Recent content by SpinFlop

This is covered very nicely in chapter 2.7 of James Annett's book "Superconductivity, Superfluids, and Condensates". Honestly, I recommend you buy the book and read the whole thing. Like all books in the Oxford Masters Series, it is little more than a slip of a book, but presents an immense...

I recommend you read my reply to a related question in the following post: https://www.physicsforums.com/threads/what-causes-magnetism-at-the-atomic-level.958654/#post-6085949 To quote the most relevant part: 1) Local moment picture/Heisenberg model, (Insulators): Electrons are almost...

For future reference, when performing a uniform operation on arrays inside of a cell, just write a function to perform your desired task and have it operate across the cell using cellfun. For your current problem it would look like this: cellfun(@(f)f(1,2), a)

To begin, your quote from an unnamed source speaks of the "spherical symmetry of the environment" which you then took upon yourself to recast as "spherical movement of an electron" (a statement that is problematic in and of itself, but one that I am not going to address.) If we focus on what the...

As general rule, it is not very reasonable to expect that atoms, when condensed into a material, will follow a simple Hund's filling. Here is an article on the 122 pnictides that includes both theory and experiment that demonstrate the important role of correlations in establishing the...

When atoms condense into a solid, then, as TeethWhitener pointed out, the primary modes are acoustic and optical phonons. These are vibrations that move as a wave through the material, ie: they are collective modes due to long range coordination of the atoms. In fact, they are guaranteed to...

They clearly have a typo in their definition of C since there are two opening parenthesis, but only one closing. My guess is that it should be ##2(n^2 - 1)(n^2 - s^2)##. Try that and see if it fixes the problem.

As I stated above, the paper I recommended works out the potential for you cubic system in the first half. The second half then works out the energy level scheme due to splitting the orbital degeneracy of electrons placed within this crystal field. So there is no simpler model, just disregard...

What you are describing is Crystal Field Theory. There is extensive literature on this and the associated energy splitting of electron orbital degeneracy within these crystalline electric fields, ie: CEF levels. In general, the number of levels is determined by the local point group symmetry of...

In my last post I said you needed to determine the interval ##dE## that matches ##dN=1##. If you are sitting at an energy interval in between two levels it should be abundantly clear that ##dN=0##. So for ##dE<(E_{n+1} - E_{n-1})##, would you agree with the following statement? $$ dN =...

If there is no degeneracy then clearly ##dN = 1##, but if you need more detail then here it is. Your density of states is give by: $$D(E_{n}) = \frac{dN}{dE}$$ where ##N(E_{n})## is the total number of states up to energy ##E_{n}##: $$N(E_{n}) = \sum_{k=0}^n g_{k}$$ As you stated there is no...

I assume you have also determined the degeneracy of the energy levels. With that in hand it you should be ready to determine the DOS by the standard prescription.

Kittel "Intro to Solid State" has a simple exercise to walk you through demonstrating this yourself. It is problem 4 at the end of chapter 3. Assuming you have a copy, then you can work this out yourself. I will abbreviate this exercise even further to provide an immediate answer and a bit of...

Attempt at Problem 4 Part a) follows directly from the fundamental property of self adjoint operators. $$ (\hat T{\mathbf a})\cdot{\mathbf b} = {\mathbf a}\cdot(\hat T{\mathbf b})$$ where ##\hat T## is a self-adjoint linear operator and ##{\mathbf a},{\mathbf b}## are two vectors in a given...

The answer as to why we do a Taylor expansion at all is because physics is hard. Indeed, for the Lennard Jones potential we use to describe lattice dynamics exact solutions are altogether intractable. So we do what we typically do with potentials in such cases, we Taylor expand around the...