A function is given like : $ f(\mathbf{r}) = (x + y + z) \ \ \ f(|\mathbf{r}|)$
I need to covert this equation to spherical harmonic form to find out specific Eigen state.To do this, first i converted them into spherical coordinates :
$${\begin{aligned}x&=r\sin \theta \cos \varphi \\y&=r\sin \theta \sin \varphi \\z&=r\cos \theta \end{aligned}}$$
After putting these values into the main equation i get:
$$\psi (r, \theta, \phi) = r\Big( \sin \theta \frac{ e^{i \phi} + e^{- i \phi}}{2} + \sin \theta \frac{ e^{i \phi} - e^{- i \phi}}{2} + \cos \theta \Big) \ \ f(r, \theta , \phi)$$
If I want to write this equation according to the Spherical harmonic table, what happen to the function $f(r, \theta, \phi)$ ? I can write the function as $\psi$ without the function $f(r)$, which in the right side.
Could you please tell me how to write the function in the form of spherical harmonic perfectly?
Since
\begin{eqnarray} Y_1^{-1} &=& \frac{1}{2}\sqrt{\frac{3}{2\pi}}e^{-i\phi}\sin\theta \\ Y_1^{1} &=& -\frac{1}{2}\sqrt{\frac{3}{2\pi}}e^{i\phi}\sin\theta \\ Y_1^0 &=& \frac{1}{2}\sqrt{\frac{3}{\pi}}\cos\theta \end{eqnarray}
You can write
\begin{eqnarray} e^{-i\phi}\sin\theta &=& 2\sqrt{\frac{2\pi}{3}}Y_1^{-1} \\ e^{i\phi}\sin\theta &=& -2\sqrt{\frac{2\pi}{3}}Y_1^{1} \\ \cos\theta &=& 2\sqrt{\frac{\pi}{3}}Y_1^{0} \end{eqnarray}
So that
\begin{eqnarray} \frac{x + y + z}{r} &=& \frac{1}{2}\sin\theta(e^{i\phi} + e^{-i\phi}) + \frac{1}{2i}\sin\theta(e^{i\phi} - e^{-i\phi}) + \cos\theta \\ &=& \frac{1}{2}(1 - i)e^{i\phi}\sin\theta + \frac{1}{2}(1 + i)e^{-i\phi}\sin\theta + \cos\theta \\ &=& -(1 - i)\sqrt{\frac{2\pi}{3}}Y_1^{1} + (1 + i)\sqrt{\frac{2\pi}{3}}Y_1^{-1}+ 2\sqrt{\frac{\pi}{3}}Y_1^{0} \\ &=& -2e^{-i\pi/4}\sqrt{\frac{\pi}{3}}Y_1^{1} + 2e^{i\pi/4}\sqrt{\frac{\pi}{3}}Y_1^{-1}+ 2\sqrt{\frac{\pi}{3}}Y_1^{0}\\ &=& 2\sqrt{\frac{\pi}{3}}(e^{i\pi/4}Y_1^{-1} - e^{-i\pi/4}Y_1^{1} + Y_1^{0}) \end{eqnarray}
The function $f$ then becomes
$$ f(\mathbf{r}) = (x + y +z)g(r) = 2\sqrt{\frac{\pi}{3}}(e^{i\pi/4}Y_1^{-1} - e^{-i\pi/4}Y_1^{1} + Y_1^{0})rg(r) $$
Note that the factor $rg(r)$ is not represented in spherical harmonics because it does not depend on either $\theta$ or $\phi$.