Conservation

TDSH

$$ i\hbar \partial_t\left|\psi \right> = \hat{H}\left|\psi \right> $$

If $\hat{F}$ is not explicitly depend on the time $\partial_t \hat{F} = 0$,

$$ \frac{d}{dt}\left<\psi\right|\hat{F}\left|\psi\right> = (\partial_t\left<\psi\right|)\hat{F}\left|\psi\right> + \left<\psi\right|\hat{F}(\partial_t \left|\psi\right>) \\

=\left<\psi\right|\left(-\frac{\hat{H}}{i\hbar}\hat{F}+\hat{F}\frac{\hat{H}}{i\hbar}\right) \left|\psi\right> \\ =\left<\psi\right| \left( \frac{1}{i\hbar}[\hat{F},\hat{H}]\right) \left|\psi\right> $$

If the physical parameter $\left<\hat{F}\right>$ is constant of the motion, which means conserved quantities,

$$ \frac{d}{dt} \left<\hat{F}\right> = 0 \quad \Leftrightarrow \quad [\hat{F},\hat{H}]=0 $$

$[\hat{F},\hat{H}]=0$ means, two variables are compatible; can be measured simultaneously, and the quantum state can be labelled by quantum numbers, $\left|\psi\right>=\left|f, h\right>$

Unitary operator

After transform $\hat{U}$

$$ \left|\psi\right> \rightarrow \left|\psi'\right>=\hat{U}\left|\psi\right> $$

still TDSH is invariant under the transform,

$$ i\hbar \partial_t\left|\psi' \right> = \hat{H}\left|\psi' \right> \quad \Rightarrow \quad [\hat{U},\hat{H}] = 0 $$

For satisfying normalized condition,

$$ \left<\psi|\psi \right>=\left<\psi'|\psi'\right>=1 $$

Therefore, $\hat{U}$ is unitary transform,

$$ \hat{U}\hat{U}^{\dagger}=\hat{U}^{\dagger}\hat{U} = \hat{I} $$

The following form with Hermitian operator $\hat{G}$; $\hat{G}=\hat{G}^{\dagger}$

$$ \hat{U} = \hat{I} -i \frac{\epsilon}{\hbar}\hat{G} $$