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Neutrino oscillations aren't near the top of particle physics

The 2015 Nobel prize in physics went to Takaaki Kajita (University of Tokyo, 1/2) and Art McDonald (Queen's University, Ontario, Canada, 1/2), for their discovery of the neutrino oscillations – which may also be interpreted as the discovery of the neutrino mass. Congratulations!

I don't think that I know the Gentlemen in person. Of course, I know their achievements in cooking. A fajita is a flavored meat strip. And it's sensible that fajitas got the prize along with the hamburgers. McDonald's restaurants have repeatedly saved my life, for example in Santa Cruz, California. While I was at Harvard, I often walked a mile towards the Central Square, away from the hard core of the People's Republic of Cambridge which de facto bans corporations such as McDonald's. (This attitude of the pompous far left-wing bigots offends my nationalist feelings as well – McDonald's only became a big chain when it was bought by Ray Kroc, 1902-1984, a Czech American whose father was born 10 miles from my home.)

But back to the prize. It's a prize in particle physics and one that is totally legitimate but I am sure that lots of top particle physicists will be left a bit unexcited by that prize, too. For example, Nima Arkani-Hamed – who, as a Westerner, is primarily Canadian, just like McDonald – has explicitly stated that neutrino physics – something that the Fermilab has pretty much promoted to the core of the institute's activities – isn't an adequate flagship for the particle physics enterprises in a country that is as great as the U.S. I have always shared this view.

The oscillating neutrinos represent an example of a simple 3-dimensional (and perhaps 4- or 6- etc. but there's no evidence for that at this point) Hilbert space with a generic \(3\times 3\) matrix as the Hamiltonian. The general evolution is periodic – more precisely, the wave function is basically a superposition of three cosines – and "more ordinary" examples of many such oscillating quantum systems have been known since the 1920s.

Conceptually, a \(3\times 3\) Hamiltonian matrix is an extremely low-brow, unexciting thing.

Moreover, Kajita and McDonald are "bosses" of experimental teams and all the difficult questions about the "credit" in larger teams – and the role of politics and spokesmen etc. – may be discussed here, too.

When we were kids, and even relatively recently, around 2000, we would often see books that said that the neutrinos may have been massless and only upper limits on the masses could have been imposed. The neutrinos are three particle species \(\nu_e, \nu_\mu, \nu_\tau\) that are the electroweak \(SU(2)\) partners of the well-known \(e^-, \mu^-,\tau^-\). Well, only the left-handed 2-spinor in the electron, muon, tau has a new \(SU(2)\) partner so the neutrinos look chiral: all neutrinos we have observed so far are left-handed and all antineutrinos (the partners of the positron, antimuon, and antitau) are right-handed.

Because of the conservation of the angular momentum, a freely flying neutrino (which is left-handed) won't ever become an antineutrino (right-handed one). However, the neutrino and the antineutrino may actually be the same particle. If a fermion is the same thing as its antiparticle, we describe it as the so-called Majorana spinor, a Dirac spinor with an extra "reality condition" imposed on it that reduces the number of its degrees of freedom to one-half.

All the empirical data we know are compatible with the assumption that the observed neutrinos are Majorana particles. (That also follows from the scenario in which right-handed neutrinos exist but do have GUT-scale Majorana masses. In that case, the see-saw mechanism – assuming an electroweak-scale coupling between the left-handed and right-handed ones – produces tiny masses for the left-handed neutrinos that are Majorana, too.) However, there may also exist the right-handed two-spinor components that extend the known neutrino fields to a full Dirac spinor, just like in the case of the charged leptons, and the Dirac masses are those observed in the oscillation experiments. At this moment, we can't really distinguish between these two possibilities. Moreover, both possibilities – Majorana neutrinos and Dirac neutrinos – follow from comparably complete and satisfactory models of high-energy physics, grand unification, or string theory. If the right-handed neutrinos (needed for a Dirac neutrino) exist, then they are even more weakly interacting than the left-handed ones. They don't even interact through the usual weak interaction.

Some experiments have a chance to tell us whether the neutrinos are Dirac or Majorana.

