1.1.10. Linear Background Matter Profile

The equation of motion in flavor basis for arbitary matter profile \(\lambda(x)\) is

\[\begin{split}i \partial_x \begin{pmatrix} \psi_e\\ \psi_x \end{pmatrix} = \left[ \left( \frac{\lambda(x)}{2} - \frac{\omega_v}{2} \cos 2\theta_v \right) \sigma_3 + \frac{\omega_v}{2} \sin 2\theta_v \sigma_1 \right]\begin{pmatrix} \psi_e\\ \psi_x \end{pmatrix}.\end{split}\]

Introduce the T transformation

\[\begin{split}\begin{pmatrix} \psi_e\\ \psi_x \end{pmatrix} = \mathbf{T} \begin{pmatrix} \psi_{b1}\\ \psi_{b2} \end{pmatrix} = \begin{pmatrix} e^{-i \eta(x)} & 0 \\ 0 & e^{i \eta(x)} \end{pmatrix} \begin{pmatrix} \psi_{b1}\\ \psi_{b2} \end{pmatrix}.\end{split}\]

By multiplying on both sides \(\mathbf{T}^\dagger\), we get

\[\begin{split}i \partial_x \begin{pmatrix} \psi_{b1}\\ \psi_{b2} \end{pmatrix} = \frac{\omega_v}{2} \sin 2\theta_v \begin{pmatrix} 0 & e^{2i\eta(x)} \\ e^{-2i\eta(x)} & 0 \end{pmatrix}\begin{pmatrix} \psi_{b1}\\ \psi_{b2} \end{pmatrix} + \left( \frac{\lambda(x)}{2} - \frac{\omega_v}{2} \cos 2\theta_v - \partial_x \eta(x) \right) \sigma_3 \begin{pmatrix} \psi_{b1}\\ \psi_{b2} \end{pmatrix}.\end{split}\]

To remove the diagonal elements we require

\[\frac{\lambda(x)}{2} - \frac{\omega_v}{2} \cos 2\theta_v - \partial_x \eta(x) = 0,\]
\[\eta(x) = \eta(0) + \frac{1}{2}\int_0^x dx' \left( \lambda(x) - \frac{\omega_v}{2} \cos 2\theta_v \right) .\]

Perodic perturbation can be used

\[\delta \lambda(x) = A\sin (kx + \phi).\]

\(\eta(x)\) has the form

\[\begin{split}\eta(x) &= \eta(0) + \frac{1}{2} \int_0^x dx' (\lambda_0(x) + A\sin(kx+\phi) ) - \frac{1}{2}\omega_v \cos 2\theta_v x \\ & = - \frac{1}{2} \omega_v \cos 2\theta_v x - \frac{A}{2k} \cos(kx+\phi) + \frac{1}{2} g(x) {\color{grey}- \frac{1}{2}g(0) - \frac{A}{2k} \cos \phi + \eta(0) },\end{split}\]

where \(g(x)=\int dx' \lambda_0(x')\) and the grey part is usally set to zero.

The equation we are dealing with is

\[\begin{split}i \partial_x \begin{pmatrix} \psi_{b1}\\ \psi_{b2} \end{pmatrix} = \frac{\omega_v}{2} \sin 2\theta_v \begin{pmatrix} 0 & e^{2i\eta(x)} \\ e^{-2i\eta(x)} & 0 \end{pmatrix}\begin{pmatrix} \psi_{b1}\\ \psi_{b2} \end{pmatrix},\end{split}\]

where

\[\eta(x) = - \frac{1}{2} \omega_v \cos 2\theta_v x - \frac{A}{2k} \cos(kx+\phi) + \frac{1}{2} g(x) .\]

Can we solve the equation

Here is the problem. For most general \(g(x)\) we are not guaranteed to have a analycal solution of the system.

For \(g(x)\propto x\) we have demonstrated that the system is solvale.

What about \(g(x)\propto x^2\) (linear background matter profile) and \(g(x)\propto e^{x}\) (exponential background matter profile)?

