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

  • Thread starter Thread starter MathematicalPhysicist
  • Start date Start date
  • Tags Tags
    Integration Qm
MathematicalPhysicist
Science Advisor
Gold Member
Messages
4,662
Reaction score
372
Questions 3 and 4 in the attachment.


The Attempt at a Solution


3. \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)}
don't know how to proceed from here?
for question 4 I got to the integral:
\int_{0}^{\infty}\int_{-1}^{1}dcos(\theta)x^2exp(-(|x-x_A|+|x-x_B|)/a)dx
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: exp(-(x+\sqrt{x^2+R^2-2Rxcos(\theta)), but still how do I proceed from here?

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

Attachments

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

\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)}}

Remember that

d\Omega = sin\theta d\theta d\phi

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:
\frac{1}{\sqrt{r_1^2+r_2^2-2r_1r_2cos(\theta_1)}}
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

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

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:
\int_{0}^{\infty}x^2 e^{-(|x-R|+x)/a}dx
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.
 
Back
Top