Wave function normalization
A book by <PERSON> led me to believe that a quick way to check that I performed step operators properly was by observing that the "wave function should appear normalized," but I have found some issue applying this in practice and believe it is due to my misunderstanding of the underlying physics; I'm trying to understand what <PERSON> meant by that and if it applies in my case.
In his case he was observing two equivalent electrons. Let's say they are two equivalent $p$ electrons for the sake of example. (He was actually considering $d$ electrons and I can provide the specifics of that if it will help.) There are several unperturbed functions which can be described by symbols such as $(1^+ 0^+)$, in which case the first number means $m_l$ of the 1st 2p electron, the next symbol indicates its spin and so forth.
For the term $ ^1 D : M_L = 2, M_S=0: (1^+ 1^-) $ is an eigenfunction of $p^2$ configuration that is known. Using a step down operator on the angular momentum gives: $M_L=1 : (2)^{(-1/2)} [ (1^+ 0^-) - (1^- 0^+) ]$. Here I get the impression that what we observe appears normalized because squaring the coefficients gives unity. I realize one could in principle perform $\int \psi^* \psi d \tau$, but I do not suspect that is what he means.
Now if we apply the step operator again we get $M_L=0: (6)^{(-1/2)} [ (1^+ -1^-) - (1^- -1^+) + 2(0^+ 0^-) ]$. Here the square of the coefficients is decidedly unity. 346
His examples given in the book also happen to go to unity; is this just coincidence or is it going to always be true?
My specific example is as follows, starting with a $d^3$ configuration: $$\psi(L,M_L,S,M_S)=\psi(5,5,\frac{1}{2},\frac{1}{2})=(2^+,2^-,1^+)$$
Applying the lowering operators gives $\sqrt{10} \psi(5,4,\frac{1}{2},\frac{1}{2}) = -\sqrt{4} (2^+,1^+,1^+) - \sqrt{4} (2^+,1^+,1^-) + \sqrt{6}(2^+,2^-,0^+)$ and the coefficients go to unity as expected. Above you will notice that the ordering in the first and third terms has changed and an odd permutation brings about a change in sign. Also notice the first term must be equal to zero by Pauli Principle. Dividing out we get,
$$\psi(5,4,\frac{1}{2},\frac{1}{2}) = \sqrt{3/5} (2^+,2^-,0^+) - \sqrt{2/5} (2^+,1^+,1^-)$$
You'll notice that the coefficients squared sum up to 1, so all appears normalized and well. Now we apply the lowering operator again to give $\sqrt{(L-M_L+1)(L+M_L)} = \sqrt{(5-4+1)(5+4)} = \sqrt{(2)(9)} = \sqrt{18}$ times the function for $M_L=3$ $^1 H$.
Applying to the RHS using $\sqrt{(l-m_l+1)(l+m_l)}$. We are working with $d$ orbitals, therefore $l=2$. So for the case of $m_l=2$ we get $\sqrt{(2-2+1)(2+2)}=\sqrt{4}$ and for $m_l=1$ we get $\sqrt{(2-1+1)(2+1)} = \sqrt{6}$ and finally for $m_l=0$ we get $\sqrt{(2-0+1)(2+0)}=\sqrt{(3)(2)} = \sqrt{6}$. 818