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As a practicing physicist in the field, I think I'm qualified to reply. The question is a valid one, and I've heard it a few times before from other researchers. In brief, the irreversible heat dissipation, calculated from the change in the entropy (S) of some sub-system of the quantum computer, does not yield a quantity that has much practical relevance. But it does lead to interesting considerations, some of which I've written out below. Following the basic thinking behind Landauer's principle, if one considers the state of the qubits themselves, the dissipation is always zero or even negative. In the usual universal gate-based model, any computation starts with a known pure quantum state for which S = 0. The gates are unitary transformations and do not change the entropy. Non-coherent interactions generally lead to a state with nonzero positive entropy. Amusingly, this could be interpeted as using the quantum register to cool its surroundings by a tiny amount, at the cost of randomizing the output of the "computation". A more fruitful approach (which you also alluded to) relates to the "cost" of the unitary transformations themselves. Implementing high-fielity gates, necessary for useful quantum computing, requires very precise time-varying external control fields. It is correct to think that there is a thermodynamic cost to ensuring that the noise in the fields experienced by the qubits are small. For a concrete example, consider the fact that superconducting qubits are controlled with microwave signals. The output of a room-temperature microwave source has thermal noise supreimposed with the desired, sythesized signal. For simplicity, we can take the noise temperature to be 300K. (It is much higher in practice.) To use this output to drive a superconducting qubit at T = 30 mK, the power needs to be attenuated by at least a factor of 10^4. (Again, the real factor is much higher.) Hence, a lot of power is seemingly wasted in the synthesis of the control signals. I think a similar argument can be made about the lasers used to control some other kinds of qubits. This type of dissipation is relevant for hypothesized large-scale quantum computers. It doesn't lead to new deep "quantum" insights, however. First, the dissipation is independent of the state of the qubits, and simply scales linearly with the number of operations. Second, the dissipation takes place outside of the delicate qubit system. Removing any amount of entropy from (i.e., cooling) the environment surrounding the qubits is merely a difficult classical engineering problem. |