# B Neutrino change of flavour and conservation of energy

1. Mar 14, 2016

### Ngineer

Hello,

I have a simple question. Has the discovery that some neutrinos change their flavor posed any issues with conservation of energy?! How has this been solved?!

Thank you.

2. Mar 14, 2016

### Staff: Mentor

No.

The distinct masses are associated with the neutrino "mass eigenstates." The "flavor eigenstates" (e, μ, τ) are superpositions (linear combinations) of the mass eigenstates.

When a neutrino is created as one flavor and then interacts (is detected) as another flavor, if you could measure the energies and momenta precisely enough, you would find that the same mass eigenstate is involved at both ends of the process. Because of conservation of energy and momentum, the neutrino has to have the same energy and momentum, and therefore also the same mass, at both ends. If you repeat this for many neutrinos (same initial and final flavors), you would find all three masses represented, with certain probabilities.

Last edited: Mar 14, 2016
3. Mar 14, 2016

### Orodruin

Staff Emeritus
Just to add that this is exactly what happens in the quark sector (apart from the neutral meson oscillations).

4. Mar 15, 2016

### vanhees71

The full answer is more complicated and goes to the heart of the particle interpretation of relativistic QFT. Such an interpretation is unambigously possible only for (asymptotic) free mass eigenstates. A neutrino, however, cannot be created or destroyed as a mass eigenstate but only in a flavor eigenstate in weak-interaction processes. Thus, strictly speaking we don't observe neutrinos as particles, because formal external neutrino lines of a Feynman diagram represent mass eigenstates. When we say we "detect a neutrino" we mean that there was some interaction of the neutrino with material in the detector, thereby destroying a neutrino-flavor state and creating other particles, which then are detected as proper asymptotic free particles.

So in reality we always have a process, where the neutrinos occur only as internal lines in the Feynman diagram, which describes the production process of a neutrino, e.g., the decay of a muon at some accelerator and the detection of the neutrino at the (maybe far distant) detector, where e.g., a lepton (of some flavor) is measured. Of course in the production and absorption process which is between true asymptotic states energy and momentum are conserved as with any other process described by a Feynman diagram. So there is no problem with energy-momentum conservation despite the "oscillation" of neutrinos between different mass eigenstates. There's a lot of confusion about this in the literature, because neutrino oscillations are treated in a simplified manner as if there is only some plane wave neutrino oscillating between different mass eigenstates. The resulting oscillation formula which involves the differences of the squared neutrino masses, the energy of the neutrinos and the distance between creation and detection places, is however a not too bad an approximation to what's really measured as described above.

5. Mar 15, 2016

### Ngineer

6. Mar 15, 2016

### vanhees71

Last edited: Mar 15, 2016
7. Mar 15, 2016

### ShayanJ

A Feynman diagram with an internal line as long as 1 AU? Really?

Also what you say, implies that the measurement of that lepton(of some flavor) is different from the measurement of the neutrino. But that also involves an interaction, in a bubble chamber or a spark chamber or something like that. What kind of interaction do you consider to be direct enough so that the detected particle can be counted as an external line?

8. Mar 15, 2016

### vanhees71

Well, internal lines of Feynman diagrams in the literal sense are infinitely long ;-)).

Of course, you always detect particles by interactions with the detector material. The difference between a particle whose mass eigenstate you can measure and a neutrino is that for a mass eigenstate an interpretation as an asymptotically free particle makes sense.

9. Mar 15, 2016

### ShayanJ

Oh...that's confusing! Can you clarify?(I hope that doesn't mean hijacking the thread! If it does, only a reference will do.)
Unless you mean the length of an internal line has no meaning so we can imagine it to be as long as we want. Is this what you mean?
But that only works if we assume that the neutrinos have no interaction in their way from sun to earth, which doesn't seem reasonable to me!
But this isn't the same as saying that neutrinos are always internal lines and other particles can be external lines. It only means you can't interpret an external neutrino line as a particle!

Last edited: Mar 15, 2016
10. Mar 15, 2016

### vanhees71

I don't know, how you read Feynman diagrams. For me they are a special very convenient and ingenious notation for the mathematical expressions that let me calculate S-matrix elements in perturbation theory. If you read it as space-time diagrams of "elementary interaction processes" as Feynman originally did in deriving the rules for QED, then the internal lines are "infinitely long", because by definition an S-matrix element describes the situation of a scattering processes of particles that are very far (mathematically infinitely far) away from each other initial, then run together and interact and then run appart after the collision. Then I wait very long (for an infinitely long time mathematically) such that the particles are very far (infinitely far mathematically) apart again, such that you can take them as non-interacting again. These are the asymptotic free in- and out-states. Physically an infinite distance doesn't make literal sense, and what's meant is that the particles are far apart compared to the range of the interaction. To discuss the case of a theory like electrodynamics, where the interaction has no range (because the photon is massless), would hijack this thread indeed.

