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I How exactly does squaring operators (e.g. <p^2> work?

  1. Feb 27, 2016 #1
    I'm having to essentially piece together a framework of background knowledge to understand parts of a QM class in which I'm lacking prerequisites; one of the things that I noticed that really confused me was the square of the momentum operator <p>, and how that translated into the integrand of an expectation value; Squaring an operator as far as I'm as aware isn't as simple as just squaring its components, because, for the momentum operator defined as [tex] \langle p^2 \rangle = \int_{-\infty}^{\infty}\psi^*(x)p^2\psi(x) = \int_{-\infty}^{\infty}\psi^*(x)\bigg(-\hbar^2 \frac{\partial^2}{\partial x^2}\psi(x)\bigg)[/tex]
    The differential part of the operator changed, too! This isn't just squaring, what is it? [tex] \frac{d}{dx}\frac{d}{dx} \neq \frac{d^2}{dx^2} [/tex] If I'm given some arbitrary operator A and told to calculate [tex] \langle A^2\rangle [/tex] How do I know what the operator "squared" takes the form of?
  2. jcsd
  3. Feb 27, 2016 #2
    The differential part of the operator changed, too! This isn't just squaring, what is it?

    operators are supposed to operate in a space - so a A^2 must be equivalent to A(op). A(op) , a product say operating on the right....
  4. Feb 27, 2016 #3
    I don't understand what that meant; I don't have any working knowledge of spaces, regardless of what they are.
    I also have no idea of what A(op) is supposed to indicate. A is the operator, and I'm only guessing "op" is shorthand for "operator", so unless A(op) is a really unnecessarily roundabout way of saying (A)(A) or something, I don't get it.
  5. Feb 27, 2016 #4
    Alright then, thanks for the help(?).
  6. Feb 28, 2016 #5


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    Staff: Mentor

    ##p^2\psi=pp\psi=p(p\psi)##, that is, you apply ##p## to ##\psi## to get a new function ##\psi_1##, then you apply ##p## to ##\psi_1##.

    Likewise for any other operator.
  7. Mar 1, 2016 #6
    Thank you very much!
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