Just read it slowly and soak in all of the details. Kittell is very terse...my professor used to call it "condensed knowledge." Let me know if you need any help.
I learned from that book alone, so I cannot point you anywhere else. However, I would highly recommend reading through all of chapter 4 before moving on. Its only 10 pages + problems, and is needed to understand the more advanced material that you're looking for. In fact, every figure in that...
Just to clarify: are you saying that its Bravais lattice is BCC with a two-atom basis, but that there still exists a [rhombohedron] primitive cell which contains only one atom?
This is not true. The "E.S." as you call it -- though, transition state is the more appropriate term -- is never lower in energy (more stable) than the products. As a matter of fact, this is completely counterintuitive! One would think that if there were to be progress along a reaction...
Came here to say exactly this! I decided not to read through the whole thread, but to instead search each page for the words "rigged" or "triplet" and came across your post on page 5. Was this post resolved? I don't see how there would be an issue in the extended nuclear space...as Ballentine...
Please correct me if I'm wrong, but isn't BCC not considered in this scenario? Its not a simple (primitive) cubic lattice...BCC is, by definition, two interpenetrating simple cubic lattices. I believe the original question dealt with the primitive cubic lattice (one atom per unit cell), in which...
Yes, of course. I didn't properly account for the fact that inner products between spatial orbitals of the same set such as those in ##\{\psi_i^{\alpha}\}## or ##\{\psi_i^{\beta}\}## can be different, and most of the time are.
Thanks for the help! It exposed a few holes I had in notation and...
Here is my train of thought: If i=j, then we are referring to the same spin orbital. Because both sets of spatial orbitals are orthonormal w.r.t. their own elements, the inner product of any spin orbital with itself must be one. If i ##\neq## j, then we are referring to any two different spin...
Okay, I think I know what you're saying now. I was trying to start with the individual spatial orbitals and go to the spin orbitals. So then I think we can arrange these 16 possibilities into a matrix:
\begin{bmatrix}...
You're right, I'm confusing a few things. The math review for this book starts at basic vectors and gradually builds. They start with dot products in Euclidean space, but once they've introduced complex vector spaces they continue to call inner products "scalar products." This is confusing, and...
I do not mean these to be states, they're just basis vectors. When I write ##\big|\psi_{j}\rangle##, I take it to mean the scalar product of a basis vector, ##\langle j\big|##, and the wavefunction, ##\big|\psi\rangle##. This gives ##\langle j\big|\cdot \big|\psi\rangle = \big|\psi_{j}\rangle##...
Thanks! I think I have it. Does this look correct?
\int d\mathbf{r}\ \psi_{i}^{\alpha*}(\mathbf{r})\psi_{j}^{\beta}(\mathbf{r})\\
= \langle\psi^{\alpha}\|i\rangle\langle j\|\psi^{\beta}\rangle\\
= \sum_{i}\langle\psi^{\alpha}\|i\rangle\langle j \|i \rangle \langle i\|\psi^{\beta}\rangle\\
=...
Homework Statement
[/B]
Taken straight out of Szabo and Ostlund's "Quantum Chemistry" problem 2.1:
Given a set of K orthonormal spatial functions, \{\psi_{i}^{\alpha}(\mathbf{r})\}, and another set of K orthonormal functions, \{\psi_{i}^{\beta}(\mathbf{r})\}, such that the first set is not...