Do all unitary matrices belong to a one-parameter unitary group (for Stone's theorem).

Question

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 $det~U_?=-1$.

Answer 1

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 $$ \det U_t = \det \exp(iHt) = \exp (\operatorname{trace}(iHt)) = e^{it}. $$