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2015Summer.tex
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%!TEX program = pdflatex
\documentclass{tufte-handout}
%\geometry{showframe}% for debugging purposes -- displays the margins
\usepackage{amsmath}
% Set up the images/graphics package
\usepackage{graphicx}
\setkeys{Gin}{width=\linewidth,totalheight=\textheight,keepaspectratio}
\graphicspath{{graphics/}}
\title{Neutrino Oscillations in Vacuum and Matter \thanks{2015 Summer}}
\author[Lei Ma]{Lei Ma}
\date{May 15 2015} % if the \date{} command is left out, the current date will be used
% The following package makes prettier tables. We're all about the bling!
\usepackage{booktabs}
% The units package provides nice, non-stacked fractions and better spacing
% for units.
\usepackage{units}
\usepackage{ulem}
% The fancyvrb package lets us customize the formatting of verbatim
% environments. We use a slightly smaller font.
\usepackage{fancyvrb}
\fvset{fontsize=\normalsize}
% Small sections of multiple columns
\usepackage{multicol}
% code highlighting
\usepackage{listings}
% For url links
\usepackage{hyperref}
% \usepackage{minted}
% \usepackage[utf8]{inputenc}
% \usepackage[english]{babel}
% Provides paragraphs of dummy text
% These commands are used to pretty-print LaTeX commands
\newcommand{\doccmd}[1]{\texttt{\textbackslash#1}}% command name -- adds backslash automatically
\newcommand{\docopt}[1]{\ensuremath{\langle}\textrm{\textit{#1}}\ensuremath{\rangle}}% optional command argument
\newcommand{\docarg}[1]{\textrm{\textit{#1}}}% (required) command argument
\newenvironment{docspec}{\begin{quote}\noindent}{\end{quote}}% command specification environment
\newcommand{\docenv}[1]{\textsf{#1}}% environment name
\newcommand{\docpkg}[1]{\texttt{#1}}% package name
\newcommand{\doccls}[1]{\texttt{#1}}% document class name
\newcommand{\docclsopt}[1]{\texttt{#1}}% document class option name
% For quantum braket notation
\newcommand{\bra}[1]{\left\langle #1\right|}
\newcommand{\ket}[1]{\left| #1\right\rangle}
\newcommand{\braket}[2]{\langle #1 \mid #2 \rangle}
\newcommand{\avg}[1]{\left< #1 \right>}
\begin{document}
\maketitle% this prints the handout title, author, and date
\begin{abstract}
\noindent Notes for neutrino oscillations in vacuum and dense matter.
\end{abstract}
\tableofcontents
\section{Symbols and Definitions}
\begin{itemize}
\item
$\Delta = \sqrt{2} G_F n(x) $
\item
$\omega = \frac{\Delta m^2}{2E}$
\end{itemize}
\newpage
\section{Interesting Topics}
\subsection{Decoherence}
Not really decoherence but in the view of wave packet formalism different propagation eigenstates will be far away from each other.
\subsection{Vacuum Oscillations 3 Flavor}
The trick to solve the equations is to subtract the $\frac{m_1^2}{2E} \mathbf{I}$ from Hamiltonian in mass eigenbasis.
\begin{align*}
&\mathbf{H}- \frac{m_1^2}{2E}\mathbf{I} \\
=& \frac{1}{2E}\begin{pmatrix}
m_1^2 & 0 & 0 \\
0 & m_2^2 & 0\\
0 & 0 & m_3^2
\end{pmatrix} - \frac{m_1^2}{2E} \mathbf{I} \\
=& \frac{1}{2E} \begin{pmatrix}
0 & 0 & 0 \\
0 & \Delta m_{12}^2 & 0 \\
0 & 0 & \Delta m_{13}^2
\end{pmatrix}
\end{align*}
For any kind of potential, do this first then add in the potentials.
Normal hierarchy and inverted hierarchy just changes the behavior of small scale oscillations. See caption of figure \ref{vacOsc3Flavor} for details.
Invert hierarchy of $m_1$ and $m_2$ also doesn't change anything on average sense.
\subsection{Matter}
Oscillation length in calculations, equation \ref{eq:matter-oscillation-length} and \ref{overal-oscillation-length}.
Then a discussion of the energy dependence of the vacuum oscillation length is shown in figures \ref{fig:comparisonVacOscLengthMatterLength} and \ref{fig:lengthComparison}.
Matter effect can be a suppression or enhancement. See equation \ref{sin2thetaofx} and the discussion below.
A general discussion of solving the problem is located at equation \ref{solving-matter-effect-general}. There are three different ways to solve it however the one begins with vacuum mass eigenbasis Hamiltonian is the best.
To solve the solar problem, there are three stages.
\begin{itemize}
\item Region with very high number density which is above resonance: $\ket{\nu_e}$ projects onto instantaneous mass eigenstates then propagate adiabatically on these states.
\item Region around resonance: transition from a instantaneous mass eigenstate to another happens.
\item After the transition: stays on instantaneous mass eigenstates.
\end{itemize}
A general discussion is presented in subsection {\it{General Discussion for Neutrinos Interacting with Matter}}.
\subsection{Ternary Diagrams}
Hierarchy of $\Delta m_{13}$ changes the small oscillations, which can be seen clearly in ternary diagrams.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% Vacuum Oscillations %%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Vacuum Oscillations}\label{vacuum-oscillations}
Schrodinger equation is
\begin{equation}
i\partial_t \ket{\Psi} = \mathbf H \ket{\Psi},
\end{equation}
where for relativistic neutrinos, the energy is\footnote{They all have the same momentum but different mass. The thing is we assume they have the same velocity since the mass is very small. To have an idea of the velocity difference, I can calculate the distance travelled by another neutrino in the frame of one neutrino.\newline
Assuming the mass of a neutrino is 1eV with energy 10MeV, we will get a speed of $1-10^{-14}$c. This $10^{-14}$c will make a difference about $3\mu\mathrm{ m}$ in 1s.
