The Antiparticle Mix: Exploring Particle Pairs

[genloz]

Why do two particles have to be antiparticles of each other to mix?

 

[Timo]

I think you should elaborate on what you mean by "mix". My understanding and usage of the term mixing for instance does not equire that condition.

 

[Meir Achuz]

Why do two particles have to be antiparticles of each other to mix?

No. They usually have to be close in mass.
It is now believed that different types of neutrino mix.

 

[mormonator_rm]

No. They usually have to be close in mass.
It is now believed that different types of neutrino mix.

They do not necessarily need to be close in mass... its just a lot more noticeable and powerful when they are. They do have to have the same quantum numbers, though. So the neutrinos can mix with each other because they are all spin-1/2 with the same lepton number. Quarks can mix with each other because they all have spin-1/2 and baryon number plus or minus 1/3. Mesons can mix as long as they share the same spin, parity, and charge conjugation. The effects of mixing are much stronger when the mixing particles are close in mass.

 

[ZapperZ]

No. They usually have to be close in mass.
It is now believed that different types of neutrino mix.

You really should wait until the OP explain what he/she meant as "mix". It is waaay to early to introduce "neutrino mix" into this and might do more to add confusion to this thread.

Zz.

 

[genloz]

Thanks for all the comments! I have a question that states:
"For two particles, A and B, to mix, A and B have to be anti particles of one another. Why then does a neutron not mix with an antineutron?"...
I'm not sure exactly what is meant by mix but just wanted to understand why the particles have to be antiparticles of each other in the first place before tackling the question...

 

[pallidin]

Of your term "mix", are you possibly referring to annihilation?
If so, look here: http://en.wikipedia.org/wiki/Annihilation

If not, disregard my post.

 

[genloz]

no, not really... because an antineutron and a neutron would annihilate each other wouldn't they?

 

[ZapperZ]

Thanks for all the comments! I have a question that states:
"For two particles, A and B, to mix, A and B have to be anti particles of one another. Why then does a neutron not mix with an antineutron?"...
I'm not sure exactly what is meant by mix but just wanted to understand why the particles have to be antiparticles of each other in the first place before tackling the question...

Where did the question come from?

Zz.

 

[genloz]

From a subatomic physics class I'm taking...

 

[blechman]

There's something *very* wrong here! A particle will never "mix" with its antiparticle since (by definition of "antiparticle") such particles have opposite quantum numbers, and are thus distinguishable. Therefore they cannot mix!

I suggest you ask your prof for clarification on the question.

 

[mormonator_rm]

Thanks for all the comments! I have a question that states:
"For two particles, A and B, to mix, A and B have to be anti particles of one another. Why then does a neutron not mix with an antineutron?"...
I'm not sure exactly what is meant by mix but just wanted to understand why the particles have to be antiparticles of each other in the first place before tackling the question...

I would say there has been a misunderstanding here on some level. You must understand that mixing is between similar states. For example, we have the pseudoscalar (0-+) I=0 mesons $$\eta$$ and $$\eta^\prime$$. Although they would appear to correspond in mass to the $$\eta_0$$ (singlet) and $$\eta_8$$ (octet) states, they are a mixture, or linear superposition, of these flavor states. They do not correspond to $$\eta_{qq\bar}$$ (light-quark) or $$\eta_{ss\bar}$$ (hidden strangeness) flavor states, either. In fact, their deviance from these states produces the observed masses and the flavor content determined from their decay products.

 

[genloz]

is there anywhere you know of that I can find a good explanation of mixing?

 

[pallidin]

By observing the rare process of D-meson mixing, BaBar collaborators can now test the intricacies of the Standard Model. To switch from matter to antimatter, the D-meson must interact with "virtual particles," which through quantum fluctuations pop into existence for a brief moment before disappearing again. Their momentary existence is enough to spark the D-meson's transformation into an anti-D-meson. Although the BaBar detector cannot directly see these virtual particles, researchers can identify their effect by measuring the frequency of the D-meson to anti D-meson transformation. Knowing that quantity will help determine whether the Standard Model is sufficient or whether the Model must be expanded to incorporate new physics processes.

"It's too soon to know if the Standard Model is capable of fully accounting for this effect, or if new physics is required to explain the observation," said Jawahery, who is a member of the Maryland Experimental High Energy Physics group. "But in the coming weeks and months we are likely to see an abundance of new theoretical work to interpret what we've observed."


