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

[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.