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

Schrödinger Equation

The Attempt at a Solution

[/B]
This seems very simple, but I am missing a factor:

Solving the time independent Schrödinger 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.