\newline {\bf To Be Discussed!} \newline Will decoherence happen due to this? For high energy neutrinos this won't be a problem however for low energy neutrinos this will definitely cause a problem for the wave function approach.Because the different mass eigenstates will become decoherent gradually along the path.\newline
A estimation of the decoherence length is\begin{equation*}
l_{\mathrm{coh}}=\frac{v_g}{\Delta v_g}\sigma.
\end{equation*}
To obtain the relation,
\begin{align*}
\Delta x &= \lvert v_1 - v_2 \rvert t_{\mathrm{coh}}\\
\frac{\hbar c}{\Delta E} & = \lvert \frac{m_1^2}{2E_1^2} - \frac{m_2^2}{2E_2^2} \rvert t_{\mathrm{coh}} \\
\frac{\hbar c}{\Delta E} & = \frac{1}{2E}\lvert \Delta m_{12}^2 \rvert t_{\mathrm{coh}}
\end{align*}
}
\begin{align*}
\mathbf H^m &= \begin{pmatrix}\sqrt{p^2 + m_1^2} & 0 & 0 \\ 0& \sqrt{p^2 + m_2^2} & 0 \\ 0 & 0 & \sqrt{p^2 + m_3^2} \end{pmatrix},
\end{align*}
in which the energy terms are simplified using the relativistic condition
\begin{align}
\sqrt{p^2+m_i^2} & = p\sqrt{1 + \frac{m_i^2}{p^2}} \\
&\approx p(1 + \frac{1}{2} \frac{m_i^2}{p^2}).
\end{align}
In general the flavor eigenstates are the mixing of the mass eigenstates with a unitary matrix $\mathbf U$, that is
\begin{equation}
\ket{\nu_{\alpha}} = U_{\alpha i} \ket{\nu_i},
\end{equation}
where the $\alpha$s are indices for flavor states while the $i$s are indices for mass eigenstates.
To find out the equation of motion for flavor states, plugin in the initary tranformation,
\begin{equation}
i U_{\alpha i} \partial_t \ket{\nu_i} = U_{\alpha i} H^m_{ij} \ket{\nu_j}.
\end{equation}
I use index ${}^m$ for representation of Hamiltonian in mass eigenstates. Applying the unitary condition of the transformation,
\begin{equation}
\mathbf I = \mathbf {U^\dagger} \mathbf U,
\end{equation}
I get
\begin{equation}
i U_{\alpha i} \partial_t \ket{\nu_i} = U_{\alpha i} H^m_{i j} {U^\dagger_{j\beta}} U_{\beta k} \ket{\nu_k},
\end{equation}
which is simplified to
\begin{equation}
i \partial_t \ket{\nu_\alpha} = H^f_{\alpha \beta} \ket{\nu_{\beta}},
\end{equation}
since the transformation is time independent.
The new Hamiltonian in the representations of flavor eigenstates reads
\begin{equation}
H^f_{\alpha\beta} = U^\dagger_{\alpha i} H^m_{ij} U_{j\beta}.
\end{equation}
\subsection{Survival Probability}
The neutrino states at any time can be written as
\begin{equation}
\ket{\Psi(t)} = X_1 \ket{\nu_1 } e^{-i E_1 t}+ X_2 \ket{ \nu_2 } e^{-i E_2 t},
\end{equation}
where $X_1$ and $X_2$ are the initial conditions which are determined using the neutrino initial states.
Survival probalility is the squrare of the projection on an flavor eigenstate,
\begin{equation}
P_{\alpha}(t) = \lvert \braket{\nu_{\alpha}}{\Psi(t)} \rvert^2.
\end{equation}
The calculation of this expression requires our knowledge of the relation between mass eigenstates and flavor eigenstates which we have already found out.
Recall that the transformation between flavor and mass states is
\begin{equation}
\ket{\nu_i} = U^{-1}_{i\alpha} \ket{\nu_\alpha},
\end{equation}
which leads to the inner product of mass eigenstates and flavor eigenstates,
\begin{align}
\braket{\nu_\alpha}{\nu_i} &= \bra{\nu_\alpha} U^{-1}_{i\beta} \ket{\nu_\beta} \\
& = U^{-1}_{i\beta}\delta_{\alpha\beta} \\
& = U^{-1}_{i\alpha}.
\end{align}
The survival probability becomes
\begin{align*}
P_\alpha (t) &= \lvert \braket{\nu_\alpha}{ X_1 \ket{\nu_1 } e^{-i E_1 t} X_2 \ket{ \nu_2 } e^{-i E_2 t} } \rvert^2 \\
& = \lvert X_1 e^{-i E_1 t} \braket{\nu_\alpha}{\ket{\nu_1} } + X_2 e^{-i E_2 t} \braket{ \nu_\alpha }{ \nu_2 } \rvert^2 \\
& = \lvert \sum_i X_i e^{-i E_i t} U^{-1}_{i \alpha} \rvert ^2 \\
& = \sum_i X_1^* e^{iE_i t} U^{\dagger *}_{i\alpha} \sum_i X_i e^{-i E_i t} U^\dagger_{i \alpha} \\
& = \lvert X_1 \rvert^2 U^{\dagger *}_{1\alpha} U^\dagger_{1\alpha} + \lvert X_2 \rvert^2 U^{\dagger *}_{2\alpha} U^\dagger_{2\alpha} + X_1^* X_2 U^{\dagger *}_{1\alpha} U^\dagger_{2\alpha} e^{i E_1 t - i E_2 t} + X_2^* X_1 U^{\dagger *}_{2\alpha} U^\dagger_{1\alpha} e^{i E_2 t - i E_1 t}
\end{align*}
$U^{\dagger *}_{i\alpha}$ stands for the $i$th row and the $\alpha$th column of the matrix $U^{\dagger *}$.
\subsection{Two Flavor States}
For 2 flavor neutrinos the Hamiltonian in the representation of propagation states,
\begin{equation*}
\mathbf H = \begin{pmatrix}
E_1 & 0 \\
0 & E_2
\end{pmatrix}
= \begin{pmatrix}
p_1 + \frac{1}{2}\frac{m_1^2}{p_1} & 0 \\
0 & p_2 + \frac{1}{2}\frac{m_1^2}{p_2}
\end{pmatrix}.
