# Homework Help: Integration (related to QM).

1. Dec 18, 2009

### MathematicalPhysicist

Questions 3 and 4 in the attachment.

3. 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 seperation 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?

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

#### Attached Files:

• ###### HW9.pdf
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Last edited: Dec 18, 2009
2. Dec 18, 2009

### MathematicalPhysicist

I don't see why my first tex code doesn't showup.

3. Dec 18, 2009

### phsopher

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.

4. Dec 19, 2009

### MathematicalPhysicist

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 cancelled cause they contain powers greater than 1 of 1/r1, though I believe this result should be exact.

Last edited: Dec 19, 2009
5. Dec 19, 2009

### phsopher

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: Dec 19, 2009
6. Dec 19, 2009

### MathematicalPhysicist

OK, thanks I solved this.

Btw can you help me with my second question?

7. Dec 19, 2009

### phsopher

Can't really think of a way right now, so let me know if you solve it/get the the solution.

8. Dec 20, 2009

### MathematicalPhysicist

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 seperation 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.

9. Dec 20, 2009

### MathematicalPhysicist

Never mind, I found the answer I was looking for in QM-vol 2 of Cohen Tannoudji from pages: 1170-1175.