I'm trying to learn about the physics of ion traps, and my understanding is that an ion can be moved from a ground state to an excited state by stimulation with a laser pulse. More specifically, this is a Raman transition: the ion absorbs an incoming photon and reaches a "virtual state", then releases a photon and decays back down to either the initial state or the excited state.

The emitted photon is thus entangled with the qubit. To maintain coherence, it needs to fit into a coherent state such as a laser state.

However, everything I've read about this kind of scattering states that the outgoing photon has a random direction (which I assume means a superposition of directions until the outgoing photons start interacting with the environment). If this happens, then the outgoing photon is no longer part of the coherent laser state (which is mostly propagated in the original direction). All outgoing photons will be correlated with the qubit, and so when they interact with the environment, they will decohere the qubit.

It seems like the only way to maintain coherence would be if the outgoing photons had the same momentum (but why would this happen?) or if there was so much scattering that there was effectively a coherent scattered state that was only weakly entangled with the trapped ion. Is any of this accurate?

  • $\begingroup$ Why does the emitted photon have to be entangled with the qubit? If there is a 100% chance of emission then the presence or absence of the photon is not entangled with the qubit. It is true that the momentum kick would be entangled with the qubit... let's say a standard shift is by ~meV and something like ytterbium has mass ~$10^{-25}$kg gives a momentum kick around $10^{-25}$kg m/s or velocity kick ~30m/s, so that's nonnegligible, but maybe the trap holds the ion in place and so the photons just become correlated with the trap lasers? $\endgroup$ Dec 11, 2021 at 16:12
  • $\begingroup$ Hmmm, okay, so the trap lasers become entangled with the qubit, but if they are in a sufficiently coherent laser state, then the entanglement is negligible? $\endgroup$
    – Sam Jaques
    Dec 12, 2021 at 17:26
  • $\begingroup$ Maybe, I'm not sure to be honest. In general the entanglement is negligible when the light is bright enough (eg if you use strong coherent states in the Jaynes Cummings model), so that could also answer the question directly about the Raman scattering, but I am not certain if the amount of light scattered is enough to still be classical. I'm still hoping for a better answer here $\endgroup$ Dec 12, 2021 at 22:31

1 Answer 1


I think you are basically getting at "spontaneous photon scattering". This is the dominant error source in ion trap quantum computing which use Raman transitions for gate operations. The effect does not appear in direct drive transitions, such as those with microwaves or two photon quadrupolar transitions.

The idea is similar to what you've described. After the initial virtual transition, the subsequent final photon emission can either be aligned along the laser direction (stimulated) or along a random direction (spontaneous). However, the reason it is predominantly along the laser direction is essentially the same physics as that of stimulated emission: the pre-existing coherent state of the second laser makes it more probable to "emit" a photon along that direction. Nonetheless, there is still a chance of emiting a photon along a random direction (i.e. coupling to the vacuum $n=0$ state of the photon field rather than the laser). This effect gives rise to photon scattering gate errors.

You can read more about this type of error here: https://journals.aps.org/pra/abstract/10.1103/PhysRevA.75.042329


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