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02:40
@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...
^ PG(3,2). Raytraced using POV-Ray
 
1 hour later…
03:57
the poisson bracket is giving commutator vibes
 
1 hour later…
05:12
Hi
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?
 
1 hour later…
06:15
morning
@Allie yes! a "naive" idea of quantization is to replace the Poisson bracket by commutators (with some factors)
 
2 hours later…
08:01
@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~
08:17
@naturallyInconsistent thanks
 
1 hour later…
09:28
Hi, can you kindly tell me whether if I got it completely wrong and help me understand Noether's theorem ?
@naturallyInconsistent I don't fully understand it either. But a 600-cell is en.wikipedia.org/wiki/600-cell
> 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.
09:50
Ah. He has that anim as a GIF on his site. Here's the simpler one for n=13
 
1 hour later…
10:58
@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
@TobiasFünke Why naive?
11:36
@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
Oh yeah I forgot about that lol even though it literally came up for me like a week ago
@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.
 
2 hours later…
13:51
are these fourier transforms or no?
@SillyGoose Discrete FT
but i don't get why we do not sum over all $\vec{k}$
other resources actually do seem to sum over all $\vec{k}$
i mean if you are going to discrete fourier transform on a lattice you should sum over all reciprocal lattice vectors
@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
if it is the usual notion of fourier transform you should sum over all values of the reciprocal lattice
this is the case in discrete fourier transform $\hat{f}(\chi_g)=\langle f, \chi_g \rangle$ where $\chi_g$ is the analogue of $\vec{k}$ here
You should sum over all $k=2\pi n/N, n\in\mathbb{Z}$
But this is the discretization coming from the PBC, which is "internal" to the 1BZ
14:01
@SignorFeynman but just because some vectors are equivalent does not mean they give nonzero contribution to the fourier transform
You should check the definition of the inverse discrete FT here
As you can see in eq. $(2)$ they sum over $k=0... N-1$
This is a sum over the first BZ in the physical context
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
$N$ points--->$k=\frac{2\pi n}{N}$ by PBC----> You have $N$ points in the 1BZ
yes but not N points in the whole crystal is my point
also this is solyom vol.1 which does not constrain $\vec{k}$ as you wrote
This looks more like a Fourier series than a discrete FT
Let me check Solyom
14:16
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 $\hat{f}(\chi) = \sum_{g \in \mathbb{Z}^3} f(g)\overline{\chi(g)}$
@SillyGoose Check appendix C.1.2, that's what I am talking about
where $\chi$ is a character of $\mathbb{Z}^3$ and clearly the sum is over all values of $\mathbb{Z}^3$
See what I mean? Right after eq. $(C.1.38)$
>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
@MoreAnonymous hey.
14:29
@SignorFeynman i still don't get why this means we can completely ignore them
if I have a sum $\sum_k f(k) e^{ikR}$, then this is like $f(k_1) e^{ik_1 R} + ...$
if there exists $k_r = k_1 + G$ where $G$ is a reciprocal lattice vector, then to this sum I will get the term $2f(k_1) e^{ik_1 R}$
What Solyom is saying is not that you ignore them; he means that the original sum is already constrained
but he is only summing over 1BZ $\vec{k}$ vectors in (C.1.37)
Yes, that's what I mean: it's the definition of the discrete FT. You sum over the 1BZ from the beginning
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
14:56
@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?
a few sources describe it as a fourier transform
Well, that probably depends on what the value range you allow for the $\vec R$ is
if $\vec R$ is drawn from the position space version of the Brillouin zone, it's a genuine (discrete) FT
if $\vec R$ is drawn from the entire unbounded position space lattice, it's not
15:31
@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 $k$
15:48
Is a question about (the basics of) qubits better for PSE or QCSE?
@ACuriousMind what would it be called if $\vec{R}$ is drawn from the entire (finite) position space lattice?
nothing in particular :P
this is how it is described in ashmerm.
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...
I can't decide what is stronger between my appreciation for your handwriting and my inability to read it :P
15:58
I'm one of those people who can't read other people's handwritings
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!
then it possibly obstructs high characters in the next line :P
That's why I was taught to just write in every second line when writing on a math grid like that
I just write wherever without respecting lines
GR
16:06
@SignorFeynman depends on the question, lol. but here should be OK, I suppose
if the question is about "computing", then it should be fine on QC too
hm wait actually a bravais lattice is not really a group right
so how can pontrygin duality be applied here
@SillyGoose What do you mean it's not a group? The sum of two lattice vectors is a lattice vector.
Why subset?
16:19
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
oh the proof of orthogonality of characters i followed hinged on the group structure (and abelianness)
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
@ACuriousMind indeed. it is essentially just a choice of a suitable orthonormal basis on a suitable Hilbert space
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
16:33
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 $a_{k}^\dagger a_k + a_{k+q}^\dagger a_{k'-q}^\dagger a_{k}a_{k'} + ...$?
Do you think that it's wrong to avoid adding details that someone who can answer the question will already know?
where those c/a operators are periodic in the reciprocal lattice, so we don't have to worry about the modulo reciprocal lattice vectors
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
@SignorFeynman I think it's always best to define all your notation
even if you just say "those are the usual XYZ operators"
I will do that, but in the limits of reason. Yeah, like that ^
16:39
yes, perhaps linking e.g. to Wikipedia or so
posts should be useful for a broader audience, and notations/conventions might and usually do differ in the same field
17:01
I have 2 questions:

