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What is the mathematical intuition behind operator embedding?

  1. Mar 8, 2013 #1

    jshrager

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    Can someone explain to me the mathematical intuition that motivates the embedding of quantum operators between the conjugate wave function and the (non-conjugated) wave function? That is, we write: [itex]\Psi[/itex][itex]^{*}[/itex][itex]\hat{H}[/itex][itex]\Psi[/itex], that is: [itex]\Psi[/itex][itex]^{*}[/itex]([itex]\hat{H}[/itex][itex]\Psi[/itex]), so that [itex]\hat{H}[/itex] operates on [itex]\Psi[/itex] (not [itex]\Psi[/itex][itex]^{*}[/itex]) and THEN this result is multiplied by [itex]\Psi[/itex][itex]^{*}[/itex]. Okay, fine, but WHY? What is the mathematical intuition behind this way of formulating quantum observables? What intuitively does it do to multiply the conjugate of [itex]\Psi[/itex] (i.e., [itex]\Psi[/itex][itex]^{*}[/itex]) by a modified (i.e., operated-upon) [itex]\Psi[/itex]?
     
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  3. Mar 9, 2013 #2

    dextercioby

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    It's not intuition, it's a necessity from the fact that the norm of a wavefunction must be real, so that to get from a complex number to a real number, you need the complex conjugate.
     
  4. Mar 9, 2013 #3

    jshrager

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    Okay, but then why don't we apply the operator to both the wave function and it's conjugate? That is, why isn't it necessary to do this: ([itex]\hat{H}[/itex][itex]\Psi^{*}[/itex])([itex]\hat{H}[/itex][itex]\Psi[/itex])? That would seem to make more sense. What amazes me is that we don't seem to have to turn ourselves inside out to take account of NOT operating on both the wave function and its conjugate, but it still works. This could only be true (to me...perhaps stupidly) if the equivalent of what I have just written is already embedded in the operator. No? (Yes, I know that it works the way we have learned to do it and that that should be enough for me...and it is. I'm just mathematically curious!)
     
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