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.