# Probability density

1. May 5, 2004

### gnome

I want to "show that the classical probability density describing a particle in an infinite square well of dimension L is P(x) = 1/L."

I know that classically, the particle bounces back and forth with constant kinetic energy and at constant speed, so at any given time it is equally likely to be found at any location in the well. It seems intuitively obvious that the probability that the particle will be between 0 and .25L, for example, at any particular moment, would be 1/4. Between .25L and .75L the probability must be 1/2.

But how do I show formally that the probability density function is 1/L?

2. May 5, 2004

### Staff: Mentor

I don't know what you mean by formally, but that won't stop me from stating the obvious. You know that the probability density must be a constant. And that its integral from 0 to L must equal 1. QED.

3. May 5, 2004

### gnome

I guess what I had in mind was something along the lines of the step-by-step proofs we've been doing in analysis of algorithms and AI (propositional & predicate logic).

The trouble was that I had only a vague notion of what a probability density function is, so I didn't know exactly what to work towards. I finally found this:

"A probability density function is a function defined on a continuous interval so that the area under the curve (and above the x-axis) described by the function is unity (meaning equal to one ( = 1))."

So, given that and your comment, is it correct to say that if one can show that the area under f(x) = 1/L from 0 to L equals 1, then it is true that f(x) is a probability density function for x in the region 0 - L?

4. May 6, 2004

### Staff: Mentor

I won't be any help to you there!
Yes, you would have shown that it meets the criteria for being a probability density function. If that's all you need to show. But that won't show that it's the correct probability density funtion. (That has to be done based on physical arguments, I would think. Our physical assumption is that the density function is a constant.)

5. May 6, 2004

### Gokul43201

Staff Emeritus
There may be a "physical" argument for a uniform density. Here's my attempt : Inside the well, the potential is a constant, so the probability density must have translational invariance inside the well. In other words, there's nothing that makes any spot inside the well special, so all points must be treated identically. So, symmetry requires that the density be constant.