Source: http://www.newsdesk.umd.edu/scitech/release.cfm?ArticleID=1409

 

[arivero]

The question is right if you add context; it is referring not to any abstract elementary particle mixing but to mixing in the sense of Kaon mixing or, as pointed above, D or B mixings. In the case of Kaon, it is K and Kbar, this is down-antistrange and antidown-strange, which obviously are antiparticles one of another. I'd not say it is a prerrequisite: what you need is to have same charges and nearby masses, and obviously it works here.

So the question stands: why the mixing does not apply to neutron?

 

[arivero]

Now, for the answer using the internet: you go to the particle data group website:
http://pdglive.lbl.gov
and then you select the neutron
http://pdglive.lbl.gov/Rsummary.brl?nodein=S017&fsizein=1&sub=Yr&return=BXXX005
and you will see that there is actually a bound for neutron antineutron mising

Mean n n-oscillation time >8.6 × 10 7 s , CL=90%

if you click in the number, you come to the actual data
http://pdglive.lbl.gov/popupblockdata.brl?nodein=S017NAN&fsizein=1
which starts telling that

A test of ΔB=2 baryon number nonconservation. MOHAPATRA 1980 and MOHAPATRA 1989 discuss the theoretical motivations for looking for n n oscillations. DOVER 1983 and DOVER 1985 give phenomenological analyses.

If your university has a good bunch of paid e-journals, you can even click in the links to go to the papers.

Anyway, here you have the answer: Baryons have baryonic number +1, thus antibaryions -1, and the mixing violates the conservation of barionic number. On the contrary, mesons have baryonic number 0.

I was thinking that, giving the mechanism of kaon mixing where the remixed wavefunctions are almost eigenstates of CP, a wrong answer that your teacher could score some points -because it proofs some knowledge- is to claim that the neutron state is already an eigenstate of CP. It is wrong in two ways: it is tautological, and anyway CP is not a preserved symmetry of the standard model.

Next question could be: why Barionic number, or at least B-L, is a preserved quantity in the standard model?

 

[samalkhaiat]

The question is right if you add context; it is referring not to any abstract elementary particle mixing but to mixing in the sense of Kaon mixing or, as pointed above, D or B mixings. In the case of Kaon, it is K and Kbar, this is down-antistrange and antidown-strange, which obviously are antiparticles one of another. I'd not say it is a prerrequisite: what you need is to have same charges and nearby masses, and obviously it works here.

In general, particles mix to produce eigenstates of the symmetry-breaking-part of the Hamiltonian.
We know that isospin and strangness conservation are violated by the weak interaction Hamiltonian. Thus there is no reason why $K^{0}$ and $\bar{K}^{0}$ should not be able to transform into each other through the weak interaction. This is in fact possible via the intermediate 2-pion state
$$K^{0} \rightarrow (\pi^{-} + \pi^{+}) \rightarrow \bar{K}^{0}$$

Thus, one can say that Kions mix to produce the weak-hamiltoian (short & long) eigenstates

$$K^{0}_{L} = 2^{-1/2}( K^{0} + \bar{K}^{0})$$
$$K^{0}_{S} = 2^{-1/2}( K^{0} - \bar{K}^{0})$$

So the question stands: why the mixing does not apply to neutron?

Not realy, It seems the baryon number is an absolutely conserved quantity, therefore a neutron can never transform into an antineutron, i.e., mixing can never occur.

regards

sam

 

[samalkhaiat]

.

Next question could be: why Barionic number, or at least B-L, is a preserved quantity in the standard model?

The symmetry group of SM is SU(3)XSU(2)XU(1), The baryon number (like the electric charge) is the Noether number corresponding to the exact (nubroken) global U(1) part.

sam

 

[arivero]

The symmetry group of SM is SU(3)XSU(2)XU(1), The baryon number (like the electric charge) is the Noether number corresponding to the exact (nubroken) global U(1) part.

sam

Fine! I never remember these arguments . And it is worse when it comes for instance to argue for things as R-symmetry for fermions and spartners or similar beasts.

Now let's pray that the original poster will care to come here to read the complete answer to his question. Probably he left

(EDITED(remark): praying answered in the next answer !)

 

[genloz]

Thanks very much! Thats been really reallyhelpful...

 

[Urvabara]

Hi!

.
The symmetry group of SM is SU(3)XSU(2)XU(1), The baryon number (like the electric charge) is the Noether number corresponding to the exact (nubroken) global U(1) part.

I would like to learn this physics stuff. Please, list here all the easily accessible and valuable particle physics articles you can find/remember. There are some very good ones that are pedagogical and easy to approach.

