Need help with expectation value

[begyu85]

Homework Statement

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. <x\cdot x>, <x\cdot y>, ..., <z\cdot z>.

Homework Equations

In spherical coordinates the x,y,z components are

x = \sin(\theta)\cdot\cos(\phi)
y= \sin(\theta)\cdot\sin(\phi)
z = \cos(\theta)


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

The Attempt at a Solution

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

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

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

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

But the solution of the above integral is not 1/3.

 

[chaoseverlasting]

You don't 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 don't know if youre calculating the same expectation value, but in QM, <f>=\int_{a}^{b}f(r)*f'*f(r)'dr,

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,

f(r)^2\int_{a}^{b}f'dr

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