However, when we focus our attention on the left-handed neutrinos that do interact weakly (which exclude antineutrinos), we may still talk about their masses (we don't know whether they're Majorana or Dirac masses). Imagine that we create a neutrino with the well-defined 3-momentum \(\vec p\). It may be an electron neutrino, a muon neutrino, a tau neutrino, or a superposition: the wave function really has 3 complex components.\[

\psi = \pmatrix{ \psi_{\nu,e}\\ \psi_{\nu,\mu} \\ \psi_{\nu,\tau} }

\] How does this state evolve in time? Well, the energy \(H=p^0\) is given by the relativistic expression \[

H = \sqrt{|\vec p|^2 + m^2 } = \dots

\] Now, the term \(m^2\) isn't really a universal constant because the masses of different flavors \(\nu_e,\nu_\mu,\nu_\tau\) may be (and, as we know, are) different. You may see that \(m^2\) is a Hermitian \(3\times 3\) matrix on the space of the three flavors. Let's assume that the neutrino moves almost by the speed of light which is equivalent to \[

|\vec p|^2 \gg m^2

\] i.e. the rest masses are negligible relatively to the momentum which is meant to say that the momentum is greater than any of the three eigenvalues of \(m^2\). If that's so, we may use the Taylor expansions to approximate \(H\) as\[

\dots = |\vec p| \cdot \sqrt{1 + m^2 / |\vec p|^2 } \approx |\vec p| + \frac{m^2}{2 |\vec p |} + \dots

\] The remaining terms are negligible in the limit. We've agreed that the initial state had a well-defined \(\vec p\) so the first, big term is proportional to the unit matrix and such terms in the Hamiltonian don't affect any observable quantities. Overall shifts of the energy don't cause or modify any oscillations etc. So it's only the second, smaller term – the greatest term in the expansion that isn't proportional to the unit matrix – which matters.\[

H_{\rm relevant} = \frac{m^2}{2|\vec p|}

\] Note that the momentum in the denominator is no longer squared; one of the copies has cancelled. This term in the Hamiltonian \(H\) is proportional to the \(3\times 3\) matrix of the squared masses. There is a coefficient which is a \(c\)-number.

It's clear how the wave function will oscillate. You may find the three eigenstates of \(m^2\) in the three-dimensional Hilbert space of the three neutrino flavors and write the most general solution of the time-dependent Schrödinger's equation. The eigenstates of \(m^2\) won't necessarily be the \(SU(2)\) partners of \(e^-, \mu^-,\tau^-\) – the latter three are defined as mass eigenstates in the three-dimensional space of charged leptons.

In fact, you may be pretty much certain that they won't be the partners. You need some \(U(3)\) transformation to switch from the neutrino mass eigenstates to the \(SU(2)\) partners of the charged lepton mass eigenstates – an analogy of the CKM matrix.

Fine. You write the most general solution of the time-dependent Schrödinger's equation and you may also determine the coefficient from the initial state. If you start with a mass eigenstate and a well-defined \(\vec p\), the state will be constant, up to an overall phase. But if you start e.g. with \(\nu_e\) which is not a mass eigenstate but a general combination of neutrinos' mass eigenstates, such an initial state will oscillate into superpositions of \(\nu_e,\nu_\mu, \nu_\tau\) and back. Back and forth. You may measure how quickly the neutrinos of one flavor become the other flavor(s) before they return (almost) back.

The rate of the oscillations will depend on the differences of the eigenvalues of \(m^2\). Note that the relevant term in the Hamiltonian also contained the factor of \(1/2|\vec p|\) which means that very high-energy (high-momentum) neutrinos will oscillate more slowly – and the oscillations will therefore manifest themselves as longer wavelengths. Let me emphasize that only the difference of the eigenvalues of \(m^2\) may be measured by the simple oscillations. If all values of \(m^2\) are shifted by the same constant, then, as long as the constant is low enough, the oscillation experiments will be insensitive to that.

OK, in various regimes, the 3-dimensional oscillations effectively reduce to various 2-dimensional oscillations which are responsible for the "main effect". In 1998, Kajita et al. at Kamiokande (Japan) observed a shortage of muon neutrinos in the atmosphere that resulted from "atmospheric neutrino oscillations", mostly \(\nu_\mu \leftrightarrow \nu_\tau\). In 2001, McDonald et al. at SNO (Ontario, Canada) observed that a part of the electron neutrinos from the Sun became muon and tau neutrinos (\(\nu_e\leftrightarrow \nu_\mu,\nu_\tau\)).