1.1.10.1. Linear Background Matter Profile

Equation Solved

It turns out that we can solve the system

\[\begin{split}i\partial_x \begin{pmatrix} \psi_{b1}\\ \psi_{b2} \end{pmatrix} = A_s \begin{pmatrix} 0 & e^{if x^2} \\ e^{-if x^2} & 0 \end{pmatrix}\begin{pmatrix} \psi_{b1}\\ \psi_{b2} \end{pmatrix}.\end{split}\]

The solution involves hypergeometric functions. For initial condition

\[\begin{split}\begin{pmatrix} \psi_{b1}(0)\\ \psi_{b2}(0) \end{pmatrix} = \begin{pmatrix} 1\\ 0 \end{pmatrix},\end{split}\]

we have the solution

\[\begin{split}\begin{pmatrix} \psi_{b1}(x)\\ \psi_{b2}(x) \end{pmatrix} = \begin{pmatrix} {}_1F_1 \left( \frac{i A_s^2}{4f};\frac{1}{2}; i f x^2 \right) \\ -i A_s x e^{-i f x^2} {}_1F_1 \left( \frac{i A_s^2}{4f}+1;\frac{3}{2}; i f x^2 \right) \end{pmatrix},\end{split}\]

where \({}_1F_1(a;b;z)\) is hypergeometric function .

We choose the background matter profile to be

\[\lambda_0(x) = a x + b,\]

while the perturbation to be

\[\delta \lambda(x) = A\sin (kx + \phi).\]

\(\eta(x)\) is solved

\[\begin{split}g(x) & = \frac{1}{2} ax^2+b \\ \eta(x) &= - \frac{1}{2} \omega_v \cos 2\theta_v x - \frac{A}{2k} \cos (kx + \phi) + \frac{1}{2} \left( \frac{1}{2}a x^2 + b x \right),\end{split}\]

so that

\[e^{2i\eta(x)} = e^{-i \omega_v \cos 2\theta_v x} e^{i ( ax^2/2+ bx )} e^{-i\left( \frac{A}{k} \cos (kx+\phi) \right)}.\]

Jacobi-Anger Expansion

\[e^{-i\left( \frac{A}{k} \cos (kx+\phi) \right)} = \sum_{n=-\infty}^{\infty} (-i)^n J_n\left(\frac{A}{k}\right) e^{in(kx+\phi)}.\]

Collect terms

\[\begin{split}e^{2i\eta(x)} &= \sum_{n=-\infty}^{\infty} (-i)^n J_n\left(\frac{A}{k}\right) e^{in(kx+\phi) + \frac{1}{2}ax^2 +bx - \omega_v \cos 2\theta_v x } \\ &= \sum_{n=-\infty}^{\infty} (-i)^n J_n\left(\frac{A}{k}\right) e^{ i \left( \frac{1}{2} a \left( x + \frac{b_n}{a} \right)^2 - \frac{b_n^2}{2a} + n\phi \right) } \\ & = \sum_{n=-\infty}^{\infty} J_n\left(\frac{A}{k}\right) e^{ i \left( \frac{1}{2} a \left( x + \frac{b_n}{a} \right)^2 - \frac{b_n^2}{2a} + n\phi + n\pi \right) } \\ & = \sum_{n=-\infty}^{\infty} J_n\left(\frac{A}{k}\right) e^{ i \left( \frac{1}{2} a \left( x + \frac{b_n}{a} \right)^2 - \frac{b_n^2}{2a} + n\phi' \right) } ,\end{split}\]

where

\[\begin{split}b_n &= b + nk - \omega_v \cos 2\theta_v,\\ \phi'&= \phi + \pi,\end{split}\]

and we have used

\[(-i)^n = e^{i n\pi}.\]

The equation to be sovled becomes

\[\begin{split}i\partial_x \begin{pmatrix} \psi_{b1}\\ \psi_{b2} \end{pmatrix} = \sum_{n=-\infty}^\infty \frac{\omega_v}{2}\sin 2\theta_v J_n\left( \frac{A}{k} \right) \begin{pmatrix} 0 & e^{i\left( \frac{1}{2} a \left( x + \frac{b_n}{a} \right)^2 - \frac{b_n^2}{2a} + n\phi' \right) } \\ e^{-i\left( \frac{1}{2} a \left( x + \frac{b_n}{a} \right)^2 - \frac{b_n^2}{2a} + n\phi' \right)} & 0 \end{pmatrix}\begin{pmatrix} \psi_{b1}\\ \psi_{b2} \end{pmatrix}.\end{split}\]

