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Proof: Operators with same expectation value

  1. Jun 23, 2011 #1
    Given some state [itex]\left|\psi\right\rangle[/itex], and two operators [itex]\hat{A}[/itex] and [itex]\hat{B}[/itex], how do you prove that if [itex]\langle\psi|\hat{A}|\psi\rangle = \langle\psi|\hat{B}| \psi\rangle[/itex] then [itex]\hat{A} = \hat{B}[/itex] ?
  2. jcsd
  3. Jun 23, 2011 #2


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    Is it true for every state?
  4. Jun 23, 2011 #3
    Proving [tex]\hat{A}[/tex] is the same as [tex]\hat{B}[/tex] depends on the state of [tex]\psi[/tex]. You've made some notation [tex]<\psi|\psi>[/tex] is a probability amplitude. Your operators are fine, but if they are equal, it depends on the state of [tex]|\psi>[/tex]. Note though [tex]\mathcal{O} \cdot x \ne x \cdot \mathcal{O}[/tex].
  5. Jun 23, 2011 #4


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    Well, you can't; you have the equality of the images for only one vector in the common domain. You must have much more than that, of course.
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