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.

 

[Architeuthis Dux]

It appears that you have made a mistake in your integration or in your understanding of the problem. The expectation value of a quantity is defined as the average value of that quantity over all possible outcomes, weighted by their probabilities. In this case, since the vector components are uniformly distributed, each component has an equal probability of being any value within its allowed range. Therefore, the expectation value of <x\cdot x> (or any other component product) should be the average value of that product over all possible values of x.

To calculate this, you can use the integral you have written, but with the probability density function (PDF) included:

<x\cdot x> = \int_{0}^{\pi}d\theta\int_{0}^{2\pi}d\phi~~[\sin(\theta)\cdot\cos(\phi)]^{2} \cdot f(x) = \frac{1}{4\pi} \int_{0}^{\pi}d\theta\int_{0}^{2\pi}d\phi~~[\sin(\theta)\cdot\cos(\phi)]^{2}

where the PDF, f(x), is equal to 1/(4\pi) for a uniform distribution over the surface of a sphere. Evaluating this integral gives the correct result of 1/3.

In general, when calculating expectation values, it is important to include the appropriate PDF in the integration to account for the weighting of different outcomes. Additionally, it may be helpful to check your work using a different method or by considering a simpler case where the answer is known.