I do not mean popular particle physics text. I mean publications and learning materials for undergraduate and graduate students in physics. For example, article that demonstrates group theory intuitively for particle physicist...

PS. There are lots of articles in arxiv.org, but it is not very easy to find the best ones.

Thanks...

 

[mormonator_rm]

The question is right if you add context; it is referring not to any abstract elementary particle mixing but to mixing in the sense of Kaon mixing or, as pointed above, D or B mixings. In the case of Kaon, it is K and Kbar, this is down-antistrange and antidown-strange, which obviously are antiparticles one of another. I'd not say it is a prerrequisite: what you need is to have same charges and nearby masses, and obviously it works here.

So the question stands: why the mixing does not apply to neutron?

You are right to say that the electric charges must be equal for mixing to occur. However, the masses do not have to be close for mixing to occur; the mixing will have the same matrix element either way, but the effect of the mixing will appear much weaker as the pure-state masses are further apart. In other words, in a particular quantum number, let's say vectors ($$J^{PC}=1^{--}$$) just for an example, we may have the $$\omega$$, $$\phi$$, and $$J/\psi$$ mesons in the $$I=0$$ isospin channel. If these all mixed with the same value for the mixing matrix element, then the light $$\omega$$ at about 770 MeV would mix most strongly with the $$\phi$$ meson at about 1020 MeV. The mixings between $$\phi$$ and $$J/\psi$$ would be much weaker since their mass difference is much greater, on the order of 8 times larger than the value of $$M_{\phi}-M_{\omega}$$. The mixing between $$\omega$$ and $$J/\psi$$ is slightly weaker yet, since the mass difference is 9 times larger than $$M_{\phi}-M_{\omega}$$. Even though the matrix mixing element is the same size in each case, the actual mass shift from the mixing with respect to $$J/\psi$$ is much smaller for all mesons involved, since the mixing eigenvalues fall to the standard $$\omega - \phi$$ mixing eigenvalues as $$M_{J/\psi}$$ approaches infinity. This is observed in the behavior of the mixing mass eigenvalue equation;

$$\lambda^2 = (M_{J/\psi}^2+M_{\omega}^2)/2 \pm \sqrt{(M_{J/\psi}^2-M_{\omega}^2)^2+4M_{mix}^4})/2$$

where either $$\omega$$ or $$\phi$$ could be in the place of $$\omega$$ in the equation.

As for the final question of "why doesn't the neutron mix?", you must understand that you cannot mix baryons with their anti-baryons. This is because baryons have baryon number b=1, and anti-baryons have b=-1. You may ask, "why then does it work for mesons?" It works for mesons because mesons and anti-mesons all have baryon number b=0. Thus, mesons may mix with their antiparticles (if they have the same charge), but baryons may not mix with their antiparticles.

 

[arivero]

You are right to say that the electric charges must be equal for mixing to occur. However, the masses do not have to be close for mixing to occur; the mixing will have the same matrix element either way, but the effect of the mixing will appear much weaker as the pure-state masses are further apart.

Well, that was the point, the effect is a lot weaker, in relative terms if you wish.

In any case I must acknowledge that when reformulating the question of the original poster I was also thinking in a completely different case of mixing, those of two states via tunnel effect. In such case, only when the classical energy levels (the equivalent of mass, in the analogy) are the same you have access to some powerful calculational tools; I am thinking instantons as in Coleman lectures. In the case of tunneling between asymmetrical wells, where the quantum levels are different but not very, the instanton contribution becomes a mixing instead of a mass splitting.

 

[mormonator_rm]

Next question could be: why Barionic number, or at least B-L, is a preserved quantity in the standard model?

Well, if you think about it, baryon number conservation should be an absolute in the standard model IF quarks and leptons are fundamental particles. Now, here is something interesting I ought to point out, but please, please take this with a "grain of salt";

B-L conservation is an inherent prediction not just of supersymmetric models, but also of rishon model. In rishon models, baryon number and lepton number are linked to the types and numbers of rishons in each fermion. If a baryon number of -1/3 is given to $$T$$ and $${V\bar}$$, while a baryon number of +1/3 is assigned to $${T\bar}$$ and $$V$$, then the quarks each have B=+1/3, antiquarks each have B=-1/3, leptons and antineutrinos each have B=-1, and antileptons and neutrinos each have B=+1. Note that the baryon number for each of the leptons and neutrinos is the opposite of its lepton number. Because of this, B-L conservation in a rishon model is really just the result of making sure that rishons are produced in rishon-antirishon pairs, which is natural in any model of particle physics. In essence, you could take a $$\Delta^-$$ baryon composed of three down quarks (3 times $$TVV$$), and observe it anomalously decay, by rearrangement, directly into an electron ($$TTT$$) and two electron-antineutrinos (2 times $$VVV$$). The initial state had B=+1, and the final state had L=-1. B-L = 1-0 initial state led to B-L = 0+1 for the final state, so B-L was indeed conserved!

