Is mass really necessary for particle oscillations?

Question

I am a beginner in particle physics. Recently I am studying neutrino oscillations and first I came to know that particles have to have non-zero mass to oscillate between each other. I thought that it is so because if a particle (relativistic particle) doesn't have mass then the time period of its oscillations will become infinite i.e., "no oscillation". But later, when I looked at the mathematics of neutrino oscillations, I got to know that it is only the mass square square difference which is essential for this oscillation phenomenon and oscillation is possible if one of the two particles has zero mass. So now, I am confused whether the mass is really necessary? If it is not (as we know from mathematics), How can we think about the oscillations of a massless particle??

Answer 1

What is oscillating is the probability density distributions for the various flavors, i.e a feynman diagram must exist which will give the probability of finding an electron neutrino in a muon neutrino beam.

In the approximation where only two neutrinos participate in the oscillation, the probability of oscillation follows a simple pattern:

The blue curve shows the probability of the original neutrino retaining its identity. The red curve shows the probability of conversion to the other neutrino. The maximum probability of conversion is equal to sin^2(2θ). The frequency of the oscillation is controlled by Δm^2.

At least one neutrino must have a mass for the oscillations to make mathematical sense.

It is only at interaction level, which measures the flavor, that these oscillations in the relative flavor content can be seen.

Suppose the beam has muon neutrinos, so starting with a definite flavor and mass

I am not good at drawing . The image wants to convey that the muon neutrino line that has a probability of turning into an electron neutrino line is off mass shell, i.e. virtual. Energy and momentum are balanced at the vertices where the real on shell particles come out, in this case a proton and an electron.

Answer 2

Short version

Oscillations are a mere consequence of the evolution of a quantum superposition of mass eigenstates when those mass eigenstate differ from the flavour eigenstates, whereas the particular dependence in $\Delta m^2$ comes from the fact that neutrino masses are so small that they are ultra-relativistic.

Longer version $\newcommand{\ket}[2][]{\left\lvert#2 \right\rangle_{#1}}\renewcommand{\vec}[1]{\mathbf{#1}}\newcommand{\braket}[3][]{\left\langle #2 \rvert #3 \right\rangle_{#1}}$

The starting point is that a neutrino mass eigenstate $i$ of energy $E_i$ and momentum $\vec{p}_i$ has a wave function

$$\ket[M]{\nu_i(t)}=\exp i\left(E_i t - \vec{p}_i.\vec{x}\right)\tag{1}$$

at time $t$ and position $\vec{x}$. Nothing special about neutrinos here: this is true for any mass eigenstate of any particle.

The second key point is that neutrinos are produced in a definite flavour $f=e,\mu,\tau$

$$\ket{\nu_f}=\sum_i U_{fi}\ket[M]{\nu_i} \tag{2}$$

which is a superposition of mass eigenstates. The matrix $U$ is unitary, but not the trivial identity matrix. $U$ is the equivalent in the neutrino sector of the Cabibbo–Kobayashi–Maskawa (CKM) matrix in the quark sector.

Finally we look at the probability of transition

$$P_{f\to f'}=\left|\braket{\nu_{f'}(t)}{\nu_f}\right|^2 \tag{3}$$

and (1), (2), and (3) results in that probability to show oscillations versus the propagation length. Now your question was partially why the wavelengths of the oscillations are proportional to the $\Delta m_{ij}^2 = m_i^2 - m_j^2$. That is a consequence of taking the ultra-relativistic limit

$$p_i = \left(E_i^2 - m_i^2\right)^{-\frac{1}{2}}\approx E_i+\frac{m_i^2}{2E_f}$$

If you work out the "interference" terms in (3), you will see you get sines of terms proportional to $\Delta m_i^2$. If the neutrinos were not ultra-relativistic, the wavelengths would not be that simple but that would not conceptually change anything else.