Homework 3 solutions link.

https://drive.google.com/file/d/0B_GIlXrjJVn4aGZWMlppVGdLcmc/edit?usp=sharing

As Eric pointed out, there is a mistake in the solution to problem 7 (and problem 8). For 7, the theta integrand is actually  $sin^3 (\theta)$, which integrates to 4/3. This changes the answer from 1.5a to 3a. It also changes the answer for problem 8 to $2 \sqrt2 a$, which is about 2.8a, i.e., a little less than 3a.



If you take a close you you can see that non-zero angular integrals, in our hybrid state calculations, come from terms of the form: $x^2 sin\theta$, $y^2 sin\theta$ or $z^2 sin\theta$. The $sin\theta$ comes from the Jacobian and the powers of x, y or z come from the state and the thing we are taking the expectation value of, respectively. (All the angular dependence of the state is in the x, y or z term, right?) For both the x^2 or y^2 cases the $\theta$ integrand will be $sin^3(\theta)$ , so that is a common integral (4/3) and the $\phi$ integrand will be $cos^2 \phi$ or $sin^2\phi$, either of which integrate to pi. For the z case the $\theta$ integrand will be $cos^2\theta sin \theta$ (which integrates to 2/3) and the $\phi$ integrand will be 1, which, of course, integrates to $2 \pi$.

The expectation, based on symmetry considerations, that the integral of $x \psi_{21x} \psi_{200}$ should be the same as the integral of $z \psi_{21z} \psi_{200}$ is thus borne out.