# Using uncertainty principle to find minimum Kinetic Energy expectation value

1. Oct 8, 2012

### AluminumFalc

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. Oct 11, 2012

### AluminumFalc

No one can help?