Neutrino Oscillation: Solving for x,t

[DeldotB]

Homework Statement

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 $k_i= \sqrt{2m_iE/ \hbar^2}$

Homework Equations

Schrodinger Equation

The Attempt at a Solution

[/B]
This seems very simple, but I am 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!

 

[mfb]

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