Struggling with Integrals in QM? Here's Some Help!

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Homework Help Overview

The discussion revolves around integrals related to quantum mechanics, specifically addressing questions involving multiple integrals and their evaluation in the context of quantum systems. Participants are exploring the mathematical intricacies involved in these integrals.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to evaluate integrals involving angular coordinates and exponential functions but expresses uncertainty about how to proceed after initial transformations. Some participants point out missing elements in the mathematical expressions and suggest clarifications regarding the setup of the integrals. Others discuss the implications of specific assumptions made about the coordinates and the behavior of the integrands.

Discussion Status

There is an ongoing exploration of different approaches to the integrals, with some participants providing guidance on simplifying the expressions and suggesting substitutions. However, there is no explicit consensus on the best method to proceed, and multiple interpretations of the problems are being considered.

Contextual Notes

Participants are working under the constraints of homework assignments, which may limit the information they can share. There are indications of reliance on specific texts for further clarification, suggesting that some foundational knowledge may be assumed or missing in the discussion.

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Questions 3 and 4 in the attachment.


The Attempt at a Solution


3. [tex]\int d\omega_1 d\omega_2 /|r1-r2|=(2\pi)^2 \int_{0}^{\pi} d\theta_1 \int_{0}^{\pi} d\theta_2 \frac{1}{\sqrt{r_1^2+r_2^2-2r_1r_2cos(\theta_1+\theta_2)}[/tex]
don't know how to proceed from here?
for question 4 I got to the integral:
[tex]\int_{0}^{\infty}\int_{-1}^{1}dcos(\theta)x^2exp(-(|x-x_A|+|x-x_B|)/a)dx[/tex]
Now I can assume that x_A is at the origin and x_B=Rx, where R is the separation between the two atoms, i.e the exponenet becomes: [tex]exp(-(x+\sqrt{x^2+R^2-2Rxcos(\theta))[/tex], but still how do I proceed from here?

Thanks in advance.
here's the attachment in case the link doesn't show.
 

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Last edited:
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I don't see why my first tex code doesn't showup.
 
There is a curly bracket missing at the end.

[tex]\int d\omega_1 d\omega_2 /|r1-r2|=(2\pi)^2 \int_{0}^{\pi} d\theta_1 \int_{0}^{\pi} d\theta_2 \frac{1}{\sqrt{r_1^2+r_2^2-2r_1r_2cos(\theta_1+\theta_2)}}[/tex]

Remember that

[tex]d\Omega = sin\theta d\theta d\phi[/tex]

Also, why do you have a sum of the angles inside cosine? Choose coordinates such taht z-axis is along one of the vectors.
 
So it's should be something like this:
[tex]\frac{1}{\sqrt{r_1^2+r_2^2-2r_1r_2cos(\theta_1)}}[/tex]
Now if r1>r2, by expandig this by taylor series, I get that the first term is 1/r1, and the other terms depend on cosines of theta1, which get terminated with the integration.

Can someone help me with the integral in question 4?

EDIT:
Only odd powers of cosines get terminated in integration, I guess that the other terms get canceled cause they contain powers greater than 1 of 1/r1, though I believe this result should be exact.
 
Last edited:
You are making this too complicated. After integrating over theta2 (which is trivial) your relult is proportional to

[tex]\int_{0}^{\pi} \frac{sin\theta_1 d\theta_1}{\sqrt{r_1^2+r_2^2-2r_1r_2cos\theta_1}}[/tex]

Which is easily integrated because the integrand is of the form f'(g)*g'. Do a substitution u=cos(theta1) if that makes it more clear.
 
Last edited:
OK, thanks I solved this.

Btw can you help me with my second question?

Thanks in advance.
 
Can't really think of a way right now, so let me know if you solve it/get the the solution.
 
I think I solved it.
What I argued is that I assume that x vector is in the direction of x-hat, and that x_A is in the origin and x_B is also in the direction of x-hat but with magnitude of R which is the separation between A and B.
I got to the next integral to solve:
[tex]\int_{0}^{\infty}x^2 e^{-(|x-R|+x)/a}dx[/tex]
which is easy to calculate.
 
Never mind, I found the answer I was looking for in QM-vol 2 of Cohen Tannoudji from pages: 1170-1175.
 

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