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!

Square of z-component of angular momentum eigenvalues

  1. Jul 23, 2013 #1
    1. The problem statement, all variables and given/known data
    I'm trying to demonstrate that if:

    $$\hat{L}_z | l, m \rangle = m \hbar | l, m \rangle$$

    Then

    $$\hat{L}_z^2 | l, m \rangle = m^2 \hbar^2 | l, m \rangle$$


    2. Relevant equations

    $$\hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2$$
    $$\hat{L}_z = -i\hbar \left [ x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right ]$$


    3. The attempt at a solution

    I'm not really sure where to start with this. I can apply the operator ##\hat{L}_z^2## to an arbitrary function ##f(x,y)##, but that gives me:

    $$\hat{L}_z^2 f(x,y) = \hbar^2\left(x^2 \frac{\partial^2}{\partial y^2} - xy \frac{\partial}{\partial y} \frac{\partial}{\partial x} -xy \frac{\partial}{\partial x} \frac{\partial}{\partial y} + y^2 \frac{\partial^2}{\partial x^2} \right)f(x,y) $$
    I've no idea if this demonstrates anything at all...
     
  2. jcsd
  3. Jul 23, 2013 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Don't think too complicated. Just apply [itex]\hat{L}_z[/itex] twice to the eigenstate. Generally one can say that almost any calculation concerning angular-momentum operators is easier in the representation-free Hilbert-space formulation than using the differential operators of the position representation.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted