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A basic assumption of various error correction threshold theorems is that the probability of an error exponentially decreases with the weight of the error.

One of the selling features of ion traps is that in a single trap, the qubits have all-to-all connectivity. For, e.g., LDPC codes, this is essential. But does this also mean that there is a much higher chance of high-weight, coherent errors (e.g., cross-talk)?

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Before going further, I think it is worth pointing out that trapped ions have any-to-any rather than all-to-all coupling. With maybe a few exceptions, no one is doing anything more than two qubit gates.

The basic idea of trapped ion quantum computing is generally as follows. Typically, you encode the qubit state in the ion's spin (hyperfine/Zeeman qubits). To perform two qubit gates, you couple the ion's spin to it's motion by using focused lasers. Without those lasers, there is virtually zero coupling between spin and motion, which is the big feature of ion trap quantum computing. In fact, this lack of coupling is what makes two qubit gates rather challenging technically - you have to use a considerable amount of laser power for a relatively long time to do gates. So the way crosstalk errors occur is actually quite local - the nonzero spot size of the lasers means that neighboring qubits will undergo unintended gate operations as well. However, since the laser intensity is gaussian in space, the effects on even next-nearest neighbors are exponentially small.

However, it is worth pointing out that gates can and do introduce motional heating if they are imperfect. The idea being that a perfect gate returns the motional state of each ion back to where it started, but imperfect gates leave a bit of motion left over. You may think that this motional heating will mess up gates coming afterwards. However, a main selling point of the Molmer-Sorenson gate with ions is that it works regardless of the initial state of the motional degrees of freedom. So while this heating does affect gate fidelity, it does not cascade and propagate errors in the way you might be wondering.

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  • $\begingroup$ You say "you couple the ion's spin to it's motion" to describe a 2 qubit gate, but such a gate needs to apply to 2 different ions, right? If those two ions are far apart, how do they communicate for the 2 qubit gate? Can that process create high-weight errors? $\endgroup$
    – Sam Jaques
    Commented Nov 7, 2022 at 11:01
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    $\begingroup$ Good question, their spins are both coupled to a vibrational mode of the ion crystal. While other ions will definitely feel this vibration as well, their spins are not coupled to the vibration unless they are exposed to lasers at the right frequency. Hence why the beam spot size must be tightly focused in these experiments. $\endgroup$
    – Dr. T. Q. Bit
    Commented Nov 7, 2022 at 11:07
  • $\begingroup$ Thanks, so maybe to condense my original line of thought: the beam spot size will not be perfect, so there is presumably a bit of erroneous vibrational coupling. That seems like it will be a "high weight" term, i.e., it will be all-to-all (but maybe that's wrong). If it is a high-weight term, then even if it's small now (say $<10^{-5}$, even), it won't get any smaller with the usual QECC. Am I on the right track? $\endgroup$
    – Sam Jaques
    Commented Nov 8, 2022 at 16:32
  • $\begingroup$ Are you talking about all-to-all or any-to-any? This type of crosstalk will only affect neighboring qubits at a given gate, not next nearest neighbors etc. $\endgroup$
    – Dr. T. Q. Bit
    Commented Nov 9, 2022 at 15:50
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    $\begingroup$ Good thoughts! In practice the way experiments are actually done it's a bit different that what your imagining. While the initial error from a single two qubit gate will leave some tiny amount of motional entanglement, you are also actively laser cooling the motional degrees of freedom before and after the gate. So in effect the motional state is measured before and after every gate (but not during) and I would say that the kind of error you're thinking of wouldn't occur. $\endgroup$
    – Dr. T. Q. Bit
    Commented Nov 10, 2022 at 1:48

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