The eigenvalues are such that the electron-muon neutrino oscillations are much slower (a smaller difference of the \(m^2\) eigenvalues) than the muon-tau (or electron-tau) oscillations (a greater difference of the \(m^2\) eigenvalues). That's why the atmosphere (a relatively short wavelength) is enough to eliminate lots of the muon neutrinos (which are the dominant kind born in the atmosphere) by turning them into tau neutrinos, but you need something comparable to the Sun-Earth distance to turn lots of electron neutrinos into the other flavors.

Moreover, the eigenstates are such that the lightest 2 neutrino mass eigenstates are both almost equal mixtures of \(\nu_e,\nu_\mu\) – the 12-mixing angle is almost exactly 45 degrees (maximum mixing) while the heaviest neutrino mass eigenstate is almost exactly \(\nu_\tau\). There's still a marginally viable possibility of the "inverse hierarchy" – the lightest neutrino mass eigenstate would be closest to \(\nu_\tau\), the partner of the heaviest charged lepton – but this possibility leads to some other implications and some experiments in 2014 or 2015 actually indicated that the "straight" hierarchy seems more likely at this point.

The 13-element of the mass matrix – something pretty much turning \(\nu_e\) to \(\nu_\tau\) directly, if you wish – was only measured (and seen to be nonzero) some three years ago or so. The value roughly agreed with some predictions that were done in F-theory, a non-perturbative approach to type IIB string theory.

When I was (and similarly, lots of particle physicists were) getting more familiar with the renormalization group and grand unified theories and similar things, I was getting increasingly certain that there was no reason for the neutrino masses to be exactly zero which really means that they won't be exactly zero. At least, some higher-dimension operators have to exist that make them massive.

So the very fact that the neutrinos are massive – and oscillating – is pretty much guaranteed from a theorist's viewpoint. The rest is a "straightforward" measurement. The experiments are fun – it's hard to detect neutrinos because they're so weakly interacting – but in some sense, this whole machinery is straightforward and conceptually trivial. You just measure the elements of some \(3\times 3\) matrix.

It's much more interesting to look at models that actually try to calculate the neutrino mass matrix etc. But those theorists are unlikely to get a Nobel prize for quite some time. Or maybe I will be proven wrong?

The entries of the neutrino mass matrix were unknown up to recently and they may be measured. So this is a great example of the "old-fashioned empirical science". But I think it's fair to say that the measurement of the neutrino mass matrix is also a great example how boring and conceptually shallow issues you run into if you insist on the "empirical rooting" of your particle physics too dogmatically.

If the U.S. decided to turn their particle physics into an industry that measures the neutrino mass matrix ever more accurately, they would return to the end of the 19th century when Lord Kelvin said that physics was over and the only remaining task was to measure some parameters more accurately than before. Well, modern physics wasn't even getting started. And even today, lots of breakthroughs that are comparable to the 20th century breakthroughs are clearly still ahead of us. In the 19th century, they actually observed millions of parameters they had no chance to explain at that time – like all the thousands if not millions of frequencies of spectral lines of elements and compounds – and a revolution in science (quantum mechanics) had to occur before they actually understood those things.

We no longer have millions of observable unexplained numbers (aside from those that we are confident to have followed from random events in the history) but we still have some 30 physically meaningful parameters in the Standard Model (now I do count the neutrino mass matrix although the neutrino masses are often said "not to belong" to the Standard Model) which means that some extra layers of explanations are probably still waiting for us.

It's OK for the Nobel committee to be giving prizes only to "totally obviously empirically proven" discoveries such as the nonzero neutrino masses. But it would be wrong for a physicist to be affected by these conventions of the Nobel committee. The truly important discoveries are probably lying elsewhere. I feel that not much has changed even in neutrino physics, let alone physics, since the time when I wrote a term paper for Glennys Farrar's course at Rutgers in 1998 – and it was at the time when the first Nobel-winning experiment was just getting published.

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