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

by Job Stancil   Last Updated August 13, 2019 18:20 PM

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 $$det~U_?=-1$$.

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#### Answers 1

It is indeed true that every unitary matrix $$U$$ is an element of some one-parameter subgroup. One can say that this is a consequence of the fact that the unitary matrices form a compact Lie group.

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 $$t = \pi$$.

Omnomnomnom
August 13, 2019 18:14 PM

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