# Neutrino Oscillation

Tags:
1. Nov 30, 2015

### DeldotB

1. The problem statement, all variables and given/known data

Suppose that two neutrinos are created in the sun - call the states $|{ \nu_1}\rangle$ and $|{ \nu_2}\rangle$.

(Among many other things) I am asked to show that once the neutrinos have propigated a distance x after a time t, the states satisfy:

$|{ \nu_1}(x,t)\rangle = e^{i \phi_1} | \nu_1(0,0) \rangle$
$|{ \nu_2}(x,t)\rangle = e^{i \phi_2} | \nu_2(0,0) \rangle$

Where $\phi_{1,2} = k_ix-E_it/ \hbar$ where $k_i= \sqrt{2m_iE/ \hbar^2}$
2. Relevant equations

Schrodinger Equation

3. The attempt at a solution

This seems very simple, but im missing a factor:

Solving the time independent schrodinger equation yields: $| \nu_1 (0,0) \rangle = e^{-ikx}$ where $k= \sqrt{2mE/ \hbar^2}$.

Tagging on time dependence yields: $| \nu_1 (t) \rangle = e^{-ikx} e^{-iEt/ \hbar}$= $e^{-iEt/ \hbar} | \nu_1 (0,0) \rangle$.

So my question is: I tagged on the time dependence factor (from solving the time dependent s.e) and I got $e^{-iEt/ \hbar} | \nu_1 (0,0) \rangle$. But the problem states after the neutrinos have propigated a distance x after a time t. But isn't the "distance x" tied up in $| \nu_1 (0,0) \rangle = e^{-ikx}$ ?
Why are the solutions of the form $|{ \nu_1}(x,t)\rangle = e^{i (k_ix-E_it/ \hbar)} | \nu_1(0,0) \rangle$ instead of just $|{ \nu_1}(x,t)\rangle = e^{iE_it/ \hbar} | \nu_1(0,0) \rangle$?

I hope this makes sense. Thanks in advance!

2. Dec 1, 2015

### Staff: Mentor

$| \nu_1(\color{red}{0},0) \rangle$ has x=0, it does not depend on x.
You calculated $| \nu_1(x,0) \rangle$ which is something different.