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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?

glS
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WishBeLeibniz
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    I believe the way you propose is quite standard. For a more general understanding of exponentiation, maybe look for an introductory course on Lie groups and Lie algebras, but for simple question like this, elementary linear algebra should do. – Peter Franek Jan 01 '15 at 09:23
  • It is not directly applicable to your context, but suppose that $U^t$ is known. For example when $A$ was already given, then $U^t$ is given by $e^{iAt}$. Taking the derivative gives $A = \frac{1}{i} \left. \frac{d}{dt} \right|_0U^t$. The interpretation of this is that any continuous time evolution of unitary operators is automatically of the form $e^{iAt}$. This is the content of Stone's Theorem. – abcdef Jun 13 '19 at 21:56

2 Answers2

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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 $P(U)=H$.

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

  2. 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. $e^{iH}=U$.

  • Thanks. What do you mean "$H$ is a polynomial in $U$"? Is it another way to say that "the $\theta_j$ are not unique"? – WishBeLeibniz Jan 26 '16 at 18:55
  • I got that $R^UR = \exp(idiag(\theta_j))$, but how this yields $U=\exp(iRdiag(\theta_j)R^)$? – DH K Nov 15 '24 at 05:42
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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.

Eric Sakk
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    @ Eric Sakk , your reasoning is not correct. You don't know $H$, then you don't know $V$. You must directly diagonalize $U$. –  Jan 16 '16 at 10:43