The h Bar

Silly Goose

01:21

does anyone know what equations 3.30 and 3.33 are actually supposed to be? There seems to be a web of typos/errors in both expressions.

Silly Goose

01:33

i think these are actually the wanted expressions...

ACuriousMind

06:45

@qwerty sure, if you look at hypersurfaces or other submanifolds, then spacetime is the ambient space for them

qwerty

good morning! or... hey-delberg :D

ACuriousMind

Ooof :D

naturallyInconsistent

oh that's such a groan

ACuriousMind

07:01

@SillyGoose there isn't really enough context here to tell

naturallyInconsistent

@ACuriousMind We are always doomed to have to ask for more context. But there is sufficient information to tell that the wacky fowl is much more likely to be wrong.

ACuriousMind

Dunno, could easily both be correct in some sense, especially with some implicit conventions since the angle is modulo pi/2

qwerty

07:31

maybe it's cos I'm very tired... but I'm wondering what peoples' personal attitudes to physics is, whether you would feel the same motivation or pleasure in doing physics if all your output, thoughts etc were never shared with other humans? if you could not discuss or teach it, and only work alone. I suppose you could draw a parallel with writers of novels who never intend to have or have their works read or published for example.

Slereah

People want ambient spacetimes because they want to believe that spacetime is really flat fundamentally

And all the curved space is a trick

qwerty

your comment reminds me of farewell ict

(complex numbers are definitely less friendly to me than a simple minus sign)

handan_toddler

07:53

how 'bout a simple ± sign

naturallyInconsistent

@handan_toddler the latest few videos, on the advanced parts, are great. I'd buy the book just to see what I can steal. But it is soooo awkward to go the crazy lengths just to avoid calculus.

handan_toddler

08:10

He calls going through those crazy lengths "french cooking."

naturallyInconsistent

08:26

I know. One really has to wonder if French cooking is even useful to instruction

Ryder Rude

08:43

hi

Slereah

I do French cooking every day

naturallyInconsistent

sure, but do your students learn better from directly sending them to do French cooking, or do they learn better from something more normal?

Ryder Rude

The Nature of Reality: A Dialogue Between a Buddhist Scholar and a Theoretical Physicist

►

Slereah

Oh I mean actual french cooking

naturallyInconsistent

I dont see why it would not apply in the actual case too

@RyderRude @ACuriousMind

Tobias Fünke

08:55

morning

qwerty

heya

Ryder Rude

@naturallyInconsistent ??

hi

Signor Feynman

10:03

buongiorno, hbar

qwerty

buongiorno, signore

Signor Feynman

G'day, mate

Another day of junctions awaits me

qwerty

buona fortuna~

Signor Feynman

I will need that :'(

Tobias Fünke

10:28

:d

Ryder Rude

Carroll says we should explore the outer universe and the Buddhist scholar says we should look inside our own mind

i am going to be doing the latter

Ryder Rude

13:20

this question has unfairly been closed. Please consider re opening

-1

Q: If the entire universe is governed by QM, and only by QM, would that imply MWI?

Edit: Based on @WillO's very relevant comment, I want to clarify what I mean by "fully governed by QM" - it means governed only by the laws of quantum mechanics, as opposed to classical physics. It effectively means "if classical systems don't exist in the universe". If those logical steps below ...

quantum-interpretations

i was writing an answer when it was closed

ACuriousMind

17:40

@qwerty No, definitely not. The rather solitary experience I had towards the end of my master was one of the reasons I didn't stay in academia. Doesn't mean there's no interest in doing physics on my own but definitely there are other people who have much higher intrinsic motivation for that than I do

naturallyInconsistent

18:15

you arent going to do something about the above?

ACuriousMind

@naturallyInconsistent it's not from one of the grifter channels I explicitly forbade, it was just one link and not spamming, it's fine

naturallyInconsistent

even if it is explicitly religious?

ACuriousMind

@Slereah What do you mena? It remains curved even if you embed things - intrinsic curvature doesn't go away. Or do you mean people want to believe in some flat ambient space as the "real" spacetime? Is that that popular?

@naturallyInconsistent not really relevant as long as it's not a blatant attempt at proselytizing or something like that

Slereah

18:40

@ACuriousMind it is popular with the layman :p

ACuriousMind

I'm not sure the average layperson has any opinion on the concept of spacetime :P

Slereah

Well you know

The pop science crowd

When they read about GR that is usually the image they have in their head

Is there a general notion to associate a topological space to a concrete sheaf, similarly to the D-topology for diffeological spaces?

[i do not mean the etale space]

ACuriousMind

@Slereah You can take $C^0(-,\mathbb{R})$

for compact Hausdorff spaces this is an equivalence of categories (or something like that) between them and Banach algebras

Dannyu NDos

A few days ago, I realized how powerful Einstein summation is, in NumPy.

Slereah

Yeah there are a few but like

Is there a general process for it?

ACuriousMind

18:54

I'm not sure what's not general about $C^0(-,\mathbb{R})$

Slereah

Like the D topology process seems to not rely exceedingly on anything about Cartesian categories

ACuriousMind

is the problem that there's no reason $\mathbb{R}$ should be special if you just look at topological spaces?

Dannyu NDos

@ACuriousMind In the compact-open topology, right?

Slereah

Yeah the process for D topologies is just depending on the sheaf itself

It's just the final topology on the "plots", ie the morphisms you obtain via the yoneda embedding

I guess that could just be enough but I find it weird that there doesn't seem to be any generic name for it?

