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!

Z-operator acting on an angular momentum quantum state

  1. Oct 19, 2011 #1
    1. The problem statement, all variables and given/known data

    I need to show that < n l m | z | n l m > = 0 for all states | n l m>

    2. Relevent Equations:

    L^2 = Lx^2 + Ly^2 + Lz ^2
    Lx = yp(z) - zp(y)
    Ly = zp(x) - xp(z)
    Lz = xp(y) - yp(x)
    L+/- = Lx +/- iLy

    3. The attempt at a solution

    I really don't know where to begin because z is not an eigenfuntion of | n l m> (and if it was this equation would not be 0 anyways). My intuition tells me that I need to somehow represent z as a function of the operators L^2, Lz, and maybe L+/-. But I can't seem to isolate z. Maybe I'm looking at this problem the wrong way. Is there some fundamental theorem that would show that this equation is true?
     
  2. jcsd
  3. Oct 19, 2011 #2
    There is certainly a way to use the algebra, but it's actually eaasier if you look at the wavefunctions in position space.
    The wavefunction is basically a Laguerre Polynomial with respect to r times a spherical harmonics.
    The spherical harmonics is basically a complex exponential of phi times an associated Legendre polynomial with respect to [itex]\cos \theta[/itex] which is [itex] \hat{z}[/itex].
    Now, check the first recurrence relation here: http://en.wikipedia.org/wiki/Associated_Legendre_polynomials#Recurrence_formula
    and you will see that multiplying with [itex] \hat{z}=\cos \theta [/itex], which is the argument of the Legendre polynomial, gives you [itex]P_{l+1}^m(\cos \theta)[/itex] and [itex]P_{l-1}^m(\cos \theta)[/itex] (with awkward coefficients), so effectively you'll get |n l+1 m> and |n l-1 m>.
     
  4. Oct 19, 2011 #3
    Thanks, that was a big help
     
  5. Oct 19, 2011 #4
    My pleasure ;)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook