1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Quamtum mechanics question

  1. Mar 26, 2005 #1
    Ahoy hoy...I'm having some trouble understanding exactly where the momentum operator comes from. The momentum operator is P=-ih/(2*pi)*d/dx
    I know that according to the DeBroglie relation p=kh/(2*pi)
    and in the first chapter of my book we introduce the operator K=-id/dx
    which is hermitian (which is necessary for getting real eigenvalues). So they say in the book that the P opertor is just P=hK/(2*pi)...but I don't see why. Is there some sort of relation between the wave # k and the operator K?
    When I asked my professor all he said was something about the units of K being right, but I don't even see that.
    Any help would be greatly appreciated.
     
  2. jcsd
  3. Mar 27, 2005 #2

    Galileo

    User Avatar
    Science Advisor
    Homework Helper

    I don't know the argument used in the book, but if you consider a plane wave (state with a well defined momentum):

    [tex]\psi(x)=Ae^{ikx}[/tex], then this is an eigenstate of the momentum operator with eigenvalue [itex]p=\hbar k[/itex].
    Does that clear it up a bit?
     
  4. Mar 27, 2005 #3

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    What book do you use...?I bet it's not a good one.A proof that in the coordinate representation the momentum operator (for a general axis 'i') [tex] \hat{P}_{i} [/tex] has the form

    [tex] \hat{P}_{i}=-i\hbar\frac{\partial}{\partial x_{i}} [/tex] *

    would use 3 things:
    1.De Broglie's relation [itex] \vec{p}=\hbar\vec{k} [/itex]
    2.Parseval identity;
    3.Fourier transformation of the wavefunction.

    Daniel.
    ------------------------------------------------------------
    * valid in the Schrödinger picture in Dirac's formulation.
     
  5. Mar 27, 2005 #4

    StatusX

    User Avatar
    Homework Helper

    You can derive the momentum operator by calling the change in time of the position expectation value the "velocity":

    [tex]<p> = m \frac{d}{dt} \int \Psi^*(x,t) \ x \ \Psi(x,t) \ dx [/tex]

    This can be put into the form of an operator between the wavefunction and its conjugate. You'll need to first move the time derivative inside the integral, use the schroedinger equation to get [itex]d\Psi(x,t)/dt[/itex], and integrate by parts a couple times, getting rid of the boundary term since the wavefunction must go to 0 at infinity. You'll end up with:

    [tex]<p> = \int \Psi^*(x,t) (-i \hbar )\frac{d\Psi(x,t)}{dx} \ \ dx [/tex]
     
  6. Mar 27, 2005 #5
    Thanks everyone, I think that cleared things up for me. It's an undergrad class and the book I'm using is "principles of quantum mechanics" by R.Shankar. It's probably not the book, it's most likely me :cry:
    Does anyone have any suggestions of a good book to compliment the one I have already?
     
  7. Mar 27, 2005 #6

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    I thought Shankar's book is pretty good.It's advanced,compared,let's say to an introductory text like Griffiths or Blokhintsev,but pretty good on the ensemble...

    Cohen-Tannoudji is very calculative,though the structure of the material is not appealing with those "Compléments".If u like more Messiah's book,be my guest.

    Daniel.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Quamtum mechanics question
  1. Mechanics Question (Replies: 1)

  2. Mechanics Question (Replies: 2)

  3. Mechanics question (Replies: 8)

  4. Mechanics question (Replies: 22)

  5. Mechanics Question (Replies: 6)

Loading...