Antiparticles of Standard Model gauge bosons

[PeterDonis]

From a recent thread:

All elementary bosons other than the W boson are their own anti-particle.

Is this true of gluons? Doesn't the color charge invert under CPT? (For example, a red-antigreen gluon's antiparticle would be a green-antired antigluon.)

 

[Drakkith]

Is this true of gluons? Doesn't the color charge invert under CPT? (For example, a red-antigreen gluon's antiparticle would be a green-antired antigluon.)

Honestly I don't know. I somehow forgot gluons had color charge when I posted earlier.

 

[PeterDonis]

Honestly I don't know.

On thinking about it, the case of gluons may be weirder than I had imagined it was. The example I gave was a red-antigreen gluon, which under CPT inversion would (I think) become a green-antired antigluon. But there is also a green-antired gluon. Is that the same as a "green-antired antigluon"? If so, then the distinction between "gluons" and "antigluons" isn't as clear-cut as I had implied in my previous post. But if not, then there must be some other quantum number that is different in a green-antired gluon, vs. a green-antired antigluon, and I have no idea what it would be.

 

[PeterDonis]

The W boson consists of a W+ and W- boson, which are a each others anti-particle,

On thinking it over some more, I'm not sure this is actually true. The two W bosons usually referred to are actually not the "fundamental" bosons of the weak force; they appear after electroweak symmetry is broken (along with the Z and the photon $\gamma$). The "fundamental" electroweak bosons are usually given as $W^1$, $W^2$, $W^3$, and $B$, and the "usual" bosons are linear combinations of these, as follows:

$$
W^+ = \frac{1}{\sqrt{2}} \left( W^1 + i W^2 \right)
$$

$$
W^- = \frac{1}{\sqrt{2}} \left( W^1 - i W^2 \right)
$$

$$
Z = \cos \theta W^3 - \sin \theta B
$$

$$
\gamma = \sin \theta W^3 + \cos \theta B
$$

where $\theta$ is the Weinberg angle. None of these are really CPT conjugates of each other; I think they all work out to be CPT conjugates of themselves, which would make them all their own "antiparticles", like the photon. (The gluons, because of the color charge, wouldn't exactly be CPT conjugates of themselves, but they wouldn't occur in nice particle-antiparticle pairs either.)

There's probably a slicker way of expressing all this in terms of group representations; I think the magic words might be something like "real" representations (particles which are their own antiparticles) vs. "complex" representations (distinct particles and antiparticles).

 

[mfb]

It is more complicated with gluons. There are 8 different gluons, but the way to list them is a bit arbitrary, so the question "which one is the antiparticle of which one" gets arbitrary as well. I don't think that this topic is the right place to go into more detail, however. "The antiparticles of gluons are gluons as well" is true in all cases.

 

[Drakkith]

None of these are really CPT conjugates of each other; I think they all work out to be CPT conjugates of themselves, which would make them all their own "antiparticles", like the photon. (The gluons, because of the color charge, wouldn't exactly be CPT conjugates of themselves, but they wouldn't occur in nice particle-antiparticle pairs either.)

Considering this thread is labeled B, would it be correct at this level to say the W^- and W⁺ are the antiparticle of each other even if it's more complicated when you get into the details?

 

[PeterDonis]

would it be correct at this level to say the W- and W+ are the antiparticle of each other even if it's more complicated when you get into the details?

I'm not sure, because I'm not sure that saying they are antiparticles is correct even at the basic level. The fact that they have opposite electric charges and equal masses is not enough by itself; they would have to bear a specific relationship to each other at the QFT level and I'm not sure whether they do or not.

 

[Drakkith]

I'm not sure, because I'm not sure that saying they are antiparticles is correct even at the basic level. The fact that they have opposite electric charges and equal masses is not enough by itself; they would have to bear a specific relationship to each other at the QFT level and I'm not sure whether they do or not.

Hmmm. I can find a lot of sources that say that the W+ and W- are antiparticles of each other, but they aren't exactly high-quality, detailed sources except for possibly the last source.

