Neutrino Oscillation Survival Probability

  • Thread starter Dahaka14
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  • #1
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Homework Statement


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]

Homework 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]


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:.
 

Answers and Replies

  • #2
George Jones
Staff Emeritus
Science Advisor
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[tex]
&=\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
[/tex]

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.
 
  • #3
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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.
 
  • #4
George Jones
Staff Emeritus
Science Advisor
Gold Member
7,494
1,231
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?
 
  • #5
73
0
Thanks a lot!
 

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