Is exponential of GUE random matrix Haar random? Consider the matrix exponential map $H \mapsto e^{i H t}$ acting on the Gaussian unitary ensemble (GUE) of Hermitian matrices. 
I would expect that for large $t$, the resulting measure on the unitary group approaches Haar measure -- is that right? Is there a simple heuristic argument showing that the limiting distribution is unitary-translation-invariant?
I imagine that some much larger class of distributions on the Lie algebra might also yield the Haar measure under exponentiation for large $t$?   (Perhaps for other Lie groups / Lie algebras as well?) And if so, I'm also curious which distribution converges most quickly to Haar measure (for some fixed normalization of the size of $H$).
 A: If you exponentiate the GUE ensemble of Hermitian matrices, $H \mapsto e^{i H t}$, then at large times, you actually don't converge to the Haar measure (AKA the measure of the circular unitary ensemble, or CUE) on the unitary group.  The intuition is simple: when you exponentiate, the distribution of eigenvectors stays the same (and matches that of the CUE), but the eigenvalues become $\lambda_i \mapsto e^{i \lambda_i t}$, and at large times, these eigenvalues circle around the unit circle at different speeds and ultimately mix uniformly.  So the limiting distribution should have eigenvalues which are i.i.d. and uniformly distributed over the unit circle, whereas the eigenvalues of the CUE are not independent and actually exhibit "eigenvalue repulsion."  
This limiting distribution on the unitary group should occur somewhat generally, independent of the initial distribution of Hermitian matrices.  Here's the sketch of a more rigorous derivation.  Consider an ensemble Hermitian matrices, with joint density of eigenvalues $p(\lambda_1,...,\lambda_n)$.  Crucially, assume that such a continuous density exists. We are interested in the new (pushforward) measure under the exponential map $\lambda_i \mapsto e^{i \lambda_i t}$.  Using change of variables and recognizing that the exponential is a many-to-one map, we can calculate new the joint density of eigenvalues.  Writing the density in terms of phases $\theta \in [0,2\pi)$ (angles on the unit circle), you find
$$\rho_t(\theta_1,...,\theta_n) = \sum_{ \lambda_1,...,\lambda_n \text{ s.t. } e^{i \lambda_i t} = e^{i \theta_i}} t^{-n} p(\lambda_1,...,\lambda_n),$$ where the factor of $t^{-n}$ comes from the Jacobian $\left\vert \frac{d \theta_i}{d \lambda_j}\right\vert$.  Taking $t$ to be a large integer $T$, we have 
$$\rho_T(\theta_1,...,\theta_n) = T^{-n} \sum_{m=1}^T \sum_{a_1,...,a_n \in \mathbb{Z}} p\left(\frac{\theta_1+2 \pi m}{T}+2\pi a_1,...,\frac{\theta_1+2 \pi m}{T} + 2\pi a_n\right)$$
$$ \approx T^{-n} \sum_{a_1,...,a_n \in \mathbb{Z}} \frac{T^n}{(2\pi)^n}\int_{2\pi a_1}^{2\pi(a_1+1)} d\lambda_1 \cdots \int_{2\pi a_n}^{2\pi(a_n+1)} d\lambda_n \; p(\lambda_1,..., \lambda_n)$$  $$= (2\pi)^{-n}\int_{-\infty}^{\infty} p(\lambda_1,...,\lambda_n) d\lambda_1 ... d\lambda_n = (2\pi)^{-n},$$
i.e. a constant density.
