Hmm, you may be right. I would assume that the low temperature is to keep the qubits from being influenced by their environment. But minimizing computational energy might be equivalent to minimizing the probability of external interaction with the system. That’s just me guessing, however.
Again, I’m really no QC expert or even physicist for that matter.
It’s confusing because it’s incorrect. Quantum computers do not perform parallel computation. Performing that many computations, according the laws of physics as we know it (see Landauer’s Principle), would consume more energy than the mass of the sun.
So the transition operator in the continuous time case might be thought of as a matrix of differentials. Instead of s mapping to Ts, it instead maps to Tsdt. Then we can integrate over the time domain to get the total change.
I was actually a bit concerned about the notation I used, as it isn’t clear that the probability distribution/density is a function of time.