# Proof that any unitary can be written as $U=e^{-iH}$ with $H$ Hamiltonian with bounded norm

I am looking for a proof that any unitary matrix can be written as:

$$U = e^{-iH}$$

where $$H$$ is some Hamiltonian with bounded norm. That is $$||H||_{2} = O(1).$$

Since $$U$$ is a normal matrix, the spectral theorem applies, i.e. we can write $$U=\sum_n\lambda_n|\lambda_n\rangle\langle\lambda_n|,$$ where $$\lambda_n$$ are the eigenvalues, and $$|\lambda_n\rangle$$ are the eigenvectors. Moreover, since $$UU^\dagger=I$$, we know that $$|\lambda_n|^2=1$$, and thus we can write $$\lambda_n=e^{-i\theta_n}$$ for $$\theta_n$$ in the range 0 to $$2\pi$$.
Now, let $$H=\sum_n\theta_n|\lambda_n\rangle\langle\lambda_n|.$$ Clearly, $$e^{-iH}=U.$$ Also, $$\|H\|_2$$ is the maximum singular value of $$H$$. Since all eigenvalues $$\theta_i$$ are positive, this is just the largest of the $$\theta_i$$, which, being less than $$2\pi$$, is $$O(1)$$.
• @stephanmg This comes from the definition of operator functions (Nielsen & Chuang section 2.1.8). For calculating $f(A)$ where $A$ is a normal matrix, you use the spectral decomposition and then apply $f(\cdot)$ to the eigenvalues of $A$: $f(A)=\sum_a f(a) |a\rangle \langle a|$. Here $f(\cdot) = exp(\cdot)$ and $A=-iH$, so you get $e^{-iH} = \sum_n e^{ -i \theta_n } |\lambda_n\rangle\langle\lambda_n|$ which is exactly $U$. – Attila Kun Sep 28 at 14:14