Of course you can formally treat neutrinos at external lines, but the cost is that you run into confusion precisely concerning the conservation of energy and momentum, because you are not interested in the mass eigenstates, which you cannot observe but only in the flavor eigenstates which you can observe due to interactions with the detector (you can only observe what makes some interaction with material in the detector giving a signal!). So it's not clear, how to calculate such a Feynman diagram to make physical sense. All makes perfect sense in terms of QFT of course, and there is no problem with energy-momentum conservation. See the above cited paper by Akhmedov:

http://arxiv.org/abs/1008.2077

11. Mar 15, 2016

### snorkack

So just why is it not possible to observe a neutrino interacting in a definite flavour AND mass eigenstate?
For example, in a beta decay of a long lived isotope, the electron and the daughter nucleus, assuming they are stable, can have their energy and momentum measured with arbitrary precision. The precision of measuring the energy and momentum of the mother nucleus is only limited by Heisenberg uncertainty of its lifetime, wich can be long.

If you could measure the rest mass of a neutrino, while emitted, more precisely than the difference between eigenstates, would you find the neutrino to have been emitted in a specific mass eigenstate?

12. Mar 15, 2016

### Khashishi

Does that mean neutrinos only exist as virtual particles?

13. Mar 16, 2016

### vanhees71

That's the point of any decay process, not only weak decays. The mother nucleus is not in a definite state of the total Hamiltonian, including the weak interactions.

You look at it formally in a perturbative way (with very high accuracy) by taking into account the strong interaction first, which is responsible for binding quark and gluons together to nucleons and the nucleons to a nucleus, which can be stable as far as the strong interaction is concerned. But that's not the full story but there's also the weak interaction, which couples the quarks in the nucleus to the leptons and enables them to make transitions which we observe as $\beta$ decay. Thus the mother nucleus is not in an energy eigenstate of the full Hamiltonian including the weak interaction and thus can make this decay. This $\beta$ instability means that the nucleus, stable as far as the strong interaction is concerned, is in fact a resonance state with a finite energy (or if looked at it in its restframe its mass) width.

Now the interaction determines that the decay is to an electron and an electron neutrino, i.e., a flavor eigenstate. That's just the coupling according to the weak-isospin group coming out of the electroweak standard model or, for this purpose, Fermi's effective theory of $\beta$ decay. The finite mass width of the mother nucleus makes all this consistent with energy-momentum conservation for the asymptotic initial and final states as described in my postings above. Strictly speaking also the mother nucleus cannot be an external line as any resonance. It's always formed by some creation process out of reactions with stable particles, but that's of course academic, if you just have a sample of a long-lived $\beta$ instable medium, and the usual effective description of the decay in the just explained perturbative sense is accurate enough and all you can observe anyway.

14. Mar 16, 2016

### vanhees71

Strictly speaking yes.

15. Mar 16, 2016

Staff Emeritus
Strictly speaking everything exists only as virtual particles, since real particles are defined as plane waves - going from infinity to infinity. But this is a singularly un-useful way of thinking about things. Our senses tell us the difference between real and virtual (real particles can be counted, for example), and the fact that real particles don't quite match up to the mathematical approximation of infinite plane waves shouldn't make us throw out the baby with the bathwater. There aren't frictionless surfaces or massless pulleys either.

16. Mar 16, 2016

### snorkack

How is energy conserved when a state has a short lifetime and therefore large natural width?
Say, you have an atom with a wide spectral line, say centered at 601,0 nm and 2,0 nm wide, from 600,0 nm to 602,0 nm. Further suppose that the width of the line is purely due to short lifetime of one state - the other state is longlived, with narrowly defined energy.
And now you irradiate these atoms with radiation of similar wavelength but produced from different states and with different line width. Say a spectral line centered at 601,9 nm, and 0,2 nm wide, from 601,8 nm to 602,0 nm.
So, if atoms are excited into a broad state centered at 601,0 nm with a narrow incoming radiation centered at 601,9 nm, will they reemit what came in, a narrow band centered at 601,9 nm, because energy has to be conserved by each atom severally? Or will the atoms reemit a broad band centered at 601,0 nm, because a state is a state no matter which legal way it was created, and create the missing energy from nowhere?

17. Mar 19, 2016

### vanhees71

Energy is always conserved when looking at "complete processes" as defined by S-matrix elements, i.e., the transitions between states of asymptotic free and thus stable particles. In your case of an atom the only stable state is its ground state. Now you excite this atom by, e.g., absorbing a photon, to an excited state, which lives some finite (average) lifetime and thus emits again a photon and the atom is going back to its ground state. Alltogether energy is conserved, although the excited state can have a large width in energy because it's of short lifetime as in your example.