Time Independence of the Momentum Uncertainty for a Free Particle Wave

[uxioq99]

Homework Statement

Prove that $\frac{\sigma_p}{dt} = 0$ for a freely moving wave packet in the absence of a potential. (Here, $\sigma_p$ denotes momentum uncertainty.)

Relevant Equations

$\frac{\sigma_p}{dt} = 0$

Mine is a simple question, so I shall keep development at a minimum. If a particle is moving in the absence of a potential ($V(x) = 0$), then
$\frac{\langle\hat p \rangle}{dt} = \langle -\frac{\partial V}{\partial x}\rangle=0$
will require that the momentum expectation value remains constant in time. Now, I must prove that $\langle \hat p^2 \rangle$ is also constant in time. I used the kinetic energy formula $\hat T = \frac{\hat p^2}{2m}$ to assert that $\frac{d\langle p \rangle}{dt} = 2m\frac{d\langle T\rangle}{dt}=0$ because the total kinetic energy of a freely moving particle is conserved. I justified my claim by arguing that there cannot be any work in the absence of a potential so that potential must be constant. Then, the momentum uncertainty $\sigma_p = \sqrt{\langle p^2 \rangle - \langle p \rangle^2}$ is formed from two functions that are constant in time is consequently time-invariant itself.

 

[anuttarasammyak]

It seems all right to me.　 You may check it directly from momentum wavefunction which you get from Gaussian coordinate wavefunction with growing dispersion, by Fourier transform.

 

[PeroK]

Homework Statement:: Prove that $\frac{\sigma_p}{dt} = 0$ for a freely moving wave packet in the absence of a potential. (Here, $\sigma_p$ denotes momentum uncertainty.)
Relevant Equations:: $\frac{\sigma_p}{dt} = 0$

Mine is a simple question, so I shall keep development at a minimum. If a particle is moving in the absence of a potential ($V(x) = 0$), then
$\frac{\langle\hat p \rangle}{dt} = \langle -\frac{\partial V}{\partial x}\rangle=0$
will require that the momentum expectation value remains constant in time. Now, I must prove that $\langle \hat p^2 \rangle$ is also constant in time. I used the kinetic energy formula $\hat T = \frac{\hat p^2}{2m}$ to assert that $\frac{d\langle p \rangle}{dt} = 2m\frac{d\langle T\rangle}{dt}=0$ because the total kinetic energy of a freely moving particle is conserved. I justified my claim by arguing that there cannot be any work in the absence of a potential so that potential must be constant. Then, the momentum uncertainty $\sigma_p = \sqrt{\langle p^2 \rangle - \langle p \rangle^2}$ is formed from two functions that are constant in time is consequently time-invariant itself.

I'm not totally convinced. Why can you assume conservation of KE? The concept of "work" is not very QM.

An alternative is to consider the relationship between momentum and a potential-free Hamiltonian.

 

[vanhees71]

I'd solve the time-dependent Schrödinger equation in momentum representation, given an arbitrary square-integrable $\psi(t=0,\vec{p})=\psi_0(\vec{p})$ and then think about, how to calculate $\sigma_p(t)$ with it.