@DIRAC1930 It's a big topic. Yes, people still study projective geometry, both pure & applied. I suppose there are more students of the applied side, since it's pretty important in 3D graphics. Once again, Ted Shifrin knows a lot of projective geometry.
Finite projective geometry is also an interesting topic, if you're into combinatorics.
In finite geometry, the Fano plane (named after Gino Fano) is a finite projective plane with the smallest possible number of points and lines: 7 points and 7 lines, with 3 points on every line and 3 lines through every point. These points and lines cannot exist with this pattern of incidences in Euclidean geometry, but they can be given coordinates using the finite field with two elements. The standard notation for this plane, as a member of a family of projective spaces, is PG(2, 2). Here, PG stands for "projective geometry", the first parameter is the geometric dimension (it is a plane, o...
I have read that translational symmetry implies conservation of momentum. And as much as I have understood that is this way that if we have two identical system in a co-ordinate system with their position vectors r1 and r2, as the physical laws must be the same everywhere in space because space itself is same everywhere
There must be something that is in both systems the same or unchanged or conserved. And it's momentum in this case just saying d²(mr)/dt²=0; but why can't the momentum change at a same rate over time in both cases? I mean that would be something same in both systems. So what's the problem?
Does Nother's theorem tries to say that there should be a conserved quantity that depends only on translation and has nothing to do with time and thus momentum is that only conserved quantity that is independent of time and depends on translational symmetry?
@SignorFeynman in that circuit, the grounding just means that electricity can be flow that way, bridging the loop; in complicated circuits, having that makes for us being able to omit useless lines
@PM2Ring miao miao no understand, but no need understand to star~
> It is the 4-dimensional analogue of the icosahedron, since it has five tetrahedra meeting at every edge, just as the icosahedron has five triangles meeting at every vertex.
> Caveat: Everything that follows assumes that the Schrödinger equation for the hydrogen atom is exactly correct, which of course it isn’t, since it neglects the spin of the electron and all relativistic and quantum field-theoretical corrections that, in reality, give an additional fine structure to the electron wave functions.
@PM2Ring I read that, though not the wiki for 600cell. Like, I know when it is really going over myow head. Pretty pics though
@PM2Ring This restriction was very very silly. He could easily have chosen to use the Dirac spherical harmonics. Alas, the basic version is likely way easier to work with
@DIRAC1930 The "brackets to commutators" map does not actually work consistently for all classical observables (most prominently this manifests in ordering ambiguities when you try do canonical quantization), this is called the Groenewold-van Howe no go theorem, see e.g. this answer of mine
@DebanjanBiswas Noether's theorem just says that to every continuous symmetry there is a conserved quantity. Without going to the Hamiltonian formalism there is no way to give this conserved quantity a direct physical interpretation with relation to the symmetry - but in the Hamiltonian formalism this quantity is the generator of the continuous symmetry in a well-defined sense.
There is no good way to see this without actually doing the math because it cannot be just some hand-waving or "intuitive" relationship between a symmetry and conservation, because for discrete symmetries there are no such conserved quantities, so at some point your argument has to engage with the formal difference between discrete and continuous symmetries.
@SillyGoose Careful, what do you think "all $k$ are?
