12.1.1. Quantum Master Equation
Projection Technique
First of all define a diagonalizing operator \(\hat D\) which just keeps the diagonal elements and simply drops the off diagonal elements. We see that \(1-\hat D\) will eliminate all diagonal elements.
We can define the diagonalized density matrix as \(\hat \rho_d = \hat D \hat \rho\) and off-diagonalized density matrix as \(\hat \rho_{od} = (1-\hat D)\hat \rho\). As an application,
\[\hat \rho = \hat \rho_d + \hat \rho_{od} .\]
Starting from the von Neumann equation,
\[i\hbar \partial_t \hat \rho = \left[\hat H, \hat \rho \right] .\]
By using the Liouville operator,
\[\partial_t \hat \rho = -i \hat L \hat \rho .\]
Apply \(\hat D\) and \(1-\hat D\) to the von Neumann equation,
\[\begin{split}\partial_t \hat \rho_d & = -i \hat D \hat L \hat \rho \\
\partial_t \hat \rho _{od} & = -i (1 - \hat D) \hat L \hat \rho .\end{split}\]
Use the relation that \(\hat \rho = \hat \rho_d + \hat \rho_{od}\), we have
\[\begin{split}\partial_t \hat \rho_d & = -i \hat D \hat L \hat \rho_d - i \hat D \hat L \hat \rho _ {od} \\
\partial_t \hat \rho _{od} & = - i (1 - \hat D) \hat L \hat \rho _ d - i (1 - \hat D) \hat L \hat \rho_{od} .\end{split}\]
Solve the second equation using Green function technique,
\[\hat \rho_{od} = e^{-i(1-\hat D)\hat L t} + \int_0^t dt' e^{-i(1-\hat D) \hat L (t-t')}(-i(1-\hat D)\hat L \hat \rho_d(t')) .\]
Hint
Recall that the solution for
\[\dot y + \alpha y = f\]
is
\[y = e^{-\alpha t} y(0) + \int_0^t dt' e^{-\alpha (t-t')} f(t') .\]
Insert this solution to the equation of \(\hat \rho_d\),
\[{\color{red}\partial_t \hat \rho_d = - i\hat D\hat L \hat \rho_d - \hat D\hat L \int_0^t dt' e^{-i(1-\hat D) \hat L (t-t')}(1-\hat D)\hat L \hat \rho_d(t')} {\color{blue} - i \hat D \hat L e^{-i(1-\hat D)\hat L t} \rho_{od}(0) }.\]
What happened to the blue term? It disapears when we apply the initial random phase condition.
When it happens we get our closed master equation for \(\hat \rho_d\), which is an equation for the probability.
In our case of neutrinos, random phase condition is not really needed since we usually deal with the situation that electron neutrinos are appearant only.
In details, we have such a density matrix,
\[\begin{split}\mathbf\rho = \begin{pmatrix}\rho_{aa} &\rho_{ab} \\ \rho_{ba} & \rho_{bb}\end{pmatrix} .\end{split}\]
The quantum master equation we would like to use is
\[\partial_t \hat \rho_d = - i\hat D\hat L \hat \rho_d - \hat D\hat L \int_0^t dt' e^{-i(1-\hat D) \hat L (t-t')}(1-\hat D)\hat L \hat \rho_d(t') .\]
12.1.2. Vacuum Oscillation Master Equation
Using this projection method, one can find out the master equation for vacuum oscillations.
Pauli Matrices
We will use Pauli matrices in the following part. Here is a review of them.
Pauli Matrices,
\[\begin{split}\sigma_1 &= \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\
\sigma_2 & = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \\
\sigma_3 & = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.\end{split}\]
Commutation Relations,
\[\begin{split}[\sigma_1,\sigma_2] &= 2i \sigma_3 \\
[\sigma_2,\sigma_3] &= 2i \sigma_1 \\
[\sigma_3,\sigma_1] &= 2i \sigma_2.\end{split}\]
The general form is
\[[\sigma_i,\sigma_j] &= 2i \epsilon_{ijk} \sigma_k.\]
All the Pauli matrices plus identity form a complate basis for 2 by 2 matrices. Vacuum oscillation Hamiltonian is
\[\begin{split}\mathbf H &\to \frac{\delta^2m}{4E} \begin{pmatrix} -\cos 2\theta & \sin 2 \theta \\ \sin 2\theta & \cos 2\theta \end{pmatrix} \\
& \equiv \begin{pmatrix} -c & s \\ s & c \end{pmatrix}\\
& = -c \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} + s\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\
& = -c \mathbf{\sigma_3} + s \mathbf{ \sigma_1},\end{split}\]
where \(c\equiv \frac{\delta^2 m}{4E}\cos 2 \theta\) and similarly for s.
