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Two-Flavor Neutrino Oscillations

Neutrino oscillations — the fact that a neutrino created as an electron neutrino can later be detected as a muon neutrino — were confirmed experimentally in 1998 (Super-Kamiokande) and earned the 2015 Nobel Prize in Physics. The key insight is that the flavor eigenstates are not the same as the mass eigenstates. Here we derive the survival probability from scratch in the two-flavor case.


The two bases#

The free-particle Hamiltonian has mass eigenstates ∣ν1⟩,∣ν2⟩|\nu_1\rangle, |\nu_2\rangle with masses m1,m2m_1, m_2. Weak interactions, however, couple to flavor eigenstates ∣νe⟩,∣νμ⟩|\nu_e\rangle, |\nu_\mu\rangle. These two bases are related by a rotation through the mixing angle θ\theta:

∣νe⟩=cos⁡θ ∣ν1⟩−sin⁡θ ∣ν2⟩|\nu_e\rangle = \cos\theta\,|\nu_1\rangle - \sin\theta\,|\nu_2\rangle

∣νμ⟩=sin⁡θ ∣ν1⟩+cos⁡θ ∣ν2⟩|\nu_\mu\rangle = \sin\theta\,|\nu_1\rangle + \cos\theta\,|\nu_2\rangle

or in matrix form:

(∣νe⟩∣νμ⟩)=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)(∣ν1⟩∣ν2⟩)\begin{pmatrix}|\nu_e\rangle \\ |\nu_\mu\rangle\end{pmatrix} = \begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}\begin{pmatrix}|\nu_1\rangle \\ |\nu_2\rangle\end{pmatrix}

A general neutrino state can be written in either basis:

∣Ψ⟩=c1∣ν1⟩+c2∣ν2⟩=ce∣νe⟩+cμ∣νμ⟩|\Psi\rangle = c_1|\nu_1\rangle + c_2|\nu_2\rangle = c_e|\nu_e\rangle + c_\mu|\nu_\mu\rangle

with normalization ∣c1∣2+∣c2∣2=1|c_1|^2 + |c_2|^2 = 1 and ∣ce∣2+∣cμ∣2=1|c_e|^2 + |c_\mu|^2 = 1.


Time evolution#

Since ∣ν1⟩,∣ν2⟩|\nu_1\rangle, |\nu_2\rangle are energy eigenstates, their amplitudes evolve simply:

c1(t)=c1(0) e−iE1t/ℏ,c2(t)=c2(0) e−iE2t/ℏc_1(t) = c_1(0)\,e^{-iE_1 t/\hbar}, \qquad c_2(t) = c_2(0)\,e^{-iE_2 t/\hbar}

The flavor amplitudes at time tt are then:

(ce(t)cμ(t))=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)(c1(0) e−iE1t/ℏc2(0) e−iE2t/ℏ)\begin{pmatrix}c_e(t) \\ c_\mu(t)\end{pmatrix} = \begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}\begin{pmatrix}c_1(0)\,e^{-iE_1 t/\hbar} \\ c_2(0)\,e^{-iE_2 t/\hbar}\end{pmatrix}

Initial condition: start as a pure νe\nu_e#

Set ce(0)=1c_e(0) = 1, cμ(0)=0c_\mu(0) = 0. Inverting the rotation matrix gives the initial mass-basis amplitudes:

c1(0)=cos⁡θ,c2(0)=−sin⁡θc_1(0) = \cos\theta, \qquad c_2(0) = -\sin\theta

Substituting:

(ce(t)cμ(t))=(cos⁡θ−sin⁡θsin⁡θcos⁡θ)(cos⁡θ  e−iE1t/ℏ−sin⁡θ  e−iE2t/ℏ)\begin{pmatrix}c_e(t) \\ c_\mu(t)\end{pmatrix} = \begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{pmatrix}\begin{pmatrix}\cos\theta\; e^{-iE_1 t/\hbar} \\ -\sin\theta\; e^{-iE_2 t/\hbar}\end{pmatrix}

This gives:

ce(t)=cos⁡2θ  e−iE1t/ℏ+sin⁡2θ  e−iE2t/ℏc_e(t) = \cos^2\theta\; e^{-iE_1 t/\hbar} + \sin^2\theta\; e^{-iE_2 t/\hbar}

cμ(t)=sin⁡θcos⁡θ(e−iE1t/ℏ−e−iE2t/ℏ)c_\mu(t) = \sin\theta\cos\theta\left(e^{-iE_1 t/\hbar} - e^{-iE_2 t/\hbar}\right)


