Just to update with the answer:
In the case of diamond, there's a two atom basis. In this atom basis, we can think of the orbitals as the bonding (filled) and anti bonding (empty) orbitals, which are the ones to give rise to their respective bands. In the case of hydrogen, there's just one atom...
Just found around that each sigma sp3 bond splits into a bonding and an anti bonding state, the former being full and viceversa. The bonding states correspond to the therefore valence bands and the anti bonding to the conductions band. I've got no problem with that, but it seems I have a...
Homework Statement
I am asked to discuss the band structure of diamond. I saw the band structure of diamond has 4 filled valence bands and then 4 conduction bands. Silicon, the same.
Homework Equations
---
The Attempt at a Solution
I'm feeling really silly because I don't understand why it is...
@Charles Link Let me ask it to the point: is ##\left(\frac{\partial u}{\partial u}\right)_T=0## (assuming ##u=cRT##)??
I think that could kind of make sense, though I had never considered it. Maybe knowing that I may avoid future problems (but I can't promise it!).
I know keeping T, v constants forbid me from changing the energy, but still I applied the definition correctly, right? The thing is, I know you are right: "the variation of pressure with respect to energy, keeping v and T constants" makes no sense because it is physically impossible. However, it...
The problem I seem to have is the following. Say I want the variation of P with respect to u keeping the v and T constants. Starting from the ideal gas law: $$\left(\frac{\partial P}{\partial u}\right)_{T,v}=0$$ since T and v re constants.
However, if I start from ##P=\frac{u}{cv}##, I get...
Hi all.
Suppose I have the ideal gas law $$P=\frac{RT}{v}$$If I'm asked about the partial derivative of P with respect to molar energy ##u##, I may think "derivative of P keeping other quantities (whatever those are) constant", so from the formula above I get $$\frac{\partial P}{\partial...
Say we have a Lagrange function with one multiplier a times a constrain. I minimize and solve the system to find a. I now add another constrain to the same system multiplied by the constant b. Is the value of a the same or can it change?
I don't understand why materials with low surface energy are hydrophobic and viceversa. All I can find are quick phenomenological explanations that don't quite deal with the physical (microscopic) process going on.
Could anyone provide a good microscopic picture of why it is that way? What's...
Yup, I meant number operator for a fermionic state (number of particles equals either 0 or 1). So ##\exp({\hat{n}})=\sum{\frac{\hat{n}^k}{k!}}=\sum{\frac{\hat{n}}{k!}}=\hat{n}\sum{\frac{1}{k!}}=\hat{n} e ##?
What the title says. Acting on a fermionic state with the number operator to a power is like acting with the fermionic operator itself. Does this allow us to define ## \hat{n}^k=\hat{n} ##? Or is there any picky mathematical reason not to do so?
Since there are no bonds at the other side of the surface, external layers of solids are usually closer to the next layer. This process is called relaxation. (Example in picture a here).
However, at a lecture I attended the other day it was mentioned that some surfaces present expansion...
If I have a Hamiltonian diagonal by blocks (H1 0; 0 H2), where H1 and H2 are square matrices, is the density matrix also diagonal by blocks in the same way?
Hi. I'm taking a look at some lectures by Charles Kane, and he uses this simple model of polyacetylene (1D chain of atoms with alternating bonds which give alternating hopping amplitudes) [view attached image].
There are two types of polyacetylene topologically inequivalent. They both give the...
Yes.
That's what I understood, but I was wondering if there was a simpler way, or at least a more polished formalism to state it. For many body systems that could easily get out of hand, right?
I don't understand what you mean by this, PeterDonis.
Sorry again for the notation. I'll take some time to learn the PF Latex notation.
PeterDonis, you were right: 3 energy states, plus spin degeneration, so 6 total states, and you got them right. And I proposed a combination of 2 states of 2 electrons each. However, I should have said that they...
Hi there. Excuse that I write it all in standard text but I don't know how to write it otherwise. I'll get right to the question and try to see if I understand the theory from the answer.
Suppose we have a system of two electrons, one spin up and one spin down in a system with 3 possible...
Hi.
I'll be doing a master's degree in nanophysics and working on electron transport in arrays of qubits.
I don't know anything (or barely) about the second quantization and would like a book which covers it, and on condensed matter overall.
So far I've been told about Bruus&Flensberg's...
Yes, actually just thought about that after posting it but forgot to clarify it in this thread. Thanks for stating it, though. So scalar fields are just functions of coordinates that are invariant under the set of transformations you consider?
My first option was to do it abroad as well, maybe the UK or Germany, too. I'm glad yours was great. I'm kind of assuming that coming from Greece, which is one of the PIGS just like Spain... the university was much better at Germany; am I right? I did an Erasmus year in Nottingham University and...
Hi. I'm just graduating and will be doing a Master's next year, but as far as I know (which isn't too far), there are always a lot of options for a PhD out there, specially if you have worked with someone who can recommend you. I'm Spanish, so perhaps it's a bit different somewhere else.
Either...
What I meant, @haushofer, is whether a scalar field would be any function on the position (event) coordinates that returns a scalar, so that when you transform the space on which it acts, you also transform the field.
Example: f(x,y)=x+y
Now y'=y; x'=2x; f'(x',y')=x'/2+y'
For clearance, what I...
Oh, I see. So scalar fields are those defined as invariant, not anything that returns a scalar value based on position. Would any function of coordinates (the current density wasn't) be then a scalar field?
In a semiconductor, the Fermi level sits in the bandgap, whereas in a metal, it does inside a band (this is, actually, what drives their behaviour). Even different orbitals (different bands) can be at the Fermi level for metals. This high density of states is what allows the "electron sea".
Hi. This question most probably shows my lack of understanding on the topic: why are scalar fields Lorentz invariant?
Imagine a field T(x) [x is a vector; I just don't know how to write it, sorry] that tells us the temperature in each point of a room. We make a rotation in the room and now...