Why Are Antiparticles Essential for Particle Mixes in Quantum Field Theory?

[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:

<br /> \frac{c}{M^2}\overline{Q}\gamma^\mu Q\overline{Q}\gamma_\mu L<br />

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! ;-)

 

[mormonator_rm]

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:

<br /> \frac{c}{M^2}\overline{Q}\gamma^\mu Q\overline{Q}\gamma_\mu L<br />

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?

Actually, that proton decay could be slightly restricted. If you think about it;

<br /> p \rightarrow \pi^0 + e^+<br />

is most likely to occur with \pi^0 in either a pure d\bar{d} or pure u\bar{u} state, since the basic interaction has to fall from either;

<br /> d + (u + u) \rightarrow d + (\bar{d} + e^+)<br />

or;

<br /> u + (u + d) \rightarrow u +(\bar{u} + e^+)<br />

Now, in the standard model we can probably say that this arrangement allows for equal amplitude of decay in either of these directions based on the near flavor symmetry of up and down, meaning that the decay width can be calculated without any flavor limitations. This is because \pi^0 = \frac{u\bar{u}-d\bar{d}}{\sqrt{2}}. This works if quarks and electrons are fundamental particles.

However, in the regime of Rishon Model, this problem takes on a different shape. When Rishons are factored in, the possibility of;

<br /> u + (u + d) \rightarrow u +(\bar{u} + e^+)<br />

falls relative to;

<br /> d + (u + u) \rightarrow d + (\bar{d} + e^+)<br />

because the latter requires only rearrangement of the initial Rishons, whereas the first requires a V\overline{V} annihilation followed by T\overline{T} production, which could reduce its probability greatly.

If the Standard Model is absolutely correct, and quarks and leptons are fundamental, then we should see the decay p \rightarrow \pi^0 + e^+ at the conventionally calculated rate. If Rishon Model is true, then we should observe p \rightarrow \pi^0 + e^+ at about half the rate prescribed by the Standard Model calculations.

 

[blechman]

Cute. But the nice thing about an operator analysis (what I was doing) is that it is model-independent (Rishon or Chiral quarks or whatever). Such an operator exists since it can't be forbidden by gauge symmetries, and you can get very far from that. However, the best you can do typically is order-of-magnitude. You say the difference between SM and R is a factor of 2 - such precision is far greater than what I was going for: I'm looking for factors of 10!

 

[mormonator_rm]

Cute. But the nice thing about an operator analysis (what I was doing) is that it is model-independent (Rishon or Chiral quarks or whatever). Such an operator exists since it can't be forbidden by gauge symmetries, and you can get very far from that. However, the best you can do typically is order-of-magnitude. You say the difference between SM and R is a factor of 2 - such precision is far greater than what I was going for: I'm looking for factors of 10!

Understood, and I think the apparent non-observation of proton decay could easily be construed as evidence that the aforementioned dimension-four terms allowing non-coservation of B and L seperately (as long as B-L is still conserved) probably do not exist (whether they do at all in the first place, after suppression). The dimension-six term from the SM looks a lot more plausible because of this, and my conjecture of the Rishon vs SM difference already assumed that only the dimension-six term survives. The factor of 2 difference was just based on the non-symmetrical flavor choice available if quarks and leptons are indeed composites of Rishons, and did not take into account any other changes in decay behavior or scaling. But if the extremely long life-time of the proton in the SM wasn't long enough to prevent us from seeing a proton decay in our lifetime, then perhaps Rishon model, if true, would keep us from seeing that event for even longer.

You are right, though, in that the operator analysis is more fundamental to the problem than deciding on a model. I am not very well practiced in operator analysis, so I try not to delve too deeply into it right now; ultimately, it will lead us to the point where we will be at the right order of magnitude, and from there a model selection could be appropriate.

I should try to learn more about operator analysis. I am trying to think where I have seen it all done before, maybe one of my old textbooks would shed some light on it.

 

[blechman]

I should try to learn more about operator analysis. I am trying to think where I have seen it all done before, maybe one of my old textbooks would shed some light on it.

