# Free particle, classical vs quantum mechanics

1. May 4, 2013

### Cogswell

1. The problem statement, all variables and given/known data
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.

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

3. 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 $\hbar k$

2. May 4, 2013

### TSny

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

3. May 5, 2013

### 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.