Non-existence of time operator in quantum mechanics
On the not Hermitian point.

Let $\hat{F}$ be a non-stationary observable (but does not depend explicitly on time $t$, hence $\partial \hat{F}/\partial t = 0)$. in the Heisenberg picture it satisfies the Heisenberg equation of motion for $\hat{F}$, namely
$$i \frac{d \hat{F}}{dt} = [\hat{F},\hat{H}] \tag{1}$$.

Now assume that whenever $d \hat{F}/dt \neq 0$ then it has an inverse
$$(d \hat{F}/dt)^{-1} = i([\hat{F},\hat{H}])^{-1}\tag{2}$$
Now, if
\begin{align}
[(d \hat{F}/dt)^{-1}, \hat{H}] &amp;= 0 \\
\implies [([\hat{F},\hat{H}])^{-1},\hat{H}] &amp;=0. 
\end{align}
Now introduce a Time operator, $\hat{T}$ as
\begin{align}
\hat{T} &amp;= \frac{1}{2}\left(\hat{F}\left(\frac{d \hat{F}}{dt}\right)^{-1} + \left(\frac{d \hat{F}}{dt}\right)^{-1} \hat{F}\right) \\
&amp;=\frac{1}{2}\left(\hat{F} ([\hat{F},\hat{H}])^{-1}+([\hat{F},\hat{H}])^{-1}\hat{F}\right)\tag{3}
\end{align}
Now using equations $(1)$ and $(2)$ into $(3)$ one finds that
$$[\hat{H}, \hat{T}] = i \tag{4}$$

That is, $\hat{T}$ and $\hat{H}$ form a canonically conjugate pair. Owing to the fact that the Hamiltonian spectrum is bounded from below, the time operator $\hat{T}$ is not self adjoint, in other words, not Hermitian.