Why is the expectation value of an observable what it is (the formula)

[gothloli]

Homework Statement

I really do not understand why the expectation value of an observable such as position is
<x> = $\int\Psi*(x)\Psi$

Homework Equations

If Q is an operator then
<Q> = = $\int\Psi*(Q)\Psi$
cn = <f,$\Psi>$

The Attempt at a Solution

What I understand this is saying is that since x is a linear transformation and $\Psi$ is an eigenvector, by x$\Psi$ would be a vector in position space. Then taking the expectation value is like taking the inner product of $\Psi$, and a position eigenfunction. But why would that give an average value? What I vaguely understand is that $\int\Psi^{2}$, is the probability density and multiplying operator by probability would give the average value. But I'm still confused as connecting linear algebra with probability and quantum, I'm having a hard time with that.

 

[mfb]

It is like a weighted average, where the weight appears via the wave function in the integral.
It just works this way. Why? Ask nature.

 

[vanhees71]

It's important to understand this. It's the key of the probabilistic meaning of the quantum mechanical state. An observable is represented by a self-adjoint operator in Hilbert space, and there's a complete set of (generalized) eigenfunctions $u_j(x)$ of this operator. The (generatlized) eigen values (or better "spectral values") of the operator are the possible values the observable can take, and the probability to measure the $j$th eigenvalue, given the system is prepared in a state described by the normalized wave function $\psi$, is given by
$$P_j=\left |\int \mathrm{d}^3 x u_j^*(x) \psi(x) \right|^2.$$
The expectation value is then given by
$$\langle O \rangle = \sum_j o_j P_j.$$
Using the completeness of the (generalized) wave functions you can then easily show that you can write this expectation value in the way given in your question.

 

[gothloli]

Ok, I understand, that makes sense.