The first BZ contains all sensible momenta; the other zones are just "shifted" (by a reciprocal lattice vector) BZ, which thus contains the same momenta
but they are FTing a function that is actually only defined on a lattice of $N$ points
i mean the fourier transform of a Bloch function should not just be fourier transforming it in the 1BZ. it should literally be FTing it in the entire reciprocal lattice, then perhaps this resulting quantity can be written in terms of just the 1BZ
i mean if we are using pontrygin duality, then for a lattice like $\mathbb{Z}^3$ we have by definition that the fourier transform is $\hat{f}(\chi) = \sum_{g \in \mathbb{Z}^3} f(g)\overline{\chi(g)}$
>Since $\hat{f}(\boldsymbol{k})=\hat{f}(\boldsymbol{k}+\boldsymbol{G})$, for any vector $\boldsymbol{G}$ of the reciprocal lattice, it is sufficient to consider one vector $\boldsymbol{k}$ in each set of equivalent vectors
But i don't see why that makes sense as a definition
i mean we should be able to use the actual fourier transform and then deduce what happens for functions defined only on lattice sites
to my eye it should be $f(\vec{R}) = \sum_{\text{all } \vec{k}} \hat{f}(\vec{k})e^{i\vec{k}\cdot \vec{R}} = \sum_{1\text{BZ}} g_\vec{k} \hat{f}(\vec{k})e^{i\vec{k}\cdot \vec{R}}$
where $g_\vec{k}$ counts the cardinality of each equivalence class of allowed crystal momentum
@SillyGoose From someone with no real condensed matter knowledge I'm a bit confused what the point of contention is here: This is merely a definition of something called "Wannier state". There is no claim that these are Fourier transforms (and as the Italian Feynman says, this is closer to a DFT). What's the problem?
@SillyGoose Well, to spare you a fair share of headaches, I can tell you that in the CM literature (and in the physics literature), there is the tendency to call Fourier transform about anything that maps to the reciprocal space
Be it a series, a discrete FT
After all, as you already mentioned, it's always Pontryagin duality. What really changes is the domain of $R$ and $k$
the first bit is just writing a periodic function in terms of its legitimate fourier series (i think).
then seemingly, we are just using fourier's trick (or whatever its called) to extract the fourier coefficients. we only need to sum (or integrate) over the first brillioun zone because it's redundant to do otherwise with respect to extracting the coefficients. this is my understanding now...
but this suggests to me that the reason for defining the wannier functions using sums over the 1BZ is to avoid divergent integrals in the infinite volume limit
because we very well could in the finite case sum over all $k$ and just normalize by a factor of however many unit cells there are in the finite volume.
the handwriting is fine, but why are all the math symbol shifted upwards? The lower parts of things like $f$ and $\psi$ are supposed to go below the base line!
er wait i mean for a finite lattice it is not a group
i guess for an infinite lattice it is some $\mathbb{Z}^d$
but for a finite lattice with periodic boundary conditions...then we can get back to a $\mathbb{Z} / n\mathbb{Z} \times \mathbb{Z} / n_1 \mathbb{Z} \times ...$
you don't really need the "periodic boundary conditions", the group structure doesn't really play much of a role
If you have a set of $n$ points, then you can talk about the functions on it (sets of $n$ numbers) and you can do DFT (the Pontryagin-Fourier transform for $\mathbb{Z}/n\mathbb{Z}$) on it to get a different bunch of $n$ numbers
The group structure mostly just gives the Haar measure in this context, but for n points the Haar measure is the same obvious counting measure you can also get a million other ways
when I mean is that if you put down the abstract lens for a moment all the DFT is doing is send n numbers to n other numbers
of course you need to prove that this doesn't lose information but you only need properties of the complex exponential (or the sines/cosines) for that, you can do that without abstract group theory
it's neat that there is a general theory of Pontryagin duality that fits both the DFT and the usual FT and Fourier series and some more exotic transforms into the same general theory but really you don't gain anything from that description over just writing down the formulae in my opinion
especially not if it leads to you wondering "how is the lattice a group" while everyone else just does the computation and moves on - it's not less rigorous just because it's not phrased in the most abstract way in this case :P
well i like the proof of orthogonality of characters because it means ive already proved all of the orthogonality relations without having to do any of the concrete computations (if i can identify the underlyign abelian group) :)
also, so (1) a bravais lattice translational symmetric hamiltonian conserved crystal momentum modulo reciprocal lattice vectors. (2) does this mean that all such hamiltonians look like $a_{k}^\dagger a_k + a_{k+q}^\dagger a_{k'-q}^\dagger a_{k}a_{k'} + ...$?
