@naturallyInconsistent but what does this even mean tho? so i have my conduction electrons in my metal and each electron is just existing throughout the entire lattice??
@Relativisticcucumber Okay. Those correspond to free particles. Free=momentum (operator) is conserved=space translation invariance
Now, the case of the lattice is a little different. An electron (let's consider only one) in a periodic lattice will be in a less symmetric situation than a free electron
But nonetheless symmetric
As we have discussed other times, what I'm talking about is the content of Bloch's theorem. The content of the theorem is intuitively explained by comparing to the free particle case
@SillyGoose I think in (intro) books indeed most often the free electron case is discussed, but some also discuss the case for an electron in a solid (e.g. in the semi-classical framework)
$$\text{Electron in a lattice}\to\text{Bloch waves}$$
And the idea is (Bloch's theorem) that, just like the setting is a less symmetric (no continuous translation invariance) version of the free particle case, the Bloch waves are a modified version of the plane waves, as you know plane waves modulated by a periodic function
@TobiasFünke you are correct. The only problem I have is that in a lecture we take things which are necessary and related to the exercise sheet, mandatory to be solved, otherwise I cannot enter the exam. So, I will have to read the lecture notes, and then the notes in the book, which correspond to the topic of discussion
in class
And if I were to look for it in the book, I would just jump from x chapter to y, to z, leaving all the theory in between
@qwerty Lol, he's not the kind of person. He could risk making a typo if he also had to distract to drive
@ACuriousMind Not gonna lie, using the information I had on your workplace, I thought you were walking down the hallway of your house at the time of writing
Yeah, I got confused because there is really no distinction in Italian. You have specific words for various types of residence, but I just call everything "casa"
the distinction is very strong in German - if I told someone we were going to "my house" and then we end up at this apartment building instead of a house in which only I (and family or roommates) live, they would be very confused
Incidentally, I'm not very precise with daily language. For most of my life I have called hoods "hats" (of course in Italian) because the words are similar
I heard that so many times already... and I even think I shouted it --without even owning a car :d
@imbAF Let me finish our discussion by recommending Nolting's many-body book, first chapter, if you want to learn second quantization (including things like Fock space, the symmetrization postulate and so on). It is available in English and German. And as I've said some hours ago, I think it is a quite good book; especially the first chapter
As a matter of fact, not so long ago I had the exact same situation with an Italian friend of mine. It was only raining a little bit ("nieseln" we say), but he was making drama lol
@TobiasFünke I'll reply here. Probably I didn't really understand the ISW back then, as I'd just been introduced to QM. Basically I was led by the idea that you could just do an infinite energy shift, making the problem equivalent, but it wasn't very evident to me back then that in reality all thosd infinities are just a neat way to state what the support of the wavefunction is
@TobiasFünke well, the whole fact that people learn about QM in infinite dimensional settings where most of the things they do are just formal and require complicated functional analysis to back it up isn't good either
As ACM says, you could learn QM without infinite dimensional spaces, e.g. with spins
@Mr.Feynman well... but you need infinite-dimensional spaces in QM. One must just be aware of possible pitfalls, "paradoxes" etc., which almost always arise when one tries to carry concepts from finite-dimensional to infinite-dimensional spaces and alike...
Well yeah, of course the basic axioms of QM work on finite-dimensional spaces. I agree with that, and it also makes sense to start with this setting, I guess. But you cannot avoid infinite-dimensional spaces at all
Damn, I had to read an article about etymology because of a caps letter
@TobiasFünke of course you do need infinite-dimensional spaces, both for physical relevance and also for motivating the introduction of QM, given that only infinite l-dimensional spaces are the result of the quantization of a classical theory. Finite dimensional ones are useful to understand the machinery, though
We consider a 1D chain of N particles. We use the variational principle to find the state of lowest energy. The trial state considered is a Bloch state. Now we consider boundary conditions such that the wave function of the electron in the N-1 position is equal to that of the 0th. I want to show 2a)
So I consider $T_{R_1}c_{\vec k,j}^\dagger=e^{ika}c_{\vec k,j}^\dagger$. This is the end of my calculation. I am giving the result only
Now, would it be accurate to say that because of the periodic boundary conditions
$k=\frac{2\pi n}{Na}$ where n=1,2,3...N-1
?
If so, how would things look like if I were to not consider the fact that $T_R\phi_0=\pji_{N-1}$ ?
First of all $T_{R_1}\psi(\vec x - \vec R_j)=\phi(\vec x - R_{j-1})$
So essentially as you use the translation operator once, you consider the wave function of the electron in the j+1 position, but because x doesn't change while j-->j+1
the potential felt by the electron is the same as the potential it feels from the j-1 atom
would you agree with this?
$T_{R_1}|\psi_j\rangle=|\psi_{j-1}\rangle$
$T_{R_1}c_{j,\sigma}^\dagger|0\rangle=c_{j-1,\sigma}^\dagger|0\rangle$ From this last eq. one can tell that:
Cross checking? Well, if you get that it is indeed an eigenstate of $T_R$ plus if it is really an eigenstate of $H$, as the next exercise suggests...then chances are good you did the correct thing
i have constructed an argument that seems wrong $tr([A,B]) = tr(AB-BA) = tr(AB)-tr(BA) = tr(AB) - tr(AB) = 0$ via linearity and cyclicity of trace. what is going awry here?
@naturallyInconsistent im more of a functionalist myself
