Neutrino Oscillation Survival Probability

[Dahaka14]

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
1-\sin^{2}2\theta\sin^{2}\left(\frac{\Delta m^{2}t}{4p}\right)

Homework Equations

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

The Attempt at a Solution

\begin{align*}<br /> P_{e\rightarrow\nu_{e}}=\langle\nu_{e}|\psi(t)\rangle&amp;=\langle\nu_{e}|\nu_{e}\rangle e^{-iEt/\hbar}=\left|<br /> \left(<br /> \begin{array}{ccc}<br /> \cos\theta &amp; \sin\theta<br /> \end{array} \right)<br /> \left(<br /> \begin{array}{ccc}<br /> \cos\theta e^{-iE_{1}t/\hbar} \\<br /> \sin\theta e^{-iE_{2}t/\hbar}<br /> \end{array} \right)<br /> \right|^{2} \\<br /> &amp;=|\cos^{2}\theta e^{-iE_{1}t/\hbar}+\sin^{2}\theta e^{-iE_{2}t/\hbar}|^{2} \\<br /> &amp;=|e^{-iE_{1}t/\hbar}(\cos^{2}\theta+\sin^{2}\theta e^{-(iE_{2}-E_{1})t/\hbar})|^{2} \\<br /> &amp;=(\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}) \\<br /> &amp;=\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 \\<br /> &amp;=\sin^{2}2\theta\cos\frac{\Delta m^{2}t}{2p}+\cos^{4}\theta+\sin^{4}\theta \\<br /> &amp;=...? \\<br /> &amp;=1-\sin^{2}2\theta\sin^{2}\left(\frac{\Delta m^{2}t}{4p}\right)<br /> \end{align*}
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 .

 

[George Jones]

<br /> &amp;=\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<br />

Shouldn't

\frac{1}{2}\sin^{2}2\theta

be

\left( \frac{1}{2}\sin2\theta \right)^2 ?

Then, I think it works.

 

[Dahaka14]

Okay, so now I have
\begin{align*}<br /> &amp;=\frac{1}{2}\sin^{2}2\theta\cos\frac{\Delta m^{2}t}{2p}+\cos^{4}\theta+\sin^{4}\theta \\<br /> &amp;=\frac{2\sin^{2}2\theta\cos\frac{\Delta m^{2}t}{2p}}{4}+\frac{3+\cos4\theta}{4}<br /> \end{align*}

I'm sorry if it might be obvious, but I can't see it. I've just looked at it for too long.

 

[George Jones]

There could be more than one way to show this. Here's one way: what does \left( \cos^2 \theta + \sin^2 \theta \right)^2 equal?

 

[Dahaka14]

Thanks a lot!