\end{equation*}
The equation of motion in matrix form is
\begin{equation}
i\partial_t \begin{pmatrix}
\nu_1 \\ \nu_2 \end{pmatrix} = \begin{pmatrix}
p_1 + \frac{1}{2}\frac{m_1^2}{p_1} & 0 \\
0 & p_2 + \frac{1}{2}\frac{m_1^2}{p_2}
\end{pmatrix} \begin{pmatrix}
\nu_1 \\ \nu_2 \end{pmatrix}
\end{equation}
The flavor eigenstate is a mixing of the propagation eigenstates,
\begin{equation}
\begin{pmatrix}
\nu_a \\ \nu_b \end{pmatrix} =
\begin{pmatrix} \cos\theta_v & \sin\theta_v \\ -\sin\theta_v & \cos\theta_v
\end{pmatrix} \begin{pmatrix} \nu_1 \\ \nu_2
\end{pmatrix}
\end{equation}
Denote the rotation matrix using $\mathbf U$, the transofmation can be written as
\begin{equation}
\ket{\nu_{\alpha}} = \mathbf U_{\alpha i} \ket{\nu_i},
\end{equation}
where $\alpha$ is for the flavor eigenstates and $i$ is for the mass eigenstates.
The survival probability has been derived in previous section, which is the projection of propagation states onto flavor states.
For arbitary initial condition,
\begin{align}
\Psi(t=0)= A \ket{\nu_a} + B \ket{\nu_b},
\end{align}
which can be rewritten into a matrix form,
\begin{align}
\Psi(t=0) = \begin{pmatrix}
A & B
\end{pmatrix}\begin{pmatrix}
\nu_a \\
\nu_b
\end{pmatrix}
\end{align}
To write down the projection, the relation
\begin{align}
\begin{pmatrix}
\nu_a \\
\nu_b
\end{pmatrix} = \begin{pmatrix}
\cos\theta_v & \sin\theta_v \\ -\sin\theta_v & \cos\theta_v
\end{pmatrix} \begin{pmatrix} \nu_1 \\ \nu_2
\end{pmatrix}
\end{align}
is needed. BTW, the inverse transformation is the transpose of $\mathbf U$ since $\mathbf U$ is unitary, thus we have the relation,
\begin{align}
\begin{pmatrix}
\nu_1 \\
\nu_2
\end{pmatrix} = \begin{pmatrix}
\cos\theta_v & -\sin\theta_v \\
\sin\theta_v & \cos\theta_v
\end{pmatrix} \begin{pmatrix}
\nu_a \\
\nu_b
\end{pmatrix}
\end{align}
Thus in the state can be written as
\begin{align}
\Psi(t=0) = \begin{pmatrix}
A & B
\end{pmatrix} \begin{pmatrix}
\cos\theta_v & \sin\theta_v \\
-\sin\theta_v & \cos\theta_v
\end{pmatrix}\begin{pmatrix}
\nu_1 \\
\nu_2
\end{pmatrix}.
\end{align}
At any $t$, the state is
\begin{align}
\Psi(t) &= \begin{pmatrix}
A \cos\theta_v - B \sin\theta_v & A\sin\theta_v + B \cos\theta_v
\end{pmatrix} \begin{pmatrix}
\nu_1 e^{-i E_1 t}\\
\nu_2 e^{-i E_2 t}
\end{pmatrix} \\
& = \begin{pmatrix}
(A \cos\theta_v - B \sin\theta_v) e^{-iE_1 t} & (A\sin\theta_v + B \cos\theta_v
) e^{-iE_2t}\end{pmatrix} \begin{pmatrix}
\nu_1 \\
\nu_2
\end{pmatrix}
\end{align}
The survival probability which is projection on a flavor state is written as
\begin{equation}
P(\nu_\alpha,t) = \lvert \braket{\nu_\alpha}{\Psi(t)} \rvert^2.
\end{equation}
The survival amplitude for $\nu_a$ is
\begin{align*}
&\phantom{=}\braket{\nu_a}{\Psi(t)} \\
=& \bra{\nu_a}\left( ( A \cos\theta_v - B \sin\theta_v) e^{-iE_1 t} \ket{\nu_1} + (A\sin\theta_v + B \cos\theta_v
) e^{-i E_2t} \ket{\nu_2} \right) \\
=& \left ( \cos\theta_v \bra{\nu_1} + \sin\theta_v \bra{\nu_2} \right) \left( ( A \cos\theta_v - B \sin\theta_v) e^{-iE_1 t} \ket{\nu_1} + (A\sin\theta_v + B \cos\theta_v
) e^{-i E_2 t} \ket{\nu_2} \right )
\end{align*}
This is simple since the transformation matrix is real.
Applying the condition that the propagation eigenstates are orthonormal, the survival probability is
\begin{align*}
P(\nu_a,t) & = \lvert \braket{\nu_a}{\Psi(t)}\rvert^2 \\
& = \lvert \cos\theta_v (A\cos\theta_v -B\sin\theta_v) e^{-iE_1t} + \sin\theta_v(A\sin\theta_v + B \cos\theta_v) e^{-iE_2t} \rvert^2\\
& =\lvert ( A \cos^2\theta_v - B \sin\theta_v \cos\theta_v ) e^{-iE_1t} + ( A \sin^2\theta_v + B \sin\theta_v \cos\theta_v )e^{-iE_2t} \rvert^2
\end{align*}
In a special limit that $E_1=E_2=E$, the probability becomes
\begin{equation}
P(\nu_a,t) = \lvert A\rvert^2
\end{equation}
which is the same as initial probability since there is no mixing at all.
