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

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.

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mfb
Mentor
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
Gold Member
2019 Award
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.

Ok, I understand, that makes sense.