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

In summary, the discussion was about a step in John Baez's derivation of the Energy-time Uncertainty relation, where he uses the solution to the Schrödinger equation, U(t)=exp(-itH/ħ), to show that <[H,A]> is equal to i\hbar times the time derivative of <ψ, Aψ> at t=0. The confusion arose because the solution was mistakenly written as U(t)=[H,A], but it was clarified that they are not the same. Instead, U(t) can be used to calculate the time derivative of <A>, which follows classical solutions.
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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")

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

1 person
"< [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).

1 person
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$\hbar$d|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.

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.

1 person
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:
$\frac{d<A>}{dt}= \frac{d<\psi(t)|A|\psi(t)>}{dt}= \frac{d<\psi_{0}|e^{iHt/\hbar}Ae^{-iHt/\hbar}|\psi_{0}>}{dt}=\frac{d<\psi_{0}|A(t)|\psi_{0}>}{dt}= \frac{d<A(t)>}{dt}$

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.

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

## What is the goal of deriving E-t uncertainty?

The goal of deriving E-t uncertainty is to understand the relationship between the uncertainty in energy (E) and the uncertainty in time (t) in a given system or experiment. This can provide valuable insights into the accuracy and precision of measurements and calculations in the field of physics.

## What are the key steps in deriving E-t uncertainty?

The key steps in deriving E-t uncertainty involve analyzing the equations and principles of uncertainty in energy and time, such as Heisenberg's uncertainty principle and the principle of energy conservation. This is often done through mathematical calculations and manipulations to determine the relationship between E and t.

## Why is the step in derivation of E-t uncertainty not clear?

The step in derivation of E-t uncertainty may not be clear due to the complexity of the concepts and equations involved. It may also be because of a lack of understanding or familiarity with the mathematical techniques used in the derivation process. Further explanation or clarification may be needed to fully understand this step.

## How can I improve my understanding of the derivation of E-t uncertainty?

To improve your understanding of the derivation of E-t uncertainty, you can review the fundamental principles and equations involved in uncertainty, such as the uncertainty principle and the concept of uncertainty in measurement. You can also seek out additional resources, such as textbooks or online tutorials, to gain a deeper understanding of the derivation process.

## What are the practical applications of E-t uncertainty?

E-t uncertainty has practical applications in various fields, such as quantum mechanics, nuclear physics, and particle physics. It can help scientists and researchers understand the limitations and accuracy of their measurements and calculations, and can also aid in the development of new technologies and theories in these fields.

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