There are two kinds of initial conditions.
\begin{itemize}
\item
The neutrinos are all in $\nu_a$ state initially, which means $A=1, B=0$. The survival probability simplifies to
\begin{align*}
P(\nu_a,t) & = \lvert \cos^2\theta_v e^{-iE_1t} + \sin^2\theta_v e^{-iE_2t} \rvert^2 \\
& = \lvert \cos^2\theta_v e^{-i(E_1 - E_2) t} + \sin^2\theta_v \rvert^2
\end{align*}
As we have already discussed, $E_1-E_2 = \frac{m_1^2 - m_2^2}{2p}$ assuming the neutrinos have the same momentum. \footnote{And here is a question.} Using the notation $\Delta \, m^2 = m_1^2 - m_2^2$ and the approximation that $E\approx p$, the survival probability can be rewritten as
\begin{align*}
P(\nu_a,t) & = \cos^4\theta_v + \sin^4\theta_v + \cos^2\theta_v \sin^2\theta_v \left( e^{-i \Delta m^2 t/E} + e^{i \Delta m^2 t/E} \right) \\
& = 1 - 2\cos^2\theta_v \sin^2\theta_v + 2\cos^2\theta_v \sin^2\theta_v \cos\left( \frac{\Delta m^2 t}{2E}\right) \\
& = 1 - 2\cos^2\theta_v \sin^2\theta_v \left( 1 - \cos\left( \frac{\Delta m^2 t}{2E} \right) \right) \\
& = 1 - 4 \cos^2\theta_v \sin^2\theta_v \sin^2 \left( \frac{\Delta m^2 t}{4E} \right) \\
& = 1 - \sin^2(2\theta_v) \sin^2 \left( \frac{\Delta m^2 t}{4E} \right)
\end{align*}
We always assuming that in the region of interest, all neutrinos are travelling with the same speed, i.e., the speed of light $c=1$. \footnote{which is not true obviously} Time is related to distance, $L = t$. Survival probability at distance $L$ is
\begin{equation}
P(\nu_a, L) = 1 - \sin^2 (2\theta_v) \sin^2 \left( \frac{\Delta m^2 L}{4E} \right)
\end{equation}
\item
The neutrinos are all in $\nu_b$ state initially. Equivalently, we have $A=0, B = 1$. Survival probability is
\begin{align*}
P(\nu_a,t) & = \lvert -\sin\theta_v \cos\theta_v e^{-iE_1t} + \sin\theta_v \cos\theta_v e^{-iE_2t} \rvert^2 \\
& = \sin^2\theta_v \cos^2\theta_v \lvert e^{-i(E_1 - E_2)t} - 1 \rvert^2 \\
& = \sin^2\theta_v \cos^2\theta_v \left( 1 +1 - e^{-i \Delta m^2 t/2E} - e^{i \Delta m^2 t/2E} \right) \\
& = 2 \sin^2\theta_v \cos^2\theta_v \left( 1- \cos\left( \frac{\Delta m^2 t}{2E} \right) \right) \\
& = \sin^2 ( 2\theta_v ) \sin^2\left( \frac{\Delta m^2 t}{4E} \right) \\
& = \sin^2 ( 2\theta_v ) \sin^2\left( \frac{\Delta m^2 L}{4E} \right)
\end{align*}
\end{itemize}
\subsection{Flavor Basis}
This problem can be solved using density matrix $\rho$ and Von Neumann equation
\begin{equation}
i \partial_t \rho = [H,\rho].
\end{equation}
The initial condition for this equation is
\begin{align*}
\rho(t=0) &= (A \ket{\nu_a} + B \ket{\nu_b} )(A^* \bra{\nu_a} + B^* \bra{\nu_b} ) \\
& = A A^* \ket{\nu_a}\bra{\nu_a} + B B^* \ket{\nu_b}\bra{\nu_b} + A B^* \ket{\nu_a}\bra{\nu_b} + A^* B \ket{\nu_b} \bra{\nu_a} .
\end{align*}
To calculate the propagation of the states, we need the Hamiltonian matrix in flavor basis.
This can be done by finding out how the Hamiltonian matrix transforms from one basis to another.
Using propagation basis,
\begin{equation}
i\partial_t \ket{\Psi_p} = H_p \ket{\Psi_p}.
\end{equation}
The states are $\ket{\Psi} = \mathbf{U} \ket{\Psi_p} $ in flavor basis, which means we could plug in $\ket{\Psi_p} = \mathbf{U^T}\ket{\Psi}$.
\begin{equation*}
i\partial_t \mathbf{U^T} \ket{\Psi} = H_p \mathbf{U^T} \ket{\Psi}.
\end{equation*}
Since $\mathbf{U}\mathbf{U^T}=\mathbf{I}$, we have a clean result by multiplying through the equation by $\mathbf{U}$.
\begin{equation*}
i\partial_t \ket{\Psi} = \mathbf{U} H_p \mathbf{U^T} \ket{\Psi}.
\end{equation*}
So we define $H = \mathbf{U}H_p \mathbf{U^T}$ as the Hamiltonian matrix in flavor basis, which is
\begin{equation}
H = \left(p + \frac{m_1^2+m_2^2}{4p} \right)\mathbf I - \frac{1}{4p}\begin{pmatrix} - \Delta m^2 \cos 2\theta & \Delta^2 m \sin 2\theta \\ \Delta m^2 \sin 2\theta & \Delta^2 m\cos 2\theta \end{pmatrix}.
