Neutrino Oscillation

  • #1
117
7

Homework Statement



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]

Homework Equations



Schrodinger Equation

The Attempt at a Solution


[/B]
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!
 

Answers and Replies

  • #2
35,714
12,293
##| \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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