1.3.2. Halo Effect - Finite Difference¶

1.3.2.1.1. Toy Model¶

The toy model we use is neutrino-only, right forward beam and reflection.

For single angle neutrino-only emission, this model can also work for the tilted emission simply by rescale the neutrino self-interaction by \(\cos\alpha\) where \(\alpha\) the angle of forward and backward beams.

1.3.2.1.2. Antineutrinos or Spectrum¶

The simple model would be two beams model with left and right beams having different energies.

1.3.2.1.3. Model with Collective Oscillations¶

Introduce multiangles and left-right asymmetries to the system.

1.3.2.2.1. Convention I Used in Code¶

For any tracelesstwo by two matrix with real diagonal elements such as density matrix,

we define a corresponding array s = { a, b ,c }. Then the density matrix is

1.3.2.2.2. Convention II: Polarization Vector¶

Polarization vector \(\vec P\) is

so that vec P = { b, -c , a }.

P[0] = s[1]
P[1] = -s[2]
P[2] = s[0]

1.3.2.3.1. Euler Method¶

We start from the equation of motion

where we denote

Calcuation of the commutator shows that

from which we conclude that

Then we discretize each equations using Euler method.

1.3.2.3.2. Evolution Operator¶

The same calculation can be achieved by using the evolution operator \(U(t) = \exp (-i H t)\).

Since the Hamiltonian in our simple calcuations is always two by two, the exact form of the algorithm can be written down exactly.

The evolution operator itself is

where we defined \(h = \sqrt{ h_1^2+h_2^2+h_3^2 }\) for short.

To obtain a simpler form, we can write all matrices as Pauli matrices.

For the purpose of formalism, we denote density matrix as \(\rho = P_i \sigma_i\). The density matrix at \(t+\Delta t\) is

where \(c=\cos( h \Delta t )\), \(s = \sin(h\Delta t)\), \(u_{1,3} =-i h_{1,3}/h\), \(u_2 = i h_2/h\), and \(h=\sqrt{h_1^2 + h_2^2 + h_3^2}\). The equation is simplified if we redefine

Then we obtain the equation

1.3.2.4.1. Single Neutrino Forward then Backward¶

1.3.2.4.2. Single Neutrino Simultaneous¶

PROGRAM START
Halo Problem Forward and Backward:
Total number of iterations: 100
Size of rhos: 1000
Range: 1.000000
Step size: 0.001000
Save Steps: 2
Total clock time: 0.070386
Clock time for 1000 iterations: 0.70409
PROGRAM END

PROGRAM START
Halo Problem Forward and Backward:
Total number of iterations: 100
Size of rhos: 1000
Range: 1.000000
Step size: 0.001000
Save Steps: 2
Total clock time: 0.018305
Clock time for 1000 iterations: 0.18335
PROGRAM END

The result shows that

For 20000 total iteractions. Calculation is within range 0 to 1 with step size 1e-5. A total of 100 steps are exported. So each step indicates iteration of 200 times.

GIF: Step size \(10^{-5}\)

We immediately spot trouble here. The calculation doesn’t reach equilibrium. I calculated using larger iterations (100000 iterations) and it doesn’t converge either.

I checked the convergence.

Comparing step size 1e-4 and 1e-5, with Ntop = 20000 total iterations and 20 outputs. Each time increment in the plot indicates 20000/20=1000 iterations.

Compare step sizes \(10^{-5}\) and \(10^{-6}\).

Comparing step size 1e-6 and 1e-5, with Ntop = 20000 total iterations and 20 outputs. Each time increment in the plot indicates 20000/20=1000 iterations.