Find Expectation Value for Particle Moving in N Steps of Length L

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SUMMARY

The expectation value for a particle moving in N steps of length L can be calculated using the normalized probability density function p(θ) = (2/π)cos²(θ/2). For a single step, the expectation value is derived as = (L²/2). To find the expectation value for N steps, one must consider the contributions from each step, which requires integrating the probability density over the defined angles.

PREREQUISITES
  • Understanding of probability density functions
  • Familiarity with polar coordinates and angular integration
  • Knowledge of expectation values in statistical mechanics
  • Basic proficiency in calculus, particularly integration techniques
NEXT STEPS
  • Research how to compute expectation values for multiple independent steps in random walks
  • Study the properties of normalized probability density functions
  • Learn about angular integration techniques in polar coordinates
  • Explore applications of expectation values in statistical physics
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Students and researchers in physics, particularly those focusing on statistical mechanics and random walks, will benefit from this discussion.

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Homework Statement


A particle moves in a sequence of steps of length L. The polar angle [tex]\theta[/tex] for each step is taken from the (normalized) probability density [tex]p(\theta)[/tex]. The azimuthal angle is uniformly distributed. Suppose the particle makes N steps.
My question is how do I find the expectation value (say [tex]<z^2>[/tex] for example).

Homework Equations


Usually for a probability density p(x) we have
[tex]<x^m>=\int x^m p(x) dx[/tex].

The Attempt at a Solution


I think that I can get the values for one step. eg.
[tex]<z^2>=\int_0^\pi (Lcos(\theta))^2p(\theta)d\theta={L^{2}\over 2}[/tex]
Note: the density [tex]p(\theta)[/tex] is normalized.
I just don't know how to treat N steps. Do I just multiply the one-step result by N?
 
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What is [tex]p(\theta)[/tex]? Is it given?
 
Oh ya. Sorry. It is
[tex]p(\theta) ={2 \over \pi}cos^2({\theta \over 2})[/tex]
 

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