From here at PF: https://www.physicsforums.com/threads/do-gauge-bosons-have-anti-particles.520402/#post-3444123

2. Yes, all gauge bosons have antiparticles. The photon is its own antiparticle. The W+ and W− are antiparticles of each other. The Z is its own antiparticle. A "red x anti-blue" gluon is the anti-particle of an "anti-red x blue" gluon, and similar for the other colors. The two diagonal gluons are each their own anti-particle.

From here: http://www.whillyard.com/science-pages/weak-nuclear-force.html

The W- boson is the anti-particle of the W+, while the neutral Z boson is its own anti-particle.

From here: http://www.benbest.com/science/standard.html

W bosons can be positive (W+) or negative (W−), each being the antiparticle of the other.

From here: https://books.google.com/books?id=ELkIp2-Po3YC&pg=PA12&lpg=PA12&dq=antiparticle+of+w%2B+boson&source=bl&ots=gAulGIYalR&sig=xVHCGYzCT5ZTtkqGqnBtNnMWk4k&hl=en&sa=X&ved=0CDkQ6AEwBTgKahUKEwiPv5vuoarHAhVQfpIKHdhnDhs#v=onepage&q=antiparticle%20of%20w%2B%20boson&f=false

The W boson contains the W⁺ and W^-, and they are each other's anti-particle.

I couldn't find any sources saying otherwise in my quick search, but if you find any, please let me know.

 

[PeterDonis]

Here is a source that describes the "fundamental" particles (i.e., before electroweak symmetry breaking) in the standard model:

http://math.ucr.edu/home/baez/qg-spring2003/elementary/

Note, first, that in the table of particles, the gauge bosons don't have "and their antiparticles" noted, while the other particles (Higgs and fermions) all do.

Second, this note near the bottom of the page:

The representations of SU(3) × SU(2) × U(1) are representations on real vector spaces for the gauge bosons, but on complex vector spaces for the Higgs boson and fermions.

For any complex vector space

there is a conjugate vector space [PLAIN]http://math.ucr.edu/home/baez/qg-spring2003/elementary/img83.gif, which we get by making

into a complex vector space in a different way, such that multiplication by $i$ now acts as what used to be multiplication by $-i$. Similarly, for any complex representation of a group on http://math.ucr.edu/home/baez/qg-spring2003/elementary/img82.gif we get a representation of this group on [PLAIN]http://math.ucr.edu/home/baez/qg-spring2003/elementary/img83.gif, called the conjugate representation.

In physics, conjugate representations describe antiparticles. In the above chart, wherever we write "and its antiparticle", we mean that in addition to the particle corresponding to the representation shown, there is also a particle corresponding to the complex conjugate of this representation.

This is what I was dimly remembering when I posted earlier about real vs. complex representations. This clearly implies that all of the gauge bosons, since they are real representations, are their own antiparticles.

Now I just need to dig enough into the QFT behind all this to understand in more detail what it means.

 

[Drakkith]

This clearly implies that all of the gauge bosons, since they are real representations, are their own antiparticles.

I'll take your word for it. I might as well be reading braille with a wooden hand.

Now I just need to dig enough into the QFT behind all this to understand in more detail what it means.

I think I'll wait to do that until after I take calc 2, EM theory, and the rest of my classes.

 

[PeterDonis]

Here is a source that describes the "fundamental" particles (i.e., before electroweak symmetry breaking) in the standard model

The "before electroweak symmetry breaking" is key, though. If you look at the expressions I posted earlier in the thread for the W+, W-, Z, and photon in terms of W1, W2, W3 and B, it's clear that the W+ and W- are complex conjugates of each other, whereas the Z and photon are both real. So it looks like electroweak symmetry breaking changes the representations of the electroweak gauge bosons, so that W+ and W- become antiparticles even though all of the electroweak bosons before symmetry breaking are their own antiparticles. So I now agree that W+ and W- are antiparticles.

I still don't see how this affects the gluons, though; they are clearly real representations in the Baez article, and electroweak symmetry breaking doesn't change them at all, so it would seem that they are all their own antiparticles.

 

[Vanadium 50]

The W+ and W- are antiparticles of each other. The w1-w2-w3 representation pre-symmetry breaking is different than W+ and W- because there are extra fields (the w3) and missing fields (the Higgs). Here the w1's antiparticle is a linear combination of w2 and w3. This combination is determined by the SU(2) algebra of the theory.

