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genloz
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Why do two particles have to be antiparticles of each other to mix?
No. They usually have to be close in mass.genloz said:Why do two particles have to be antiparticles of each other to mix?
clem said:No. They usually have to be close in mass.
It is now believed that different types of neutrino mix.
clem said:No. They usually have to be close in mass.
It is now believed that different types of neutrino mix.
genloz said: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...
genloz said: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...
if you click in the number, you come to the actual dataMean n n-oscillation time >8.6 × 10 7 s , CL=90%
If your university has a good bunch of paid e-journals, you can even click in the links to go to the papers.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.
arivero said: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 [itex]K^{0}[/itex] and [itex]\bar{K}^{0}[/itex] should not be able to transform into each other through the weak interaction. This is in fact possible via the intermediate 2-pion state
[tex] K^{0} \rightarrow (\pi^{-} + \pi^{+}) \rightarrow \bar{K}^{0}[/tex]
Thus, one can say that Kions mix to produce the weak-hamiltoian (short & long) eigenstates
[tex]K^{0}_{L} = 2^{-1/2}( K^{0} + \bar{K}^{0})[/tex]
[tex]K^{0}_{S} = 2^{-1/2}( K^{0} - \bar{K}^{0})[/tex]
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
arivero said:.
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
samalkhaiat said: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
samalkhaiat said:arivero said:.
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...
arivero said: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?
mormonator_rm said: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.
arivero said:Next question could be: why Barionic number, or at least B-L, is a preserved quantity in the standard model?
Urvabara said: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...
samalkhaiat said: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...
samalkhaiat said: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
blechman said: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.
arivero said: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?
arivero said: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 said: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).
blechman said: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 [itex]M\sim M_{\rm GUT}[/itex].
Does that help?
arivero said: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.