Z-operator acting on an angular momentum quantum state

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The discussion centers on demonstrating that the expectation value of the z-operator, < n l m | z | n l m >, equals zero for all quantum states | n l m>. The participant expresses confusion about starting the problem since z is not an eigenfunction of | n l m>. They suggest that z might be represented using angular momentum operators L^2, Lz, and L±, but struggle to isolate z. The conversation highlights the importance of wavefunctions, specifically noting that the spherical harmonics relate to the z-component through associated Legendre polynomials. Ultimately, the solution involves recognizing how the multiplication with the z-component leads to transitions between different angular momentum states.
QuantumKyle
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Homework Statement



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

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?
 
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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 \cos \theta which is \hat{z}.
Now, check the first recurrence relation here: http://en.wikipedia.org/wiki/Associated_Legendre_polynomials#Recurrence_formula
and you will see that multiplying with \hat{z}=\cos \theta, which is the argument of the Legendre polynomial, gives you P_{l+1}^m(\cos \theta) and P_{l-1}^m(\cos \theta) (with awkward coefficients), so effectively you'll get |n l+1 m> and |n l-1 m>.
 
Thanks, that was a big help
 
My pleasure ;)
 

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