Free particle, classical vs quantum mechanics

[Cogswell]

Homework Statement

Consider first a free particle (Potential energy zero everywhere). When the particle at a given time is prepared in a state ## \psi (x) ## it has <x> = 0 and <p> = 0.
The particle is now prepared in
## \Psi (x, t=0) = \psi (x) e^{ikx} ##

1. Give <p> at time t=0.

2. It can be shown in quantum mechanics that <p> is independent of time for a free particle, in anology to Newton's first law in classical mechanics. Give <x> as a function of time.

Homework Equations

## \displaystyle <p> = \int ^{ \infty} _{- \infty} \Psi (x,t)^* \left( \dfrac{\hbar}{i} \dfrac{ \partial}{\partial x} \right) \Psi (x,t) dx ##

The Attempt at a Solution

So I put Psi into the integral and get:

## \displaystyle <p> = \dfrac{\hbar}{i} \int ^{ \infty} _{- \infty} \psi (x)^* e^{-ikx} \left( \dfrac{ \partial}{\partial x} \right) \psi (x) e^{ikx} dx ##

## \displaystyle <p> = \dfrac{\hbar}{i} \int ^{ \infty} _{- \infty} \psi (x)^* e^{-ikx} \left[ \dfrac{d \psi (x)}{dx} e^{ikx} + ik \psi (x) e^{ikx} \right] dx ##

The exponentials cancel out and I get:

## \displaystyle <p> = \dfrac{\hbar}{i} \int ^{ \infty} _{- \infty} \psi (x)^* \dfrac{d \psi (x)}{dx} dx + \dfrac{ik \hbar}{i} \int ^{ \infty} _{- \infty} \psi (x)^* \psi (x) dx ##


So, for a normalised distribution, the area under the curve wil be 1, and so the second integral will give me ## \hbar k##
What about the first integral?

 

[TSny]

## \displaystyle <p> = \dfrac{\hbar}{i} \int ^{ \infty} _{- \infty} \psi (x)^* \dfrac{d \psi (x)}{dx} dx + \dfrac{ik \hbar}{i} \int ^{ \infty} _{- \infty} \psi (x)^* \psi (x) dx ##


So, for a normalised distribution, the area under the curve wil be 1, and so the second integral will give me ## \hbar k##
What about the first integral?

Can you relate the first integral to some property of the state ##\psi (x)## that was specified at the beginning of the problem statement?

 

[Cogswell]

Oh right silly me, the first integral is just the expectation value of the momentum of the function ## \psi (x) ## which is given to us as 0.