Timeline for For any unitary $U = e^{iA}$, question about the matrix element $A_{pq}$ of matrix $A$ and its functional derivative

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Jul 24, 2023 at 17:31	comment	added	Quillo		@JonMegan it means that typically the Hilbert space is much larger and, possibly, uncontrollable (it is infinite-dimensional). However, there may be a finite number $N$ of (orthogonal) states that can be reached ("prepared") via experimental techniques. Therefore, the analysis is restricted to this subspace (or to spin systems that are naturally finite dimensional.. or an ensemble of Q-bits).
Jul 24, 2023 at 17:26	comment	added	Jon Megan		I also believe that this assumption is not well-explained. I have thought about it deeper and now it at least makes sense that every matrix element of $A$ should be independent with respect to $C(t)$. This is because we produce every $2^N$ computational basis states: $\lvert 00 \cdots 00 \rangle, \dotsc, \lvert 11 \cdots 11\rangle$. Since this is a (sub)set of basis for any $U$, it kind of makes sense that every element matrix of $U$ can be independently addressable w.r.t $C(t)$, and so thus $A$ ($A$ is hermitian, so matrix element of $A$ must be directly related to $U = e^{iA}$).
Jul 24, 2023 at 17:21	comment	added	Jon Megan		Thanks you for your comment. I am not sure if I understand the first paragraph properly. What do you mean by "𝑁-dimensional Hilbert space is just the "controllable" portion of a possibly larger system"?
Jul 24, 2023 at 11:57	history	edited	Quillo	CC BY-SA 4.0	new links
Jul 24, 2023 at 11:41	history	answered	Quillo	CC BY-SA 4.0