# Normalization constant for orbital wave functions

1. Oct 8, 2011

### Kaiten7

Suppose I have a wavefunction

ψ(r1, r2)= (∅1s(r1) ∅1p(r2) - ∅1s(r2) ∅1p(r1))

And I know that ∅1s(r1) and ∅1p(r1) are normalized. How would I go about finding the normalization constant for ψ(r1, r2)?

Everywhere I look just whips out a $\frac{1}{\sqrt{2}}$ out of nowhere:

http://en.wikipedia.org/wiki/Slater_determinant
http://farside.ph.utexas.edu/teaching/qmech/lectures/node59.html
http://vergil.chemistry.gatech.edu/notes/intro_estruc/intro_estruc.pdf

Are a few examples

A few of those seem to mention something about orthonormal orbitals, but their definition seems to rely on Dirac notation, which I'm not that familiar with. That also means it was infuriating to find this other thread https://www.physicsforums.com/showthread.php?t=178292, that looks like would've really helped me had I understood what the second poster said.

Currently I'm exactly where the first poster is, with

$\frac{1}{N²}$ = $\int$(∅1s(r1) ∅1p(r2) - 2 $\int$ ∅1s(r1) ∅1p(r2) ∅1s(r2) ∅1p(r1) + $\int$ ∅1p(r2) ∅1s(r2

As I mentioned, I think I know that the first and third terms must equal 1 for some reason, and the middle one equal 0, but I don't exactly know why. Any help?

2. Oct 8, 2011

### DiracRules

Dirac's notation is "simply" (for our purpose)

$<a|b>=\int_{\Omega} a^*b\, d\Omega$, where a-star is the complex conjugate.

If two functions are orthonormal (in QM meaning) means that $<a|b>=1$ if a=b, 0 otherwise.

Now, since the function we are dealing with are orthonormal, you expression simplifies because $N^2=<\psi(x_1,x_2)|\psi(x_1,x_2)>=<\chi_1(x_1)\chi_2(x_2)-\chi_1(x_2)\chi_2(x_1)|\chi_1(x_1)\chi_2(x_2)-\chi_1(x_2)\chi_2(x_1)>=\int \left[\chi_1(x_1)^*\chi_2(x_2)^*-\chi_1(x_2)^*\chi_2(x_1)^*\right]\cdot\left[\chi_1(x_1)\chi_2(x_2)-\chi_1(x_2)\chi_2(x_1)\right] d\Omega$

From now on, it's only the application of integration rules
Dirac's formalism is more straightforward, so I'll use it, I hope you'll understand. Otherwise, tell me and I'll re-write all the stuff :
$N^2=<\psi(x_1,x_2)|\psi(x_1,x_2)>=<\chi_1(x_1)\chi_2(x_2)-\chi_1(x_2)\chi_2(x_1)|\chi_1(x_1)\chi_2(x_2)-\chi_1(x_2)\chi_2(x_1)>=$
$<\chi_1(x_1)\chi_2(x_2)|\chi_1(x_1)\chi_2(x_2)-\chi_1(x_2)\chi_2(x_1)>-<\chi_1(x_2)\chi_2(x_1)|\chi_1(x_1)\chi_2(x_2)-\chi_1(x_2)\chi_2(x_1)>=$
$<\chi_1(x_1)\chi_2(x_2)|\chi_1(x_1)\chi_2(x_2)>-<\chi_1(x_1)\chi_2(x_2)|\chi_1(x_2)\chi_2(x_1)>-<\chi_1(x_2)\chi_2(x_1)|\chi_1(x_1)\chi_2(x_2)+<\chi_1(x_2)\chi_2(x_1)|\chi_1(x_2) \chi_2(x_1)>$
In the first and in the last terms the same terms appear at the left and the right of the |, like <a|a>. Because of orthonormality, they both account for 1.
The middle terms have different terms, like <a|b> and <b|a>, so, again, for orthonormality they account for 0.
So
$N^2<\chi_1(x_1)\chi_2(x_2)|\chi_1(x_1)\chi_2(x_2)>-<\chi_1(x_1)\chi_2(x_2)|\chi_1(x_2)\chi_2(x_1)>-<\chi_1(x_2)\chi_2(x_1)|\chi_1(x_1)\chi_2(x_2)+<\chi_1(x_2)\chi_2(x_1)|\chi_1(x_2)\chi_2(x_1)>=1-0-0+1=2$
So the normalization constant is $1/N=\frac{1}{\sqrt{2}}$

Hope it is clearer =)

3. Oct 8, 2011

### Kaiten7

It is! I think I got it now, thanks!