Solving More Wave Integrals: Proving p_x = - 2^7 \frac{a}{3^5}

[bigevil]

Homework Statement

$$p_x = - \int\int\int \sqrt{\frac{3}{32{\pi}^2}} {sin}^3 \theta cos \phi e^{i\phi} \sqrt{\frac{1}{8a^6}} \frac{2r^4 e^{-3r/2a}}{a\sqrt{3}} dr d\theta d\phi$$

We are to prove that $$p_x = - 2^7 \frac{a}{3^5}$$

Homework Equations

The constants collapse to
$$\frac{1}{8a^4\pi}$$

I have already combined the coefficients for spherical coordinates.

The Attempt at a Solution

With respect to r,

$$\int_0^{\infty} r^4 e^{-3r/2a} dr = 4.3.2.1 (- \frac{2a}{3})^5 = 6. - 2^7 \frac{a^5}{3^5}$$

With respect to $$\theta$$,

$$\int_0^{\pi} {sin}^3 \theta d\theta = \frac{4}{3}$$ (trust me)

With respect to $$\phi$$

$$\int_0^{2\pi} cos\phi e^{i\phi} d\phi = \int_0^{2\phi} \frac{1}{2} {cos}^2 \phi + \frac{1}{2} + icos\phi sin\phi d\phi = \pi$$

This gives us $$p_x = 2^7 \frac{a}{3^5}$$

So where did my minus go?? I Don't think there's a big problem with the integration but not sure if I've made any careless mistakes.

 

[Dick]

If you are claiming the r integral is negative, how can that be? The integrand is positive. If a<0 then the integral is divergent. Where are you getting these horrible integrals?

 

[bigevil]

It's from Mathematical Methods for Physics and Engineering. The question actually cites two wavefunctions for the position of a hydrogen electron and then asks you to find the dipole matrix element $$p_x$$ between the two. Whatever the case, I have combined the two wavefunctions and the volume element in spherical coordinates, which gives you the monstrosity above.

Well, this integral could be worse, at least the variables are separable.

I was thinking my r-integral was wrong too, but I'm not sure where.

a isn't <0 because it's the Bohr radius. Thanks for your help Dick.