I mean, I'm writing a Hamiltonian and there would be several parameters or operators with quite a standard notation, I don't want to make an useless wall of text
@User198 But if you are fishing for the equivalent of expressing the evolution of a state in terms of the Poisson bracket with that second question, you just need to turn the state into an operator - its density matrix
@User198 For a direct comparison to the classical phase space formulation you can then go to the Wigner function, where the evolution equation turns into not the Poisson bracket, but the Moyal bracket.
in any case, you might want to search for "symmetric tridiagonal matrix constant off-diagonal elements" or so
...but perhaps one can conclude some things about the eigenvectors of the Hamiltonian without diagonalizing it. For example, have you tried to see if the Hamiltonian commutes with a suitable parity operator?
@Allie Oh Goldstein is a wonderful book,I personally went through his first five chapters(essentially lagrangian formulation and rigid body dynamics) and have used landau for Hamiltonian formulation since I was tight on time lol
@Allie Also check out sunil golwala's notes on classical mechanics, they are pretty neat detailed
@ACuriousMind And than the Moyal equation is ${\displaystyle {\frac {\partial W}{\partial t}}=-\{\{W,H\}\}=-\{W,H\}+O(\hbar ^{2})}$. And as $\hbar \rightarrow 0$ I should get the Hamiltonian flow: ${\displaystyle {\frac {df}{d t}}=\{{f,H\}}}$
Is the minus sign in the Moyal bracket expansion correct?
@User198 The minus sign is because states evolve with a minus sign compared to observables, that's also the case in the classical Liouville equation for a classical phase space density.
On that Liouville equation Wiki page, there is the equation ${\displaystyle {\frac {\partial \rho }{\partial t}}=\{\,H,\rho \,\}}$.
But the general thing in Ham. mechanics is: ${\displaystyle {\frac {df }{d t}}=\{\,f,H \,\}}$
I am confused now, why is it for $\rho$ the Poisson bracket is in the "wrong" order. Is that because of that what you said: "states evolve with a minus sign compared to observables" also? That is that?
@User198 Yes; that's Liouville's theorem (at least one reading of it). The equation you cite is for classical observables; the phase space density is not an observable in this sense.
In quantum mechanics it's easy to see by switching between the Schrödinger and Heisenberg pictures: A time-evolving expectation value of an observable $A$ for a state $\rho$ is $\mathrm{Tr}(UAU^\dagger \rho)$. You can say $A$ evolves as $A(t) = UAU^\dagger$ or that $\rho$ evolves as $\rho(t) = U^\dagger \rho U$ (cyclicity of the trace). The different order of the $U$ and $U^\dagger$ correspond to the different sign in the infinitesimal equation
(the order might be exactly the other way around, I'm not checking signs :P)
they're calling it "ground state" because when you go to the quantum version this will turn out to be the VEV, i.e the expectation value in the ground state
@imbAF From the screenshots it's impossible to tell whether it's a book, an article, lecture notes, whatever, so I said "resource" because it covers all of them. If that's the level of nitpicking you apply to the texts you read you'll never get through them. Try to be a bit more open to people not using exactly the phrasing you'd expect them to use.
And the lecturer did not bother to emphasize the fact that we consider CFT. Which led me to believe that it is not so important. But obviously it is since in the book he based his notes on, clearly puts an emphasis on that
Oh, you're right. I stopped at the first equation. The decomposition uses the same notation as the usual one of Higgs QED :P
@ACuriousMind at least this is a genuine symmetry breaking :P
I found another "version" yesterday and there was a comment of yours beneath, by the way. That answer said that breaking the global part of the gauge symmetry (about which we talked last time in the chat, the center of the gauge group) amounts to breaking the gauge symmetry, so it shouldn't be considered
And your comment said what you told me last time about not considering the center
@imbAF Oh sorry, I thought I had deleted that message. I meant that using GF (which are VEV of time ordered products of fields) you can find the observables
@imbAF but what is the difference between the value of a scalar and a non-scalar field under a Lorentz transformation? You don't need to know the specific transformation of the non-scalar field.
