I am an undergraduate student in Physics, I have a basic understanding of Particle Physics and Quantum Mechanics but none whatsoever of Quantum Field Theory.

I know that Neutrino mixing requires neutrinos to be massive (but why? Physically, couldn't neutrinos mix if they were massless?), and that their mass is usually estimated to be lower than an upper threshold.

But mathematically, does the Standard Model actually predict an upper limit on the neutrino mass, or does it just say that they are massless? In the former case, what is it stopping it from predicting a lower limit? In the latter case, so is it wrong?


3 Answers 3


Your question is addressed in this paper. The Standard Model as is can accomodate massive neutrinos but if the neutrinos have a mass, and no right handed neutrinos are added, the model becomes non-renormalisable. Adding right handed neutrinos fixes this.

The Standard Model doesn't make any predictions of neutrino mass, but then it doesn't predict any of the fermion masses. The masses of leptons and quarks are input parameters.


The Standard Model doesn't 'say' anything. It is a model, that we built, so we have to tell it something first for it to know about said thing.


There was no need for the neutrions to be massive when the model was built, so no mass terms were put into it's Lagrangian.

Then, some years later, these neutrino osciallation experiments were recorded.


As a side note, the experiments were actually looking for proton decays, but ended up finding none of them, and instead finding neutrino oscillations.


So the model builders went back to the drawing board, did some maths, and came up with a description of the oscillations. They found that the probability to measure a neutrino oscillation goes as

$$ P \propto \sin^2\left( \Delta m \right) $$

where $\Delta m$ is the difference in the mass of the two neutrinos (the one that it 'used to be' and the one that it oscillates to).

So we see that in order for this probability to be non-zero, we must have $\Delta m$ is non-zero.

But in order for $\Delta m$ to be non-zero the neutrino masses have to be non-zero to begin with!

The difference between zero and zero is certainly not non-zero!

So this is why they concluded that, in fact, the neutrinos must be massive, albeit it very slim.

Then they 'told' this information to the Standard Model, and the Standard Model now has non-massless neutrinos.


So in brief, the probability for an oscillation depends on the (difference in the) mass(es) of the neutrinos, which is what lets us conclude that they are not massless. Furthermore the Standard Model cannot 'explain' anything. It only repeats what we tell it (though one must admit that it does a fairly good job of it too!).

  • $\begingroup$ I may get lost somewhere, but where do you get the two zeros in "The difference between zero and zero is certainly not non-zero!"? $\endgroup$
    – Ooker
    Commented Oct 19, 2015 at 12:54
  • $\begingroup$ @Ooker That is, if neutrinos were massless, the the difference in their masses would be $$ 0 - 0 = 0 \ne a $$ with $a$ nonzero. $\endgroup$
    – Flint72
    Commented Nov 15, 2015 at 17:48

As could be quickly found from a wikipedia search, the 'classical' Standard Model of particle physics indeed predicts massless neutrino's. Therefore, the experimental evidence of neutrino oscillations is a strong indication that the Standard Model is missing some important physics.

This is not a huge problem (it's an exciting challenge, though!), since the Standard Model doesn't claim 'absolute truth'. It has been known for quite a while that the Standard Model is just an effective theory which it works very well in most situations that are relevant on the energy scales that are accessible to us right now, but does not claim to be a theory of everything (which it couldn't anyways, since it's missing gravity). Another wikipedia article lists some possible explanations for the observed neutrino mass and oscillations

  • $\begingroup$ I thought that the masslessness of neutrinos was an input variable of the SM, rather than a prediction (but I'm not sure). With the evidence of neutrino oscillations it appears that neutrinos do have mass, and so people are looking for valid extensions of the SM to incorporate this fact. $\endgroup$
    – Hunter
    Commented May 18, 2014 at 14:15
  • $\begingroup$ @Hunter To be honest, I'm not sure either now that you mention it! I just took the wikipedia quote 'Experiments indicate that neutrinos have mass, which the classic Standard Model did not allow' at face value... $\endgroup$
    – Danu
    Commented May 18, 2014 at 14:23
  • $\begingroup$ Ok, I guess we will we have to wait for someone with more knowledge to give us a conclusive answer ;). $\endgroup$
    – Hunter
    Commented May 18, 2014 at 14:26
  • $\begingroup$ The particular symmetries of the standard model are only proper for massless neutrinos. There are "one step more complicated" symmetries which have massive neutrinos but then you have to make a choice (Majorana or Fermi) that they had no basis to make at the time and the evidence on the ground was that any mass was very, very small... $\endgroup$ Commented May 18, 2014 at 17:51
  • 1
    $\begingroup$ @user23873 There has been (recently) a strong theoretical preference for Majorana neutrinos on the basis of the consequences that arise in leptogenesis allowing one to explain much of the matter-antimatter asymmetry. However EXO, KamLAND-Xen and friends are rapidly excluding the theoretically favored part of the parameter space, so things may get interesting... $\endgroup$ Commented May 18, 2014 at 18:34

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