Liouville Operator
Liouville operator in quantum mechanics is
\[\hat L = [H, *],\]
where the asterisk is the slot for an operator.
In the case of vacuum oscillation, we can calculate the following results,
\[\begin{split}\hat L \sigma_1 &= [H, \sigma_1] = -2ic\sigma_2 \\
\hat L \sigma_2 &= [H, \sigma_2] = 2ic\sigma_1 + 2is\sigma_3.\end{split}\]
Notice that \(\sigma_3\) has diagonal terms only. It will dispear when we apply \(1-\mathscr D\) which removes the diagonal elements, i.e.,
\[\begin{split}(1-\mathscr D)\hat L \sigma_1 &= -2ic\sigma_2 \\
(1-\mathscr D)\hat L \sigma_2 &= 2ic\sigma_1.\end{split}\]
Diagonalized density matrix \(\rho_d=\mathrm {diag}(\rho_1,\rho_2)\) is
\[\begin{split}\mathrm {\rho_d} &= \begin{pmatrix} \rho_1 & 0 \\ 0 & \rho_2 \end{pmatrix} \\
& = \frac{1}{2} \left(\begin{pmatrix} \rho_1 -\rho_2 & 0 \\ 0 & \rho_2 -\rho_1 \end{pmatrix} + \begin{pmatrix} \rho_1+\rho_2 & 0 \\ 0 & \rho_1 + \rho_2 \end{pmatrix} \right) \\
& = \frac{1}{2}\left( (\rho_1-\rho_2)\sigma_3 + (\rho_1+\rho_2)\mathbf I \right)\end{split}\]
Note
Actually \(\rho_1+\rho_2=1\) for such a system. We’ll see the proof of this later.
Apply \((1-\mathscr D)\hat L\) we get
\[\begin{split}(1-\mathscr D)\hat L \rho_d &= i s (\rho_2-\rho_1) \sigma_2,\\
\mathscr D \hat L \rho_d & = -\frac{1}{2}c (\rho_1+\rho_2)\sigma_3.\end{split}\]
Exponential Operator
Exponential operator is understood when series expansion is done,
\[e^{\hat A} = \hat I + \hat A + \frac{1}{2!}{\hat A}^2 + \frac{1}{3!} {\hat A}^3 +\cdots\]
Recall that the master equation is
\[\begin{split}\partial_t \rho_d(t) &= - i \mathscr D \hat L \rho_d - \mathscr D\hat L \int_0^t dt' e^{-i(1-\mathscr D)\hat L (t-t')} (1-\mathscr D) \hat L \hat \rho_d(t') \\
& = \frac{1}{2}ic(\rho_1+\rho_2)\sigma_3 - \mathscr D\hat L \int_0^t dt' \left( i s (\rho_2-\rho_1) e^{-i(1-\mathscr D)\hat L (t-t')} \sigma_2 \right) \\\end{split}\]
So we need to calculate
\[\begin{split}e^{-i(1-\mathscr D)\hat L (t-t')} \sigma_2 &= \left[1 -i(1-\mathscr D)\hat L (t-t') + \frac{1}{2} (-i(1-\mathscr D)\hat L (t-t') )^2 + \frac{1}{3!}(-i(1-\mathscr D)\hat L (t-t') )^3 + \cdots \right]\sigma_2\\
&\equiv T_0 + T_1 + \frac{1}{2} T_2 + \frac{1}{3!}T_3 + \cdots .\end{split}\]
We will calculate it term by term and find the pattern.
