# Are neutrinos just different states of a single particle, in essence?

When Neutrino oscillation phenomenon is analysed quantum mechanically, it is said that,

\begin{align} |\nu_e \rangle &= \cos(\theta) | \nu_1 \rangle + \sin(\theta) |\nu_2 \rangle,\\ |\nu_\mu \rangle &= -\sin(\theta )| \nu_1 \rangle + \cos(\theta) | \nu_2 \rangle,\end{align} where $$| \nu_1 \rangle, | \nu_2 \rangle$$ are energy eigenkets.

After this point, using some machinery about how the energy eigenkets evolve in time, it has been found that a $$\nu_e$$ turns into $$\nu_\mu$$ with some probability having an oscillatory characteristic.

However, when I first saw this analysis, I stumbled for a couple of minutes because before that point I always imagined in my mind that "we have a particle, and it has a state $$|\alpha \rangle$$ ". However, the above argument suggest that all of the neutrinos, in a sense, are just different states of a single particle, so we gave different names to the different states of a some "particle" or "thing".

For example, in Stern-Gerlach experiment, we could have named silver atoms that are going up as up-silver, and the others as down-silver; if we were to do that, we would have named different states of a single atom.

Question:

In the case of Neutrino oscillation phenomenon, up to what degree the mental picture outlined above would be correct? Where would it fail? Is there any other such phenomena?

On both sides of your two equations, there are two distinct particles: The r.h.s. takes place in the "mass basis", where you identify particles $$\lvert \nu_1\rangle$$ and $$\lvert \nu_2\rangle$$. The l.h.s. lives in the "lepton basis", where you identify particles $$\lvert \nu_\mu\rangle$$ and $$\lvert \nu_\mathrm{e}\rangle$$.

These are just two different choices of basis for the space of neutrinos. The mass basis is natural when you look at stationary states, but the lepton basis is natural when you want to talk about interactions with other particles - it is the $$\lvert \nu_\mu\rangle$$ that participates in weak interactions involving muons, not one of the mass basis states.

On a more general level, a particle state is just another quantum state. And quantum states can always be expressed as superpositions of other states. I could define two "electrophotons" as the superpositions $$\lvert \gamma\rangle \pm \lvert \mathrm{e}\rangle$$ of a photon and an electron state and re-express all electron and photon states in terms of electrophotons. It just wouldn't be very useful.

Maybe it helps to see that this notion of re-evaluating our notions of the "best" choice of basis for a particle states is not exclusive to neutrinos: When we think about electrons and how the weak interaction only couples to states with a particular chirality, we also find that the notion of the "mass basis" electron can be expressed as the superposition of electron states with a definite chirality. It is the mass basis electron that we detect, but the chiral states that participate in the weak interaction (or not). For more on that particular case, see this answer of mine.

In the end, quantum states are simply not amenable to picturing them with our classical intuition. You're surprised that one can express a particle state as a superposition mostly because your ontology of "a particle" is intuitively classical - a little ball that sits in space. But that's not how quantum states work - state vectors can superpose in a way that classical states cannot.

• About your last paragraph, maybe I'm missing some point, but what made me confused initially was the fact that we are not defining the state of some particle, we are defining the particle's itself. – Our Jan 23 '19 at 21:34

Defining what a particle is and what not is a complicated question. It is basically convention.

In one particle quantum mechanics you have a wavefunction $$\psi(x)$$ which basically tells you what is the probability to find the particle at position $$x$$. In quantum field theory you have instead a "wavefunction" $$\psi\{A^\mu(x)\}$$ which gives you the probability that the field has the classical configuration $$A^\mu$$(x) (For example $$A^\mu$$ could be the electromagnetic field). Now you choose one of those "wavefunctions" $$|\psi\{A^\mu(x)\}\rangle$$ and simply define it as a particle. In principle you basically could define whatever you want as a particle.

In reality it is convenient to define them as eigenstates of the total momentum operator (i.e. such that your particle has a definite momentum) and of the Hamiltonian (such that the particle has a definite energy). That means that the particle states you define depend on the Hamiltonian!

Now go back to neutrino oscillations. It is due to a mass term in the Hamiltonian. If you exclude this from your theory you have a Hamiltonian which only has a kinetic energy term. Then you define the neutrinos as eigenstates (your $$|\nu_e\rangle$$ and $$|\nu_\mu\rangle$$) of this Hamiltonian. Now you add a mass term to the Hamiltonian. Since the Hamiltonian is different you will also have different states ($$|\nu_1\rangle$$ and $$|\nu_2\rangle$$). These are just superpositions of the old states. And now you simply define them as your new particles.

The questions whether or not neutrinos are just different states of the same particle or not is a philosophical one. You can say that all the particles are just different states of one particle. This procedure described above is not restricted to neutrinos. For example in QED you start with a "plain" electron and a "plain" photon. When you add the electromagnetic interaction you find that the real observable electron is actually a superposition of your old electron with photons. So you could also say that electron and photon are just different states of a particle.

Edit: Before I come to your question about neutrino going into gluon I forgot to mention something. To define particle states you normally take into account not only Hamiltonian and momentum operator, but also all other conserved quantities (like charge, color, spin, ... whatever). This means your particles have a definite energy, momentum, charge, color, spin, ... This means you can only built superpositions which satisfy this. For example a superposition of a neutrino (which has no color) with a gluon (which has color) would not give a state with definite color (and its forbidden therefore). However it is possible (and I am pretty sure that this is realized in nature) to have a superposition of a neutrino and two gluons (where the gluons have opposite color, such that the total color vanishes.)

Ok now why don't we see a neutrino going into a gluon? The reason is that the mixing is super super super small. Do you know Feynman diagrams? Imagine a diagram where a neutrino is coming in and a neutrino is going out. Anything that can happen in between will mix into your state. We want a neutrino and two gluons. For example $$\nu \to e^- + W^+$$ and then $$W^+ \to u + \bar{u}$$, then the $$u$$'s are emitting gluons, then $$u + \bar{u} \to W^+$$ and finally $$e^- + W^+ \to \nu$$ again. However there are many interaction vertices involved here and with each interaction in the diagram there is a lower probability of the process to occur. Therefore $$\nu \to g$$ is vanishingly small while for example $$\nu_e \to \nu_\mu$$ (which only needs one interaction vertex) has a much higher probability.

So typically particles mix when they have a direct interaction between them. Note that this is also the case for the electron + photon mixing because electrons interact directly with photons.

• I don't agree with the statement, that what a particle is, is convention (certain details are, like what is the particle and the antiparticle... but generally speaking it is incorrect). You are ignoring the core of QFT which is symmetries and representation theory. A particle is commonly understood as a localized excitation of a field that fulfils certain symmetry requirements. Comments on Wigner's Classification are a must here. This answer must be improved. – ohneVal Jan 23 '19 at 13:28
• About "You can say that all the particles are just different states of one particle"; however, for example, we observe an electron neutrino is turning into an muon neutrino, but not into a, say, gluon, so in that sense, should there be some groups s.t an element of one group cannot turn into another one ? – Our Jan 23 '19 at 14:02