Can you be more specific about where you are having a hard time? Is it the change of coordinate system?
Also, when asking a question like this, it's quite reasonable to mention that you're setting h_p=h=h_s=0. Makes it easier to relate your equations to those in the paper.
Well, I'm not sure how sensitive DFT is to these things or how the specific material is supposed to behave, but in other simulations one often needs to pay attention to boundary conditions (you can check if you have an even-odd effect when changing cell size) and to finite size effects. For the...
The main issue with your first question is that there isn't really such a model. Or rather, one can only write something like H_{tot}=H_e + H_I + H_{int} where H_{tot} is the total Hamiltonian, H_{e} is the electron Hamiltonian which contains the kinetic energy of each electron and all...
Yeah, it has to do with stacking. There are several notations, of which I'm most used the ABC one myself. However, I think this is Ramsdell notation, in which H=hexagonal and the number two would signify the number of layers. Hence 2H should correspond to AB stacking, which agrees with the...
The average \langle S_j^z \rangle is certainly changed by the external magnetic field - consider the extreme case with an antiferromagnetic J and a strong field along some axis. This will change the mean field from zero in the case of no field to its maximal value in the case of a strong field...
I believe this tends to be called the "Ising model on a complete graph". See chapter 3 in these lecture notes. Since the atomic orbitals tend to be quite localized (few papers even consider next next nearest neighbor interactions) this case tends to be more popular in mathematics (judging by...
If you are asking about the technique rather than the detailed physical processes, note that photolithography works essentially the same way, but is usually much better explained. That wikipedia page illustrates it quite nicely with that figure. Electron beam lithography uses some other...
The surface states tend to have them though, see e.g. Joel Moore's article.
I am also not quite sure what the question is actually about. If the Fermi energy is at (or close to) the Dirac point, then the low-energy states will have a linear dispersion relation. This is a clear signal, and quite...
Actually, the author of those notes probably just switched to a complex field
E_V=\sqrt{\frac{\epsilon_0}{2}}E + i\frac{B}{\sqrt{2\mu_0}}
in which case the energy density comes out as
u=\int |E_V|^2 d^3 r = \int \left( \frac{\epsilon_0}{2}E^2 + \frac{B^2}{2\mu_0} \right) d^3 r
as it should.
Hi all,
This is all in the context of interaction between (two-level) atoms and an electromagnetic field, basically the Wigner-Weisskopf model. In particular, I tried to derive the value of the atom-field interaction constant and show that it satisfied...
By the way, I don't know why I dragged up the whole even function bit. That doesn't really help.
I see. Yeah, I'm not sure about how to do that. But didn't we use dE=\nabla E \cdot \mathbf{k} to remove one of the momentum dimensions already? That kind of suggests that it might be simplest to...
Ah, ok. Yeah, that Taylor series (unless you do the full infinite sum, in which case you wouldn't lose information) is mainly useful near the origin. But the other solutions are for k_x =\pm \pi/a, right? And |\nabla \epsilon| is an even function, isn't it?
Yeah, it doesn't make sense if you think of i,j as vectors. The upshot is that you don't have to - the indices are just supposed to label all sites. As long as your system has finite length, width, height (and so on, if you want to) - all you need is one number. It doesn't need to have anything...
For your and others' future reference, this review on the experimental status of the Luttinger liquids was posted on arXiv yesterday: http://arxiv.org/abs/1308.2731
The review is relatively short, but Giamarchi (the author) covers more tests than I was aware of anyway, and provides references...
Yes, there are no electronic differences between isotopes. The chemical difference is due to the increased mass, making reactions and transport slower. (This effect becomes negligible for heavier elements, by the way.) Practically speaking, isotopes above tritium have too short half-lives to be...
I haven't seen it outside that book or the article by Lamb they cite, but then again, it does seem like the kind of detail any book except for Bethe & Salpeter would hide under the rug. Except for the case of Hydrogen and possibly Helium, it really is a small effect.
Did you read the...
What would you consider a definite proof? The Luttinger liquid does a good job for e.g. carbon nanotubes and edge states of fractional quantum Hall systems, but it is a linearised theory which can only deal with low energy excitations. As it is a limit, there will clearly be cases when it does...
Is your question whether the Fermi liquid is a state with an order different from the Fermi gas, or whether you can have phase transitions in the Fermi liquid? If it is the latter, then yes, you can have instabilities that lead to magnetic ordering or pairing. If it is the first, I think the...
Hi,
Reading http://rmp.aps.org/abstract/RMP/v70/i4/p1039_1 I've run into a notational question of atomic p orbitals. The authors use symbols like p_\sigma and p_\pi. From their fig. 3 (attached), they do look like p_x and p_z orbitals, respectively, rather than anything close to σ or π bonds...
Where does it say that? That is quite misleading.
Basically, one can use the traditional hand waving argument: a travelling electron will perceive electric fields as magnetic fields (which the electron spin can couple to). In the simplest case, a lone Hydrogen atom, the field is entirely due...
See http://physics.nist.gov/PhysRefData/ASD/Html/lineshelp.html#OUTWAVELENGTH
The Ritz wavelengths are calculated from the difference in the energy levels involved in the transition, whereas the observed values are the experimental data. The accuracy of the Ritz wavelengths obviously depends...
The spin vector (if we allow ourselves to use the semi-classical picture, in which spin forms a vector) has a fixed lenght for any particle (that is, the electron spin 1/2 doesn't change), so it's not really a dimension in that (infinite) sense. You could add a number of "spin dimensions" equal...