# 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 $(1^2+(-1)^2+2^2)/6=1$ ? – Ramashalanka Jun 30 '11 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. – Chris Jun 30 '11 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. – David Z Jun 30 '11 at 17:00
• @Ramashalanka; Yes, you're right. I'll delete my answer (and probably this comment, eventually). – Carl Brannen Jul 2 '11 at 4:36
• @all; nice puzzle. – Carl Brannen Jul 2 '11 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. – Peter Morgan Jun 30 '11 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? – Ramashalanka Jun 30 '11 at 22:35
• that's very interesting! @Ramashalanka, I hope it is something silly with my algebra: I have posted more details on that. – Chris Jun 30 '11 at 23:18
• @Carl: +1 well spotted. – Ramashalanka Jul 2 '11 at 8:29