除了 Godric Seer 的建议，您还可以对指数使用有理逼近，例如

$e^z = \frac{1 + z/2}{1 - z/2} + \mathcal{O}(z^3)$,

为矩阵指数设计更准确的近似值：

$e^{-itH} \approx (I + itH)^{-1}(I - itH)$.

这种近似具有作为酉算子的优点，因此在精确算术中$\|w\|$ remains unchanged. Of course you have to watch out in floating point arithmetic if the condition number of $H$ is very large.

The paper 19 Dubious Ways to Compute the Exponential of a Matrix is worth a read, and if you want to dig deeper, Saad's book is both very readable and very comprehensive.

By definition

$e^{idtH} = \sum\limits_{k=0}^\infty \frac{1}{k!}\left( i dt H\right)^k$

which is easily approximated by truncating the sum after a number of terms. Ideally you would want to use only two terms so that

$e^{i dt H} \approx 1 + i dt H$

What that means is that $\frac{1}{2}(i dt H)^2$ must have a small effect on the solution. Assuming $H$ has eigenvalues $\lambda_i$, then you need $dt^2 \lambda_i^2 << 1$ for every eigenvalue. As long as you choose a small enough $dt$ that this is true, you can avoid the matrix exponential with only the first two terms of the sum.

Is H time-dependent? If not, can not you just diagonalize H, and then express your initial vector "v" as a linear combination of the eigenvectors of H, and then propagate those?

If H is time-dependent, another technique (in addition to those already mentioned) would be to use a split operator technique like the Trotter decomposition.