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Expectation value of operator derivation

  1. Feb 15, 2015 #1
    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. jcsd
  3. Feb 15, 2015 #2

    jfizzix

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

    [itex]\langle A\rangle\equiv \sum_{i}P(a_{i}) a_{i}[/itex]

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

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

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

    With a little algebra, this becomes:
    [itex]\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[/itex].
     
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