 

[PeterDonis]

Here the w1's antiparticle is a linear combination of w2 and w3.

According to the Baez article, as I'm reading it, the W1, W2, W3, and W0 (which I called B) are all their own antiparticles; they are all real representations. The W+ and W- are each other's antiparticles because the way their representations are constructed as linear combinations of W1 and W2 makes them complex conjugates of each other. The Z and photon are real linear combinations of W3 and W0/B, so they are both their own antiparticles.

Also, just for completeness since you mentioned the Higgs, the Higgs representation, before symmetry breaking, contains the H+ and H0 and their antiparticles, which I'll call H- and H0'. After symmetry breaking, as I understand it, the H+ is "eaten" by the W+ to give it mass, the H- is "eaten" by the W- to give it mass, some linear combination of H0 and H0' is "eaten" by the Z to give it mass, and the remaining degree of freedom (which will be some other linear combination of H0 and H0') is the Higgs particle the LHC is detecting.

 

[Drakkith]

If you look at the expressions I posted earlier in the thread for the W+, W-, Z, and photon in terms of W1, W2, W3 and B, it's clear that the W+ and W- are complex conjugates of each other, whereas the Z and photon are both real.

Why is one particle being the complex conjugate of the other important? (vs 'being real')

 

[Avodyne]

Antiparticles are only defined when there is a conserved U(1) symmetry. Then, if there is a particle which is charged under the symmetry, there must be another particle with the same mass with equal and opposite charge: this is the "antiparticle". In the Standard Model, the U(1)'s that are conventionally used to define antiparticles are electric charge, baryon number, and lepton number. With this convention, the W- is the antiparticle of the W+, and all other gauge bosons (including gluons) are their own antiparticles (because they are all neutral under the three U(1)'s).

But if we think of the Standard Model before spontaneous symmetry breaking, then we have the hyperchage U(1) instead of the electromagnetic U(1). Then it is more natural to use hyper charge to define antiparticles. With this definition, the W1, W2, and W3 bosons are their own antiparticles.

 

[samalkhaiat]

The gauge fields of any gauge group transform by the adjoint map, i.e., you belong to the real adjoint representation of the group. For example the $W^{\mu}_{a}$ of $SU(2)$ belong to the representation space $[3] = [\bar{3}]$ and the gluons $G^{\mu}_{a}$ of $SU_{c}(3)$ belong to the real adjoint representation $[8] = [\bar{8}]$. Charge conjugation operator acts not on the real gauge fields but on the complex fields which represent the gauge bosons $W^{\mu}_{a} \tau^{a}$.

 

[PeterDonis]

Why is one particle being the complex conjugate of the other important? (vs 'being real')

Because that's how the distinction between "having a distinct particle and antiparticle" vs. "a particle being its own antiparticle" appears in the math.

 

[Vanadium 50]

According to the Baez article, as I'm reading it, the W1, W2, W3, and W0 (which I called B) are all their own antiparticles

They can't be. Can we agree that a necessary condition for A and B to be antiparticles is that there is a combination A+B that has the same quantum numbers as the vacuum? So an e+ and e- can be antiparticles because they can combine in a neutral spin singlet, which has the same quantum numbers as the vacuum. (They can also combine to form a spin triplet, which does not, but that doesn't matter to this argument. An electron and a proton cannot be antiparticles, because while they can combine in a neutral spin singlet, this combination has net baryon number and net lepton number, unlike the vacuum.

The w1, w2 and w3 form an SU(2) triplet. Combining two elements produces a singlet, a triplet and a quintuplet. That's just the SU(2) algebra.

If w1 was its own antiparticle, the w1w1 combination needs to be a singlet. Since which field we call "w1" is arbitrary (that's what SU(2) symmetry means, after all), that means w2w2 and w3w3 have to be singlets as well. That's three singlets, but we know there is only one. So we know that can't be the answer.

We actually know a bit more: we know w1w1, w2w2 and w3w3 are all in the quintuplet. So they are manifestly not singlets.