1.) Can the Poisson brackets {f,H} be equated to the Lie derivative of f when f is flowing along the Hamiltonian vector field X_H ?

2.) What is the commutator [psi, H] where psi is the wave function and H is the hamiltonian operator. Is such a thing defined?
@User198 1. Yes, $L_{X_H} f = \{f,H\}$. 2. No, that makes no sense, since $\psi$ is not an operator.
I have done the deed
Wow, 2 fast upvotes :P
17:20
@ACuriousMind Thank you.
@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
Or put $\psi( \hat{x})$ in lol
Can the Schrodinger equation be written using Lie derivatives?
@SignorFeynman I might be mistaken, but the Hamiltonian resembles the form of the Wannier Stark Ladder, IIRC
@ACuriousMind That was sort of what I was looking for. Thank you.
@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.
see eq. 1.10. do I miss something or is this a very similar problem?
I just remember Mike Stone here posted this under a question to which I've answered. Hope this helps a bit
...but I could be wrong here, by e.g. missing some subtleties in their notation or so
The Moyal bracket has an interesting history where its publication was delayed or something because of Dirac
hieeee
17:32
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?
Hi Allie
hi
im nervous about grad decisions
i need to just forget about them
:/ you should be noticed anytime soon, no?
@TobiasFünke I don't know what it is, but mine is very similar to a tight binding Hamiltonian
@SignorFeynman yeah, but here you have a non-constant diagonal
Yes, that was the "similar" :P
17:37
yeah within the next week or 2, if i got in
ugh i know i need to just distract myself
@SignorFeynman that of course makes everything non-trivial :p
I think this is the last straw, my friend
@Allie yeah... read some more Hamiltonian mechanics or DFT :p
Once I get rid of qubits, my life will be good again
Quoth the QRaven: Nevermore!
17:39
is goldstein a good book for hamiltonian/lagrangian?
It's a very standard reference for classical mech
i liked the introduction given by Taylor but i probably want to go a bit deeper
18:02
@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
2nd chap does a good job
18:30
a classic mathematical approach to classical mechanics is arnold's "mathematical methods of classical mechanics"
@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.
18:49
@ACuriousMind Thank you.
i think i might take a break today
One might say that in classical mechanics the "fluid" is incompressible, whilst in quantum mechanics the "fluid" is compressible. :)
@User198 it's not just that - the Wigner function isn't a probability distribution in the classical sense, it can be negative
19:08
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.
@ACuriousMind Thank you.
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)
@ACuriousMind Thanks.
19:50
Wrong steps, right results. Always lol D:
Next Feynman is insane Feynman
Take it as me speaking to myself:P
20:17
well...I guess it is the second best "outcome", no? :)
what is a constant field in QFT?
no particle number change?
@imbAF without more context, it's impossible to tell
In the context of sponatenous symmetry breaking
We consider as an example the complex scalar field
with an interaction term
you're not considering a "constant field"
you're considering a field with non-zero VEV
it says we consider a constant $\phi(x)$ such that the potential term in the hamiltonian
becomes minimal
20:33
Your screenshot explicitly says you're considering $\phi$ as a classical field
so why are you asking about constant fields "in QFT"?
Because further down the line
he quantizes it
And
The meaning of a constant field in the classical case should be obvious
in the classical case it is
But at some point
and I'm sure in that context the replacement of the classical constant value with a constant VEV in the quantum case will be discussed
Since when, $\phi$ which can be the representation of CFT or QFT
is now, suddenly the ground state representative
And since you are talking about ground state
I don't believe you are in CFT
20:37
As long as you're talking about a literal constant value of the field you're still treating it classically
yeah but what is the ground state
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
phi, if classical takes values
you have to have a bit of patience with these expositions
VEV of what?
@ACuriousMind For the last 3 days I am reading this part
it's very odd
20:39
@imbAF of the field operator; if you don't understand what I mean then you simply haven't read far enough yet
Idk what the right word would be to describe it
(or you're following a horrible resource :P)
you mean book ?
@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.
Oh, gauge symmetry breaking, great
20:44
I simply wasn't sure about what were you talking
And also, it's implied that SSB is ONLY present for degenerate ground state
Quite the statement
doesn't explain why isn't the case if there is for example an excited degenerate state
It's quite delicate if the symmetry is a gauge symmetry though
@imbAF in QFT you are interested in VEV
Not EEV
why?
I mean, look at everything you compute in standard QFT, you have vacuum sandwiches everywhere
@SignorFeynman technically so far none of the screenshots are about gauge symmetries, it's just a $\phi^4$ theory with ordinary symmetries
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
20:47
(of course this will be used as a toy model to then do the Higgs mechanism muddling the waters, but that's how it goes :P)
@SignorFeynman I notice that. But doesn't sound like a strong argument. But what do I know about QFT anyways
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
5 years have been D:
Do you make a distinction whether the gauge symmetry breaking is a global or a local one ?
@imbAF I think the strong argument is that VEV are enough to generate all the physics
what does it mean to generate all the physics?
20:52
@imbAF only local symmetry is gauge symmetry
VEV just happens to be present when you calculate matrix elements
but that's it really
@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
One more thing
@SignorFeynman at least I am consistent :)
I am consistent in saying you're consistent :P
The only time I denied such consistency was when I pointed out the time of "happy new year, nerds"!
20:58
@RyderRude needs to come back
what is the system when one talks about a quantized field? the field excitations and space time?
And why does invariance under LT and translations imply a scalar field?
@imbAF How would a non-zero VEV of a non-scalar field behave under Lorentz transformations?
I don't know
depends how the field behaves under LTs?
@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.
@DIRAC1930 he sent a message today
21:09
A scalar physical quantity is invariant under a LT. I suppose the same can be said for a 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.
@ACuriousMind I don't understand. If the VEV for a non-scalar happens to NOT be zero, that's something that is the case.
I don't understand
@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.
 
