For any unitary matrix $U$, there exists a Hermitian matrix $H$ such that $U=e^{iH}$.
Is the above statement true? (I guess not)
If not, how "many" unitary matrixes can the expression $U=e^{iH}$ cover?
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$\begingroup$ it is nice question $\endgroup$ – zeraoulia rafik Jul 12 '15 at 17:07
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$\begingroup$ Check out here math.stackexchange.com/questions/1087375/… $\endgroup$ – A.Γ. Jul 12 '15 at 17:27
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$\begingroup$ Yes. en.wikipedia.org/wiki/Spectral_theorem#Normal_matrices $\endgroup$ – David C. Ullrich Jul 12 '15 at 17:28
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The statement is true. Any unitary matrix $U$ can be written as $U=e^{iH}$ for some $H$. However, $H$ is not unique, since taking $H'=H+2\pi I$, we have $e^{iH'}=e^{iH}=U$.
To see why the statement is true, simply write $U$ in its diagonal form (we can do this because of Spectral Theorem): $$U=VDV^\dagger$$ where $D=\text{diag}(e^{i\theta_1},e^{i\theta_2},\ldots)$, where $\theta_j\in\mathbb{R}$, and $e^{i\theta_j}$ are eigenvalues of $U$ which by definition of unitarity must be complex numbers with modulus 1. Then one choice of $H$ is $$H=\frac{1}{i}\ln U = \frac{1}{i}V \ln(D)V^\dagger = V\text{diag}(\theta_1,\theta_2,\ldots)V^\dagger$$ It isn't hard to check $e^{iH}=U$, and $H^\dagger = H$ is Hermitian since $\theta_j$ are real.