The symmetry conditions requires that there be an equal amount of w2 and w3 in the antiparticle of w1. I think a fair conclusion (although not the one I would draw) is that the w1 has no antiparticle, as its antiparticle is an admixture of multiple fields.

A key point touched upon earlier is that in U(1), charge is a number. Finding the anticharge is easy. In non-Abelian theories, like SU(2), charge is a matrix. The anticharge is the inverse of this matrix, but the matrix is not guaranteed to be invertible, and even if it is, the resulting matrix is not guaranteed to be an element of that representation of the group. In the SU(2) case, it's not. In SU(3) color, it is: so each of the eight gluons has another member of the octet that is its own antiparticle.

 

[PeterDonis]

Can we agree that a necessary condition for A and B to be antiparticles is that there is a combination A+B that has the same quantum numbers as the vacuum?

This and the rest of your post seems plausible, but it's leading to answers that contradict what I'm getting from other sources (including other posters in this thread as well as the Baez article), so I'm going to defer further comment until I've dug into this more.

 

[fzero]

You can work out the antiparticles from some group theory and perhaps the Lagrangian. In a general case, we would first diagonalize the mass matrix. If the particles are massless, then we just have to consider the kinetic terms. These will be obtained from the quadratic Casimir, so they naturally take the form $\partial (\phi^a)^\dagger \partial \phi^a$. For a real representation, we could just treat $\phi^a$ as real fields, so for an appropriate quantization $\phi^a$ will destroy particles and create antiparticles. For a more general treatment, we can consider the weights of the representation.

For the example of the unbroken $SU(2)$, we have massless gauge fields $W^a$ in the adjoint. We can choose a parameterization of the weights of the representation (roots for the adjoint), so there is a zero-vector corresponding to $W^3$ and the root $\alpha$ which we can associate with $W^- = W^1 - i W^2$. Since the corresponding generator satisfies $E_\alpha^\dagger = E_{-\alpha}$, $W^+$ has opposite charge to $W^-$. We conclude that $W^3$ is its own antiparticle and $W^\pm$ are antiparticles of each other. We could repeat the argument for the gluons and the roots would tell us how to associate particles to antiparticles.

For a representation $R$ different from the adjoint, we again have weights $w_i$. The antiparticles are elements of the conjugate representation $\bar{R}$ and have weights $-w_i$, so they are in 1-1 correspondence with the particles.

 

[PeterDonis]

These will be obtained from the quadratic Casimir, so they naturally take the form $\partial (\phi^a)^\dagger \partial \phi^a$. For a real representation, we could just treat $\phi^a$ as real fields, so for an appropriate quantization $\phi^a$ will destroy particles and create antiparticles.

If the representation is real, that means $(\phi^a)^\dagger = \phi^a$, correct?

For the example of the unbroken $SU(2)$, we have massless gauge fields $W^a$ in the adjoint.

These are real fields, correct? The Baez article lists them that way (and he explicitly contrasts that with the complex fields listed for the Higgs and the fermions).

We can choose a parameterization of the weights of the representation (roots for the adjoint)

This amounts to changing which fields we are looking at, correct? In other words, instead of looking at $W^1$, $W^2$, and $W^3$, all of which are real, we construct two new fields, $W^+ = W^1 + i W^2$ and $W^- = W^1 - i W^2$, which are manifestly complex and are conjugates of each other.

We conclude that $W^3$ is its own antiparticle and $W^\pm$ are antiparticles of each other.

From the above, this appears to depend on choosing to look at $W^\pm$ and $W^3$ instead of $W^1$, $W^2$, and $W^3$, correct?

 

[fzero]

If the representation is real, that means $(\phi^a)^\dagger = \phi^a$, correct?

These are real fields, correct? The Baez article lists them that way (and he explicitly contrasts that with the complex fields listed for the Higgs and the fermions).

Real representation means that the generators $\lambda_a$ for this representation are real. But this also means that we can choose $(\phi^a)^\dagger = \phi^a$ so the mode expansion only contains a single pair $a_k, a_k^\dagger$ and the field is real.

This amounts to changing which fields we are looking at, correct? In other words, instead of looking at $W^1$, $W^2$, and $W^3$, all of which are real, we construct two new fields, $W^+ = W^1 + i W^2$ and $W^- = W^1 - i W^2$, which are manifestly complex and are conjugates of each other.

