Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    mathman

    User Avatar
    Science Advisor
    Gold Member

    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

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proof: Operators with same expectation value
Loading...