Of course, this does hinge on whether you feel that rishon models have any merit in the first place... but it is interestingly simple...

 

[nrqed]

Hi!



I would like to learn this physics stuff. Please, list here all the easily accessible and valuable particle physics articles you can find/remember. There are some very good ones that are pedagogical and easy to approach.

I do not mean popular particle physics text. I mean publications and learning materials for undergraduate and graduate students in physics. For example, article that demonstrates group theory intuitively for particle physicist...

PS. There are lots of articles in arxiv.org, but it is not very easy to find the best ones.

Thanks...

at the undergraduate level, I would suggest to get started by reading Griffiths' book on Particle Physics. A very good introductory text.

 

[blechman]

I would like to learn this physics stuff. Please, list here all the easily accessible and valuable particle physics articles you can find/remember. There are some very good ones that are pedagogical and easy to approach.

I do not mean popular particle physics text. I mean publications and learning materials for undergraduate and graduate students in physics. For example, article that demonstrates group theory intuitively for particle physicist...

When I was a grad student I used Bigi and Sanda's textbook on CP Violation, an excellent text for the advanced student/researcher. But I would not recommend it for anyone who has not already mastered QM at the Sakurai/Landau-Lif****z level, and has fully worked through Peskin-Schroeder or the equivalent.

As far as group theory goes: check out Georgi's text "Lie Algebras for Particle Physics". There's also a very famous Phys Report by Slansky - but this is definitely only for those who actually use these things in detail (it's more like a user-manual than an explanation).

 

[blechman]

Wow, I can't even say the guy's name!

Hey - Landau's poor colleague deserves to have his name published, don't you agree?! If there's a moderator out there, what's the deal with that?

 

[blechman]

The symmetry group of SM is SU(3)XSU(2)XU(1), The baryon number (like the electric charge) is the Noether number corresponding to the exact (nubroken) global U(1) part.

sam

Well, that's not entirely right. Baryon number is only sensitive to the quarks - the leptons have hypercharge but no baryon number. Furthermore (and for this reason), B is an anomalous symmetry in the standard model, so it's broken by nonperturbative effects (that are exponentially tiny, of course). The hypercharge gauge symmetry (the U(1) above) is NOT anomalous. Neither is B-L (baryon minus lepton number).

To answer arivero's question: baryon number is simply an accidental symmetry of the standard model. You write down all the allowed operators by the gauge symmetries, and you stop after dimension 4 (to keep things renormalizable). Then you find that you have this accidental symmetry. You expect higher-dimension operators to violate this symmetry, and they do. Extensions of the renormalizable part of the SM have dimension 5 and 6 operators that permit proton decay and other B and L violating processes. Since they're higher than dimension 4 they are suppressed by a scale, and proton decay searches push that scale to around $10^{16}$ GeV, which also just so happens to be the GUT scale where the couplings unify (yay!)

 

[Urvabara]

Thanks guys!

 

[arivero]

You write down all the allowed operators by the gauge symmetries, and you stop after dimension 4 (to keep things renormalizable). Then you find that you have this accidental symmetry. You expect higher-dimension operators to violate this symmetry, and they do. Extensions of the renormalizable part of the SM have dimension 5 and 6 operators that permit proton decay and other B and L violating processes.

It could be useful to look more slowly at this B, L, B-L stuff.

quarks had barionic number B=1/3, lepton number 0.
leptons had B=0, L=1

thus B-L for quarks is 1/3, for leptons is -1.
for antiquarks we had -1/3, for antileptons +1.

so a B-L preserving interaction could change 3 quarks into an antilepton. For instance a proton could disintegrate into a single positron. Is it? the B number changes from 1 to 0 but the L number changes from 0 to -1, so B-L goes from from (1-0) to (0-(-1)) so it does not change.