The survival probability#

The probability of still detecting a νe\nu_e at time tt is P(νe→νe)=∣ce(t)∣2P(\nu_e \to \nu_e) = |c_e(t)|^2. Computing this:

∣ce∣2=cos⁡4θ+sin⁡4θ+2sin⁡2θcos⁡2θcos⁡ ⁣((E2−E1)tℏ)|c_e|^2 = \cos^4\theta + \sin^4\theta + 2\sin^2\theta\cos^2\theta\cos\!\left(\frac{(E_2 - E_1)t}{\hbar}\right)

Using cos⁡4θ+sin⁡4θ=1−2sin⁡2θcos⁡2θ=1−12sin⁡2(2θ)\cos^4\theta + \sin^4\theta = 1 - 2\sin^2\theta\cos^2\theta = 1 - \frac{1}{2}\sin^2(2\theta) and the identity 2sin⁡2θcos⁡2θ=12sin⁡2(2θ)2\sin^2\theta\cos^2\theta = \frac{1}{2}\sin^2(2\theta):

∣ce∣2=1−sin⁡2(2θ)sin⁡2 ⁣((E2−E1)t2ℏ)|c_e|^2 = 1 - \sin^2(2\theta)\sin^2\!\left(\frac{(E_2-E_1)t}{2\hbar}\right)

Relativistic energy approximation#

Neutrinos are ultra-relativistic: Ei≫mic2E_i \gg m_i c^2. For fixed momentum pp:

Ei=p2c2+mi2c4≈pc(1+mi2c22p2)E_i = \sqrt{p^2c^2 + m_i^2c^4} \approx pc\left(1 + \frac{m_i^2 c^2}{2p^2}\right)

So:

E2−E1≈(m22−m12)c42E=Δm2c42EE_2 - E_1 \approx \frac{(m_2^2 - m_1^2)c^4}{2E} = \frac{\Delta m^2 c^4}{2E}

where E≈pcE \approx pc is the common energy. With L=ctL = ct (distance traveled):

(E2−E1)t2ℏ=Δm2c4 L4Eℏc\frac{(E_2 - E_1)t}{2\hbar} = \frac{\Delta m^2 c^4\, L}{4E\hbar c}

Putting it all together:

P(νe→νe)=1−sin⁡2(2θ)sin⁡2 ⁣(Δm2c4 L4Eℏc)\boxed{P(\nu_e \to \nu_e) = 1 - \sin^2(2\theta)\sin^2\!\left(\frac{\Delta m^2 c^4\, L}{4E\hbar c}\right)}


Conceptual questions#

Why do oscillations imply neutrinos have mass?#

Look at the argument of the sin⁡2\sin^2: it contains Δm2=m22−m12\Delta m^2 = m_2^2 - m_1^2. If both neutrinos were massless, Δm2=0\Delta m^2 = 0 and the probability would be identically 1 — no oscillation. More precisely:

  • Oscillations require two different phases e−iE1t/ℏe^{-iE_1 t/\hbar} and e−iE2t/ℏe^{-iE_2 t/\hbar} to interfere.
  • For massless particles, Ei=pcE_i = pc for all ii, so both phases are identical and they never interfere.
  • A non-trivial oscillation pattern therefore requires m1≠m2m_1 \neq m_2, and at least one must be non-zero.

Note: oscillations only constrain mass differences, not absolute masses. This is why neutrino oscillation experiments cannot tell us the absolute mass scale — only that Δm2≠0\Delta m^2 \neq 0.

Three flavors: how many must be massive?#

In the three-flavor case there are three mass eigenstates ∣ν1⟩,∣ν2⟩,∣ν3⟩|\nu_1\rangle, |\nu_2\rangle, |\nu_3\rangle and three flavor eigenstates ∣νe⟩,∣νμ⟩,∣ντ⟩|\nu_e\rangle, |\nu_\mu\rangle, |\nu_\tau\rangle. Oscillations between all three flavor pairs have been observed.

If only one mass eigenstate were massive (say m1≠0m_1 \neq 0, m2=m3=0m_2 = m_3 = 0), we would have Δm232=0\Delta m^2_{23} = 0, meaning no oscillation between the states that mix with ∣ν2⟩|\nu_2\rangle and ∣ν3⟩|\nu_3\rangle. Since oscillations between all flavors are observed, we need Δm122≠0\Delta m^2_{12} \neq 0 and Δm232≠0\Delta m^2_{23} \neq 0, which requires at least two massive eigenstates.

Two-Flavor Neutrino Oscillations
https://rohankulkarni.me/posts/notes/neutrino-oscillations/
Author
Rohan Kulkarni
Published at
2023-07-20
License
CC BY-NC-SA 4.0
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