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I need to verify that the following Hydrogen atom wave function $$\Psi(x,y,z;t=0)\equiv\frac{4}{(2a)^{3/2}}\left[\frac{1}{\sqrt{4\pi}}e^{-r/a}+A\frac{r}{a}e^{-r/(2a)}\left(-iY_1^{+1}+Y_1^{-1}+\sqrt{7}Y_1^0\right)\right]$$ is normalized for $A=1/(12\sqrt{6})$.

I know that this implies showing that the inner product of $\Psi$ with itself equals 1, though getting there has proven to be a challenge.

I have tried both plugging in the definitions of the spherical harmonics and solving the integral directly (this leads to a very large number of terms) and by substituting in $R_{nl}$ radial functions (but this leaves the first term as a radial function on its own).

It feels like the second method would be the intended way to approach the problem but I could really use a tip in the right direction.

Edit: I forgot that $Y_0^0=\sqrt{1/4\pi}$. Plugging this into the first term and massaging the coefficients to obtain $R_{nl}$'s for both terms allows for a representation of $\Psi$ in terms of Hydrogen $\psi_{nlm}$ wave functions.

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  • $\begingroup$ Use orthonormality of the spherical harmonics. That will kill six cross terms and simplify three more. You can get rid of four more terms because the $\phi$ integral of a spherical harmonic with $m\ne 0$ vanishes. That leaves three integrals to evaluate: two of them are independent of $\theta$ and $\phi$ and the other one is a simple one over the angles; that leaves just the r integrations which are also simple. $\endgroup$ – NickD Apr 20 '17 at 5:40
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Orthogonality of the spherical harmonics will eliminate the cross-terms : $\int d\Omega Y_{\ell m}Y_{\ell m'}^* =\delta_{\ell\ell'}\delta_{mm'}$ so your term with spherical harmonics, upon $\int d\Omega$, will give you $9\vert A\vert^2 r^2 e^{-r/a}/a^2$.

You are rapidly left with $$ \frac{8}{a^3} \int_0^\infty dr\, r^2 \left(e^{-2a/r} + \vert A\vert^2 \frac{r^2}{a^2}e^{-r/a}9 \right) $$ where the $1/4\pi$ factor has been eliminated using $\int d\Omega=4\pi$. The rest in integration by parts.


Edit This question alreay has your $\psi$ in the form $$ \psi(r,\theta,\phi)=\sum_{n\ell m}c_{n\ell m} \psi_{n\ell m}(r,\theta,\phi) $$ so normalization amounts to verifying that $\sum_{n\ell m}\vert c_{n\ell m}\vert^2=1$ since the solutions $\psi_{n\ell m}(r,\theta,\phi)$ are orthogonal under integration

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Multiply this with its complex conjugate of same function and integrate over all space ($r\theta$ and $\phi$). Using normalization property of spherical harmonics and perform integration of $r$ coordinates $0$ to $∞$.

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