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I've written the following proof:

Let $A,B$ be observables (Hermitian operators). define $ ΔA:=A-<A> $. the uncertainty principle is: $$ ⟨(ΔA)^2⟩⟨(ΔB)^2⟩ ≥|⟨[A,B]⟩|^2/4 $$
The Heisenberg equation for $A$ gives: $$ i d/dt ⟨(ΔA)^2 ⟩=⟨[A^2,H]⟩-2⟨A⟩⟨[A,H]⟩ $$ so: $$i d/dt⟨(ΔA)^2⟩⟨(ΔB)^2⟩ = (⟨[A^2,H]⟩-2⟨A⟩⟨[A,H]⟩)⟨(ΔB)^2⟩ + A↔B $$ define $$ C^2:=-i[A^2-2⟨A⟩A,H] , D^2:=-i[B^2-2⟨B⟩B,H] $$ which are Hermitian (are observables and have square roots). using the uncertainty principle for $(C,B)$ and $(D,A)$, we get: $$ d/dt ⟨(ΔA)^2 ⟩⟨(ΔB)^2 ⟩≥|⟨[C,B]⟩|^2/4+|⟨[D,A]⟩|^2/4+⟨C⟩^2⟨(ΔB)^2 ⟩+⟨(ΔA)^2 ⟩⟨D⟩^2 $$
zero is assumed if $A$ or $B$ are constants of motion. so: $$ d/dt ⟨(ΔA)^2 ⟩⟨(ΔB)^2 ⟩≥0 $$ but this is wrong (check out Does Heisenberg's uncertainty under time evolution always grow?). I'm having trouble figuring out what's exactly wrong with my proof? does it only apply when $C,D$ are positive definite?

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  • $\begingroup$ I suspect the issue is that $C^2$ being Hermitian does not imply that $C$ is Hermitian. $\endgroup$ Commented Sep 6 at 20:08
  • $\begingroup$ @J.Delaney is it when C^2 is positive definite? (using the square root of its spectrum in C's spectral decomposition) $\endgroup$ Commented Sep 7 at 5:48
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    $\begingroup$ Yes. But of course assuming that $C^2$ is positive definite is the same as assuming that $d/dt\langle(\Delta A)^2\rangle >0$, which makes the result a bit trivial. $\endgroup$ Commented Sep 7 at 9:16
  • $\begingroup$ @J.Delaney thanks. $\endgroup$ Commented Sep 7 at 12:38

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