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Using uncertainty principle to find minimum Kinetic Energy expectation value

  1. Oct 8, 2012 #1
    1. The problem statement, all variables and given/known data
    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##.


    2. Relevant equations
    $$ΔxΔp≥\frac{1}{2} \hbar$$
    $$\left\langle\hat{T}\right\rangle = \left\langle\frac{\hat{p}^2}{2m}\right\rangle = \frac{\left\langle\hat{p}^2\right\rangle}{2m}$$
    $$v = \left\langle\hat{p}/m\right\rangle$$
    $$Δp = \sqrt{\left\langle\hat{p}^2\right\rangle - \left\langle\hat{p}\right\rangle^2}$$


    3. 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}##.
     
  2. jcsd
  3. Oct 11, 2012 #2
    No one can help?
     
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