# Decoherence in a Penning trap

• I
Malamala

Twigg, gentzen and asd789

Gold Member
Sorry for the slow and short reply.

I'm not 100% sure how to calculate the decoherence rate, but I can tell you that the blackbody radiation shifts the energy difference ##E_e - E_g## by ##\delta E = (\alpha_e - \alpha_g) \| \vec{E}_{BBR} \|^2## where ##\alpha_{e,g}## is the polarizability of the ground (excited) state (at a particular frequency). This is for Stark shifted qubit states. There is an analogous expression for Zeeman states. You can get ##\| \vec{E}_{BBR} \|^2## from Planck's law for blackbody radiation (spectral irradiance) and your trap's geometry.

My gut feeling is that you can get the decoherence rate by taking the variance of this blackbody frequency shift (##\gamma = \sigma_\nu##, same idea as when you calculate the coherence time of a laser from bandwidth). Thus, the decoherence rate would bee $$\gamma = |\alpha_e - \alpha_g| \sqrt{\langle \| \vec{E}_{BBR} \|^4 \rangle - \langle \| \vec{E}_{BBR} \|^2 \rangle^2}$$ The quadratic term can be calculated from the blackbody partition function by looking at the expectation value of energy squared (just as you do when you calculate energy fluctuations in an ideal gas from the heat capacity).

Does that make sense?