2
$\begingroup$

Is some amount proton decay necessary in the standard model or is it possible for the proton lifetime to be infinite?

$\endgroup$
1

1 Answer 1

3
$\begingroup$

To answer your question, let me first concentrate on the first part "Is some amount proton decay necessary in the standard model (SM)?".

==> Well, proton decay or any processes are a consequence of a particular model, since so far proton decay is not observed experimentally in nature so it is certainly not necessary to incorporate in the SM . The SM and many extensions of it have proton decay, but the rate of the decay can vary significantly depending on the details of the model.

Let me now move to the second part of the question *** "is it possible for the proton lifetime to be infinite (in SM)?" ***.

==> Probably you already know that the SM has a $U_{B}(1)$ global symmetry, where $B$ is the baryon number. The consequence of this symmetry is, the lightest baryon, which is the proton is absolutely stable and this is why proton lifetime is infinity.

But note that, this is valid at the classical level. When quantum effects are taken in to account, this is not true. Quantum effects do break the baryon number conservation of the SM. This is done by the non-perturbative instanton effects. For a technical details see for example High Energy Behavior of Baryon and Lepton Number Violating Scattering Amplitudes and Breakdown of Unitarity in the Standard Model by Olivier Espinosa (pdf). At the quantum level the baryonic current is not conserved and one can compute to show that such non-perturbative processes has lead to both baryon and lepton number violations ($\Delta B=\Delta L =3$; L=lepton number) with a rate that is suppressed by a factor of $\sim 10^{-173}$.

So the bottom-line is, in the SM, at the classical level, proton is absolutely stable. But proton can decay due to quantum effects but the life time is extremely big but not infinity.

$\endgroup$
9
  • $\begingroup$ That's approximately what I expected; the question then is: How big? $\endgroup$
    – Schroeder
    Commented Aug 15, 2016 at 17:20
  • $\begingroup$ I already mentioned about the huge suppression factor, but now you are asking the exact number for proton lifetime. So I guess you already know how to compute lifetime. Lets assume you do. So, Proton decay is caused by dimension 6 operators that are made from the combination of 3 quarks and a lepton from the SM. Due to this in calculating the amplitude of processes that lead to proton decay, Yukawa couplings come into play. $\endgroup$
    – SAS
    Commented Aug 15, 2016 at 18:27
  • 1
    $\begingroup$ Which means one needs to know the details of the Yukawa coupling matrices to calculate precisely the proton lifetime. You must know that in the SM, Yukawa couplings are completely arbitrary parameters and this is why you can not calculate it. But if you assume that the Yukawa sector has similar structure like other theories where these matrices are known, only then you can compare the two results. And this extra suppression factor then will tell you how long would be the proton decay compared to other models. $\endgroup$
    – SAS
    Commented Aug 15, 2016 at 18:27
  • $\begingroup$ Remember, I am avoiding many technicalities, things are not as simple as I am trying to make them. $\endgroup$
    – SAS
    Commented Aug 15, 2016 at 18:27
  • $\begingroup$ The link has rotted. Could you please post a durable reference? $\endgroup$
    – Schroeder
    Commented Jul 25, 2017 at 19:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.