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Quantum mechanics operator manipulation

  1. Apr 4, 2013 #1
    1. The problem statement, all variables and given/known data
    consider operator defined as [itex]\hat{O_A} = \hat{A} -<\hat{A}> [/itex]
    show that [itex](ΔA)^2=<\hat{O_A}^2>[/itex]

    2. Relevant equations
    [itex](ΔA)^2=<\hat{A}^2>-<\hat{A}>^2[/itex]


    3. The attempt at a solution
    [itex](ΔA)^2=<\hat{A}^2>-<\hat{A}>^2[/itex]
    [itex] = <\hat{A}^2> - (\hat{A} -\hat{O_A})^2 [/itex]
    [itex] = <\hat{A}^2> - \hat{A}^2 + 2\hat{A}<\hat{O_A}> - \hat{O_A}^2 [/itex]

    or

    [itex]\hat{O_A} = \hat{A} -<\hat{A}>[/itex]

    [itex]\hat{O_A}^2 = (\hat{A} -<\hat{A}>)^2[/itex]

    [itex] \hat{O_A}^2 = \hat{A}^2-2\hat{A}<\hat{A}>+\hat{A}^2 [/itex]

    But I don't know how to convert the operator [itex]\hat{O_A} [/itex] into the expectation value [itex] <\hat{O_A}> [/itex]...?
     
  2. jcsd
  3. Apr 4, 2013 #2

    George Jones

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    What happens when the expectation of both sides of the last line is taken?
     
  4. Apr 4, 2013 #3
    Does this happen:

    [itex] <\hat{O_A}^2> = <\hat{A}^2>-2<\hat{A}><\hat{A}>+<\hat{A}>^2 [/itex]

    [itex] <\hat{O_A}^2> = <\hat{A}^2>-2<\hat{A}>^2+<\hat{A}>^2 [/itex]

    [itex] <\hat{O_A}^2> = (ΔA)^2 [/itex]....?

    I am not sure what happens when you take the expectation value of a term that is already an expectation value, does it just remains unchanged?
     
  5. Apr 4, 2013 #4

    vela

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    An expectation value is just a number, so yes, it remains unchanged.
     
  6. Apr 5, 2013 #5
    Oh yes, of course. Thank you.
     
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