Time-energy uncertainty and derivative of an operator

[peripatein]

Homework Statement

I would appreciate feedback on the following two problems:
(1) For a given operator A with no explicit time dependence I am asked to show that d/dt(e^At)=A(e^At)
(2) A free wave packet of width Δx is traveling at a constant velocity v₀=p₀/m. I am asked to estimate the time-energy uncertainty relation.

Homework Equations

The Attempt at a Solution

(1) Is this as straightforward as it appears to be? Probably isn't, right? Is this to be solved using Ehrenfest's theorem? Or how?
(2) I can easily show that ΔEΔt ~ ħ but is this what I am plainly asked, or am I rather asked to arrive at a more exact result (i.e. find the proportionality coefficient)?

 

[jfizzix]

To take the time derivative of e^{At} with respect to t, you can take the power series expansion of e^{At}, differentiate term by term, and re-exponentiate, since what you get left should be A times another exponential.

For the second problem, it looks like they're asking you to calculate the uncertainty product \Delta E\Delta t from the given information. Then you can see for yourself how much larger than \frac{\hbar}{2} it is.

 

[peripatein]

For the second problem, how may I go about it then? Is it necessary to write the general wave function for the wave packet and then find the variances?

 

[jfizzix]

I would think so, if you know what the wavefunction of the wave packet is. I can't say off the top of my head what that would be, though.

 

[peripatein]

The question, as formulated, does not provide any additional information other than that in #1. Any ideas?

 

[jfizzix]

I'm guessing that there is a standard formula for a free wave packet of given \Delta x and p_{0}, to which the problem is referring.

Failing that, the best I can think of is to find a limit to the momentum uncertainty with \Delta p\geq \frac{\hbar}{2 \Delta x}.

Then, knowing p_{0} and a bound for \Delta p, you can use that E=\frac{p^{2}}{2m} and the propagation of uncertainty

(\Delta E)^{2}\approx (\frac{\partial E}{\partial p})^{2}(\Delta p)^{2}

Getting an estimate for the time uncertainty is a bit trickier (since time is not an observable), but you can say that

\Delta t = \frac{\Delta x}{(\frac{d <x>}{dt})}

That's about all I can say on the matter, though.