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The electro-weak force is known to contain a chiral anomaly that breaks $B+L$ conservation. In other words, it allows for the sum of baryons and leptons to change, but still conserves the difference between the two. This means that the standard model could have a channel for protons to decay, for example into a pion and a positron. Does anyone know what the total proton decay rate through standard model channels is?

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  • $\begingroup$ Are you asking for the experimental results or a theoretical calculation? $\endgroup$ Commented Oct 26, 2016 at 3:15
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    $\begingroup$ Theoretical calculation is preferred - I know that experimental results are presently only upper limits on the rate. Also, I'm asking specifically about the standard model, not GUT extensions. $\endgroup$ Commented Oct 26, 2016 at 3:19
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    $\begingroup$ Proton decay is prohibited in the SM. The only B number violating process is the sphaleron process (which preserves B-L), but that process by its very nature can't lead to proton decay. BSM GUT models in contract, routinely predict proton decay and a routinely ruled out be strict experimental bounds ruling out such processes. $\endgroup$
    – ohwilleke
    Commented Oct 26, 2016 at 5:08

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Electroweak instantons violate baryon number (and lepton number) by three units (all three generations participate in the 't Hooft vertex). This is explained in 't Hooft's original paper. As a result, the proton is absolutely stable in the standard model. The lightest baryonic state that is unstable to decay into leptons is $^3$He. The deuteron is unstable with regard to decay into an anti-proton and leptons.

The rate is proportional to $[\exp(-8\pi^2/g_w^2)]^2$, which is much smaller than the rates for proton decay that have been discussed in extensions of the standard model. Note that the decay $^3\mathrm{He}\to$ leptons involves virtual $(b,t)$ quarks, and the rate contains extra powers of $g_w$ in the pre-exponent (which does not matter much, given that the exponent is already very big).

Just to give a rough number, the lifetime is a typical weak decay lifetime (say, $10^{-8}$ sec), multiplied by the instanton factor $$ \tau = \tau_w \exp(16\pi^2/g_w^2)=\tau_w\exp(4\pi\cdot 137\cdot\sin^2\theta_W) = \tau_w\cdot 10^{187}\sim 10^{180}\, sec \sim 10^{173}\, yrs $$ where I have neglected many pre-exponential factors which can be calculated, in principle, in the standard model.

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    $\begingroup$ Could I ask you to oomph this answer up a bit with (i) references for the stability of the proton in the SM, (ii) details and references for the decay of ${}^3$He, and (iii) an estimate for the SM half-life of ${}^3$He in years? (or e.g. as multiples of the current age of the universe.) I'm happy to bounty this up once it's eligible. $\endgroup$ Commented Oct 26, 2016 at 13:37
  • $\begingroup$ For reference: the Hubble time is, to 1 significant figure, $4\times 10^{14} \operatorname{sec}$. $\endgroup$ Commented Nov 14, 2016 at 0:49
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    $\begingroup$ Typo correction : the Hubble time is, to 1 significant figure, 5×10^17 sec. $\endgroup$ Commented Sep 21, 2018 at 17:55
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As far as I know, the standard model is assumed to have vanishing anomalies, i.e. that the proton does not decay in the standard model. See page 5 in this reference.

You are asking for this calculation. I do not know if one can keep calling it "the standard model".

Here is a strong statement, at the end of chapter 7.3.1:

Thus all possible anomalies cancel for every generation of the standard model. If in one generation a quark (or any other particle) were missing, one would get non-vanishing anomalies (not for SU(3)SU(3)SU(3), but for the three other combinations)

This was for the unbroken phase, but it continuous to make the same statement for the broken phase.

So the answer is that there should be an extension of the standard model to study B+L conservation effects.

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    $\begingroup$ They mean vanishing gauge anomalies. B+L is indeed anomalous. $\endgroup$
    – Thomas
    Commented Oct 26, 2016 at 4:49
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    $\begingroup$ @Thomas so, can you give a link where a proton decay is allowed within the standard model? The link I give is about anomalies in general. $\endgroup$
    – anna v
    Commented Oct 26, 2016 at 5:13
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    $\begingroup$ When people say that the SM is anomaly free they refer to gauge anomalies (color is trivially anomaly free, so they check SU(2)xU(1)). However, in the SM B+L is not gauged so it can (and indeed it does) have an anomaly. $\endgroup$
    – Thomas
    Commented Oct 26, 2016 at 18:18
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    $\begingroup$ 't Hooft studies a theory in which V-A is gauged, and there is a charm quark. Obviously, in the SM V-A is gauged (SU(2)_W), and the charm quark exists ('t Hooft cares about charm because he assumes that all fermions are weak doublets, and before the discovery of charm there was no particle to complete the weak doublet that contains the strange quark). 't Hooft did not know about (b,t), so in 1976 he concluded that in what we now call the standard model B violation by two units is allowed. We now know that B is violated by three units. $\endgroup$
    – Thomas
    Commented Oct 26, 2016 at 18:23
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    $\begingroup$ Regarding $\pi^0\to 2\gamma$: There is a subtlety. QCD has global flavor symmetries, which are anomalous, hence the triangle anomaly that contributes to neutral pion decay. But in the SM these flavor currents are gauged, and there are no gauge anomalies. How can that be? At low energies the QCD flavor anomalies get represented by WZ terms that describe anomalous pion interactions, and in the SM these terms add to the flavor anomalies from leptons so that the total flavor anomaly cancels. $\endgroup$
    – Thomas
    Commented Oct 27, 2016 at 4:06
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There have been several attempts to measure proton decay. So far, all have been unsuccessful. Various calculations give estimates ranging from $10^{30}$ to $10^{36}$ years.

Knowing the sensitivity of the experiments, we can set limits for the proton half-life. The current best measurements indicate it is $10^{34}$ years or more. For example, a 2014 publication from the Super-Kamiokande neutron detector in Japan gives a minimum of $5.9 × 10^{33}$ years.

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    $\begingroup$ Those are all predictions based on standard model extensions, not the predicted standard model decay rate. $\endgroup$ Commented Oct 26, 2016 at 3:21
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    $\begingroup$ Did you mean neutrino detector? $\endgroup$
    – Jens
    Commented Oct 26, 2016 at 11:27
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    $\begingroup$ -1 for not reading the question. $\endgroup$ Commented Oct 26, 2016 at 13:39

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