What I do not get is the dimensionality of the operator. For instance, such disintegration could be mediated by a gauge "diquark boson" having colour in 3* and charge +1/3: in a first step two quarks u and d collide to create this boson, and then it is absorbed by the extant u quark, which mutates into a positron. But what is the dimension of this boson operator?

Another point is that this operator had changed the number of particles, or if you prefer had converted a particle into an antiparticle. That seems to violate T, doesn't it? Is it possible to keep B-L and violate B without violate T? At a first glance, it seems it isn't.

 

[blechman]

What I do not get is the dimensionality of the operator. For instance, such disintegration could be mediated by a gauge "diquark boson" having colour in 3* and charge +1/3: in a first step two quarks u and d collide to create this boson, and then it is absorbed by the extant u quark, which mutates into a positron. But what is the dimension of this boson operator?

What, precisely, is this "gauge diquark boson"? There's no such object living in the standard model. The canonical example of proton decay in the standard model comes from the dimension-6 operator:

$$\frac{c}{M^2}\overline{Q}\gamma^\mu Q\overline{Q}\gamma_\mu L$$

where Q, L are the quark and lepton weak-isospin doublets; c is an order one constant and M is the scale of new physics (the cutoff). This operator would destroy a u and d quark, and create a lepton and an antiquark, mediating the decay p --> pi+e. This operator does not violate any gauge symmetries, and therefore it is allowed; it violates B and L, but not B-L. But it is a 4-quark operator, and is therefore dimension 6 (each fermion has dimension 3/2 in 4D spacetime).

The example you gave would involve the existence of a new gauge symmetry. This is the kind of things that happens in GUT models. But when you break the GUT symmetry and integrate out all of the heavy extra gauge bosons, you end up with operators like the one I wrote down, with $M\sim M_{\rm GUT}$.

Does that help?

 

[blechman]

Another point is that this operator had changed the number of particles, or if you prefer had converted a particle into an antiparticle. That seems to violate T, doesn't it? Is it possible to keep B-L and violate B without violate T? At a first glance, it seems it isn't.

Sorry, I just noticed this second question. You haven't violated fermion number, so I'm not sure why T must be violated. Changing a quark to an antiquark doesn't violate T, since a quark moving forward in time is the same as an antiquark moving backward in time (T is anti-unitary). Of course, T is violated in the SM, but that's a different story.

 

[arivero]

Sorry, I just noticed this second question. You haven't violated fermion number, so I'm not sure why T must be violated.

I was thinking the whole process to disintegrate the proton: I start with three quarks, then I obtain an antiquark and two quarks (and the gauge boson of a beyond SM group), them an antiquark, quark and a lepton (because the gauge boson is absorbed by the extant quark) and finally a single lepton. Pretty sure I have three fermions at the start and only one at the end.

Changing a quark to an antiquark doesn't violate T, since a quark moving forward in time is the same as an antiquark moving backward in time (T is anti-unitary).

Hmm perhaps I was wrong here, I was under the impression that if the antiquark was moving forward too, then some violation could be invoked, but I did not stop to think technicaly, sorry.

 

[arivero]

But when you break the GUT symmetry and integrate out all of the heavy extra gauge bosons, you end up with operators like the one I wrote down, with $M\sim M_{\rm GUT}$.

Does that help?

Indeed it does. It seems that my problem was that I keep looking at GUT and the dimension counting is done, as you have shown, within the standard model. It seems that to jump from one to another is as jumping from fermi four-interaction to gauge SU(2) W bosons.

 

[blechman]

I was thinking the whole process to disintegrate the proton: I start with three quarks, then I obtain an antiquark and two quarks (and the gauge boson of a beyond SM group), them an antiquark, quark and a lepton (because the gauge boson is absorbed by the extant quark) and finally a single lepton. Pretty sure I have three fermions at the start and only one at the end.

I'm confusing myself: the total fermion number of the proton decay process does not change (one fermion in, one fermion out) but the quark number and lepton number does change. Nevertheless, I still don't think this violates T. The dimension-6 operator above certainly does not, as is easy to see by just doing the transformation.

As to dimension-counting: yes, this is why proton decay is such a strong constraint in GUT models and not the standard model. In the SM, proton decay can only occur through dimension-6 operators. However, in GUT models, they occur through two dimension-4 operators, and this could potentially be VERY bad since it might not be suppressed. But the point is: even if you didn't have a GUT model, you should still expect the dimension-6 operator to exist in the SM since it's not forbidden by the gauge symmetry; but it's small due to the $M^2$ suppression.

Your analogy to Fermi theory is exactly correct! ;-)