Recent content by mcas

\omega= \frac{2\pi c}{\lambda}=\frac{2 \cdot \pi \cdot3\cdot 10^8}{500 \cdot 10^{-9}}\approx 3.77\cdot 10^{15} Hz Which is still less than \omega_p so R=1.

The Drude model was the only one we've discussed during the lecture, so I would expect it is used in the solution. Thank you for clarifying this! I'll need to ask my tutor whether we should use something different that wasn't covered yet.

Note: for some reason frequency on this lecture is indicated by \omega. I wanted to calculate the reflectance using one of these equations that were given to us during the lecture: R=1 where \omega < \omega_p...

It's a problem formulated by my professor. After you post I contacted him and indeed, he made a mistake in the homework statement, as the last operator \hat{c}_{-k \downarrow} should be conjugated for the expected value to be one. In the original statement, the value will be 0. Thank you!

\langle \phi_0| \hat{c}_{-k \downarrow} \hat{c}_{k \uparrow}\hat{c}^\dagger_{k \uparrow}\hat{c}_{-k \downarrow}|\phi_0\rangle = \\ \langle \phi_0| - \hat{c}_{k \uparrow} \hat{c}_{-k \downarrow} \hat{c}^\dagger_{k \uparrow}\hat{c}_{-k \downarrow}|\phi_0\rangle = \\ \langle \phi_0| \hat{c}_{k...

We have a one dimensional lattice with a lattice constant equal to a (I'm omitting vector notation because we are in 1D). The reciprocal lattice vector is k_n=n\frac{2 \pi}{a}. So to get the nearest neighbour approximation I need to sum over k = -\frac{2 \pi}{a}, 0, \frac{2 \pi}{a}. If I...

That's it! Thank you so much once again.

Ok, I plugged in the numbers and I might still need a little help. Using the result from the answers in c), I came to a conclusion that the reduced mass (I'm going to denote it by \mu must be equal to \mu = 9.2 \cdot 10^{-25} \ kg because F= kx = k\cdot 1 = \mu \omega_0^2 so we get 23 = 25...

The last one got me confused because the lecturer just waved their hands saying "well you know it's the atoms". But that explains a lot. Thank you! Really, I needed some guidance 😅 Have a nice day!

I got these values from the attached figure: n_{st}=2.4 and n_{\infty}=1.3. Ok, now the explaining. I define a complex dielectric function (permittivity) as \widetilde{\varepsilon} = \varepsilon_1 + i \varepsilon_2 So \varepsilon_1 is the real part of the dielectric function and \varepsilon_2...

a) I managed to obtain some results that are roughly around what is given in the answers. Because \varepsilon_{st} and \varepsilon_{\infty} are values of \varepsilon_{1}, I used this approximation: n\approx \frac{1}{\sqrt{2}} (\varepsilon_{1}+\sqrt{\varepsilon_{1}^2})^{1/2} -> \varepsilon_{1} =...

The limit itself is pretty easy to calculate ##lim_{T->0} \ lim_{\mu->\epsilon_F} \ (e^{\frac{(\epsilon_F - \mu)}{kT}}+1)^{-1} = \frac{1}{2}## But I'm very confused about changing ##\epsilon_\vec{k}## to ##\epsilon_F##. Why do we do this?

I have a problem with finding the energy of an electron in an FCC lattice using the weak potential method. We did that for a one-dimensional lattice during class, and I know that there was a double degeneration at the boundaries of the first Brillouin Zone. However, I'm not sure what...

Thank you! Now I just need to express ##\frac {\partial n_i} {\partial n_j}## as ##\delta_{ij}## and it's proven.

Thank you! That looks much better now. Now I get $$\frac{\partial E}{\partial n_j}=\sum_{i=1}^{\alpha} [\frac{\partial E_j^{(p)}}{\partial n_i}n_i ]$$ And I'm still confused as to what ##\frac{\partial E}{\partial n_j}## is and how to deal with this term.