\end{equation}
The derivation of this is
\begin{align*}
\mathbf H_{\alpha} & = \mathbf U \hat H_j \mathbf U^T \\
& = \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} \left( p \mathbf I + \frac{1}{2p}\begin{pmatrix} m_1^2 & 0 \\ 0 & m_2^2 \end{pmatrix} \right) \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} \\
& = p \mathbf I + \frac{1}{2p} \begin{pmatrix} \cos^2\theta m_1^2 + \sin^2\theta m_2^2 & -\sin\theta\cos\theta m_1^2 + \sin\theta\cos\theta m_2^2 \\ -\sin\theta\cos\theta m_1^2 + \sin\theta\cos\theta m_2^2 & \sin^2\theta m_1^2 + \cos^2\theta m_2^2 \end{pmatrix} \\
& = p \mathbf I + \frac{1}{2p} \begin{pmatrix} m_1^2 - \Delta m^2 \sin^2\theta & -\frac{1}{2}\sin 2\theta \Delta m^2 \\ -\frac{1}{2}\sin 2\theta \Delta m^2 & m_2^2+ \Delta m^2 \sin^2\theta \end{pmatrix} \\
& = p \mathbf I + \frac{1}{2p} \left( \frac{1}{2}(m_1^2+m_2^2) \mathbf I - \frac{1}{2}\begin{pmatrix} - \Delta m^2 \cos 2\theta & \Delta^2 m \sin 2\theta \\ \Delta m^2 \sin 2\theta & \Delta^2 m\cos 2\theta \end{pmatrix} \right) \\
& = \left(p + \frac{m_1^2+m_2^2}{4p} \right)\mathbf I - \frac{1}{4p}\begin{pmatrix} - \Delta m^2 \cos 2\theta & \Delta^2 m \sin 2\theta \\ \Delta m^2 \sin 2\theta & \Delta^2 m\cos 2\theta \end{pmatrix}
\end{align*}
where $\Delta m^2 = m_1^2 - m_2^2$.
Since identity matrix only shifts the eigenvalues we are only interested in the second term, thus the Hamiltonian we are going to use is
\begin{equation}
H = \frac{\Delta m^2}{4E} \begin{pmatrix}
\cos 2\theta & - \sin 2\theta \\ - \sin 2\theta & - \cos 2\theta
\end{pmatrix}.
\end{equation}
The equation of motion becomes
\begin{equation}
i \partial_t \begin{pmatrix}
u(t) \\ v(t)
\end{pmatrix} = \frac{\Delta m^2}{4E} \begin{pmatrix}
\cos 2\theta & - \sin 2\theta \\ - \sin 2\theta & - \cos 2\theta
\end{pmatrix} \begin{pmatrix}
u(t) \\ v(t)
\end{pmatrix}
\end{equation}
To solve this we need the eigenvalues and eigenvectors of the Hamiltonian matrix.
\subsection{An Example of Survival Probability}
Suppose the neutrinos are prepared in electron flavor initially, the survival probability of electron flavor neutrinos is calculated using the result I get previously.
Electron neutrinos are the lighter ones, then I have ${}_a = {}_e$ and denote ${}_b={}_x$. \footnote{In the small mixing angle limit,
\begin{equation*}
\begin{pmatrix}\nu_e \\ \nu_x\end{pmatrix} \to \begin{pmatrix} 1 & \theta \\ -\theta & 1 \end{pmatrix} \begin{pmatrix}\nu_1 \\ \nu_2\end{pmatrix}
\end{equation*}
which is very close to an identity matrix. This implies that electron neutrino is more like mass eigenstate $\nu_1$ . By $\nu_1$ we mean the state with energy $\frac{ \delta m^2 }{4E}$ in vacuum.
}
In fact the dynamics of the system is very easily solved without dive into the math. Suppose we have $\ket{\nu_e}$ initially, which is
\begin{equation*}
\Psi(x=0)=\ket{\nu_e} = \cos \theta_v \ket{\nu_1} - \sin \theta_v \ket{\nu_2},
\end{equation*}
the state of the system at distance $x$ is directly written down
\begin{align*}
\Psi(x) &= \cos \theta_v \ket{\nu_1} e^{-i E_1 x} - \sin \theta_v \ket{\nu_2} e^{-i E_2 x} \\
&= e^{-i E_1 x}( \cos \theta_v \ket{\nu_1} - \sin \theta_v \ket{\nu_2} e^{i(E_1 - E_2) x}).
\end{align*}
Since a global phase doesn't change the detection, we write the state as
\begin{equation*}
\Psi(x) = \cos \theta_v \ket{\nu_1} - \sin \theta_v \ket{\nu_2} e^{i(E_1 - E_2) x} .
\end{equation*}
Notice that the period of the expression is
\begin{equation*}
l_v = \frac{2\pi}{E_1 - E_2} = - \frac{4\pi E}{\Delta m_{12}}.
\end{equation*}
Then the state becomes
\begin{equation*}
\Psi(x) = \cos \theta_v \ket{\nu_1} - \sin \theta_v \ket{\nu_2} e^{i2\pi x/l_v} .
\end{equation*}
The survival probability for electron neutrinos is
\begin{align*}
P(\nu_e,L) &= 1-\sin^2(2\theta_v)\sin^2\left( \frac{\Delta m^2 L}{4E} \right) \\
&= 1- \frac{1}{2}\sin^2 2\theta_v \left(1- \cos\left( \frac{2\pi x}{l_v} \right) \right)
\end{align*}
\subsection{Another Method for 2 Flavor Oscillations}
To solve a set of first order differential equations, I need the determinant of coefficient matrix. For 2 flavor neutrino oscillations,
\begin{equation*}
\partial_t \begin{pmatrix}
u(t) \\ v(t)
\end{pmatrix} = - i \frac{\Delta m^2}{4E} \begin{pmatrix}
\cos 2\theta & - \sin 2\theta \\ - \sin 2\theta & - \cos 2\theta
\end{pmatrix} \begin{pmatrix}
u(t) \\ v(t)
\end{pmatrix}.
\end{equation*}
To find the solutions I need the eigenvalues $\lambda$. I need to find the determinant
\begin{align*}
&\det \left( - i\frac{\Delta m^2}{4E} \begin{pmatrix}
\cos 2\theta & - \sin 2\theta \\ - \sin 2\theta & - \cos 2\theta
\end{pmatrix} - \lambda \mathbf{I} \right) \\
=& \begin{vmatrix}
-i \frac{\Delta m^2}{4E} \cos 2\theta - \lambda & i \frac{\Delta m^2}{4E} \sin 2\theta \\
i \frac{\Delta m^2}{4E} \sin 2\theta & i \frac{\Delta m^2}{4E} \cos 2\theta - \lambda
\end{vmatrix} .
