Are bosons always their own antiparticle

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I'm wondering about the fundamental difference between fermions and bosons. Isn't it true that bosons are always their own antiparticle but fermions are not? But then I wonder about the Weak bosons, W+, W-, Z0. All these are bosons, right? But isn't the W+ the antiparticle of the W-. Or is it that the W+ is its own "antiparticle" in the weak, isospin charge?

I'm thinking this can be answered, yes or no. But some follow-up questions might require higher math. So I make this an "A" thread. Thanks.
 

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  • #2
Vanadium 50
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I'm thinking this can be answered, yes or no
It can. The answer is no.
 
  • #3
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It can. The answer is no.
What bosons are not its own antiparticle, at least with respect to the charge in which they are considered bosons? Or what fermions are their own antiparticle? Thanks.
 
  • #4
PeterDonis
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What bosons are not its own antiparticle, at least with respect to the charge in which they are considered bosons?
I don't know what you mean by "with respect to the charge in which they are considered bosons".

In the Standard Model, the ##W^+## and ##W^-## bosons are antiparticles of each other. All of the other bosons are their own antiparticles. (At least, I think that's right, but I'm not completely sure about the gluons.)

what fermions are their own antiparticle?
In the Standard Model, there aren't any.
 
  • #5
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Well, I understand that there is a thing called charge conjugation. IIRC, that means all charges of every force are reversed. The Weak force particles have isospin charge and electric charge. And the quarks have color charge, isospin charge, and electric charge. If you were to charge conjugate a W+ boson, then one would have to reverse the electric charge and the isospin charge. The quarks would have to reverse the color, isospin and electric charge. So when I say, "with respect to the charge in which they are considered bosons", for the W+, I mean if we only reverse the isospin charge. The W+ is not a boson wrt the electric charge, but is it a boson wrt the isospin charge? If you reversed only its isospin charge, is it the same thing? Thanks.
 
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PeterDonis
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I understand that there is a thing called charge conjugation. IIRC, that means all charges of every force are reversed.
Charge conjugation is the operation of taking particles to their antiparticles. If the particles have a charge, their antiparticles will have opposite charge, so charge conjugation does reverse all charges. But that's really a side effect of taking particles to antiparticles.

If you were to charge conjugate a W+ boson, then one would have to reverse the electric charge and the isospin charge.
Yes. And that's exactly what happens: the ##W^+## and ##W^-## have opposite electric charge and opposite weak isospin.

The quarks would have to reverse the color, isospin and electric charge.
Yes.

The W+ is not a boson wrt the electric charge
There is no such thing as "a boson with respect to a particular charge". There are just bosons and fermions. Being a boson or a fermion has nothing to do with charge. The ##W^+## is a boson.
 
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OK. Is there a name for reversing only the isospin charge, or for reversing only the color charge? I like the name charge conjugation in the context of my studies. Can I say charge conjugate isospin, or color charge conjugation? Would everyone understand the limited scope of those terms?
 
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PeterDonis
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Is there a name for reversing only the isospin charge, or for reversing only the color charge?
No, because there is nothing physically meaningful about doing that.
 
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No, because there is nothing physically meaningful about doing that.
So every quark that has opposite color charge must also have opposite electric charge? That doesn't sound right.
 
  • #10
PeterDonis
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So every quark that has opposite color charge must also have opposite electric charge?
No. But that's not what a hypothetical operation "change the color charge but not the electric charge" does. We have some things to clear up here.

First, you used the term "reversing" isospin or color charge, but neither of those charges are a simple plus/minus like electric charge is. Weak isospin is a continuous variation in SU(2): the "up and "down" weak isospin that is usually talked about is just a set of particular basis vectors in the SU(2) vector space. So, for example, the electron and the electron neutrino are the "up" and "down" weak isospin states of the first family of leptons in the Standard Model. (Actually it's the first family of left-handed leptons, but we'll leave out parity for now.) But there are also weak isospin states that are, for example, ##1 / \sqrt{2}## of an electron plus ##1 / \sqrt{2}## of an electron neutrino. Similarly, an "up" quark and a "down" quark are the "up" and "down" weak isospin states of the first family of (left-handed) quarks in the Standard Model; but there are also weak isospin states that are, for example, ##1 / \sqrt{2}## of an up quark plus ##1 / \sqrt{2}## of a down quark.