\[T_0 = \sigma_2\]
\[\begin{split}T_1 &= -i(1-\mathscr D)\hat L (t-t') \sigma_2 \\
& = 2c\sigma_1 (t-t')\end{split}\]
\[\begin{split}T_2 & = -i(1-\mathscr D)\hat L (t-t') (2c\sigma_1 (t-t')) \\
& = -i(t-t')^2 2c(-2ic\sigma_2) \\
& = - 2^2 c^2 (t-t')^2 \sigma_2\end{split}\]
\[\begin{split}T_3 & = -i(1-\mathscr D)\hat L (t-t') (- 4c^2 (t-t')^2 \sigma_2) \\
& = -2^3c^3(t-t')^3\sigma_1\end{split}\]
\[\begin{split}T_4 & = -i(1-\mathscr D)\hat L (t-t') (-2^3c^3(t-t')^3\sigma_1) \\
& = -i (t-t') (-2^3 c^3(t-t')^3) (-2ic\sigma_2) \\
& = 2^4 c^4 (t-t')^4 \sigma_2\end{split}\]
\[\begin{split}T_5 & = -i(1-\mathscr D)\hat L (t-t') 2^4 c^4 (t-t')^4 \sigma_2 \\
& = -i(t-t')2^4 c^4 (t-t')^4 2ic\sigma_1 \\
& = 2^5c^5 (t-t')^5 \sigma_1\end{split}\]
Carry on this calculation we can infer that
\[e^{-i(1-\mathscr D)\hat L (t-t')} \sigma_2 &= \sigma_2 + 2c\sigma_1 (t-t') + \frac{1}{2}(- 2^2 c^2 (t-t')^2 \sigma_2) + \frac{1}{3!}(-2^3c^3(t-t')^3\sigma_1) + \frac{1}{4!} 2^4 c^4 (t-t')^4 \sigma_2 + \frac{1}{5!} 2^5c^5 (t-t')^5 \sigma_1 + \cdots\]
Taylor Series
Taylor series of \(\sin x\) and \(\cos x\) around \(x=0\) are
\[\begin{split}\sin x &= x - \frac{1}{3!} x^3 + \frac{1}{5!}x^5 + \cdots \\
\cos x & = 1 -\frac{1}{2!} x^2 + \frac{1}{4!} x^4 + \cdots .\end{split}\]
Now we see that
\[e^{-i(1-\mathscr D)\hat L (t-t')} \sigma_2 &= \sigma_1 \sin(M) + \sigma_2 \cos(M),\]
where \(M\equiv 2 c (t-t')\).
The master equation we need is
\[\begin{split}\partial_t \rho_d(t) &= \frac{1}{2}ic(\rho_1+\rho_2)\sigma_3 - \mathscr D \hat L \int_0^t dt' i s (\rho_2-\rho_1) \left(\sigma_1 \sin(2c(t-t')) + \sigma_2 \cos(2c(t-t'))\right) \\
& = \frac{1}{2}ic(\rho_1+\rho_2)\sigma_3 - \mathscr D \hat L i s \int_0^t dt' (\rho_2-\rho_1) \left(\sigma_2 \cos(2c(t-t'))\right) \\
& = \frac{1}{2}ic(\rho_1+\rho_2)\sigma_3 - i s G(t) \mathscr D \hat L \sigma_2 \\
& = \frac{1}{2}ic(\rho_1+\rho_2)\sigma_3 + 2 s^2 G(t) \sigma_3 \\
& = \frac{1}{2}ic(\rho_1+\rho_2)\sigma_3 + 2 s^2 \int_0^t dt' (\rho_2-\rho_1) \sigma_3 \cos(2c(t-t')) \\
& = \frac{1}{2}ic(\rho_1+\rho_2)\sigma_3 + 2 s^2 \int_0^t dt' \left( -2\rho_d(t') + (\rho_1+\rho_2) \mathbf I \right) \cos(2c(t-t')) \\
& = \frac{1}{2}ic\sigma_3 + 2 s^2 \int_0^t dt' \left( -2\rho_d(t') + \mathbf I \right) \cos(2c(t-t'))\end{split}\]
In the calculation, \(G=\int_0^t dt'(\rho_2-\rho_1)\cos(2c(t-t'))\).
Plug in the definitions of c and s, we have
\[\partial_t\rho_d(t) = \frac{\omega}{4} i \cos 2\theta_v \sigma_3 + \frac{\omega^2 \sin^2 2\theta_v }{2} \int_0^t dt' (\mathbf I - 2 \rho_d(t'))\cos \left( 2 \cos 2\theta_v (t - t') \right).\]
This result shows that the equations of different elements are decoupled. This is very important.
What to Do?
I don’t see anything good about this method. What to do next? I can predict that it’s also won’t cost a lot to solve the MSW effect. But what’s the point? These problems are not very hard to solve even using wave function method.
I am just leaving this result here and move on to other topics.