# The free particle

E92M3

## Homework Statement

$$\psi(x,0)=Ae^{-ax^2}$$
Normalize and find:
$$\psi(x,t)$$

## Homework Equations

$$1=\int_{-\infty}^\infty\psi^*\psi dx$$

## The Attempt at a Solution

$$1=A^2\int_{-\infty}^\infty e^{-2ax^2} dx$$
let:
$$u=x\sqrt{2a}$$

$$1=A^2\frac{1}{\sqrt{2a}}\int_{-\infty}^\inftye^{-u^2} du=A^2\sqrt{\frac{\pi}{2a}}$$
Therefore:
$$A=(\frac{2a}{\pi})^{1/4}$$

$$\psi(x,t)=(\frac{2a}{\pi})^{1/4}e^{-2ax^2-i\frac{\hbar k}{2m}t}$$
This is too simple to be the right answer. I think I'm missing the point of the question. Please point me to the right direction.

Homework Helper
Gold Member
$$A=(\frac{2a}{\pi})^{1/4}$$

$$\psi(x,t)=(\frac{2a}{\pi})^{1/4}e^{-2ax^2-i\frac{\hbar k}{2m}t}$$
This is too simple to be the right answer. I think I'm missing the point of the question. Please point me to the right direction.

Your value of $A$ is correct; which leaves you with

$$\psi(x,0)=\left(\frac{2a}{\pi}\right)^{1/4}e^{-ax^2}$$

How did you go from that, to your equation for $\psi(x,t)$?