If you studied chapter 9, then you know that free electron bands are parabolic and the weak potential introduces gaps at the BZ boundary. If these gaps are large, the electrons will remain in the same band in extended BZ. If the band gap are small, the electrons may end up in a higher band if...
Part of this we discussed in another thread. An electric field will drive the electrons through the band in k space. When they reach a zone boundary and there is a band gap, they get reflected to the opposite zone boundary. However, if there is no gap, the dynamics will be non-adiabatic and they...
Applying a spatially constant electric field to a charged particle will increase its momentum and k value.
Including another gauge, namely V=eU=eEx, makes the spatial dependence explicit.
Yes, the hamiltonian is periodic in k with period G. The question, how slow A(t) has to change with time is difficult and may differ for different states. It will usually hold if ## \langle i | (\partial H(t)/\partial t | j \rangle << | E_i(t)-E_j(t)| ## for all states i and j in question.
Well, the wavefunction of an electron in a periodic potential is of Bloch form: ##\psi_{kn}(x)=u_{kn}(x)\exp(ikx)##, where ##u_{kn}(x)## has the periodicity of the lattice.
The hamiltonian, which has the ##u_{kn}## as eigenfunctions is
##H=\frac{1}{2m}(p-ik)^2+U(x)##, i.e., ##k## appears as a...
Strangely enough, for at least 50 years now, no theoretical chemist believes any more in the geometry being dictated by hybridization. Yet teachers don't stop parroting this old lore.
I would not consider f0, f7 and f14 as exceptions. Binding energy increases and size of f orbitals decreases continuously with increasing nuclear charge. So evidently f0 and f14 are extremal. After f7, orbitals have to be occupied with two electrons, which costs more energy.
Also, while...
Wannier functions are a kind of localized functions which evidently can also be calculated in a LCAO approach. So I don't think this is an alternative method of calculation, but rather representation. Note that the full wavefunction in Hartree Fock is invariant under arbitrary unitary...
There are many factors influencing the precise electron configuration of the lanthanides and actinides, but there are some trends. The f-electrons are bad in screening each other from the nuclear charge, hence, the more f-electrons in an atom, the more tightly they are bound. On the other hand...
You are right! Of course it is possible to have the following two reactions
##\mathrm{OCl^- \rightarrow Cl^- + \frac{1}{2} O_2}## and
##\mathrm{H_2O_2 \rightarrow \frac{1}{2} O_2 + H_2O}##,
which somehow autocatalyse each other.
A possibility to distinguish the two possibilities is to use...
Let me add this:
The most classical approximation to a many electron problem is via a Hartree product wavefunction. To get exchange symmetry, you have to consider all permutations and end up with a Slater matrix. This is the most general wavefunction describing somehow non-correlated electrons...
d and f orbitals are orbital in inner shells (n-1 and n-2, respectively), hence they are more localized in the inner of the atoms. The precise size depends also on the filling of the sub-shells. electrons in d- and f- orbitals hardly shield each other from the nuclear charge. Hence, as the...