A step in derivation of E-t uncertainty not clear

[nomadreid]

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,
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\hbar 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/\hbar) ψ
Thus
<[H, A]> = i\hbar 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/\hbar) ψ>
to
i\hbar d<A>/dt
(i.e., i\hbar 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")

 

[ChrisVer]

why did you write [H,A]=U?

 

[kith]

"< [H,A] > = <ψ, [H,A] ψ>
is i\hbar 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/\hbar) ψ
Thus
<[H, A]> = i\hbar d<A>/dt"

Yeah, I agree that this is confusing. You can compute d/dt <A> = d/dt <ψ|A|ψ> directly by using the product rule and applying the Schrödinger equation twice.

Instead of the Schrödinger equation, you can also use U(t). If you do, you will see what he wanted to tell you. But you may have a misconception regarding U(t). U(t) is not [H,A]. U(t) is exp(-itH/ħ).

 

[nomadreid]

First, a big thanks to both ChrisVer and kith. Oddly enough, ChrisVer's first reply no longer shows up. (Thankfully I copied it before it disappeared.)
kith's first paragraph sketches the straightforward way, and ChrisVer's disappeared reply spelled the calculation out. This way was very enlightening, and I understood it. In this derivation there was no reason, as far as I can see, to use the solution to the Schrödinger equation, but just to use the original form of the equation H |ψ> = i\hbard|ψ>/dt.
However, kith wrote that another way (and apparently the way that John Baez meant) was in using the solution. (ChrisVer asked me why I wrote U = [H,A] -- the short answer, as kith pointed out, was that I made a mistake.) However, I am still puzzled how the solution could be used (except of course of working backwards to the original equation). If kith or anyone has the patience to spell that point out for me, it would be a great help.

 

[kith]

Well, instead of using the Schrödinger eqaution to substitute d/dt |ψ(t)> you can also calculate d/dt |ψ(t)> = d/dt (U(t)|ψ>) = (d/dt U(t))|ψ> by using the known form of U(t).

The time evolution operator U(t) can be derived from the Schrödinger equation and vice versa so in this sense, both are equally fundamental. Also note that in the case of time-dependent Hamiltonians the form of U(t) is more complicated than what we have been talking about so far.

 

[ChrisVer]

My first post was deleted, because I mistyped something in it. Unfortunately I said that |psi> is independent of time, whereas I think the correct thing I should have stated would be that I used bracket of |psi(0)> instead (so it can go into the time derivative without problem)...
On the other hand let me:
\frac{d&lt;A&gt;}{dt}= \frac{d&lt;\psi(t)|A|\psi(t)&gt;}{dt}= \frac{d&lt;\psi_{0}|e^{iHt/\hbar}Ae^{-iHt/\hbar}|\psi_{0}&gt;}{dt}=\frac{d&lt;\psi_{0}|A(t)|\psi_{0}&gt;}{dt}= \frac{d&lt;A(t)&gt;}{dt}

where then you can proceed by evaluating the Heisenberg's equation, shown in wiki:
http://en.wikipedia.org/wiki/Heisenberg_picture

which equation? the evaluated one?
The mean values follow classical solutions. So for example you can try to calculate how the mean value of \hat{x} evolves with time for the Harmonic Oscillator, and see that it's the classical solution.

 

[nomadreid]

Many thanks to both kith and ChrisVer. The derivation is much clearer now.