Photon and axion as two mixing states

You need complex exponentials, a two-by-two symmetric matrix and its eigenvalues. Natural units are used throughout, with the reduced Planck constant and the speed of light set to one, so energies, inverse lengths and masses are all measured in electronvolts. At the end the answers are converted to kiloparsecs.

The equation for one polarization

Let the beam travel along $z$ with energy $E$. Only the magnetic field component across the beam, $B_\perp$, can mix the photon with the axion, and only the photon polarization along that component takes part. Call its amplitude $A_\parallel$ and the axion amplitude $a$. The other polarization, $A_\perp$, only picks up a phase.

For energies far above the axion mass and the plasma frequency, the second-order wave equation reduces to a first-order one. In the sign convention used on this site,

with three entries, each an inverse length:

The off-diagonal $\Delta_{a\gamma}$ is the mixing. The diagonal entries say how far each state's phase drifts from that of a massless wave. The axion drifts because of its mass $m_a$. The photon drifts because free electrons in the plasma give it an effective mass equal to the plasma frequency $\omega_{\rm pl}$. The same matrix, with the other polarization added, appears as eq. (5) of Meyer, Horns and Raue, 2013 and in the appendix of Mirizzi, Raffelt and Serpico, 2007. Both also carry a vacuum-polarization term that is negligible for keV photons in microgauss fields, and this chapter drops it.

Solving it by rotating the basis

A constant symmetric matrix has two real eigenvalues and orthogonal eigenvectors. Write the mixing angle $\theta$ through

The eigenvalues are $\lambda_\pm=\tfrac12(\Delta_{\rm pl}+\Delta_a)\pm\tfrac12\Delta_{\rm osc}$, where

In the rotated basis each component simply gains a phase $\mathrm e^{i\lambda_\pm z}$. Start with a pure photon, $A_\parallel=1$ and $a=0$. Rotate into the eigenbasis, let each part evolve, and rotate back. The axion amplitude is

Its squared magnitude is the conversion probability. Using $\lvert\mathrm e^{ix}-\mathrm e^{iy}\rvert^2=4\sin^2\tfrac{x-y}{2}$ and $\sin2\theta=2\Delta_{a\gamma}/\Delta_{\rm osc}$,

The common phase from $\Delta_{\rm pl}+\Delta_a$ cancels. Only the difference between the two states and the mixing between them matter.

Four limits to check

Set the coupling or the field to zero. Then $\Delta_{a\gamma}=0$ and the probability vanishes at every length and energy, as it must.

Make the path short, $\Delta_{\rm osc}L\ll1$. The sine is close to its argument and the mismatch drops out:

This is why the probability grows as the square of field times length at first.

Tune the axion mass to the plasma frequency. The diagonal entries become equal, the amplitude factor is one, and $P_{\gamma\to a}=\sin^2(\Delta_{a\gamma}L)$. The energy has dropped out completely. The next chapter uses this resonance.

Finally, photon survival plus axion fraction always add to one, because the evolution only rotates and rephases the two amplitudes.

From electronvolts to kiloparsecs

The units need two conversions. One tesla is $195.35\,\mathrm{eV^2}$ in these units, so one microgauss, $10^{-10}$ tesla, is $1.9535\times10^{-8}\,\mathrm{eV^2}$. One kiloparsec is $1.5637\times10^{26}\,\mathrm{eV^{-1}}$. With $g_{a\gamma}=10^{-11}\,\mathrm{GeV^{-1}}=10^{-20}\,\mathrm{eV^{-1}}$ and $B_\perp=1\,\mu\mathrm G$,

Multiplying by the number of inverse electronvolts in one kiloparsec turns this into

The mixing term scales linearly with both the coupling and the field. We derived these conversions from CODATA constants and checked the tesla value a second way, through the critical field $m_e^2/e$.

Worked example: one uniform field

Take $g_{a\gamma}=10^{-11}\,\mathrm{GeV^{-1}}$, $B_\perp=5\,\mu\mathrm G$, electron density $10^{-2.5}\,\mathrm{cm^{-3}}$, $m_a=10^{-13}\,\mathrm{eV}$, a 96 kpc path and $E=6.6\,\mathrm{keV}$. The three entries are

Then $\Delta_{\rm osc}=0.161\,\mathrm{kpc^{-1}}$, so one full cycle takes $2\pi/\Delta_{\rm osc}=39.0$ kpc. The amplitude factor is $(0.1527/0.1612)^2=0.898$. The phase over the path is $\Delta_{\rm osc}L/2=7.74$ rad, and $\sin^2(7.74)=0.987$. A photon polarized along the field converts with probability

Unpolarized light is an equal mixture of the two polarizations, and only one of them mixes. Its survival is $1-P_{\gamma\to a}/2=0.557$. At 2, 4 and 9 keV the same path gives survival 0.777, 0.718 and 0.570, so the dimming changes with energy.

Lab tasks

1. Predict. Choose the uniform-field preset. Before moving anything, use the formula to say whether survival at 6.6 keV is above or below one half.
2. Change one condition. Drag the energy cursor from 0.5 to 10 keV. Note where the solid curve has its minima.
3. Compare. In uniform mode the dashed reference is the closed formula above. Check that the two curves lie on top of each other at every energy.
4. Explain. Lower the coupling slider to an exponent of minus twelve. The dips shrink to a few percent and their positions move. Use the amplitude factor to explain why: once the mixing is small next to the mismatch, the largest possible conversion falls as the square of the coupling.

Exercises

1. In the short-path limit, by what factor does conversion change if the coupling doubles and the field halves?
2. For which mismatch $\Delta_{\rm pl}-\Delta_a$ is the largest possible conversion exactly one half?
3. Unpolarized light crosses one uniform field. Show that the surviving light has linear polarization degree $P_{\gamma\to a}/(2-P_{\gamma\to a})$, and evaluate it for the worked example.

Try it in the Lab: Choose Oscillation in a uniform field and drag the energy cursor.