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

  • Thread starter gothloli
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  • #1
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Homework Statement


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

Homework Equations


If Q is an operator then
<Q> = = [itex]\int\Psi*(Q)\Psi[/itex]
cn = <f,[itex]\Psi>[/itex]


The Attempt at a Solution


What I understand this is saying is that since x is a linear transformation and [itex]\Psi[/itex] is an eigenvector, by x[itex]\Psi[/itex] would be a vector in position space. Then taking the expectation value is like taking the inner product of [itex]\Psi[/itex], and a position eigenfunction. But why would that give an average value? What I vaguely understand is that [itex]\int\Psi^{2}[/itex], 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.
 

Answers and Replies

  • #2
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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.
 
  • #3
vanhees71
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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 [itex]u_j(x)[/itex] 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 [itex]j[/itex]th eigenvalue, given the system is prepared in a state described by the normalized wave function [itex]\psi[/itex], is given by
[tex]P_j=\left |\int \mathrm{d}^3 x u_j^*(x) \psi(x) \right|^2.[/tex]
The expectation value is then given by
[tex]\langle O \rangle = \sum_j o_j P_j.[/tex]
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.
 
  • #4
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Ok, I understand, that makes sense.
 

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