Equations of motion for position/momentum uncertainty

[superiority]

Homework Statement

A particle is moving freely in the x direction.
Find the initially allowed (i.e. at t=0) values of $\langle x^2 \rangle$ and $\langle p^2 \rangle$, and find equations of motion for $\Delta x$ and $\Delta p$,

Homework Equations

$$\frac{d}{dt}\langle A\rangle = \frac{1}{i\hbar}\langle [A, H]\rangle + \left\langle \frac{\partial A}{\partial t} \right\rangle$$
$$\frac{d}{dt} \langle x\rangle = \frac{1}{m} \langle p\rangle$$
$$\frac{d}{dt} \langle -\nabla V(\mathbf{r})\rangle$$
$$\Delta A = \sqrt{\langle x^2 \rangle - \langle x\rangle ^2}$$

The Attempt at a Solution

I've gotten $\frac{d}{dt} \langle x^2 \rangle = \frac{1}{m} \langle xp + px\rangle = \frac{1}{m} \langle 2xp - i\hbar \rangle$, but I'm not sure how this helps me with the initial value. Do I integrate it with respect to time? If so, how? I know that <x> and <p> are both initially zero. Is <xp> zero at t=0 as well?

As for the equations for $\Delta x$ and $\Delta p$, I'm imagining that I find the time derivative of each uncertainty in terms of the expectation values of position/position squared and momentum/momentum squared, then integrate with respect to time. I'm not sure that I'm on the right track, though, and every time I try to start, I get stuck. Help?

 

[mark726]

To find the initially allowed values of <x^2> and <p^2>, you can use the fact that the particle is moving freely in the x direction. This means that there is no external force acting on the particle, and thus the potential energy V(x) is constant. Since the particle is in a state of constant energy, we can use the time-independent Schrodinger equation to find the allowed values of <x^2> and <p^2>. This equation is given by:

\hat{H}\psi(x) = E\psi(x)

where \hat{H} is the Hamiltonian operator, \psi(x) is the wavefunction, and E is the energy of the particle. In the case of a free particle, the Hamiltonian is given by:

\hat{H} = \frac{\hat{p}^2}{2m}

where \hat{p} is the momentum operator and m is the mass of the particle. Plugging this into the time-independent Schrodinger equation, we get:

\frac{\hat{p}^2}{2m}\psi(x) = E\psi(x)

Using the momentum operator \hat{p} = -i\hbar\frac{\partial}{\partial x} and rearranging, we get:

\frac{\partial^2 \psi(x)}{\partial x^2} = -\frac{2mE}{\hbar^2}\psi(x)

This is a second-order differential equation with solutions of the form:

\psi(x) = Ae^{ikx} + Be^{-ikx}

where k = \sqrt{\frac{2mE}{\hbar^2}}. Since the particle is moving freely in the x direction, we can assume that the wavefunction is a plane wave, meaning that B = 0. We can also normalize the wavefunction to get the allowed values of <x^2> and <p^2>. The normalized wavefunction is given by:

\psi(x) = \frac{1}{\sqrt{2\pi}}e^{ikx}

The allowed values of <x^2> and <p^2> can then be found by integrating \psi^*x^2\psi and \psi^*p^2\psi over all space. This gives us:

<x^2> = \frac{\hbar^2}{2