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Quantum Mechanics}Angular Momentum

Recall the angular momentum eigen-equations, L^2 \psi = \hbar^2 l(l+1) \psi and L_z \psi = m\hbar \psi.

The problem wants L^2 \psi = 6\hbar^2 \psi and L_z \psi = -\hbar \psi. Matching coefficients with the above equations, one finds that

l(l+1) = 6 and m=-1. Solving, one finds that l=2,-3. Any one of the spherical harmonics with Y^{-1}_{2} or Y^{-1}_{-3} would work. So, since the second spherical harmonic isn't listed, take choice (B).

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Alternate Solutions
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2009-07-03 15:08:49
How did you know to use 6? The problem said to use 3S and there are 2 electrons.

3 = 2s+1, which can be solved for spin. But where does the 6 come from?
2011-09-03 06:26:43
The eigenvalues of \L^2 are \l(l+1). So one can solve \l(l+1) to get l=2 or l=-3. That why the 6 is given.
2011-09-03 06:30:19
The eigenvalues of \ L^2 are . So one can solve l(l+1)=6 to get l=2 or l=-3. That why the 6 is given.
2007-10-02 11:34:01
A BIG correction to Yosun in here. That is regarding the possibility of having the orbital quantum number l=-3. This is never possible since l is defined to be ALWAYS POSITIVE and in the range of 0,1,2,...,n-1; where n is the principal quantum number. So it is not as if the choice of Y^{-1}_{-3} is not part of the multiple choice, but even if it was, you should always know l can't have negative values.
2008-10-20 03:23:08
2010-11-10 02:47:36
I had an oh shit moment. Whew!
2013-09-15 05:49:30
Hey Yosun, please revise this.

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