\end{align*}
By defining $\lambda' = \lambda/(-i \Delta m^2 / 4E)$, the determinant is
\begin{align*}
- \left( \frac{\Delta m^2}{4E} \right)^2 ( (\cos 2\theta - \lambda')(-\cos 2\theta - \lambda') - \sin 2\theta \sin 2\theta ) .
\end{align*}
The eigenvalues are the solutions to
\begin{align*}
- \left( \frac{\Delta m^2}{4E} \right)^2 ( (\cos 2\theta - \lambda')(-\cos 2\theta - \lambda') - \sin^2 2\theta \sin 2\theta ) =0 ,
\end{align*}
whose solution is
\begin{equation*}
\lambda' = \pm 1.
\end{equation*}
With the solutions
\begin{equation*}
\lambda = \pm i \frac{\Delta m^2}{4E},
\end{equation*}
the eigenvectors can also be solved.
\begin{align*}
\begin{pmatrix}
\cos 2\theta - 1 & - \sin 2\theta \\ - \sin 2\theta & - \cos 2\theta -1
\end{pmatrix} \begin{pmatrix}
\eta_1 \\ \eta_2
\end{pmatrix} = \begin{pmatrix}
0 \\ 0
\end{pmatrix}
\end{align*}
gives us $\eta_2 = -\tan \theta \eta_1$, which means the eigenvectors are
\begin{equation}
\begin{pmatrix}
1 \\ -\tan\theta
\end{pmatrix} , \begin{pmatrix}
1 \\ \cot \theta
\end{pmatrix}.
\end{equation}
The general solution of the first order differential equations is
\begin{align*}
\begin{pmatrix}
1 \\ -\tan\theta
\end{pmatrix} e^{-i \Delta m^2 t/ 4E } \\
\begin{pmatrix}
1 \\ \cot \theta
\end{pmatrix} e^{i \Delta m^2 t/ 4E }.
\end{align*}
Initial condition is
\begin{equation*}
\begin{pmatrix}
1 \\ 0
\end{pmatrix},
\end{equation*}
and it determines the final solution
\begin{align*}
& \cos^2\theta \begin{pmatrix}
1 \\ -\tan\theta
\end{pmatrix} e^{-i \Delta m^2 t/ 4E } + \sin^2\theta
\begin{pmatrix}
1 \\ \cot \theta
\end{pmatrix} e^{i \Delta m^2 t/ 4E } \\
= & \begin{pmatrix}
\cos^2\theta \\ -\sin\theta \cos\theta
\end{pmatrix} e^{-i \Delta m^2 t/ 4E } +
\begin{pmatrix}
\sin^2\theta \\ \sin\theta \cos \theta
\end{pmatrix} e^{i \Delta m^2 t/ 4E }
\end{align*}
The survival probability of electron neutrino is
\begin{align*}
P &= \lvert \cos^2\theta e^{-i \Delta m^2 t/4E} + \sin^2\theta e^{i\Delta m^2 t/4E} \rvert^2 \\
& = \lvert \cos^2 \theta e^{-i \Delta m^2 t/2E} + \sin^2 \theta \rvert ^2 ,
\end{align*}
which gets back to the result we had using the previous method.
This problem can also be solved using numerical methods. Here is a comparison between this analytical result and a numerical result.
\begin{marginfigure}
\includegraphics{assets/vacuumOsc}
\caption{They overlap on all the range completely.}
\end{marginfigure}
\subsection{Three Flavor States}
For three flavor neutrinos, the oscillations matrix is 3 by 3 which is called the PMNS matrix.
\begin{equation*}
\mathbf U = \begin{pmatrix}
U_{11} & U_{12} & U_{13} \\
U_{21} & U_{22} & U_{23} \\
U_{31} & U_{32} & U_{33}
\end{pmatrix},
\end{equation*}
which acts on the mass eigenstates to give the flavor eigenstates, i.e.,
\begin{equation*}
\ket{\nu_\alpha}= \mathbf{U}\ket{\nu_i}.
\end{equation*}
where
\begin{equation*}
\mathbf{U} = \begin{pmatrix}
U_{e1} & U_{e2} & U_{e3} \\
U_{\mu 1} & U_{\mu 2} & U_{\mu 3}\\
U_{\tau 1} & U_{\tau 2} & U_{\tau 3}
\end{pmatrix}.
\end{equation*}
In general, a rotation for 3D with a CP violation phase $\delta$ is
% A new command that scales down the size of the matrix
\newcommand\scalemath[2]{\scalebox{#1}{\mbox{\ensuremath{\displaystyle #2}}}}
\begin{equation*}
\mathbf{U} = \left(
\scalemath{0.8}{
\begin{array}{ccc}
\cos \left(\theta _{12}\right) \cos \left(\theta _{13}\right) & \cos \left(\theta _{13}\right) \sin \left(\theta _{12}\right) & e^{-i \delta _{\text{CP}}} \sin \left(\theta _{13}\right) \\
-\cos \left(\theta _{23}\right) \sin \left(\theta _{12}\right)-e^{i \delta _{\text{CP}}} \cos \left(\theta _{12}\right) \sin \left(\theta _{13}\right) \sin \left(\theta _{23}\right) & \cos \left(\theta _{12}\right) \cos \left(\theta _{23}\right)-e^{i \delta _{\text{CP}}} \sin \left(\theta _{12}\right) \sin \left(\theta _{13}\right) \sin \left(\theta _{23}\right) & \cos \left(\theta _{13}\right) \sin \left(\theta _{23}\right) \\
\sin \left(\theta _{12}\right) \sin \left(\theta _{23}\right)-e^{i \delta _{\text{CP}}} \cos \left(\theta _{12}\right) \cos \left(\theta _{23}\right) \sin \left(\theta _{13}\right) & -e^{i \delta _{\text{CP}}} \cos \left(\theta _{23}\right) \sin \left(\theta _{12}\right) \sin \left(\theta _{13}\right)-\cos \left(\theta _{12}\right) \sin \left(\theta _{23}\right) & \cos \left(\theta _{13}\right) \cos \left(\theta _{23}\right) \\
\end{array}
}
\right).
