I am interested in mathematically proving the time-energy uncertainty relation by Mandelstamm-Tamm in a system seen in the Schrödinger-Picture (time-dependent states but time-independent operators). Let this system be prepared in the state $|\Psi(t)\rangle$.
Let $A$ be a time-independent operator (i.e. not explicitly depending on t) and let $H$ be the Hamiltonian of the system. I want to prove that
$$\Delta H\frac{\Delta A}{|\frac{d}{dt}\langle A(t)\rangle|}\geq\frac{\hbar}{2}$$ where $\Delta A:=\sqrt{\langle A^2\rangle-\langle A\rangle^2}$ denotes the standard deviation of $A$ (and for $H$ analogously). I use the following Hamiltonian: $$H=\frac{p^2}{2m}+V(x)=-\frac{\hbar}{2m}\nabla^2+V(x)$$ where $p=-i\hbar\nabla$ denotes the impulse operator, $x$ the impulse operator and $V:\mathbb{R}^3\to\mathbb{R}$ a given potential.
I wish to use the general uncertainty principle, which states that $$\Delta H\Delta A\geq\frac{1}{2}|\langle[H,A]\rangle|=\frac{1}{2}|\langle\Psi(t)|[H,A]|\Psi(t)\rangle|=\,\stackrel{?}{\cdots}\,=\frac{\hbar}{2}\left|\frac{d}{dt}\langle\Psi(t)|A|\Psi(t)\rangle\right|$$ from which we could deduce the desired by dividing by $\frac{\hbar}{2}|\frac{d}{dt}\langle A\rangle|$ on both sides.
I have problems at the question mark and I cannot figure out how to move on with $$[H,A]|\Psi(t)\rangle=(HA-AH)|\Psi(t)\rangle=\frac{\hbar}{2m}(A\nabla^2|\Psi(t)\rangle-\nabla^2A|\Psi(t)\rangle)$$ since there is not time derivative $\frac{d}{dt}$ involved here. Did I use the wrong Hamiltonian or did I oversee something?
EDIT: I forgot to mention that $A(t):=U^\dagger AU$ with $U=U(t)=\exp\left(-\frac{i}{\hbar}Ht\right)$ (I want $H$ to be time-independent).
EDIT 2: I was pointed to the relation $$i\hbar\frac{d}{dt}A(t)=[A(t),H]$$ but I am not sure if I can use this in my system and I also tried to use this to find \begin{align*} \Delta A\Delta H&\geq\frac{1}{2}|\langle\Psi(t)|[H,A]|\Psi(t)\rangle|=\frac{1}{2}|\langle\Psi(t)|\left(-i\hbar\frac{d}{dt}A(t)\right)|\Psi(t)\rangle| \\ &= \frac{\hbar}{2}|\langle\Psi(t)|\frac{d}{dt}A(t)|\Psi(t)\rangle|\stackrel{?}{=}\frac{\hbar}{2}\left|\frac{d}{dt}\langle\Psi(t)|A(t)|\Psi(t)\rangle\right| \end{align*} where I am not sure about interchanging $\frac{d}{dt}$ with $\langle\cdot\rangle$ in the last step.
