- #1
AluminumFalc
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Homework Statement
Assume that a particle travels with a certain known (average) velocity ##v = \left\langle\hat{p}/m\right\rangle##. You know it's position with an uncertainty ##Δx##. Use the uncertainty principle to determine the least possible value for the article's kinetic energy expectation value ##\left\langle\hat{T}\right\rangle##.
Homework Equations
$$ΔxΔp≥\frac{1}{2} \hbar$$
$$\left\langle\hat{T}\right\rangle = \left\langle\frac{\hat{p}^2}{2m}\right\rangle = \frac{\left\langle\hat{p}^2\right\rangle}{2m}$$
$$v = \left\langle\hat{p}/m\right\rangle$$
$$Δp = \sqrt{\left\langle\hat{p}^2\right\rangle - \left\langle\hat{p}\right\rangle^2}$$
The Attempt at a Solution
First I solved for ##Δp##.
$$ΔxΔp≥\frac{1}{2} \hbar$$
$$Δp≥\frac{\hbar}{2Δx}$$
Then I tried getting ##\left\langle\hat{p}^2\right\rangle## on its own
$$Δp^2=\left\langle\hat{p}^2\right\rangle - \left\langle\hat{p}\right\rangle^2$$
$$\left\langle\hat{p}^2\right\rangle = Δp^2 + \left\langle\hat{p}\right\rangle^2$$
$$\left\langle\hat{p}^2\right\rangle = Δp^2 + m^2 v^2$$
Then plugged that into ##\left\langle\hat{T}\right\rangle##
$$\left\langle\hat{T}\right\rangle = \frac{Δp^2 + m^2 v^2}{2m} \geq \frac{\frac{\hbar^2}{4Δx^2} + m^2 v^2}{2m}$$
$$\left\langle\hat{T}\right\rangle \geq \frac{\hbar^2 + 4Δx^2 m^2 v^2}{8mΔx^2}$$
However, I've always seen this written as
$$\left\langle\hat{T}\right\rangle \geq \frac{\hbar^2}{8mΔx^2}$$
According to what I found, that's the case when ##v=0##, or alternatively when ##\left\langle\hat{p}\right\rangle=0##.
Earlier, when we looked at Gaussian wave packets at ##t=0##, we found that ##\left\langle\hat{p}\right\rangle=0## because the particle is just as likely to be moving to the left as it is to the right. But in this case ##v## is defined, which is why I'm confused. And when we did problems with the time evolution of ##\left\langle\hat{p}\right\rangle##, we found that ##\left\langle\hat{p}(t)\right\rangle = \hbar k_{0}##, where ##k_{0}## is the phase shift on the Fourier transform of ##\psi(x,t)##.
I guess I'm confused about the time evolution of the momentum expectation value and how I should think about it here. I don't get how ##\left\langle\hat{p}(t)\right\rangle = \hbar k_{0}##.