\end{equation*}
However, with the CP violation phase $\delta$ this matrix is not unitary thus I'll choose $\delta=0$ for simplicity.\footnote{The effect of the phase can also be studied easily with the help of Mathematica.}
Now the matrix becomes,
\begin{equation*}
\mathbf{U} = \left(
\scalemath{0.8}{
\begin{array}{ccc}
\cos \left(\theta _{12}\right) \cos \left(\theta _{13}\right) & \cos \left(\theta _{13}\right) \sin \left(\theta _{12}\right) & \sin \left(\theta _{13}\right) \\
-\cos \left(\theta _{23}\right) \sin \left(\theta _{12}\right)-\cos \left(\theta _{12}\right) \sin \left(\theta _{13}\right) \sin \left(\theta _{23}\right) & \cos \left(\theta _{12}\right) \cos \left(\theta _{23}\right)-\sin \left(\theta _{12}\right) \sin \left(\theta _{13}\right) \sin \left(\theta _{23}\right) & \cos \left(\theta _{13}\right) \sin \left(\theta _{23}\right) \\
\sin \left(\theta _{12}\right) \sin \left(\theta _{23}\right)-\cos \left(\theta _{12}\right) \cos \left(\theta _{23}\right) \sin \left(\theta _{13}\right) & -\cos \left(\theta _{23}\right) \sin \left(\theta _{12}\right) \sin \left(\theta _{13}\right)-\cos \left(\theta _{12}\right) \sin \left(\theta _{23}\right) & \cos \left(\theta _{13}\right) \cos \left(\theta _{23}\right) \\
\end{array}
}
\right).
\end{equation*}
The survival probability is given by the same derivation as the 2 flavor example. First of all we need to find the Hamiltonian in flavor basis from the propagation Hamiltonian, which is
\begin{align*}
\mathbf{H} &= \mathbf{U} \mathbf{H_p} \mathbf{U^{-1}} \\
& =
\left(
\scalemath{0.25}{
\begin{array}{ccc}
\left(\cos ^2\left(\theta _{12}\right) m_1^2+\sin ^2\left(\theta _{12}\right) m_2^2\right) \cos ^2\left(\theta _{13}\right)+\sin ^2\left(\theta _{13}\right) m_3^2 & \frac{1}{2} \left(\cos \left(\theta _{13}\right) \cos \left(\theta _{23}\right) \sin \left(2 \theta _{12}\right) \left(m_2^2-m_1^2\right)+\sin \left(2 \theta _{13}\right) \sin \left(\theta _{23}\right) \left(-\cos ^2\left(\theta _{12}\right) m_1^2-\sin ^2\left(\theta _{12}\right) m_2^2+m_3^2\right)\right) & \frac{1}{2} \left(\cos \left(\theta _{13}\right) \sin \left(2 \theta _{12}\right) \sin \left(\theta _{23}\right) \left(m_1^2-m_2^2\right)+\cos \left(\theta _{23}\right) \sin \left(2 \theta _{13}\right) \left(-\cos ^2\left(\theta _{12}\right) m_1^2-\sin ^2\left(\theta _{12}\right) m_2^2+m_3^2\right)\right) \\
\frac{1}{2} \left(\cos \left(\theta _{13}\right) \cos \left(\theta _{23}\right) \sin \left(2 \theta _{12}\right) \left(m_2^2-m_1^2\right)+\sin \left(2 \theta _{13}\right) \sin \left(\theta _{23}\right) \left(-\cos ^2\left(\theta _{12}\right) m_1^2-\sin ^2\left(\theta _{12}\right) m_2^2+m_3^2\right)\right) & \left(\cos \left(\theta _{23}\right) \sin \left(\theta _{12}\right)+\cos \left(\theta _{12}\right) \sin \left(\theta _{13}\right) \sin \left(\theta _{23}\right)\right){}^2 m_1^2+\left(\cos \left(\theta _{12}\right) \cos \left(\theta _{23}\right)-\sin \left(\theta _{12}\right) \sin \left(\theta _{13}\right) \sin \left(\theta _{23}\right)\right){}^2 m_2^2+\cos ^2\left(\theta _{13}\right) \sin ^2\left(\theta _{23}\right) m_3^2 & \frac{1}{8} \left(4 \cos \left(2 \theta _{23}\right) \sin \left(2 \theta _{12}\right) \sin \left(\theta _{13}\right) \left(m_1^2-m_2^2\right)-2 \cos \left(2 \theta _{13}\right) \sin \left(2 \theta _{23}\right) \left(\cos ^2\left(\theta _{12}\right) m_1^2+\sin ^2\left(\theta _{12}\right) m_2^2-m_3^2\right)+\sin \left(2 \theta _{23}\right) \left(\left(3 \cos \left(2 \theta _{12}\right)-1\right) m_1^2-\left(3 \cos \left(2 \theta _{12}\right)+1\right) m_2^2+2 m_3^2\right)\right) \\
\frac{1}{2} \left(\cos \left(\theta _{13}\right) \sin \left(2 \theta _{12}\right) \sin \left(\theta _{23}\right) \left(m_1^2-m_2^2\right)+\cos \left(\theta _{23}\right) \sin \left(2 \theta _{13}\right) \left(-\cos ^2\left(\theta _{12}\right) m_1^2-\sin ^2\left(\theta _{12}\right) m_2^2+m_3^2\right)\right) & \frac{1}{8} \left(4 \cos \left(2 \theta _{23}\right) \sin \left(2 \theta _{12}\right) \sin \left(\theta _{13}\right) \left(m_1^2-m_2^2\right)-2 \cos \left(2 \theta _{13}\right) \sin \left(2 \theta _{23}\right) \left(\cos ^2\left(\theta _{12}\right) m_1^2+\sin ^2\left(\theta _{12}\right) m_2^2-m_3^2\right)+\sin \left(2 \theta _{23}\right) \left(\left(3 \cos \left(2 \theta _{12}\right)-1\right) m_1^2-\left(3 \cos \left(2 \theta _{12}\right)+1\right) m_2^2+2 m_3^2\right)\right) & \left(\cos \left(\theta _{12}\right) \cos \left(\theta _{23}\right) \sin \left(\theta _{13}\right)-\sin \left(\theta _{12}\right) \sin \left(\theta _{23}\right)\right){}^2 m_1^2+\left(\cos \left(\theta _{23}\right) \sin \left(\theta _{12}\right) \sin \left(\theta _{13}\right)+\cos \left(\theta _{12}\right) \sin \left(\theta _{23}\right)\right){}^2 m_2^2+\cos ^2\left(\theta _{13}\right) \cos ^2\left(\theta _{23}\right) m_3^2 \\
\end{array}
}
\right)
\end{align*}
where
\begin{align*}
\mathbf{H_p} = \begin{pmatrix}
m_1^2 & 0 & 0 \\
0 & m_2^2 & 0 \\
0 & 0 & m_3^2
\end{pmatrix}.
