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 $$||H||_{2} = O(1).$$


1 Answer 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 $O(1)$.

  • $\begingroup$ Why, clearly? Can you add more information. $\endgroup$
    – stephanmg
    Commented Sep 28, 2020 at 13:10
  • 3
    $\begingroup$ @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 $U$. $\endgroup$
    – Attila Kun
    Commented Sep 28, 2020 at 14:14

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