I am not sure if I should be posting this under QM or under Linear Algebra, since it appears to be an algebraic step that I do not see, and am asking the wonderful people on this forum to spell it out for me. In John Baez's derivation of the Energy-time Uncertainty relation,(adsbygoogle = window.adsbygoogle || []).push({});

http://math.ucr.edu/home/baez/uncertainty.html

there is the following step, which first I will quote and then state my question:Selecting an observable A which does not commute with H,

"< [H,A] > = <ψ, [H,A] ψ>

is i[itex]\hbar[/itex] times the time derivative at t = 0 of <ψ, A ψ>, as you can see if you note that the solution to the Schrödinger equation can be written in the form

U(t) ψ = exp(-itH/[itex]\hbar[/itex]) ψ

Thus

<[H, A]> = i[itex]\hbar[/itex] d<A>/dt"

If he is using the commutator [H,A] as the operator U(t), then I do not see how to get from

<[H, A]> =<ψ, exp(-itH/[itex]\hbar[/itex]) ψ>

to

i[itex]\hbar[/itex] d<A>/dt

(i.e., i[itex]\hbar[/itex] d<ψ, Aψ>/dt)

If he is not using the commutator as the operator, then I am lost even worse.

I would be very grateful for anyone showing me this step more explicitly. ("Derivation for Dummies")

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# A step in derivation of E-t uncertainty not clear

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