From the above, this appears to depend on choosing to look at $W^\pm$ and $W^3$ instead of $W^1$, $W^2$, and $W^3$, correct?

Yes. The advantage of doing this is that the same rules work for complex representations. Also, it is more physical, since we've respected the charges wrt to the Cartan subalgebra.

 

[George Jones]

For me "real representation" mean "representation on a real vector space", i.e., the representative operators operate on a real vector space.

Real representation means that the generators $\lambda_a$ for this representation are real.

Does this mean that the generators span a complex vector space (Lie algebra) on which there is a preferred complex structure? Or possibly that a choice a generators is used to pick out a preferred complex structure?

Without a complex structure, it is not possible to label some vectors in a vectors space over $\mathbb{C}$ as "real" and other as "imaginary".

 

[fzero]

For me "real representation" mean "representation on a real vector space", i.e., the representative operators operate on a real vector space.
Does this mean that the generators span a complex vector space (Lie algebra) on which there is a preferred complex structure? Or possibly that a choice a generators is used to pick out a preferred complex structure?

Without a complex structure, it is not possible to label some vectors in a vectors space over $\mathbb{C}$ as "real" and other as "imaginary".

I don't want to overgeneralize, so if we restrict to the compact Lie groups that are used in model building, they are all real manifolds. The adjoint representations then have real dimension equal to that of the group so we can consider them to be real vector spaces. I'm sure that this dictates a preferred choice of generators, but since the adjoint representation is so special, we'd have expected that.

There may be some more rigorous things to say, but they don't come to mind immediately.

 

[PeterDonis]

The adjoint representations then have real dimension equal to that of the group so we can consider them to be real vector spaces.

But the basis vectors of the real vector space in question (the Lie Algebra $\mathfrak{su}2$) are $W^1$, $W^2$, and $W^3$, correct? When you talk about $W^\pm$, you are talking about switching to a different vector space, a complex one, which is constructed (somehow) by taking two real fields, $W^1$ and $W^2$, and "repackaging" them as one complex field, $W^\pm$, where $W^+ = (W^-)^\dagger$. (This then leaves one remaining real field, $W^3$.) At least, that's how I'm reading your previous posts. If that's correct, then George Jones's question could be rephrased as, what picks out the particular linear combinations $W^+ = W^1 + i W^2$ and $W^- = W^1 - i W^2$ as the "right" complex structure? (It looks obvious enough to me, but that's not a very rigorous answer. )

 

[fzero]

But the basis vectors of the real vector space in question (the Lie Algebra $\mathfrak{su}2$) are $W^1$, $W^2$, and $W^3$, correct? When you talk about $W^\pm$, you are talking about switching to a different vector space, a complex one, which is constructed (somehow) by taking two real fields, $W^1$ and $W^2$, and "repackaging" them as one complex field, $W^\pm$, where $W^+ = (W^-)^\dagger$. (This then leaves one remaining real field, $W^3$.) At least, that's how I'm reading your previous posts. If that's correct, then George Jones's question could be rephrased as, what picks out the particular linear combinations $W^+ = W^1 + i W^2$ and $W^- = W^1 - i W^2$ as the "right" complex structure? (It looks obvious enough to me, but that's not a very rigorous answer. )

OK, I think I have a better way to explain it. Given an $n$-real dimensional compact group, we have a tangent space which is $\mathbb{R}^n$. This is isomorphic to the vector space of the adjoint representation. There is a canonical, real orthonormal basis here. We see that the adjoint representation is real.

In the example we're discussing, we have the adjoint representation of $SU(2)$ which is the same as the fundamental (=adjoint) representation of $SO(3)$, so we can use anti-Hermitian (though in this case really antisymmetric) 3x3 matrices. A convenient choice of generators is $(L_c)_{ab} = -\epsilon_{abc}$ (I choose the - sign since it leads to an easier convention below). As physicists doing quantum mechanics, it is often preferable to use Hermitian operators, so we stick a factor of $i$ into the commutation relations and somewhat obscure the fact that we had a real representation. Then we can take $(J_c)_{ab} = - i\epsilon_{abc}$ as generators. We will let $W^a$ be the canonical basis for this choice of generators.

