1) I know the Slater determinant. It is a many-body wave function constructed upon single-particle wave function so that the Pauli exclusion principle (or the exchange interaction) is taken care (using the property of exchanging row/column of a determinant).
2) and 3) I think, we get...
I think, hydrogen atom and one electron atom are equivalent. Now once you put one extra electron, trouble enters. Now the Schroedinger equation has two coordinates: r_1 and r_2, and due to Coulomb repulsion they are coupled, i.e. we have a term like e^2/|r_{1}-r_{2}|. So unless we get rid of...
Now the discussion has become pretty interesting.
If an approximation works, then there should be arguments for it. If we lack reasoning, then it's just a fitting formula.
I find no reasoning to extend the same hydrogen atomic orbitals (electronic probability densities) and corresponding...
I think, s,p,d orbitals were defined only for the hydrogen or hydrogen-like (alkali) atoms.
How do you determine probability densities for other atoms?
I guess, the wiki link tells about shortcoming of the Bohr's model, not about the BS model.
How can you be sure there are no other...
See, the Bohr's model as an ansatz gives correct values for the hydrogen atom spectra (apart from the fine-structure correction). And solving Schroedinger equation for the H-atom reproduces Bohr's formula. In that sense, we have a microscopic theory that supports Bohr's formula for the H-atom...
@DiracRules
Sorry. A severe objection. We can think of subshells. But they don't need to have the same geometry of the hydrogen atom s,p,d orbitals.
Moreover, transition doesn't show the nature of the orbitals. It just reflects the difference between two energy levels.
So it seems...
As far as I know, the quantum chemistry methods use Hartree-Fock or post-Hartree-Fock methods that use linear combinations of atomic orbitals and again those orbitals are treated like s, p, d kind of orbitals, which are hard to accept.
Hope you got my point. Again my question is very simple...
Actually it's the first time I hear about cuprate.
However, the fact is that the atomic configuration of an atom is not fixed, but it can change if a new configuration with lower energy is available.
It's not only about cuprates.We always find the oxidation state and then decide the...
I think, if you can answer my last question, it'll be clear.
How can one say Cu has d9 electronic state in cuprates? Or does it need to have d-state at all?
Let me put a few more questions?
1) Are there experiments that can see real electronic orbitals of H-atom? Note that the orbital...
This is a very naive question. But I think, it's an important point that has been unattempted in textbooks. The question is:
How far should one trust the Bohr-Sommerfeld model or the atomic shell theory for all elements in the periodic table?
This question generally comes in mind, since...
@DrDu
Sorry, the latex didn't work! I guess, you could follow the notation.
And I think, the Huckel model and the tight-binding model are equivalent. Hopping from one site to other site is similar to hopping from one orbital to another orbital.
Sorry. Still not clear to me. What if I write
$t_{ij}$ instead of $-t_{ij}$ in the kinetic part of the Hubbard Hamiltonian?
Does it make any difference? If yes, then what difference?
That's true. But there must be a meaning for keeping this minus sign. I have seen Hubbard's original paper. Surprisingly that paper does not have this minus sign!
If the sign is just a matter of convention, then why unnecessarily an extra minus was chosen as a convention?
Thanks.
Might be a naive question.
In many places, specially in the Hubbard model, people use a minus sign before the tight-binding hopping amplitude. What does the negative sign signify for? Any special intention?
Thanks.
I had been reading several articles on topological insulators (TI) including the Kane and Hasan's 2010 RMP. I am not very much clear about the Z_2 invariant TI. I mean, the even-odd argument proposed by Kane and Male (also argued by S. C. Zhang's group and Joel Moore's group in a different way)...