Do all unitary matrices belong to a one-parameter unitary group (for Stone's theorem). Background.  Per Stone's theorem, a one-parameter unitary matrix group $U_t$ corresponds to a Hermitian matrix $H$:
$$U_t=e^{iHt}$$
Example. The group of unitary matrices
$$U_t=
\left(
\begin{matrix}
1&0&0&0 \\ 0&1&0&0 \\ 0&0&\cos t&i\sin t \\ 0&0&i\sin t&\cos t \\
\end{matrix}
\right)
$$
corresponds to the matrix
$$H=
\left(
\begin{matrix}
0&0&0&0 \\ 0&0&0&0 \\ 0&0&0&1 \\ 0&0&1&0 \\
\end{matrix}
\right)
$$
and then for $t=\pi/2$ we have
$$
U_{\pi/2}=
\left(
\begin{matrix}
1&0&0&0 \\ 0&1&0&0 \\ 0&0&0&i \\ 0&0&i&0 \\
\end{matrix}
\right)
$$
Question. For any given unitary matrix $U$, does it belong to a one-parameter group, such that Stone's theorem applies?
I've been trying to find $H$ for the matrix below, but haven't yet managed to:
$$
U_{?}=
\left(
\begin{matrix}
1&0&0&0 \\ 0&1&0&0 \\ 0&0&0&1 \\ 0&0&1&0 \\
\end{matrix}
\right)
$$
Note that may not be relevant: it's unitary, though with $det~U_?=-1$.
 A: It is indeed true that every unitary matrix $U$ is an element of some one-parameter subgroup.  Equivalently, the map $H \mapsto \exp(iH)$ is a surjective map from the Hermitian matrices to the unitary matrices. One can say that this is a consequence of the fact that the unitary matrices form a connected, compact Lie group; if you prefer, this can also be proven as a consequence of the construction suggested below.
The $U$ which you have given can be diagonalized as
$$
U = \pmatrix{1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0} = \\
\pmatrix{1&0&0&0\\0&1&0&0\\0&0&1/\sqrt{2}&1/\sqrt{2}\\0&0&1/\sqrt{2}&-1/\sqrt{2}}
\pmatrix{1\\&1\\&&1\\&&&-1}
\pmatrix{1&0&0&0\\0&1&0&0\\0&0&1/\sqrt{2}&1/\sqrt{2}\\0&0&1/\sqrt{2}&-1/\sqrt{2}}^T
$$
so, it belongs to the one-parameter subgroup
$$
U_t = \pmatrix{1&0&0&0\\0&1&0&0\\0&0&1/\sqrt{2}&1/\sqrt{2}\\0&0&1/\sqrt{2}&-1/\sqrt{2}}
\pmatrix{1\\&1\\&&1\\&&&e^{it}}
\pmatrix{1&0&0&0\\0&1&0&0\\0&0&1/\sqrt{2}&1/\sqrt{2}\\0&0&1/\sqrt{2}&-1/\sqrt{2}}^T
\\ 
= I + \frac{e^{it} - 1}{2}\pmatrix{0&0&0&0\\0&0&0&0\\0&0&1&-1\\0&0&-1&1},
$$
which corresponds to the generator
$$
H = \pmatrix{1&0&0&0\\0&1&0&0\\0&0&1/\sqrt{2}&1/\sqrt{2}\\0&0&1/\sqrt{2}&-1/\sqrt{2}}
\pmatrix{0\\&0\\&&0\\&&&1}
\pmatrix{1&0&0&0\\0&1&0&0\\0&0&1/\sqrt{2}&1/\sqrt{2}\\0&0&1/\sqrt{2}&-1/\sqrt{2}}^T
\\ = \frac 12 \pmatrix{0&0&0&0\\0&0&0&0\\0&0&1&-1\\0&0&-1&1}.
$$
Your $U$ is attained at $t = \pi$.
As far as the determinant goes, you might find it interesting that
$$
\det U_t = \det \exp(iHt) = \exp (\operatorname{trace}(iHt)) = e^{it}.
$$
