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It is common knowledge that in neutrinos can be Dirac particles without any Majorana masses as given a mass matrix, \begin{equation} \left( \begin{array}{cc}\nu _L & \nu _R \end{array} \right) \left( \begin{array}{cc} 0 & m \\ m & 0 \end{array} \right) \left( \begin{array}{c} \nu _L \\ \nu _R \end{array} \right) \end{equation} However, I can't figure out why this model hasn't been ruled out by now just by considering the number of neutrinos hitting our detectors.

To see what I mean take the sun as an example. We know precisely how many neutrinos are leaving the sun and we know that they are all produced through the weak interaction, and so all the neutrinos leaving the sun are left-handed. If the mass matrix above is correct then the neutrinos would begin oscillating with their right-handed counterparts.

Thus when we detect neutrinos on earth (through a detector that only notices the weak interaction) only half the neutrinos should be left-handed and we should see half the signal we expect. This is of course not the case since this model isn't ruled out. What am I missing?

Ellie
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JeffDror
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  • The problem is that particles interact only with the left-handed component of the neutrino. Since interactions are the only way to "see" or measure particles, independent of whether neutrinos are Dirac or Majorana, only the left-handed component can be measured. Thus, nothing can be ruled out based on this data. – Prahar Nov 24 '14 at 14:15
  • Related. – Cosmas Zachos Feb 22 '18 at 19:24
  • @CosmasZachos Is it not true that if I create a purely left chiral Dirac fermion, after some time it will be purely right chiral? That’s my worry; why does this not affect observations? It does for other particles, as stated in the question you linked. – knzhou Feb 22 '18 at 19:44
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    The effect is covered in Bernardini 2006, and is associated to / predicated on zitterbewegung, but they all seem to agree it is infinitesimal for solar oscillations... They seem to be looking for areas where it might amount to something. It seems to be associated with delicate details of the wave packet shape of the neutrino... – Cosmas Zachos Feb 22 '18 at 20:58
  • To sum up, he Zitterbewegung superposed on the wave packet's rectilinear motion has frequency ~ 2 m (ν) ~ $10^{12}$/sec for neutrino mass of 1 meV, for the sake of argument. However, the modulation of the oscillation is resolutely minuscule, correcting unity by a negligible piece of $O((m_\nu/m_W)^2)$. – Cosmas Zachos Feb 23 '18 at 15:17
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    @CosmasZachos Yeah, after thinking about it more I came to the same conclusion! I'm now pretty sure that the highly upvoted answer below, the bountied answer, and the accepted answer are all wrong. – knzhou Feb 24 '18 at 10:09
  • @knzhou If you had the patience i lack, you might answer the linked "Related" question above... the issue is Dirac particles and their novel Zitterbewegung, and not the basically "scalar" wave packet oscillations on this page.... – Cosmas Zachos Feb 24 '18 at 16:04
  • I'm pretty confused about this issue. The discussions in two other SE answers imply that Zittbewegung does not exist (for standard definitions of the kinematic operators): https://physics.stackexchange.com/questions/19378/what-was-missing-in-diracs-argument-to-come-up-with-the-modern-interpretation-o?noredirect=1&lq=1 https://physics.stackexchange.com/questions/28672/is-zitterbewegung-an-artefact-of-single-particle-theory – Rococo Feb 28 '18 at 00:47
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    @Rococo There's no Zittbewegung but there are chirality oscillations! I'm not sure how the two issues are related. I always think about the chirality oscillations for plane waves, in which case there is no Zittbewegung since the position is not defined at all. – knzhou Mar 01 '18 at 18:38

3 Answers3

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Neutrinos interact in the Standard Model only through their left-handed component, via electroweak interactions. However, the propagating neutrinos, which are mass eigenstates, are described by a field that is a Dirac spinor, i.e. with both chiralities $$ \nu=\nu_L+\nu_R. $$ Therefore, when neutrinos are created or measured, the Dirac spinor is projected onto its left-handed component $$ \nu_L=P_L\,\nu=\frac{1-\gamma_5}{2}\nu. $$ This projection does not halve the number of counts, it just selects the interacting component of each Dirac spinor.

A reduction in the number counts is observed when oscillations among flavours are taken into account, and this is the effect measured by oscillation experiments.

mrf1g12
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  • I do see the OP's point. I guess naively from QM you think of a normalized wavefunction, $\nu = \frac{1}{\sqrt{2}}(\nu_L + \nu_R)$ giving a $\frac12$ in a probability. – innisfree Nov 24 '14 at 15:58
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    Thanks for the anser. The part that confused me was not the production of neutrinos but why they were not oscillating into right handed neutrinos which can occur as mentioned above by @innisfree. I think I understand better why that doesn't happen now as explained in my own answer. – JeffDror Nov 24 '14 at 16:46
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The point is that there is no mass difference between left-handed and right-handed neutrinos. The mass is only defined for the complete field $\nu=\nu_L+\nu_R$. For this reason, I think it's not correct to use the oscillation probability formula to compute oscillations between LH and RH neutrinos.

mrf1g12
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I think I now have an answer. My problem was that I assumed how much neutrinos oscillated depended solely on their level of mixing. With that intuition it seems that neutrinos should oscillate significantly into their right handed counterparts. However, there is more to the story. Oscillations are also dependent on the difference of masses between the mass eigenstates. If mass eigenstates are degenerate no oscillations can't take place (I still don't really have intuition to why this needs to be true...).

To see this is the case we can just recycle the well known neutrino oscillation formula. For two flavors we have the well known formula, \begin{equation} P ( \nu_L \rightarrow \nu _R ) = \sin ^2 2 \theta \sin ^2 \frac{ \Delta m ^2 L }{ 4E} \end{equation} where $\theta $ is the mixing angle (what I was more after $P(\nu_L \rightarrow \nu_L) $ is of course just $1-P(\nu_L \rightarrow \nu_R)$ in this simple case.

Maximal mixing (the situation above) corresponds to the case where $\theta=45^o$ and we indeed have $\sin^2\theta=1$, a mixing enhancement. However, there is a second contribution dependent on the difference of the masses between the eigenstates. When this is zero (as above). The oscillation probability also becomes zero.

To get a feeling for how this works I plotted the probability for the mass matrix, \begin{equation} \left(\begin{array}{cc} 1 & x \\ x & y \end{array}\right) \end{equation} This is the most general symmetric matrix up to normalization. I plotted the result as a function of $y$ (kind of equal to $\Delta m^{-1}$) and for different $x$ (kind of equal to the mixing), enter image description here

We see that at small $x $ (small mixing) and large $y$ (small mass difference) the oscillation probability vanishes. With this we see that number of measured neutrinos can put bounds on pseudo-Dirac neutrinos (ones with both Majorana and Dirac masses) but not Dirac neutrinos.

JeffDror
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    I don't think this answer is correct. If you take the argument literally, it basically implies that a Dirac mass never changes chirality, but this is clearly untrue for, e.g. electrons. The use of the neutrino oscillation formula is misleading because we are superposing positive and negative frequency solutions to the Dirac equation -- that is, the difference in frequencies is not zero, but $2m$! – knzhou Feb 24 '18 at 10:07
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    The real answer is that neutrinos are always ultrarelativistic, and in this limit the mixing angle goes to zero; this is also called 'helicity suppression'. – knzhou Feb 24 '18 at 10:08