Thanks for the 1 D argument. You write that the same argument works in $n$ dimensions, which is not absolutely clear to me since the representation of $f$ via $\nabla f$ cannot obviously be used in the same way.

@Venkataramana Thank you very much.

@Venkataramana Thanks for your comment: How do you get that identity?

Thanks for your comment: How do you get that identity?

Let us take a look at the evolution equation, with $A=-iQ$, $Q$ self-adjoint, $$ \frac{du}{dt}+(P+iQ)u=0, \quad u(0)=u_0. $$ With $u=e^{-it Q}v$, we have $e^{it Q}$ unitary and $$ e^{-it Q}\dot v- iQ e^{-it Q} v+iQe^{-it Q} v+Pe^{-it Q}v=0,\quad\text{ i.e.,} $$ $$ \dot v+e^{it Q}Pe^{-it Q}v=0. $$ In the simple case where $P$ and $Q$ commute, then $v= e^{-t P}v(0)$. With $P(t)=e^{it Q}Pe^{-it Q},$ we have $$ \Vert v(t)\Vert^2+ 2\int_0^t\Vert P^{1/2} e^{-is Q} v(s)\Vert^2 ds=\Vert v(0)\Vert^2. $$It seems that whenever $[P,iQ]\ge 0$ the situation should not be much different.

@AHusain Thanks to your comment, I have added some details in my answer.