So the operation you are calling "reverse weak isospin" is really something like "rotate weak isospin by a particular angle in a particular direction", and there are multiple different operations of this type that take "up" weak isospin into "down" weak isospin; there are multiple "directions" in which you can do the rotation in weak isospin space, and those different rotations will have different effects on states that are mixtures of "up" and "down".

Similar remarks apply to color charge, except that for color charge the vector space is SU(3) and there are three basis vectors, often referred to as "red", "green", and "blue" (though this in itself leaves out some technical points), and there are even more different ways to "rotate" in the SU(3) vector space to take particular color states to other color states.

Next, the operation of charge conjugation, i.e., taking particles to antiparticles, is not an operation of "reversing" particular properties of particles. It "reverses" which particles you call "particles" and "antiparticles". In the Standard Model, for each particle that has a distinct antiparticle, there are separate quantum fields for the two, and "charge conjugation" means switching the labels between the two quantum fields, i.e., switching which one we call the "particle" and which one we call the "antiparticle". The fact that the laws of physics remain the same when we do this is a powerful constraint on both the laws and the fields: it requires the laws to have particular properties, and it requires the fields of particles and antiparticles to be related in particular ways. (The details of all this are well beyond the scope of a "B" level thread.) When we say that charge conjugation "reverses" all of the charges, what we are really saying is that the charges on the antiparticle fields are all the reverse of the corresponding charges on the corresponding particle fields.

Note that this means that the "operation" of charge conjugation is a different kind of operation from the operation of "reverse weak isospin" or "reverse color charge" or "reverse electric charge". The latter operations are "rotations" in the vector spaces of particular fields (electron, quark, etc.). But the charge conjugation operation is an exchange of the "particle" and "antiparticle" labels we put on the fields and their vector spaces. So the fact that charge conjugation "reverses" all of the charges on the particles (in the sense given in the previous paragraph) is a separate thing from the effects that various kinds of rotations in the vector spaces have on the properties of the fields.

Taking all the above into consideration, consider, say, a red up quark. What would it mean to "reverse" (bearing in mind the complications described above) the weak isospin or color charge, without changing the electric charge?

The isospin "reverse" of a red up quark would be a red down quark. But a down quark does not have the same electric charge as an up quark; the latter is charge ##+ 2/3##, the former is charge ##- 1/3##. So, physically, you can't just "reverse" the isospin of an up quark and get a physically meaningful result. Nor could you combine this with "reversing" the electric charge, since that would give ##- 2/3##, not ##- 1/3##. The operation in electric charge space that accompanies going from "up" to "down" in weak isospin space is "subtract ##-1## from the charge". (Note that this also works for electron neutrinos and electrons; the former is the "up" weak isospin state and the latter is the "down" one, and going from "up" to "down" decreases the electric charge by ##1##.)

In other words, physically speaking, it isn't really possible to only "rotate" in weak isospin space, without "rotating" at all in electric charge space. You have to do both in concert, related in a particular way, in order to obtain a physically meaningful result. (This, btw, is a consequence of the fact that the weak and electromagnetic interactions are not really separate; they are combined into the electroweak interaction.)

As far as the Standard Model is concerned, the strong interaction is separate from the electroweak interaction, so, mathematically speaking, you can, for example, "rotate" from red to green in the strong interaction space without doing anything in electroweak space. So you could "rotate" a red up quark into a green up quark, and it would still be an up quark and would still be a physically meaningful state. The problem here is that no state that is actually physically observable can have any color charge; so there is no such thing as a red up quark or a green up quark in isolation. Such a thing would only occur bound into a meson or hadron, and you only observe the colorless meson or hadron, and you have no way of telling whether, for example, a pi meson is composed of a red up quark and an anti-red anti-down quark, or a green up quark and an anti-green anti-down quark. So the operation of "rotating" red to green does not correspond to anything you can physically observe.
 
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  • #11
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Thank you, PeterDonis, for that detailed response. That certainly gives me something to think about.
 
  • #12
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In the Standard Model, the ##W^+## and ##W^-## bosons are antiparticles of each other. All of the other bosons are their own antiparticles. (At least, I think that's right, but I'm not completely sure about the gluons.)
In the theory of superfluidity, ##^4 He## atoms are treated as point like boson particles, and you could obviously convert all the constituent elementary particles to antiparticles to get an anti-##^4 He## atom. But that's probably not the kind of thing that is meant in particle physics context.
 

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