# Particle mixing

1. Oct 25, 2007

### genloz

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

2. Oct 25, 2007

### 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.

3. Oct 25, 2007

### clem

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

4. Oct 25, 2007

### mormonator_rm

They do not necessarily need to be close in mass... its just alot 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.

5. Oct 25, 2007

### ZapperZ

Staff Emeritus
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.

6. Oct 26, 2007

### 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...

7. Oct 26, 2007

### pallidin

8. Oct 26, 2007

### genloz

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

9. Oct 26, 2007

### ZapperZ

Staff Emeritus
Where did the question come from?

Zz.

10. Oct 26, 2007

### genloz

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

11. Oct 26, 2007

### 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!

12. Oct 26, 2007

### mormonator_rm

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.

13. Oct 27, 2007

### genloz

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

14. Oct 27, 2007

### 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

15. Oct 27, 2007

### 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?

Last edited: Oct 27, 2007
16. Oct 27, 2007

### 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
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
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?

17. Oct 28, 2007

### samalkhaiat

18. Oct 28, 2007

### samalkhaiat

19. Oct 28, 2007

### arivero

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 lets pray that the original poster will care to come here to read the complete answer to his question. Probably he left