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

A question about <p>

  1. Mar 15, 2009 #1
    <p> is the integral of the product of a real function and its first derivative, multiplied by an imaginary number. But why <p> is real?
     
  2. jcsd
  3. Mar 15, 2009 #2

    Ben Niehoff

    User Avatar
    Science Advisor
    Gold Member

    If the wavefunction is purely real, then <p> = 0:

    [tex]\langle p \rangle = \int_{-\infty}^{\infty} \psi^* \left(-i\hbar \frac{d\psi}{dx} \right) \; dx[/tex]

    If [itex]\psi^* = \psi[/itex], then this can be written

    [tex]\langle p \rangle = -i\hbar \left. \int_{-\infty}^{\infty} \psi \frac{d\psi}{dx} \; dx = -i\hbar \psi(x) \right|_{-\infty}^{\infty} = 0[/tex]

    where we have used the boundary condition that the wavefunction must vanish as [itex]x \to \pm \infty[/itex]. On the other hand, if we can write

    [tex]\psi(x) = e^{ikx} R(x)[/tex]

    where R(x) is real, then

    [tex]\langle p \rangle = \int_{-\infty}^{\infty} \psi^* \left(-i\hbar \frac{d\psi}{dx} \right) \; dx = -i\hbar \int_{-\infty}^{\infty} \left( e^{-ikx} R(x) (ik) e^{ikx} R(x) + e^{-ikx} R(x) e^{ikx} \frac{dR}{dx} \right) \; dx = \hbar k \int_{-\infty}^{\infty} R(x)^2 \; dx = \hbar k[/tex]

    So, to have a nonzero momentum, the wavefunction needs a phase that varies with x.
     
  4. Mar 15, 2009 #3
    Because of hermitian conjugate
     
  5. Mar 22, 2009 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    If [itex] \hat{p} [/itex] is selfadjoint, then its expectation value in any state is real.
     
  6. Mar 22, 2009 #5
    As in the last two posts, the most elegant and general way to see that <p> comes from the fact that p is a hermitian operator.

    But if you only know about p in position space (i.e. p = -ih d/dx) then you can see that <p> is real by showing that <p> = <p>* (is this statement obvious?).

    Do this by starting with the first equation in Ben Niehoff's post. Take the complex conjugate of both sides, i's become -i's, and psi -> psi* and psi* -> psi. Then with some manipulations (including an integration by parts) you can show that this expression is identical to the formula you started with, so <p>*=<p>.

    Hope that helps a little.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook