For any unitary matrix $U$, there exists a Hermitian matrix $H$ such that $U=e^{iH}$

Question

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?

Answer 1

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.

Answer 2

Preface

I welcome those more knowledgeable to edit this post, offer comments, or post a more thorough answer than Leo's.

The answer by Leo appears well regarded. Yet, I had questions and Leo appears to be no more. So, I rewrote the proof in a way that I could (more or less) understand it. I also modified the problem statement to make the space finite dimensional. I did this because I have a feeling that if the space is of infinite dimension, then proof might not hold.

Proposition Let $U$ be a unitary matrix on a finite dimensional space $\mathbb{C}^n$. There exists a Hermitian matrix $H$ such that $U=e^{iH}$. The Hermitian matrix is not unique.

Proof

Existence

Since $U$ is unitary, therefore it is also normal. From spectral theory [1], I know that since $U$ is normal there exists a unitary matrix $V$ such that $$ U=VDV^{\dagger} $$ where $D$ is a diagonal matrix. The entries of the diagonal of $D$ are the eigenvalues of $U$. The column vectors of $V$ are the eigenvectors of $U$ and they are orthonormal. Further, the entries of $D$ need not be real; rather the eigenvalues of $U$ lie on the unit circle [2].

Since non of the eigenvalues of $U$ are zero, therefore $U$ is invertible. Further, since the spectrum of $U$ does not contain the origin and since there is a path going from the origin to infinity not crossing the spectrum of $U$ (it being finite dimensional), this implies that one can define $ln(U)$ [3].

That said, set $$H=\frac{1}{i}\ln U.\tag{1}$$
Then, $$ e^{iH} = e^{ln(U)} = U $$ and \begin{align} H &= \frac{1}{i}\ln\left( U \right) \\ &= \frac{1}{i}\ln\left( VDV^{\dagger} \right) \\ &= \frac{1}{i}\ln\left( VDV^{-1} \right) && V~\text{is unitary} \\ &= \frac{1}{i}V \ln(D)V^{-1} && [3]~\text{but I do not understand who} \\ &= \frac{1}{i}V \ln(D)V^\dagger \\ &= \frac{1}{i}V \ln\left(\operatorname{Diag}\left(e^{i\theta_1},e^{i\theta_2},\cdots\right)\right)V^\dagger \\ &= \frac{1}{i}V \left(\operatorname{Diag}\left(\ln(e^{i\theta_1}),\ln(e^{i\theta_2}),\cdots\right)\right)V^\dagger \\ &= \frac{1}{i}V \left(\operatorname{Diag}\left( ( {i\theta_1}), ( {i\theta_2}),\cdots\right)\right)V^\dagger \\ &= V \,\operatorname{Diag}\left( \theta_1 , \theta_2 ,\cdots\right) \, V^\dagger. \end{align} Since $$ V \,\operatorname{Diag}\left( \theta_1 , \theta_2 ,\cdots\right) \, V^\dagger = \left( V \,\operatorname{Diag}\left( \theta_1 , \theta_2 ,\cdots\right) \, V^\dagger\right)^\dagger, $$ then so too $$ H =H^\dagger. $$

I have identified an object $H$ that is Hermitian and such that $U=e^{iH}$. Thereby I have demonstrated existence.

Uniqueness

$H$ is not unique. I can show this by example. Taking $H'=H+2\pi I$, we have \begin{align*} e^{iH'} &= e^{i\left(H+2\pi I\right)} \\ &= e^{iH+2\pi iI } \\ &= e^{iH } e^{ 2\pi i I } \\ &= e^{iH } I \\ &= e^{iH} = U. \end{align*}

Bibliography

[1] https://en.wikipedia.org/wiki/Spectral_theorem

[2] https://en.wikipedia.org/wiki/Unitary_matrix

[3] https://en.wikipedia.org/wiki/Logarithm_of_a_matrix