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

Issue with squaring operators

  1. May 17, 2013 #1
    I'm, slowly, working through "Quantum Physics" by Stephen Gasiorowicz, second edition. On page 49, he gives the equation

    i[itex]\hbar[/itex][itex]\frac{\partial ψ(x, t)}{\partial t}[/itex] = -[itex]\frac{\hbar^{2}}{2m}[/itex][itex]\frac{\partial ^{2}ψ(x, t)}{\partial x^2}[/itex]

    He then makes the identification [itex](h/i)(\partial/\partial x) = p_{op}[/itex], which he has previously identified as the momentum operator, and rewrites the equation as

    i[itex]\hbar[/itex][itex]\frac{\partial ψ(x, t)}{\partial t}[/itex] = [itex]\frac{p^2_{op}}{2m}[/itex][itex]ψ(x, t)[/itex]

    Then he observes that the operator on the right (which I construe to mean [itex]p^2_{op}/2m[/itex]) is just the energy for a free particle. And that's where I get confused. As I understand it, the square of the momentum divided by twice the mass gives the energy, but [itex]p^2_{op}[/itex] isn't saying use the momentum squared, but apply the momentum operator twice.

    What am I missing here?

    Sylvia.
     
  2. jcsd
  3. May 17, 2013 #2
    Consider a state with a definite value of momentum. In QM, this means a state that is an eigenstate of the momentum operator. For example, a state ##\psi(x)## satisfying

    ##\hat{p} \psi(x) = k \psi(x)##

    (where ##\hat{p}## is the momentum operator) has a momentum of ##k##. Suppose we apply the momentum operator twice:

    ##\hat{p}^2 \psi(x) = \hat{p} \hat{p} \psi(x) = \hat{p} (k \psi(x)) = k \hat{p} \psi(x) = k^2 \psi(x)##.

    So an eigenstate of the operator ##\hat{p}## with eigenvalue ##k## is an eigenstate of the operator ##\hat{p}^2## with eigenvalue ##k^2##. The point is that applying the operator twice gives us exactly what we want: it squares the eigenvalue. We can legitimately think of the operator ##\hat{p}^2## as the "momentum-squared" operator.
     
  4. May 17, 2013 #3
    OK, I can follow that, thanks. He starts discussing eigenfunctions and eigentvalues a bit later. Perhaps readers are meant to take it on trust until then.

    Sylvia.
     
  5. May 18, 2013 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    The mathematical operation of <squaring> a linear operator means <applying it twice>:

    [tex] \hat{p}^2 := \hat{p}\circ\hat{p} [/tex]

    (For simplicity, let's neglect hat the momentum operator is unbounded).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook