For a spinless particle with the wavefunction
\begin{equation} \psi(x,y,z)= K(x+y+2z)\exp(-\alpha r) \end{equation} with $r=\sqrt{x^2+y^2+z^2}$ and K and $\alpha$ are real constants.

I have to calculate the probability of finding the particle in a solid angle $d\Omega$ at $\theta$ and $\phi$. I have already interpreted the wave function using the spherical harmonics and calculated the root of the expectation value of the squared orbital angular momentum, its z-component, and its probability, now all I have to do is find the probability of the function in the solid angle $d\Omega$. Can anyone give me a clue?

Edit: I have found this equation for the needed probability while searching for a solution: \begin{equation} W_{nl}(\theta,\phi)d\Omega= Y_{l}^{m*}(\theta,\phi) Y_{l}^{m}(\theta,\phi)d\Omega \int{R_{nl}(r)r^2 dr} \end{equation} I still don't know how to go from here.


closed as off-topic by Kyle Kanos, user191954, stafusa, Jon Custer, ZeroTheHero Oct 3 '18 at 23:59

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  • $\begingroup$ Do you mean probability of finding a particle at position $x$, $y$, $z$? $\endgroup$ – K_inverse Oct 3 '18 at 10:23
  • $\begingroup$ No, the problem says finding the particle in the solid angle $d\Omega$ around $\theta$ and $\phi$. $\endgroup$ – Mohamed Mossad Oct 3 '18 at 10:38

Probability of finding the particle in a certain solid angle will be a function of $\theta$ and $\phi$. The radial coordinate $r$ does not matter, since we need to find the probability in a certain direction in space, not in a patch of given finite distance. So you should integrate over $r$ (from $0$ to $\infty$).


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