# Wave function normalization

A book by C. J. Ballhausen 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 C.J.B. 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. 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 $\sqrt{(2-0+1)(2+0)}=\sqrt{(3)(2)} = \sqrt{6}$. Applying this gives

$$\sqrt{18} \psi(5,3) = \sqrt{3/5} [ \sqrt{4} (1^+, 2^-, 0^+) + \sqrt{4} (2^+,1^-,0^+) + \sqrt{6} (2^+, 2^-, -1^+)]$$
$- \sqrt{2/5} [ \sqrt{4} (1^+,1^+,1^-) + \sqrt{6} (2^+,0^+, 1^-) + \sqrt{6} (2^+,1^+,0^-) ]$

Simplification results in:

$$\sqrt{18} \psi(5,3) = \sqrt{12/5} (1^+, 2^-, 0^+) + \sqrt{12/5} (2^+,1^-,0^+) + \sqrt{18/5} (2^+, 2^-, -1^+)$$
$- \sqrt{8/5}(1^+,1^+,1^-) - \sqrt{12/5} (2^+,0^+, 1^-) - \sqrt{12/5}(2^+,1^+,0^-)$

The fourth term cannot exist by the Pauli Principle, so we have instead,

$$\sqrt{18} \psi(5,3) = \sqrt{12/5} (1^+, 2^-, 0^+) + \sqrt{12/5} (2^+,1^-,0^+)$$
$+ \sqrt{18/5} (2^+, 2^-, -1^+) - \sqrt{12/5} (2^+,0^+, 1^-) - \sqrt{12/5}(2^+,1^+,0^-)$

Now we need to fix the ordering of the first term and the fourth term to give,

$$\sqrt{18} \psi(5,3) = -\sqrt{12/5} (2^-, 1^ +, 0^+) + \sqrt{12/5} (2^+,1^-,0^+) + \sqrt{18/5} (2^+, 2^-, -1^+)$$
$+ \sqrt{12/5} (2^+,1^-, 0^+) - \sqrt{12/5}(2^+,1^+,0-)$

Now we divide thru by $\sqrt{18}$ the like terms yielding,

$$\psi(5,3) = -\sqrt{12/90} (2^-, 1^ +, 0^+) + \sqrt{12/90} (2^+,1^-,0^+) + \sqrt{18/90} (2^+, 2^-, -1^+)$$
$+ \sqrt{12/90} (2^+,1^-, 0^+) - \sqrt{12/90}(2^+,1^+,0^-)$

Our problem is that 12/90 + 12/90 +18/90 + 12/90 +12/90 =11/15, instead of 15/15. I'm sure my mistake is stupid somewhere, can somebody point out where I've gone wrong?

• Isn't the square of coefficients in the answer you wrote down for your second example $(1^2+(-1)^2+2^2)/6=1$ ? Jun 30, 2011 at 8:22
• you're right, so this isn't just coincidence? So I wonder if this means in my specific application I am just messing up something very simple. Jun 30, 2011 at 15:42
• Just as an aside, "normalized" means $\langle\psi|\psi\rangle = 1$. If you have a spatial wavefunction then that reduces to $\int\psi^*\psi\mathrm{d}\tau = 1$, but for something else like a spinor, then it's $\sum a^* a = 1$ where the $a$'s are the coefficients, as you've done in your examples. Jun 30, 2011 at 17:00
• @Ramashalanka; Yes, you're right. I'll delete my answer (and probably this comment, eventually). Jul 2, 2011 at 4:36
• @all; nice puzzle. Jul 2, 2011 at 7:46

It was just an arithmetic error:
$$\psi(5,3) = -\sqrt{12/90} (2^-, 1^+, 0^+) + \sqrt{12/90} (2^+,1^-,0^+) + \sqrt{18/90} (2^+, 2^-, -1^+)$$
$$+ \sqrt{12/90} (2^+,1^-, 0^+) - \sqrt{12/90}(2^+,1^+,0^-)$$
needs to be simplified as the second and fourth terms are the same. One has:
$$\psi(5,3) = -\sqrt{12/90} (2^-, 1^+, 0^+) + 2\sqrt{12/90} (2^+,1^-,0^+) + \sqrt{18/90} (2^+, 2^-, -1^+)$$
$$- \sqrt{12/90}(2^+,1^+,0^-)$$
which is normalized:
$$12/90 + 4(12/90) + 18/90 + 12/90 = 1.$$

• I think this isn't perfectly expressed, but it's Useful, and hopefully enough for Chris. Suppose we construct state vectors to be orthonormal; raising and lowering operators are not unitary, which is as much as to say that they do not in general transform normalized state vectors to normalized state vectors. Jun 30, 2011 at 22:20
• I don't think this is relevant to what @Chris is doing. He is applying $J_-/\sqrt{(j+m)(j-m+1)}$ (e.g. when he divides by $\sqrt{10}=\sqrt{(5+5)(5-5+1)}$), so since both $|jm\rangle$ and $|jm-1\rangle$ should be normalized, he should be fine. I think he must have an algebra error. I think the "solved for all the relevant values of $m$" is expressed poorly. We know $j$ and $m$: what are you solving for? Jun 30, 2011 at 22:35
• that's very interesting! @Ramashalanka, I hope it is something silly with my algebra: I have posted more details on that. Jun 30, 2011 at 23:18
• @Carl: +1 well spotted. Jul 2, 2011 at 8:29