Probability density of a particle

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Discussion Overview

The discussion revolves around the classical probability density of a particle confined in an infinite square well of dimension L. Participants explore the formal derivation of the probability density function, its properties, and the physical reasoning behind a uniform density within the well.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant states that the probability density function must be constant and its integral from 0 to L must equal 1.
  • Another participant seeks a formal proof and references the definition of a probability density function, emphasizing the need to show that the area under the curve equals 1.
  • A later reply confirms that demonstrating the area under f(x) = 1/L from 0 to L equals 1 would satisfy the criteria for a probability density function, but notes that this does not establish it as the correct function without physical arguments.
  • One participant proposes a physical argument for uniform density, suggesting that the constant potential inside the well implies translational invariance, leading to a constant probability density due to symmetry.

Areas of Agreement / Disagreement

Participants generally agree on the requirement for the probability density function to be constant and the integral condition, but there is no consensus on the formal proof or the necessity of physical arguments to establish correctness.

Contextual Notes

Participants express varying levels of understanding regarding the formal definition and proof of probability density functions, indicating potential gaps in assumptions or knowledge about the topic.

gnome
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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?
 
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gnome said:
But how do I show formally that the probability density function is 1/L?
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. :smile:
 
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?
 
gnome said:
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).
I won't be any help to you there! :eek:
...
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?
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.)
 
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.
 

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