# Neutrino Oscillation Survival Probability

## 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*} 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*}
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
Staff Emeritus
Gold Member
$$&=\frac{1}{2}\sin^{2}2\theta\left(\cos\frac{\Delta m^{2}t}{2p}-i\sin\frac{\Delta m^{2}t}{2p}+\cos\frac{\Delta m^{2}t}{2p}+i\sin\frac{\Delta m^{2}t}{2p}\right)+\cos^{4}\theta+\sin^{4}\theta$$

Shouldn't

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

be

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

Then, I think it works.

Okay, so now I have
\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*}

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
Staff Emeritus
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?