It's a VERY powerful tool, used all over physics. Most QFT texts discuss it, but you can probably get an even better explanation from summer-school lectures. Look around for TASI or Les Houches lectures with the title "Effective Field Theory" - there are a whole lot of them out there. Some of my favorites are those by David B. Kaplan; Ira Rothstein; Aneesh Manohar; Howard Georgi; Cliff Burgess; this list goes on and on! Each of these reviews approaches the problem from the point of view of the author's specific research, so you can check out several of them to see different applications.

Have fun!

 

[nrqed]

It's a VERY powerful tool, used all over physics. Most QFT texts discuss it, but you can probably get an even better explanation from summer-school lectures. Look around for TASI or Les Houches lectures with the title "Effective Field Theory" - there are a whole lot of them out there. Some of my favorites are those by David B. Kaplan; Ira Rothstein; Aneesh Manohar; Howard Georgi; Cliff Burgess; this list goes on and on! Each of these reviews approaches the problem from the point of view of the author's specific research, so you can check out several of them to see different applications.

Have fun!

It's neat to see that you are working on this!
I did my PhD with Peter Lepage on applications of EFTs to nonrelativistic bound states and my postdoc with Burgess. I unfortunately got discouraged with the job market and left research for teaching, hoping I would do research on the side but things got in the way Now I am trying to get back doing research.

 

[blechman]

It's neat to see that you are working on this!
I did my PhD with Peter Lepage on applications of EFTs to nonrelativistic bound states and my postdoc with Burgess. I unfortunately got discouraged with the job market and left research for teaching, hoping I would do research on the side but things got in the way Now I am trying to get back doing research.

Your username gives you away! ;-)
I've been bounced around, working on all types of effective theories and also BSM/LHC physics. I'm also getting frustrated with the job market and considering teaching positions, but I'm trying not to think about it.

Cliff's a great guy, just down the road from me. We've talked now and then.

Anyway, I won't clutter the thread with personal anecdotes. But yeah, this stuff is great! And for grad students reading this: it's one of the most useful things you can learn, since it teaches you how to think not only about physics, but any kind of quantitative analysis. You can apply these ideas (in a slightly modified form) to industry, finance, economics, politics, medicine... EFT teaches you how to think! Once you know that, the sky's the limit!

OK, I'm done now. ;-P~~~~

 

[samalkhaiat]

The canonical example of proton decay in the standard model comes from the dimension-6 operator:

<br /> \frac{c}{M^2}\overline{Q}\gamma^\mu Q\overline{Q}\gamma_\mu L<br />

Where, in the SM Lagrangian, do you find such coupling? Or, which part of the Lagrangian produces the "current" \overline{Q} \gamma_{\mu} L in your operator?
In order to have such current, the Lagrangian must contain a kitetic term of the form

i \overline{Q} \gamma_{\mu} \partial^{\mu}L

But, the SM Lagrangian does not have such kinetic term! Therefore, your operator (quoted above) does not live in the SM, i.e., you can not speak about proton decay in the SM.
The same conclusion can be reached if we study the representation of the symmetry group of the SM:

The fields in ANY Noether current must belong to the same representation (multiplet). In the SM, quarks and leptons do not occupy the same multiplet, i.e., under the action of
SU(2)XU(1), Q and L do not transform into each other. Therefore, the gauge group
SU(2)XU(1) does not allow you to write the object \overline{Q}\gamma_{\mu}L. Indeed, doublets such as (d , e)^{t} or (u , \nu)^{t} do not belong to the representation of SU(2)XU(1). This is because the product group
SU(2)XU(1) requiers all members of an SU(2) boublet to have the same weak hypercharge [the generator of U(1)];
Since Y(Q) \neq Y(L), SU(2)-doublets must be of the form (Q_{1},Q_{2})^{t}, (L_{1},L_{2})^{t} with Y(Q_{1}) = Y(Q_{2}) and Y(L_{1}) = Y(L_{2}), a doublets like (Q_{i},L_{j})^{t} do not transform correctly under SU(2)XU(1). So, within the symmetry group of the SM, your operator
( \overline{Q} \gamma^{\mu} Q)( \overline{Q} \gamma_{\mu} L)
is not allowed, i.e., the mathematical structure of the SM has no room for proton decay.
However, your current and operator are allowed beyond the SM, i.e., when the gauge group is larger than SU(3)XSU(2)XU(1) [say SU(5)]. Indeed, in this case, quarks and leptons appear in the same representation ([5],[5*] and [10]) allowing us to speak about proton decay


regards

sam

 

