2

My question is very simple: is there a fundametal difference between the CKM (responsible for quark mixing) and the PMNS matrix (responsible for lepton mixing)?

The CKM matrix allows (in charged current) mixing between quark families: for example $W^+\rightarrow u\bar{s}$.

While in the leptonic sector, a decay of type $W^+\rightarrow e^+\nu_{\mu}$ is impossible. However, when switching from $SU(2)$ eigenstates to mass eigenstates in the Lagrangian, I do not see why this decay is not allowed.

Qmechanic
  • 201,751
AxelAE
  • 273

1 Answers1

2

The CKM, PMNS matrices are mathematically absolutely analogous, except that the values and even hierarchies of all the parameters are entirely different in the two cases. (Also, we don't know whether right-handed neutrinos exist and whether the effective Majorana masses may be derived from Dirac masses or something else.) But both matrices may be reduced to three "real angles" and one complex "CP-violating phase".

And the decay $W^+\to e^+\nu_\mu$ is entirely possible if $\nu_\mu$ actually represents a mass eigenstate of the neutrino, much like the convention we use in the case of quarks.

However, the difference is that the "oscillations" between quarks that are not mass eigenstates are extremely fast, and happen at the length scale of the nuclear radius, while the oscillations between flavors of neutrinos take hundreds or kilometers or much more. The difference arises because the neutrino mass matrix entries are hugely lower than the quark masses – even the general elements of the quark mass matrix. The strange quark's mass is 150 MeV while the neutrino masses are comparable to 1 meV. That's a difference of some 11 orders of magnitude. The "in practice" difference is actually much higher because the neutrinos always move nearly by the speed of light and their oscillations are therefore slowed down by the relativistic time dilation (much greater than for the quarks).

For this reason, because of the very slow neutrino oscillations, we don't actually use the neutrino mass eigenstates in discussions about particle physics processes. By $\nu_\mu$, we really mean the $SU(2)$ partner of the left-handed muon (the middle mass eigenstate in the 3D space of the charged leptons of a given charge), and then the decay to $e^+\nu_\mu$ is strictly prohibited by the $SU(2)$ gauge symmetry.

But the process $W^+\to e^+\nu_2$ where $\nu_2$ is a mass eigenstate of the neutrinos is allowed because $\nu_2$ contains a nonzero addition of $\nu_e$, and it's this process involving $\nu_2$ that is analogous to the flavor-changing processes with quarks (controlled by the Cabibbo angle etc.).

Luboš Motl
  • 179,018
  • Modern SM school charts finally list mass eigenstates $\nu_L, \nu_M, \nu_H$ in generations, instead of the confusing non-propagating superpositions $\nu_e, \nu_\mu, \nu_\tau$, which have confused hundreds... – Cosmas Zachos Oct 18 '20 at 15:12
  • I completely disagree with the term "non-propagating". Every superposition may be propagating as well as any other superposition, that's guaranteed by the superposition principle of QM. If someone is confused by one basis more than others, it means that he or she has completely misunderstood basics of linear algebra and its role in quantum physics. – Luboš Motl Oct 19 '20 at 16:40
  • Yes they have... several, e.g. this, adduce this answer in support of their very (complete) misunderstanding. By "propagating", I'd mean something with a one-pole propagator... – Cosmas Zachos Oct 19 '20 at 17:51
  • Neutrinos have a one pole propagator, just the location of the pole is given by a matrix. – Luboš Motl Oct 21 '20 at 01:01
  • Sure, three locations in a matrix. A $\nu_\mu$ oscillates rather than propagate immutably. Obviously we don’t disagree on facts. Surely you are not arguing that that guy is not misreading you. – Cosmas Zachos Oct 21 '20 at 02:49
  • Oscillation is the same thing as the propagation of a superposition of energy eigenstates. It's really a big mistake to say that the 3-generation neutrino propagator has "several poles". That would mean that it's a product of three 1/(p^2 - m_I^2) terms. But it is not. It is a sum of such terms. Each term only has one pole so it's right to say that the sum also has one pole. You may define "propagate" in some narrower way but I still think that it is a definition appropriate for a person who doesn't understand linear algebra. – Luboš Motl Oct 22 '20 at 10:04
  • More generally, again, I don't think it is a sign of intelligence when someone insists that "a particular basis he likes" "has to be used" to express some matrix-like information. This bias is just irrational. It is a point of linear algebra - and the linearity postulate of quantum mechanics - that all bases are really equally legitimate. – Luboš Motl Oct 22 '20 at 10:05
  • Sigh... miscommunication has it’s own rewards... – Cosmas Zachos Oct 22 '20 at 10:38