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Neutrino Oscillation

  1. Nov 30, 2015 #1
    1. The problem statement, all variables and given/known data

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

    (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:

    [itex] |{ \nu_1}(x,t)\rangle = e^{i \phi_1} | \nu_1(0,0) \rangle [/itex]
    [itex] |{ \nu_2}(x,t)\rangle = e^{i \phi_2} | \nu_2(0,0) \rangle [/itex]

    Where [itex] \phi_{1,2} = k_ix-E_it/ \hbar[/itex] where [itex] k_i= \sqrt{2m_iE/ \hbar^2} [/itex]
    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: [itex] | \nu_1 (0,0) \rangle = e^{-ikx} [/itex] where [itex] k= \sqrt{2mE/ \hbar^2} [/itex].

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

    So my question is: I tagged on the time dependence factor (from solving the time dependent s.e) and I got [itex]e^{-iEt/ \hbar} | \nu_1 (0,0) \rangle [/itex]. 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 [itex] | \nu_1 (0,0) \rangle = e^{-ikx} [/itex] ?
    Why are the solutions of the form [itex] |{ \nu_1}(x,t)\rangle = e^{i (k_ix-E_it/ \hbar)} | \nu_1(0,0) \rangle [/itex] instead of just [itex] |{ \nu_1}(x,t)\rangle = e^{iE_it/ \hbar} | \nu_1(0,0) \rangle [/itex]?

    I hope this makes sense. Thanks in advance!
     
  2. jcsd
  3. Dec 1, 2015 #2

    mfb

    User Avatar
    2016 Award

    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.
     
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