Let $U$ be a unitary matrix, and let $H$ be an Hermitian matrix. I want to know if there is a $t \in\mathbb R$ such that $\exp(i t H) = U$.
A connected question is: given a set $\{g_1, g_2, ..., g_N\}$ of Hermitian matrices, does there exist a set of real parameters $\{\lambda_1,...,\lambda_N\}$ such that $\exp\left(i\sum_k \lambda_k g_k\right) = U$?
Clearly this won't always be possible, but is there some known criterion to say something about when it would or would not, for a given choice of matrices?
Has this question been studied? Does this kind of problem have a name? It looks very connected to the theory of Lie algebras/groups, but because I'm looking at a specific matrix $U$ and Hermitian $H$ (or set of hermitians), as opposite to whole algebras, it also seems to be a little bit different.
To better show the nontriviality of this kind of question, let us write the spectral decomposition of $U$ as $ U = \sum_k \lambda_k P_k$. If $U$ is nondegenerate we can write $$ \log U = \sum_k \operatorname{Log}(\lambda_k) P_k + 2\pi i\sum_k \nu_k P_k, $$ with $\nu_k \in \mathbb Z$. This expression i relatively easy to handle, because the projectors $P_k$ are fixed by the choice of $U$.
If $U$ is degenerate, let us write its spectral decomposition as $ U = \sum_k \lambda_k \sum_j P_{kj}$, where $k$ denotes the $k$-th eigenspace, and $P_{kj}$ are the corresponding trace-1 projectors. However, the choice of $\{P_{kj}\}_j$ has some freedom in it, because for any given choice of it, and any unitary rotation $R_k$ of the the degenerate eigenspace, I can write the decomposition as $$ U = \sum_k \lambda_k \sum_j R_k P_{kj} R_k^\dagger \equiv \sum_k \lambda_k \tilde{P}_{kj},$$ using the notation $\tilde{P}_{kj} \equiv R_k P_{kj} R_k^\dagger$. This implies that $$ \log U = \sum_k \operatorname{Log}(\lambda_k) P_k + 2\pi i\sum_k \nu_k \tilde{P}_{kj}, $$ from which you can hopefully better see where the problem lies: there is a lot of freedom in the way $\tilde{P}_{kj}$ are chosen, especially for highly degenerate matrices.
As a somewhat trivial example, consider the unitary (and Hermitian) 4x4 matrix $X_1 X_2 \equiv X \otimes X$, with $X$ denoting the Pauli $X$ matrix: $ X = \begin{pmatrix}0&1\\1&0\end{pmatrix}.$ Its "principal" (or a principal) logarithm is easily seen to be $$ \operatorname{Log}(X_1 X_2) = \frac{i\pi}{2}(1 - X_1 X_2).$$ However, it is also true that $$ \operatorname{Log}(X_1 X_2) = \frac{i\pi}{2}(2 - X_1 - X_2).$$ So while the "naive" logarithm uses $\{1, X_1 X_2\}$, one can find another logarithm that uses $\{1, X_1, X_2\}$ as generators. This case is trivial of course, because $X_1 X_2$ is separable and I obtain the second logarithm simply by taking the logarithm of $X_1$ and $X_2$ separately and then multiplying them together, nor I need the rotation of the degenerate space to go from one to the other.
There are less trivial cases in which it is however far less obvious thatwhether one can do something of this sort, hence my question.