\end{align*}
The next step is to define
\begin{align*}
\Delta m_{12}^2 &= m_1^2 - m_2^2 \\
\Delta m_{23}^2 &= m_2^2 - m_3^2
\end{align*}
so that they simplifies the Hermitian Hamiltonian to
\begin{equation*}
\mathbf{H} =
\begin{pmatrix}
H_{11} & H_{12} & H_{13} \\
H_{21} & H_{22} & H_{23} \\
H_{31} & H_{32} & H_{33}
\end{pmatrix},
\end{equation*}
where \footnote{Haven't being copied from Mathematica code.}
\begin{align*}
H_{11} & = \cos^2\theta_{13}( m_2^2 - \sin^2\theta \Delta m_{12}^2 ) + \sin^2\theta_{13} m_3^2
\end{align*}
\subsection{Numerical Results for Vacuum 3 Flavor Oscillations} \label{numerical-results-for-vacuum-3-flavor-oscillations}
\begin{figure}
\includegraphics{assets/vacOsc3Flavor.jpg}
\caption{The overall shape is the same however they differ on small scales.}
\label{vacOsc3Flavor}
\end{figure}
\begin{figure}
\centering
\includegraphics{assets/vacOscNormInvComp}
\caption{Comparison of normal hierarchy and inverted hierarchy.The reason that they are almost the same is that the oscillation length for $\Delta m_{13}^2$ is small thus it only changes the oscillation patterns for the small oscillations. Vacuum energy scales in normal hierarchy are
$$\omega_{12}= \frac{\Delta m_{12}^2}{2E} = 3.8\times 10^{-20}\mathrm{GeV}$$
$$\omega_{13}= \frac{\Delta m_{13}^2}{2E} = 1.7\times 10^{-18}\mathrm{GeV}$$
$$\omega_{23}= \frac{\Delta m_{23}^2}{2E} \approx \omega_{13}$$
which shows that basically only two scales and the larger one determines the small oscillation.}
\label{fig:vacOscNormInvComp}
\end{figure}
\begin{figure}
\centering
\includegraphics{assets/vacOscNormInvComp-Invert12.png}
\caption{Comparison of normal hierarchy and inverted hierarchy but with inverted $\Delta m_{12}^2$.}
\label{fig:vacOscNormInvComp-Invert12}
\end{figure}
\subsection{Ternary Diagram for Neutrino Flavor Oscillation}
Since the probability for differential flavors of neutrinos are represented in barycentric coordinates\footnote{They sum up to 1.}, a ternary plot would be nice to representation the oscillations. An example is shown in figure \ref{fig:ternaryPlot900}.
\begin{figure}
\centering
\includegraphics{assets/vacOsc3FlavorTernary900}
\caption{Ternary diagram for neutrino oscillations. The state starts from bottom left, which means that the system has only electron neutrinos. As the neutrino travels, it oscillates in curves. After one period of the beat, it reaches the far end and then oscillates backwards.}
\label{fig:ternaryPlot900}
\end{figure}
\begin{marginfigure}
\includegraphics{assets/vacOsc3FlavorTernary300}
%\caption{This plot shows the behavior of neutrino oscillations in the range [0,300].}
\end{marginfigure}
\begin{marginfigure}
\includegraphics{assets/vacOsc3FlavorTernary5000}
%\caption{This plot shows the behavior of neutrino oscillations in the range [0,5000].}
\end{marginfigure}
\begin{figure}
\centering
\includegraphics{assets/ternary/Inv-1000-1.png}
\caption{Ternary diagrram for inverted hierarchy.}
\label{fig:Inv-1000-1}
\end{figure}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% Oscillations in Matter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Oscillations in Matter}
The Hamiltonian should be determined first. We have already derived the Hamiltonian for vacuum oscillation,
\begin{equation*}
H_v=\frac{ \delta m^2 }{2E}\frac{1}{2}\begin{pmatrix} -\cos 2\theta_v & \sin 2 \theta_v \\ \sin 2\theta_v & \cos 2\theta_v \end{pmatrix},
\end{equation*}
where we would like to define a new matrix,
\begin{equation*}
\mathbf B = \frac{1}{2}\begin{pmatrix} -\cos 2\theta_v & \sin 2 \theta_v \\ \sin 2\theta_v & \cos 2\theta_v \end{pmatrix},
\end{equation*}
so that the vacuum Hamiltonian can be written as
\begin{equation*}
H_v = \frac{ \delta m^2 }{2E}\mathbf B.
\end{equation*}
The effect of matter, adds an extra term to this vacuum Hamiltonian which makes the electron population weighs more,
\begin{equation*}
H_m = \sqrt{2}G_F n_e L.
\end{equation*}
Here we have \footnote{In principle we could shift the Hamiltonian using an identity matrix $\alpha \mathbf{I}$ without change the eigenvectors. We will use an term $\frac{1}{2}\mathbf{\sigma_3} $ instead in the following sections.}
\begin{equation*}
L = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}.
\end{equation*}
Without emphasizing the self-interaction of the neutrinos, the Hamiltonian to be used is
\begin{equation}
H = H_v + H_m.
\end{equation}