not-discovered gluon etc.

| cite | improve this answer | | | | |
$\endgroup$
0
$\begingroup$

There is no 9th "gluon" in standard model, as the number of generators in the $su(N)$ algebra is $N^2 -1 = 9-1 = 8$ for $N=3$.

However, in the context of beyond standard model with gauge group $$U (3)*SU(2)*U(1)_R, $$ (note that the right-handed isospin $U(1)_R$ gauge is only applicable to the right-handed fermions) we could have a 9th "gluon", which corresponds to the $U(1)_{B-L}$ in the decomposition of $$ U(3)=SU(3)*U(1)_{B-L}. $$ The 9th "gluon" can be heuristically written as $$ \frac{r\bar{r}+b\bar{b}+g\bar{g}}{\sqrt3}. $$ So it can be regarded as the linear combination of 3 gluons in the OP's parlance. The charge of this 9th "gluon" is baryon number B (1/3 for quarks) minus lepton number L (1 for leptons).

The combined gauge group of $$ U(1)_{B-L}*U(1)_R $$ is spontaneously broken at the GUT (Majonara mass) energy scale. There are two leftovers from the GUT-scale spontaneous symmetry breaking:

  • One combination ($Z'$ boson) of the $U(1)_{B-L}$ and $U(1)_R$ gauge fields acquires GUT scale mass, thus short-ranged.
  • The other combination of the $U(1)_{B-L}$ and $U(1)_R$ constitutes the hypercharge $U(1)_Y$ gauge field, which in turn combines with the $W^0$ component of the electroweak $SU(2)$ gauge field to form the electromagnetic gauge field via the electroweak Higgs mechanism. In other words, some relic of the "9th gluon" is long-ranged, manifested as being a portion of the electromagnetic force.

Note added: This color-neutral $U(1)$ 9th gluon is also mentioned in the answer of @Luboš Motl to a similar question: Do color-neutral gluons exist?.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ I didn't understand the answer. The strong interaction is described by color SU(3), and the decomposition $3\times 3^*=8+1$ means we have an octet that transforms like an irreducible representation of SU(3) and another like a singlet of SU(3). I don't know where you get a U(3) from. I think you are mixing up flavor SU(3) with color SU(3). $\endgroup$ – mithusengupta123 Sep 5 '19 at 15:31
  • $\begingroup$ @mithusengupta123, as I mentioned in the answer, the discussion is in the context of BSM (with gauge group $U(3)*SU(2)*U(1)$). We don't have a 9th "gluon" in SM, right? $\endgroup$ – MadMax Sep 5 '19 at 15:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.