# Expectation value of operator derivation

1. Feb 15, 2015

### Goodver

Where one can find a proof of the expectation value of operator expression.

<A> = < Ψ | A | Ψ >

or

<A> = integral( Ψ* A Ψ dx )

Thanks.

2. Feb 15, 2015

### jfizzix

The expectation value of an observable is the average value of its outcome weighted according to the probability of each outcome.

$\langle A\rangle\equiv \sum_{i}P(a_{i}) a_{i}$

The observable $A$ can be expressed in terms of its eigenvalues and eigenstates as:
$\hat{A}=\sum_{i}a_{i}|a_{i}\rangle\langle a_{i}|$, where $\hat{A}|a_{i}\rangle=a_{i}|a_{i}\rangle$

The probability of a given outcome is given by the state of the system $|\psi\rangle$ and the Born rule:
$P(a_{i})=|\langle a_{i}|\psi\rangle|^{2}=\langle \psi|a_{i}\rangle\langle a_{i}|\psi\rangle$

Combining these together, we find the expectation value:
$\langle A\rangle= \sum_{i}\langle \psi|a_{i}\rangle\langle a_{i}|\psi\rangle a_{i}$

With a little algebra, this becomes:
$\langle A\rangle=\langle \psi|\big(\sum_{i}a_{i}|a_{i}\rangle\langle a_{i}|\big)|\psi\rangle =\langle \psi|\hat{A}|\psi\rangle$.

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