Prove the wave function for dxy orbital

[chem1309]

Homework Statement

"The angular part of the wave function for the dxy orbital is (√(15/∏)/4)sin^2(θ)sin(2$\phi$). Show that this expression corresponds to the dxy orbital"

Homework Equations

conversion of Cartesian to spherical coordinates:
r=√(x^2+y^2+z^2)
cosθ=z/r
tan($\phi$)=y/x

trig identity:
sin(2x)=2sinxcosx

normalization:
N^2∫ψ*ψdτ

dτ=r^2sinθdrdθd$\phi$

0≤r≤∞
0≤θ≤∏
0≤$\phi$≤2∏

The Attempt at a Solution

in Cartesian coordinates dxy is represented as simply xy. I converted xy to spherical coordinates and manipulated the equation the relevant equations to get xy=(r/2)sin^2(θ)sin(2$\phi$) as follows:

xy=rsincos$\phi$rsinθsin$\phi$
xy=rsin^2(θ)cos$\phi$sin$\phi$
xy=rsin^2(θ)sin(2$\phi$)/2

Then I tried to normalize the equation, but I ended up with

∫r^3 from 0 to ∞, which goes to ∞/does not converge

and ∫sin2$\phi$ which equal zero.

 

[vela]

I'm not sure why you think you need to normalize the wave function to show it corresponds to the d_xy orbital.

In any case, the total wave function is of the form $\psi(\vec{r}) = R_{nl}(r)Y_l^m(\theta,\phi)$. Normalization requires that
$$\int \psi^*(\vec{r})\psi(\vec{r})\,d^3\vec{r} = \int R^*_{nl}(r)R_{nl}(r)\,dr \int {Y_l^m}^*(\theta,\phi)Y_l^m(\theta,\phi)\,d\Omega = 1.$$ Does seeing this clear up your two questions?