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Need help with expectation value !

  1. Feb 29, 2008 #1
    1. The problem statement, all variables and given/known data

    I have a random, uniformly distributed vector with Cartesian components x,y,z. I should calculate the expectation value of the products of the components, e.g. [tex]<x\cdot x>, <x\cdot y>, ..., <z\cdot z>[/tex].

    2. Relevant equations

    In spherical coordinates the [tex] x,y,z[/tex] components are

    [tex]x = \sin(\theta)\cdot\cos(\phi)[/tex]
    [tex]y= \sin(\theta)\cdot\sin(\phi)[/tex]
    [tex]z = \cos(\theta)[/tex]


    [tex]<x\cdot x> =\int f(x)\cdot x^2 dx = const.\times \int x^2 dx[/tex],
    because the probability density function is constant of the uniform probability distribution.

    3. The attempt at a solution

    I think it is useful to convert the Cartesian coordinates to spherical coordinates. So for example

    [tex]<x\cdot x> = \int_{0}^{\pi}d\theta\int_{0}^{2\pi}d\phi~~[\sin(\theta)\cdot\cos(\phi)]^{2}[/tex]

    I wrote a program for this, and the solution is:

    [tex]<x\cdot x> = <y\cdot y> = <z\cdot z> = \frac{1}{3}[/tex]
    and the other terms are zero.

    But the solution of the above integral is not 1/3.
     
  2. jcsd
  3. Mar 1, 2008 #2
    You dont need to convert to spherical coordinates, the limit you're assuming is from 0 to 1 on all three axes. And the value of the integral IS 1/3, but you've forgotten to multiply by f(x).

    I dont know if youre calculating the same expectation value, but in QM, [tex]<f>=\int_{a}^{b}f(r)*f'*f(r)'dr[/tex],

    where f' is the operator, f(r) is the function (physical quantity), and f(r)' is its conjugate. If f is real only, then f(r)=f(r)' and the integral becomes,

    [tex]f(r)^2\int_{a}^{b}f'dr[/tex]

    If that is what you were trying to calculate, then you missed out by a factor of [tex]f(x)^2[/tex]. Also, [tex]\theta[/tex] is a constant, not a variable.
     
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