Furthermore, it is often more convenient to choose a basis in terms of eigenstates of the Cartan generators. Choosing $J_3$ as the Cartan generator, we find $W^3$ has eigenvalue $0$, while $W^\pm$ have eigenvalues $\pm 1$.

So the "complex structure" is forced in two places. First we wanted Hermitian generators, then we wanted the eigenstates of the Cartan generator. Had we just used real antisymmetric generators, we would have found that the eigenvalues of $L_z$ were imaginary and the eigenvectors complex.

 

[PeterDonis]

We will let $W^a$ be the canonical basis for this choice of generators.

In other words, these are $W^1$, $W^2$, and $W^3$, as listed in the Baez article.

it is often more convenient to choose a basis in terms of eigenstates of the Cartan generators. Choosing $J_3$ as the Cartan generator, we find $W^3$ has eigenvalue $0$, while $W^\pm$ have eigenvalues $\pm 1$.

In other words, we have changed basis, at least for two of the basis vectors, so this is a different representation from the one described in the Baez article.

Had we just used real antisymmetric generators, we would have found that the eigenvalues of $L_z$ were imaginary and the eigenvectors complex.

This is the part I'm not sure I follow. Earlier, you said:

we can use anti-Hermitian (though in this case really antisymmetric)

But "anti-Hermitian" presupposes a complex structure, whereas "antisymmetric" does not; it's perfectly possible to have a real antisymmetric matrix, and as I understand it, that's how the Lie algebra of SO(3) is normally viewed. On this view, the eigenvalues and eigenvectors of all three generators are real.

Of course, since you multiplied the generators by $i$, you now can get imaginary eigenvalues and complex eigenvectors; but I'm still not sure why that's required. I probably need to write out the matrices and vectors in gory detail.

 

[fzero]

But "anti-Hermitian" presupposes a complex structure, whereas "antisymmetric" does not; it's perfectly possible to have a real antisymmetric matrix, and as I understand it, that's how the Lie algebra of SO(3) is normally viewed. On this view, the eigenvalues and eigenvectors of all three generators are real.

Of course, since you multiplied the generators by $i$, you now can get imaginary eigenvalues and complex eigenvectors; but I'm still not sure why that's required. I probably need to write out the matrices and vectors in gory detail.

I think the problem is just terminology. If we were dealing with the unitary group in generality, it would be natural to choose Hermitian or anti-Hermitian generators. However, when we are dealing with the adjoint representation, the argument I gave about the tangent space makes it clear that we can choose explicitly real matrices in a natural way. However, for the unitary and orthogonal algebras, these matrices are antisymmetric with some imaginary eigenvalues.
For instance, with the choice I made,
$$ L_z = \begin{pmatrix} 0 & - 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$
If we multiply them by $i$, we get Hermitian matrices with real eigenvalues:
$$ J_z = \begin{pmatrix} 0 & - i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$

I don't think the choice matters at all from the point of view of representation theory, but it's clear that the Hermitian choice is better for doing quantum mechanics.

 

[PeterDonis]

the argument I gave about the tangent space makes it clear that we can choose explicitly real matrices in a natural way. However, for the unitary and orthogonal algebras, these matrices are antisymmetric with some imaginary eigenvalues.

Yes, having written out the matrices and eigenvalue equations explicitly, I agree. Or, to put it another way, if we insist on keeping everything real (matrix elements and eigenvalues), then we are stuck with matrices that have only one real eigenvalue, and it's zero.

 

[samalkhaiat]

But the basis vectors of the real vector space in question (the Lie Algebra $\mathfrak{su}2$) are $W^1$, $W^2$, and $W^3$, correct? When you talk about $W^\pm$, you are talking about switching to a different vector space, a complex one

