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Quantum Mechanics: Expected momentum of a real wavefunction

  1. Oct 4, 2012 #1
    1. The problem statement, all variables and given/known data
    Given the wavefunction, [itex]\psi(x) = Cx[/itex] for 0 < x < 10 and [itex]\psi(x) = 0[/itex] for all other values.

    What is the normalization constant of C? I got [itex]\sqrt{3/1000}[/itex].
    What is <x>? I got 30/4.
    What is <p>? Here is where I'm confused.


    2. Relevant equations
    [tex]\langle p \rangle = C^2 \frac{\hbar}{i} \int_0^{10} \psi^* \frac{d}{dx} \psi \, dx[/tex]

    3. The attempt at a solution

    I worked out the integral, and got [itex](3/20)(\hbar/i)[/itex].

    It's here that I can't figure out whether to leave it as that, or say it's 0, since the momentum is an observable, and must be real. I realize that the function is odd, but I'm not integrating it from -a to a, which means I can't just say it's 0 for that reason.

    [strike]PS: Sorry for my horrible usage of LaTeX.[/strike]
    Mod note: Fixed your LaTeX for you.
     
    Last edited by a moderator: Oct 4, 2012
  2. jcsd
  3. Oct 4, 2012 #2
    Try integrating by parts. Also, your notation is not good. You already pulled the C's out of the integral so it should be [itex] \langle p \rangle = C^2 \frac{\hbar}{i} \int_0^{10} x \frac{d}{dx} x \, dx [/itex]


    NVM the above ;(. Think about it. You know that <p> is real and you get a purely imaginary result. What can you conclude? ;)
     
    Last edited: Oct 4, 2012
  4. Oct 5, 2012 #3
    I conclude that I am either wrong, or that <p> = 0. I'm inclined to the former, but the latter makes sense, because if I was integrating around the origin (say, from -5 to 5 instead), it would cancel.
     
  5. Oct 5, 2012 #4
    But your integral goes from 0 to 10, so <p> = 0.
     
  6. Oct 5, 2012 #5
    Yes.
     
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