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Neutrino Oscillation Survival Probability

  1. Oct 25, 2008 #1
    1. The problem statement, all variables and given/known data
    I'm lost at how to derive the probability of a neutrino species surviving an oscillation. After performing calculations, I can't seem to get it into the nice tidy form
    [tex]1-\sin^{2}2\theta\sin^{2}\left(\frac{\Delta m^{2}t}{4p}\right)[/tex]

    2. Relevant equations
    Whatev...
    [tex]|\langle\nu_{e}|\psi(t)\rangle|^{2}[/tex]
    [tex]E_{i}=\sqrt{p^{2}+m_{i}^{2}}\approx p+\frac{m_{i}^{2}}{2p},~\text{where}~p\gg m[/tex]
    [tex]\text{and}~\Delta m^{2}=m_{2}^{2}-m_{1}^{2}[/tex]


    3. The attempt at a solution
    [tex]\begin{align*}
    P_{e\rightarrow\nu_{e}}=\langle\nu_{e}|\psi(t)\rangle&=\langle\nu_{e}|\nu_{e}\rangle e^{-iEt/\hbar}=\left|
    \left(
    \begin{array}{ccc}
    \cos\theta & \sin\theta
    \end{array} \right)
    \left(
    \begin{array}{ccc}
    \cos\theta e^{-iE_{1}t/\hbar} \\
    \sin\theta e^{-iE_{2}t/\hbar}
    \end{array} \right)
    \right|^{2} \\
    &=|\cos^{2}\theta e^{-iE_{1}t/\hbar}+\sin^{2}\theta e^{-iE_{2}t/\hbar}|^{2} \\
    &=|e^{-iE_{1}t/\hbar}(\cos^{2}\theta+\sin^{2}\theta e^{-(iE_{2}-E_{1})t/\hbar})|^{2} \\
    &=(\cos^{2}\theta+\sin^{2}\theta e^{-i(E_{2}-E_{1})t/\hbar})(\cos^{2}\theta+\sin^{2}\theta e^{i(E_{2}-E_{1})t/\hbar}) \\
    &=\frac{1}{2}\sin^{2}2\theta\left(\cos\frac{\Delta m^{2}t}{2p}-i\sin\frac{\Delta m^{2}t}{2p}+\cos\frac{\Delta m^{2}t}{2p}+i\sin\frac{\Delta m^{2}t}{2p}\right)+\cos^{4}\theta+\sin^{4}\theta \\
    &=\sin^{2}2\theta\cos\frac{\Delta m^{2}t}{2p}+\cos^{4}\theta+\sin^{4}\theta \\
    &=...? \\
    &=1-\sin^{2}2\theta\sin^{2}\left(\frac{\Delta m^{2}t}{4p}\right)
    \end{align*}[/tex]
    Can someone help me fill in the blank? It would be best if I could do it on my own, so if possible just give me hints. If it is too explicit, then just tell me I guess. But as we all know, in order for me to truly own the idea, I should only be gently pushed toward the answer :smile:.
     
  2. jcsd
  3. Oct 25, 2008 #2

    George Jones

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    Staff Emeritus
    Science Advisor
    Gold Member

    Shouldn't

    [tex]\frac{1}{2}\sin^{2}2\theta[/tex]

    be

    [tex]\left( \frac{1}{2}\sin2\theta \right)^2 ?[/tex]

    Then, I think it works.
     
  4. Oct 25, 2008 #3
    Okay, so now I have
    [tex]\begin{align*}
    &=\frac{1}{2}\sin^{2}2\theta\cos\frac{\Delta m^{2}t}{2p}+\cos^{4}\theta+\sin^{4}\theta \\
    &=\frac{2\sin^{2}2\theta\cos\frac{\Delta m^{2}t}{2p}}{4}+\frac{3+\cos4\theta}{4}
    \end{align*}[/tex]

    I'm sorry if it might be obvious, but I can't see it. I've just looked at it for too long.
     
  5. Oct 25, 2008 #4

    George Jones

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    Staff Emeritus
    Science Advisor
    Gold Member

    There could be more than one way to show this. Here's one way: what does [itex]\left( \cos^2 \theta + \sin^2 \theta \right)^2[/itex] equal?
     
  6. Oct 25, 2008 #5
    Thanks a lot!
     
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