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

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The discussion focuses on the calculation of the integral transitioning from equations (3-22) to (3-23) in quantum mechanics, specifically using spherical harmonics. A user initially struggles with the calculation but receives hints that emphasize the constant nature of the spherical harmonics and the importance of orthogonality in the integration process. After applying these hints, the user successfully derives the integral, leading to the final expression for J(r1, r2) as (1/r1) - (1/r2). The conversation also touches on the lack of a "Solved" feature in the forum, indicating that simply stating the solution suffices. Overall, the thread highlights the collaborative nature of problem-solving in quantum mechanics.
jamal_lamaj
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I don't understand how to get this result, can you help me? Thanks!!!
Capture.PNG


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".

Capture1.PNG
 
Last edited:
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PeterDonis said:
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.
 
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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.
 
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!
 
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jamal_lamaj said:
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.
 
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