1 hour later…
22:19
Did everyone do a formal course on complex analysis?
yes, but it was not obligatory for the physics curriculum
I feel like it would have been useful
The only courses from my undergraduate that were really useful were the advanced QM course and an electrodynamics course which was fantastic
Unfortunately I've forgotten nearly all of E&M
Ngl the experimental labs were quite fun lol
22:35
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 haha what the...
@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
Aaaaah, labs
What is radial move, in the picture?
22:50
@imbAF You didn't reply to his last message :P
@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)
@SignorFeynman Because I don't understand it. And I won't be stuck with something the whole time
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
I remember hungover ACM in the lab
I think you mentioned it
22:59
yes, my lab stories obviously remain the same over time :P
Missed an opportunity to brag about consistency
@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
People are inconsistent with their stories all the time! Even intentionally. The show must go on
23:14
Some more questions:

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}}.
What am I interpreting wrong?
@User198 what's your definition of "information"?
saying "information is conserved" is meaningless if you don't have a quantitative idea what of what means
@ACuriousMind :d yeah, something like that sounds familiar. but the fact that 4 years everyone just copy pasted old results is insanely funny haha
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...
@TobiasFünke oh, it was very funny and a bit sad (we essentially got the best grade for revealing the experiment didn't work like in the manual :P)
@ACuriousMind this statement is soooo annoying ("information of the universe is conserved" or so)
23:25
@User198 that's not a proper definition :P
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
@ACuriousMind I know, I just tought of it. xD Thanks anyway.
@TobiasFünke fortunately I had a Kosmos electronics kit as a child so I understood circuits ;P
hehe nice
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
23:32
oh, that sounds interesting
It was great; meanwhile I was scared of the chemistry version because almost everything in there had some hazardous warning :P
much to the slight disappoinment of my parents who are both trained chemical lab assistants :P
I am checking out the Kosmos kits on amazon ahah they look funny
I was about to ask if your parents have background in natural sciences
@ACuriousMind in some alternate universe you're interested in circuit QED
Circuit gauge theory
23:38
@TobiasFünke well, this was like 20-15 years ago, so I have no idea how their current offerings compare
but back then they were pretty awesome
@DIRAC1930 i will come back after a bit
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
@ACuriousMind yes. but also to "arouse curiosity", I think.
@qwerty hahaha
and to motivate to think "why does this work", and to teach that one can find out, eventually; yes.
23:44
> 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
I wonder how many kids would actually love science if they would be exposed to such things at an early stage... sad
@qwerty lol, I'm just talking about some copper sulfate
I know you weren't a kid in the 1950s xD
23:48
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

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