Z-operator acting on an angular momentum quantum state

3 replies · 5K views · last reply
QuantumKyle
Messages
3
Reaction score
0

Homework Statement



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

2. relevant 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

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?
 
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>.