No. The real fields $W^{\mu}_{a}, \ a = 1,2,3$ are the Cartesian components of the adjoint vector $\mathbb{W}^{\mu} \in [3]$. Since the adjoint representation $[3]$ is a real vector space of dimension equal to that of the algebra, $\mathbb{W}^{\mu}$ can be expanded in any basis of the algebra: $\mathbb{W}^{\mu} = W^{\mu}_{a} X^{a}$. Taking $X^{a} =\frac{\tau^{a}}{2}$ to be the generators matrices of the fundamental representation, we obtain the following traceless Hermitian matrix

$$\mathbb{W}^{\mu} = \frac{1}{2} \begin{pmatrix} W_{3} & W_{1} - i W_{2} \\ W_{1} + i W_{2} & - W_{3} \end{pmatrix} ^{\mu} = \frac{1}{2} \begin{pmatrix} W_{3} & \sqrt{2} W^{+} \\ \sqrt{2} W^{-} & - W_{3} \end{pmatrix} ^{\mu} .$$ Notice that the real gauge fields $W^{\mu}_{a}$ do not correspond to any physical particle. Therefore, the question about “particle-antiparticle” does not make any sense. The physical particles (i.e. the charge eigen-states or the gauge bosons) correspond to the matrix elements of the adjoint vector $\mathbb{W}^{\mu}$.

We have similar situation in flavour and colour $SU(3)$. Recall the Mesons octet $M = \phi_{a} \frac{\lambda^{a}}{2} \in [8]$ of flavour $SU(3)$

$$M = \frac{1}{2} \begin{pmatrix} \phi_{3} + \frac{\phi_{8}}{\sqrt{3}} & \phi_{1} - i \phi_{2} & \phi_{4} - i \phi_{5} \\ \phi_{1} + i \phi_{2} & - \phi_{3} + \frac{\phi_{8}}{\sqrt{3}} & \phi_{6} - i \frac{\phi_{7}}{\sqrt{3}} \\ \phi_{4} + i \phi_{5} & \phi_{6} + i \frac{\phi_{7}}{\sqrt{3}} & - \frac{2}{\sqrt{3}} \phi_{8} \end{pmatrix} = \frac{1}{2} \begin{pmatrix} \pi^{0} + \frac{\eta_{8}}{\sqrt{3}} & \sqrt{2} \pi^{+} & \sqrt{2} K^{+} \\ \sqrt{2} \pi^{-} & - \pi^{0} + \frac{\eta_{8}}{\sqrt{3}} & \sqrt{2} K^{0} \\ \sqrt{2} K^{-} & \sqrt{2} \bar{K}^{0} & - \frac{2}{\sqrt{3}} \eta_{8} \end{pmatrix} .$$ The real scalar fields $\phi_{a} , \ a = 1, 2 , …, 8$ by themselves do not represent any physical particle, so we do not ask which one is particle and which one is antiparticle. The physical particles are represented by the matrix elements of the adjoint vector $M$.

 

[stedwards]

Might I open a slightly different manner of query that might have some resolution?

It could very well be that this idea is not very good. If so, I would like to know why, in the language that even a non-particle physicist would understand.

We could represent a standard model particle $P$ as $P=a X_1 + b X_2 + c X_3$ having $SU(3)$ symmetry, requiring that $a^2+b^2+c^2=1$, 'cause there is just one particle.

Generally, for $P$ having $SU(n)$ symmetry then there are $n$ terms, right?

Given that all this is not too stupid, what is the action of the operator $CPT$ on $P$ ? I have no idea how to apply the action of $C$, $P$, and then $T$ is applied.

Say $\bar{P} = CPT(P)$. I assume that if $\bar{P} = P$, then $P$ is it's own antiparticle and for all standard model particles, $(CPT)^2 P = P$.

 

[ChrisVer]

Say P¯=CPT(P) \bar{P} = CPT(P). I assume that if P¯=P\bar{P} = P, then PP is it's own antiparticle and for all standard model particles, (CPT)2P=P(CPT)^2 P = P.

You don't need the invariance under CPT to have particle=antiparticle... In fact any particle in the Standard Model should be invariant under CPT.
For the particle= antiparticle you have the action of C alone.

 

[stedwards]

You don't need the invariance under CPT to have particle=antiparticle... In fact any particle in the Standard Model should be invariant under CPT.
For the particle= antiparticle you have the action of C alone.

Say, for example, $\frac{i}{\sqrt{3}}R + \frac{1}{\sqrt{3}}G - \frac{1}{\sqrt{3}}B$, how would you apply charge conjugation?