- #1

uxioq99

- 11

- 4

- 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.

##\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.