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

In summary, the calculation of the integral from (3-22) to (3-23) in quantum mechanics involves evaluating the transition amplitude between two states using the principles of wave functions and operators. This process typically requires applying techniques from calculus and quantum theory, such as integrating over the relevant wave functions and utilizing properties of the Hamiltonian operator. The result provides insights into the probabilities of transitions between quantum states, which are fundamental to understanding quantum dynamics.
  • #1
jamal_lamaj
3
2
TL;DR Summary
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:
Physics news on Phys.org
  • #3
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.
 
Last edited:
  • #4
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.
 
  • Like
Likes topsquark
  • #5
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!
 
Last edited:
  • Like
Likes vanhees71 and Haborix
  • #6
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.
 
  • Like
Likes vanhees71

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

What is the significance of the integral from (3-22) to (3-23) in quantum mechanics?

The integral from (3-22) to (3-23) often represents a transition amplitude or probability amplitude in quantum mechanics. It is crucial for calculating the likelihood of a particle transitioning from one state to another, which is fundamental to understanding quantum behavior.

What mathematical methods are used to evaluate the integral from (3-22) to (3-23)?

To evaluate such integrals, physicists typically use techniques from complex analysis, such as contour integration, as well as approximation methods like the saddle point method or stationary phase approximation. Numerical integration methods may also be employed when an analytical solution is not feasible.

Why is the integral from (3-22) to (3-23) often complex-valued?

In quantum mechanics, wavefunctions and their associated integrals are generally complex-valued because they encode both amplitude and phase information. This complex nature allows for the interference effects that are characteristic of quantum systems.

How does the integral from (3-22) to (3-23) relate to the Schrödinger equation?

The integral from (3-22) to (3-23) can be derived from the solutions of the Schrödinger equation. It often represents the inner product of wavefunctions or the overlap integral, which is essential for determining probabilities and expectation values in quantum systems.

Can the integral from (3-22) to (3-23) be interpreted physically?

Yes, the integral from (3-22) to (3-23) has a physical interpretation. It usually corresponds to the probability amplitude for a particle to transition between states, which, when squared, gives the probability of such a transition occurring. This is a cornerstone of quantum mechanics, linking mathematical formalism to physical observables.

Similar threads

I Confusions I have regarding Measurements in Quantum Mechanics
Replies
11
Views
2K
A Probability flux integrated over all space is mean momentum?
Replies
2
Views
2K
B Quantum entanglement and hidden variables
Replies
7
Views
2K
I Understanding No Energy Degeneracy in Sakurai's Quantum Mechanics
Replies
2
Views
1K
I How to "derive" momentum operator in position basis using STE?
Replies
3
Views
947
What Are the Dimensions of Hilbert Space Elements in Quantum Mechanics?
Replies
14
Views
1K
I Determining the Number of Points in the n-Space
Replies
16
Views
2K
I Looking For Help Understanding Bell's Theorem, Hidden Variables & QM
3
Replies
80
Views
5K
B Time-reversal of an unknown quantum state
Replies
1
Views
1K
I How are good quantum numbers related to perturbation theory?
Replies
18
Views
3K

Hot Threads

Recent Insights

Back
Top