[arivero]

. EFT teaches you how to think! Once you know that, the sky's the

With a caveat: not to let EFT become an ideology, or you can start claiming that after all every QFT is an effective one and than any accelerator experiment is not fundamental and that renormalization is not a guide for any fundamental thing at this level and then to go to GUT scale, and even then to Planck scale for fundamental truth. A sad process to follow.

 

[blechman]

Therefore, your operator (quoted above) does not live in the SM, i.e., you can not speak about proton decay in the SM.
The same conclusion can be reached if we study the representation of the symmetry group of the SM...

(1) In the formal definition of the **perturbative** SM which only keeps the renormalizable operators (with dimension <=4) you are right - there is technically no proton decay. The operator cannot be generated since there's no current, as you say.

But what you are forgetting is that B and L are ANOMALOUS! Therefore, there are instanton effects that can generate such an operator as the one I wrote down. Of course, such operators come with coefficients that are exponentially small, so there is no contradiction with experiment. However, the operator is generated, although not through perturbation theory (no feynman graphs)!

(2) It is generally accepted that the SM is not the end of physics - if nothing else, something must happen at the Planck scale to incorporate gravity into the picture. Since B and L are only accidental symmetries of the SM, there is absolutely no reason to assume that they will continue to hold past dimension-4 operators. Since the operator I wrote down is allowed by all the gauge symmetries, it must generally be there, unless there's an additional symmetry imposed beyond the SM. Indeed, in most naive extensions of the SM (SUSY, GUT, technicolor, etc) such operators are in fact generated after integrating out the new physics.

So the operator is there, formal definition of the perturbative SM aside!

 

[blechman]

Burned!

Sam, everyone: I would like to formally apologize! I have made a foolish error: the operator I wrote down isn't even gauge invariant, as Sam rightfully pointed out! I am humbled. However, it is easily fixed. What I meant to write down is:

\bar{Q}\gamma^\mu u \bar{Q}\gamma_\mu e

There! That operator is gauge invariant, and everything I have said up to this point still holds. Again, let me apologize if I've caused any confusion.

Let me also point out that the non-perturbative argument I have given was a little off-the-cuff, and although I think it's right, perhaps you shouldn't take it too seriously unless you've worked it out for yourselves.

Finally, I want to say something about generating higher-dimensional operators in the SM. Look: Sam is quite right that you cannot generate this operator perturbatively. You can see this right away if you know about "spurion analysis": B and L are symmetries of the SM (even if only accidentally), and therefore the SM cannot break those symmetries by itself. It needs something else (GUTs, SUSY, etc) to do it. This I cannot deny (non-perturbative stuff aside).

But the question you should now ask yourself is: why have we not simply written this operator down to begin with? The answer simply is that it is non-renormalizable. That's the ONLY reason we didn't write it down (since we know B and L are anomalous, using that symmetry argument is not very convincing). Historically, we thought that renormalizability was some God-like quality of good QFTs. We now understand that this is not so: there is nothing special about renormalizable operators. Therefore, there is NO reason why we should not write this operator down. Never mind where it comes from; it's allowed, and therefore it should be there!

Fortunately, since it's suppressed by the 1/M^2 coefficient, it is not very important as long as M is some very large energy scale. That's why we don't include it generally. But even though it cannot be "generated" in the renormalizable standard model, that's not a reason to forbid it. It's there, and it deserves our respect!

 

[arivero]

Historically, we thought that renormalizability was some God-like quality of good QFTs. We now understand that this is not so: there is nothing special about renormalizable operators.

Well, historically, we lived under the rule of a nonrenormalizeble operator, Fermi quartic coupling, since the mid fifties under the late sixties, and only in the early seventies a right renormalizable theory was found (and renormalizability proved by veltman and 't hoof. I think that the doctrine of Effective QFT was already strong in the late seventies, so the historicity of the blessing of renormalizability can be a myth.