Recent content by voila

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...

Then how is it that ##\left(\frac{\partial P}{\partial u}\right)_{T,v}## can give a different result depending on how I pick 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?

@hilbert2 Say you have a metal surface. How are the bondings between the metal and the liquid?

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...

@Demystifier yes I did, @stevendaryl pointed that out. Thanks everyone for the answers, I understood it.

How come?

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?