# Antiparticles of Standard Model gauge bosons

1. Aug 14, 2015

### Staff: Mentor

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.)

Last edited: Aug 15, 2015
2. Aug 14, 2015

### Staff: Mentor

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

3. Aug 14, 2015

### Staff: Mentor

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.

4. Aug 14, 2015

### Staff: Mentor

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).

5. Aug 14, 2015

### Staff: Mentor

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.

6. Aug 14, 2015

### Staff: Mentor

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?

7. Aug 14, 2015

### Staff: Mentor

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.

8. Aug 14, 2015

### Staff: Mentor

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: 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.

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.

Last edited by a moderator: May 7, 2017
9. Aug 15, 2015

### Staff: Mentor

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:

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.

Last edited by a moderator: May 7, 2017
10. Aug 15, 2015

### Staff: Mentor

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

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

11. Aug 15, 2015

### Staff: Mentor

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.

12. Aug 15, 2015

Staff Emeritus
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.

Last edited by a moderator: Aug 15, 2015
13. Aug 15, 2015

### Staff: Mentor

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.

Last edited: Aug 15, 2015
14. Aug 15, 2015

### Staff: Mentor

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

15. Aug 15, 2015

### 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.

16. Aug 15, 2015

### 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 $W^{\mu}_{a} \tau^{a}$.

17. Aug 15, 2015

### Staff: Mentor

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

18. Aug 15, 2015

Staff Emeritus
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.

19. Aug 15, 2015

### Staff: Mentor

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.

20. Aug 18, 2015

### 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.

21. Aug 18, 2015

### Staff: Mentor

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).

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?

22. Aug 18, 2015

### fzero

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.

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.

23. Aug 19, 2015

### George Jones

Staff Emeritus
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".

24. Aug 19, 2015

### fzero

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.

25. Aug 19, 2015

### Staff: Mentor

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. )