Undergrad How Is the Integral from (3-22) to (3-23) Calculated in Quantum Mechanics?

[jamal_lamaj]

TL;DR

I don't understand how to get this result, can you help me? Thanks!!!

I don't get the step from (3-22) to (3-23), can you how this integral was calculated? Thanks!
Below there is a screenshoot of (3-9). Images are taken from "Intermediate Quantum Mechanics, 3rd Edition - Bethe, Jackiw".

 

[PeterDonis]

What textbook is your excerpt from?

 

[jamal_lamaj]

What textbook is your excerpt from?

"Intermediate Quantum Mechanics, 3rd Edition - Bethe, Jackiw"
And the same calculation in also in "Quantum Mechanics of One and Two Electron - Bethe, Salpeter", but no steps included.

 

[Haborix]

I don't have the time at the moment to type it all out (and you should work through that yourself if you want to understand). But some hints:

1. Recognize $Y_{00}(\Omega_1)=\text{constant}$, so $Y^2_{00}(\Omega_1)=\text{constant}Y_{00}(\Omega_1)$
2. Substitute 3-9 into the integral and think about the integration over $\Omega_1$, keeping the orthogonality of the spherical harmonics in mind.

 

[jamal_lamaj]

Hi! Your hints really helped me: I solved it, now it's clear!
If you wanna check up my calculation, here they are:

$$
r_1>r_2
$$

$$
\begin{aligned}

J(r_1,r_2) &= \int\int\ d\Omega_1 d\Omega_2 \hspace{5pt} Y^2_{00}(\Omega_1) |Y_{lm}(\Omega_2)|^2 (\frac{1}{r_{12}}-\frac{1}{r_2}) \\

&= \int\int\ d\Omega_1 d\Omega_2 \hspace{5pt} Y^2_{00}(\Omega_1) |Y_{lm}(\Omega_2)|^2 \frac{1}{r_{12}} - \int\int\ d\Omega_1 d\Omega_2 \hspace{5pt} Y^2_{00}(\Omega_1) |Y_{lm}(\Omega_2)|^2 \frac{1}{r_2} \\

&= \int\int\ d\Omega_1 d\Omega_2 \hspace{5pt} Y^2_{00}(\Omega_1) |Y_{lm}(\Omega_2)|^2 \frac{1}{r_{12}} - \frac{1}{r_2} \int\int\ d\Omega_1 d\Omega_2 \hspace{5pt} Y^2_{00}(\Omega_1) |Y_{lm}(\Omega_2)|^2 \\

&= \int\int\ d\Omega_1 d\Omega_2 \hspace{5pt} Y^2_{00}(\Omega_1) |Y_{lm}(\Omega_2)|^2 \frac{1}{r_{12}} - \frac{1}{r_2} \int d\Omega_1 \hspace{5pt} Y^2_{00}(\Omega_1) \int d\Omega_2 \hspace{5pt} |Y_{lm}(\Omega_2)|^2 \\

&= \int\int\ d\Omega_1 d\Omega_2 \hspace{5pt} Y^2_{00}(\Omega_1) |Y_{lm}(\Omega_2)|^2 \frac{1}{r_{12}} - \frac{1}{r_2} \\

&= I(r_1,r_2) - \frac{1}{r_2} \\

\end{aligned}
$$

$$
\begin{aligned}

I(r_1,r_2) &= \int\int\ d\Omega_1 d\Omega_2 \hspace{5pt} Y^2_{00}(\Omega_1) |Y_{lm}(\Omega_2)|^2 \frac{1}{r_{12}} \\

&= \int\int\ d\Omega_1 d\Omega_2 \hspace{5pt} Y^2_{00}(\Omega_1) |Y_{lm}(\Omega_2)|^2 \frac{1}{r_1} \sum_{l=0}^{\infty} (\frac{r_2}{r_1})^l \frac{4\pi}{2l+1} \sum_{m=-l}^l Y_{lm}(\Omega_1) Y^*_{lm}(\Omega_2) \\

&= \frac{4\pi}{2\sqrt{\pi}} \frac{1}{r_1} \int\int\ d\Omega_1 d\Omega_2 \hspace{5pt} Y_{00}(\Omega_1) |Y_{lm}(\Omega_2)|^2 \sum_{l=0}^{\infty} (\frac{r_2}{r_1})^l \frac{1}{2l+1} \sum_{m=-l}^l Y_{lm}(\Omega_1) Y^*_{lm}(\Omega_2) \\

&= \frac{2\sqrt{\pi}}{r_1} \int\int\ d\Omega_1 d\Omega_2 \hspace{5pt} |Y_{lm}(\Omega_2)|^2 \sum_{l=0}^{\infty} (\frac{r_2}{r_1})^l \frac{1}{2l+1} \sum_{m=-l}^l Y_{00}(\Omega_1) Y_{lm}(\Omega_1) Y^*_{lm}(\Omega_2) \\

&= \frac{2\sqrt{\pi}}{r_1} \int\int\ d\Omega_1 d\Omega_2 \hspace{5pt} |Y_{lm}(\Omega_2)|^2 \sum_{l=0}^{\infty} (\frac{r_2}{r_1})^l \frac{1}{2l+1} \sum_{m=-l}^l Y^*_{00}(\Omega_1) Y_{lm}(\Omega_1) Y^*_{lm}(\Omega_2) \\

&= \frac{2\sqrt{\pi}}{r_1} \int\int\ d\Omega_1 d\Omega_2 \hspace{5pt} |Y_{lm}(\Omega_2)|^2 \sum_{l=0}^{\infty} (\frac{r_2}{r_1})^l \frac{1}{2l+1} \sum_{m=-l}^l Y^2_{00}(\Omega_1) \delta_{0}^{l} \delta_{0}^{m} Y^*_{lm}(\Omega_2) \\

&= \frac{2\sqrt{\pi}}{r_1} \int\int\ d\Omega_1 d\Omega_2 \hspace{5pt} |Y_{00}(\Omega_2)|^2 (\frac{r_2}{r_1})^0 \frac{1}{2 \cdot 0+1} Y^2_{00}(\Omega_1) Y^*_{00}(\Omega_2) \\

&= \frac{2\sqrt{\pi}}{r_1} \int\int\ d\Omega_1 d\Omega_2 \hspace{5pt} Y^2_{00}(\Omega_2) Y^2_{00}(\Omega_1) Y_{00}(\Omega_2) \\

&= \frac{2\sqrt{\pi}}{r_1} \frac{1}{2\sqrt{\pi}} \int\int\ d\Omega_1 d\Omega_2 \hspace{5pt} Y^2_{00}(\Omega_2) Y^2_{00}(\Omega_1) \\

&= \frac{1}{r_1} \int d\Omega_1 \hspace{5pt} Y^2_{00}(\Omega_1) \int d\Omega_2 \hspace{5pt} Y^2_{00}(\Omega_2) \\

&= \frac{1}{r_1} \\

\end{aligned}
$$

$$
\begin{aligned}

J(r_1,r_2) &= I(r_1,r_2) - \frac{1}{r_2} \\

J(r_1,r_2) &= \frac{1}{r_1} - \frac{1}{r_2} \hspace{0.5cm} \blacksquare

\end{aligned}
$$

Thanks.

P.s.
I'm quite new to the forum, can you explain me how to mark the post as "Solved"?
Bye!

 

[PeterDonis]

can you explain me how to mark the post as "Solved"?

There isn't a way to do that here. In general PF discussions aren't as simple to categorize as "Solved" vs. "Not Solved" so we don't have any such markings for them. Your post here saying you found the solution is sufficient.