With initial condition

\[\begin{split}\begin{pmatrix} \psi_{b1}(0)\\ \psi_{b2}(0) \end{pmatrix} = \begin{pmatrix} 1\\ 0 \end{pmatrix},\end{split}\]

the equation

\[\begin{split}i\partial_x \begin{pmatrix} \psi_{b1}\\ \psi_{b2} \end{pmatrix} = \frac{\omega_v}{2}\sin 2\theta_v J_n\left( \frac{A}{k} \right) \begin{pmatrix} 0 & e^{i\left( \frac{1}{2} a \left( x + \frac{b_n}{a} \right)^2 - \frac{b_n^2}{2a} + n\phi' \right) } \\ e^{-i\left( \frac{1}{2} a \left( x + \frac{b_n}{a} \right)^2 - \frac{b_n^2}{2a} + n\phi' \right)} & 0 \end{pmatrix}\begin{pmatrix} \psi_{b1}\\ \psi_{b2} \end{pmatrix}.\end{split}\]

can be solved. To save keystroke, we define

\[A_s= J_n\left(\frac{A}{k}\right)\frac{\omega_v}{2} \sin 2\theta_v.\]

We write down the solution

\[\psi_2(x) = (-1)^{1/4} \sqrt{2} A_s e^{-i \left( \frac{1}{2} a \left( x + \frac{b_n}{a} \right)^2 - \frac{b_n^2}{2a} + n\phi' \right) } b_n \left[ - H_{-1 - i A_s^2/a} (p_1 + p_2 x) F_1 \left( 1 + i \frac{A_s^2}{2a};\frac{3}{2};\frac{i b_n^2}{2a} \right) + (1+\frac{a}{b_n} x) H_{-1 - i A_s^2/a} (p_1) F_1 \left( 1 + i \frac{A_s^2}{2a};\frac{3}{2}; (p_1 + p_2 x)^2\right) \right] \bigg / \left\{ \sqrt{a} H_{-i A_s^2/a} (p_1) \left[ (-1)^{3/4} \sqrt{2a} F_1 \left( i \frac{A_s^2}{2a};\frac{1}{2};\frac{i b_n^2}{2a} \right) - b_n F_1 \left( 1 + i \frac{A_s^2}{2a};\frac{3}{2};\frac{i b_n^2}{2a} \right) \right] \right\},\]

where

\[\begin{split}p_1 & = \frac{(-1)^{1/4} b_n}{\sqrt{2a}}, \\ p_2 & = \frac{ (-1)^{1/4} \sqrt{a} }{ \sqrt{2} }.\end{split}\]

\(F_1\equiv {}_1 F_1\) is the hypergeometric function. What’s important is that we can always define

\[\begin{split}M_n & = \sqrt{a} H_{-i A_s^2/a} (p_1) \left[ (-1)^{3/4} \sqrt{2a} F_1 \left( i \frac{A_s^2}{2a};\frac{1}{2};\frac{i b_n^2}{2a} \right) - b_n F_1 \left( 1 + i \frac{A_s^2}{2a};\frac{3}{2};\frac{i b_n^2}{2a} \right) \right] \\ K_{F,n} &= F_1 \left( 1 + i \frac{A_s^2}{2a};\frac{3}{2};\frac{i b_n^2}{2a} \right) \\ K_{H,n} &= H_{-1 - i A_s^2/a} (p_1),\end{split}\]

which are all constants given the parameters and n. We can reduce the solution to

\[\psi_2 = (-1)^{1/4} \sqrt{2} A_s b_n \frac{K_{H,n} F_1 \left( 1 + i \frac{A_s^2}{2a};\frac{3}{2}; (p_1 + p_2 x)^2\right) (1+\frac{a}{b_n} x) - K_{F,n} H_{-1 - i A_s^2/a} (p_1 + p_2 x) }{ M_n } e^{-i \left( \frac{1}{2} a \left( x + \frac{b_n}{a} \right)^2 - \frac{b_n^2}{2a} + n\phi' \right) }.\]

Some points:

  1. The third argument of \(F_1\), which is denoted as \(p_1 +p_2 x\) is in fact imaginary. As a result, this value of hypergeometric function is actually dropping as \(x\) increase.
  2. In this result, the x dependent Hermite polynomial is a monotonically decreasing function.
  3. The overall phase doesn’t play a role in the transition probability.

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