Given a unitary matrix $U$, how do I find $A$ such that $U=e^{iA}$?

Question

A unitary matrix $U \in \mathbb C^{n \times n}$ can always be written in exponential form

$$U = e^{iA} \tag{1}$$

where $A$ is Hermitian. My goal is to find the Hermitian matrix $A$, given the unitary matrix $U$. I figured out a way by diagonalizing $U$, in the following form:

$$U = V^{\dagger} [e^{ia_{kk}}] V$$

Therefore, we get

$$A = V^{\dagger} [a_{kk}] V$$

Is this the standard way for finding the Hermitian matrix $A$ in equation (1)?

If I'd like to learn more about the exponentiation of unitary operators, and their general properties, what topics should I read?

Answer 1

Given $U=e^{iH}$, assume V diagonalizes H:

$e^{V^{-1} iH V} = V^{-1} e^{iH} V = V^{-1} U V$ implying that V also diagonalizes U;

hence, $V$ is easily found. Let $\alpha_i$ denote the $i^{th}$ diagonal element of $V^{-1} U V$, then $\alpha_i = e^{i \theta_i}$ where $e^{i \theta_i}$ is the $i^{th}$ diagonal element of $e^{V^{-1} iH V}$. Then, it easily follows that $\theta_i = \tan^{-1}(Im(\alpha_i)/Re(\alpha_i))$ since the eigenvalues of U lie on the unit circle.

Answer 2

I detail the way WishBeLeibniz proposes.

Since $U$ is normal, there is a unitary $R$ s.t. $R^*UR=diag(\lambda_j)$. Since $U$ is unitary, $\lambda_j=e^{i\theta_j}$ where $\theta_j\in\mathbb{R}$; then $R^*UR=\exp(idiag(\theta_j))$ and $U=\exp(iRdiag(\theta_j)R^*)$; finally, $H=Rdiag(\theta_j)R^*$ is hermitian and satisfies $U=e^{iH}$. Of course, $H$ is not unique, because the $\theta_j$ are not unique. Yet, if we choose $\theta_j=\theta_k$ when $\lambda_j=\lambda_k$, then $H$ is a polynomial in $U$.

EDIT. Answer to WishBeLeibniz. 1. Above we choose $H$ s.t. $U,H$ are diagonal in the same basis and thus $UH=HU$; if moreover we choose $\theta_j=\theta_k$ when $\lambda_j=\lambda_k$, then let $P$ be the Lagrange interpolating polynomial s.t. $P(\lambda_j)=\theta_j$; clearly $P(U)=H$.

2. $H$ is not unique because we can change $\theta_j$ with $\theta_j+2i\pi$.

3. Note that if $e^{iH}=U$, then $UH=HU$; since $H,U$ are both diagonalizable over $\mathbb{C}$, $H,U$ are simultaneously diagonalizable; thus we obtained above essentially whole set of hermitian $H$ s.t. $e^{iH}=U$.