How is energy balanced in the Primakoff effect for axions? In the presence of a strong magnetic field, (theoretical) axions are converted to high energy photons via the Primakoff effect.
How is energy balanced in this process? Is the mass of the axion being converted directly to energy, or does some energy come from the transverse magnetic field?
Assuming the energy to drive the B-field is sufficiently low, is it possible to extract energy via the outgoing photons? In current experiments (ABRACADABRA, ADMX), the ratio of the power the machine consumes to the potential signal power in the detected photons is likely many orders of magnitude.
 A: In the Primakoff process $\gamma + a \to \gamma$, energy is conserved, i.e.
$$E_0 + E_a = E_1.$$
Since $E = \hbar \omega$, this can also be written as
$$\omega_0 + \omega_a = \omega_1.$$
Most axion experiments consider a static background field, $\omega_0 = 0$, so all the energy comes from the axion field. But in some proposed experimental setups, $\omega_0$ is nonzero, so that $\omega_1$ can be increased. (Disclaimer: I helped write that paper.)
Thus, energy extraction from axion dark matter is possible in principle. The axion power absorbed in a typical resonant cavity is, in natural units,
$$P_{\text{sig}} \sim \frac{g_{a \gamma \gamma}^2 \rho_{\mathrm{DM}}}{m_a} \, (B^2 V Q).$$
For a typical cavity haloscope along the lines of ADMX, we have cavity parameters $B \sim 10 \, \mathrm{T}$, $V \sim 1 \, \mathrm{m}^3$, $Q \sim 10^6$. The local dark matter density is $\rho_{\mathrm{DM}} \approx 0.4 \, \mathrm{GeV}/\mathrm{cm}^3$ and the typical axions targted by ADMX have $g_{a \gamma \gamma} \sim 10^{-16}\, \mathrm{GeV}^{-1}$ and $m_a \sim 10^{-6} \, \mathrm{eV}$. Plugging these numbers in gives the astoundingly small power
$$P_{\text{sig}} \sim 10^{-22} \, \mathrm{W}$$
which is of course the reason these experiments are difficult.
As a fun but incredibly unrealistic final calculation, note that the rate of dark matter mass flow through the Earth is $\mu \sim \rho_{\mathrm{DM}} v R_E^2$ where $v \sim 10^{-3} c$. If you can only capture the kinetic energy, the maximum possible rate of power harvested is roughly $\mu v^2 \sim 1 \, \mathrm{GW}$, which is the output of a typical power plant. If you can capture the full mass-energy, it's $\mu c^2 \sim 10^3 \, \mathrm{TW}$, which dwarfs total human power consumption.