@imbAF Alright. So given that the vacuum is invariant under LTs, you should be able to derive that any non-zero VEV for a non-scalar field/operator violates the assumption that the vacuum is invariant.
@imbAF 1. Assume that $\langle 0\vert O\vert 0\rangle\neq 0$ for any non-Lorentz invariant operator $O$. 2. Derive a contradiction with $U(\Lambda) \lvert 0\rangle = \lvert 0\rangle$, i.e. Lorentz invariance of the vacuum.
I did a lab where it became apparent my lab partner and I were the first people in years to actually do it instead of copying old records :P
at some point the software steering the setup had been updated so the instructions in the manual didn't work anymore - but that had been like 4 years ago and we were the first people to complain that the examples in the manual didn't do what they said they did
@ACuriousMind could you help me with a commentary, regarding goldstone bosons. This is the topic. The image, uploaded above is: i.sstatic.net/19dGXjd3.jpg
@TobiasFünke the labs were fun - I also had one where the supervisor looked at his watch, shouted "oh no, my train!" and ran out of the room, leaving us to complete the experiment on our own (and the main supervisor didn't believe us at first when we told her he just ran out of the room)
I find so frustrating to think about people faking the data to get the right results in the labs. Why do they freak out? I was almost happy to have an opportunity to write what had gone wrong
the most horrible one was probably the one where we had to measure the rotational speed of a top by matching the frequency of a stroboscope to the rotation frequency - I was extremely hungover that day and staring at a stroboscope isn't really what you want to do in that state :P
And thanks to my theoretical tendencies, it happened many times. I mean, I fucked up the data of my final lab thesis because I didn't set a damn resistance right :P
@imbAF If you don't understand why non-scalar VEVs would break Lorentz invariance of the vacuum, I'm afraid there is little chance you will understand anything of the rest of symmetry breaking. The proof is literally a one-liner, go back and think about this some more.
@SignorFeynman I mean, consistency in what I think about physics is one thing, but if I were inconsistent with stories about my life it would be rather more concerniing :P
1.) Can I think of the Liouville theorem in classical mechanics as the conservation of information? Since the area (volume) is constant, no info is lost during evolution.
2.) If when doing QM, I replace the Poisson with Moyal brackets, my "area" can now shrink. So I lost some information?
But how can that be. Isn't one of the halmarks of QM that information is preserved. But it seems that it is preserved only in classical mechancis {Poisson brackets}, and not in QM {{Moyal brackets}}.
In the 1.) I was thinking of how, when doing an experiment. I can not know the initial conditions perfectly, I can know it at best within some range in the phase space dq dp. So whatever that area was, it will always stay the same size.
So I would define information as "the smallest possible area dqdp in the phase space". Or sth like that, I am not sure...
I was kind of bad, especially in more advanced lab courses. I don't know why but I really have problems in dealing with circuits lol. Luckily I had a very bright and friendly lab mate
the crowning task of that one was building a basic radio, it was pretty awesome to essentially just plug an LC circuit into a loudspeaker and hear some random radio station
the radio really got me - I was like "no way just plugging these things together just makes a radio" and then I turned it on and it indeed was a very basic radio
extremely good material for teaching kids that most of the world is, in fact, explainable
> the company’s warning was couched not in terms of health risk but rather as bad scientific practice: Removing the ore from its jar would raise the background radiation, thereby invalidating your experimental results.
@TobiasFünke Yes, the booklet that accompanied it was always insistent on me trying out what happens if I replace that one resistor with something else etc.; again, I don't know how the current versions are, but I would recommend the version I had without any reservations, it was pop science done right
or how many would love literature if they would be exposed to books in a better way... in Germany the degree of education unfortunately is extremely influenced by the education of the parents...
but if I recall correctly there is some positive trend, at least
@TobiasFünke there is so much we could do if we actually took education as a value in itself seriously and not just conceiving of school as a place where parents get to off-load their children or as some bureaucratic hindrance to getting well-paid jobs