ACuriousMind

@DannyuNDos yepp

Slereah

18:56

Are there any conditions necessary for this to work?

ACuriousMind

@Slereah oh sorry, I read that the wrong way around

I thought you wanted a sheaf for a space, but you want a space for a sheaf

Slereah

Well there is an inverse functor for the d topology :p

Nlab called them poetically dtplg and cdflg

Well adjoint, not inverse

ACuriousMind

I would say you need to generally have some set of distinguished maps (like the plots of a diffeological space), then the topology is just the coarsest topology that makes those distinguished maps continuous

you can't do it to "just a sheaf"

Slereah

The plots are apparently defined in a sheaf theoretic manner tho

ACuriousMind

sure, but it's additional data that's part of the definition of the diff. space

Slereah

19:02

For the sheaf X and site object U, every element of X(U) is a plot via the Yoneda lemma

is it?

It's just elements of the sheaf itself

ACuriousMind

ah, that's what you mean

Slereah

via $X(U) = Hom(y(U), X)$

ACuriousMind

I mean then the construction there always works at least in principle, no? If you have a concrete sheaf $D$ with points $D(\ast) = X$, you give $X$ the topology that makes all $D(U)$ (interpreted as maps $U\to X$ via the concreteness condition) continuous

Slereah

Sounds reasonable, although I beware since there are always weird corner cases you can forget :p

I find it strange that I can't find a general case for this?

since "sheaves as geometry" is a big part of their raison d'être

Maybe considering them as topological spaces is considered too simple for category theorists idk

ACuriousMind

I think it's just not very useful in general? As in, this will give you strange $X$s with weird topologies?

Slereah

19:11

I try my best to connect category theory things to simpler math :p

also I am trying to figure out how the intensive-extensive thing works

which might help with such a thing

ACuriousMind

consider that even for diffeological spaces this isn't really the topology you "want" - the plots are not supposed to be the continuous functions, they're supposed to be the smooth functions

Slereah

True, although apparently for manifolds you do get the usual "manifold topology"

John Rennie

@Relativisticcucumber metals don't have flat surfaces. They can be machines to have a flat surface, or cooled from the melt to inherit a flat surface from the liquid surface, but this isn't unique to metals. There is nothing about the electronic structure of metals that predisposes them to be flat.

ACuriousMind

well, that's just because manifolds are nice :P

Slereah

Not sure how things work out in general

As I said, the maps between Smooth and Top aren't inverses but adjoints

So presumably at some point you won't get the same thing back

I'm not sure when tho

You do get the discrete topology for the fine diffeology, also

That is how you get to embed topological spaces into Smooth I guess

> The D-topology on a discrete diffeological space is discrete, and the D-topology on an
indiscrete diffeological space is indiscrete

Well at least the extremes are fine :p

and the manifolds in the middle

ACuriousMind

19:17

also, note that for manifolds with more than one smooth structure the procedure clearly loses information

Slereah

True

Question is

ACuriousMind

i.e. this adjunction may be neat but you shouldn't treat the resulting topological space as encoding the same information

Slereah

which manifold do you get

ACuriousMind

@Slereah what do you mean?

Slereah

Oh I'm not, I just want some more pedagogical examples :p

@ACuriousMind Well if you take some manifolds of different smooth structures and map them to Top and then back to Smooth

Which one is The Manifold

The manifold invariant under the dtplg-cdflg monad

The dtplg-cdflg type, as it is called

I don't want to be mean to Saint Grothendieck but having a topological space once in a while is a bit more pedagogical than having the sheaf

ACuriousMind

19:21

that's just because you haven't drunk enough of the koolaid yet ;P

Slereah

@ACuriousMind Do you mean those potions they found in Grothendieck's shack

ACuriousMind

possibly

those are going to do something to you, for sure

and I mean the whole point of this sheaf stuff is that you believe that it's not the space that counts, it's the maps :P

if you just want nice sets with maps on them you can just stay in the category of manifolds :P

Slereah

Legends say that if you drink from them, you will understand the Topos

@ACuriousMind I want to move freely between the two :p

Get all them angles

I want the sheaf, the set, the topology, the type and the logic

My attempt at analyzing the nlab science of logic thing is slowly turning into a book

Not again

qwerty

20:21

good morning

ACuriousMind

g'evenin'

Sine of the Time

9 pm here :)

Slereah

@ACuriousMind turning Australian

qwerty

@Slereah imo, not aided by rubber sheet demos in high schools :(

Slereah

Do they even have GR in high school

i don't remember any GR in mine

qwerty

20:24

Oct 1, 2024 at 11:02, by qwerty

@naturallyInconsistent OK, but our syllabuses were messed up at the time I was doing it -- we were teaching superconductors and cooper pairs BEFORE a simple harmonic oscillator

no GR explicitly, but there was a module that to have understood properly would have required GR (I don't remember the context... maybe cosmology) and I do remember mention of a rubber sheet analogy

@Slereah my french friend picked up a slight Aussie accent during his PhD here >=] I'm corrupting your countrymen too...

Signor Feynman

@ACuriousMind Unless you already said it, what do you mean "solitary"?

@Slereah the lack of "mate" is truly dissappointing

Although, ACM has a long history of g'greetings

Indeed

ACuriousMind

@SignorFeynman I didn't have a lot of classes (meaning I hung out much less on campus) and there just weren't a lot of people to talk to about the stuff I was looking into

Slereah

Damo and Darren 1 - 'Ciggy Butt Brain'

►

It's a beautiful language

Signor Feynman

@ACuriousMind I felt that too at some point :(

qwerty

20:45

@Slereah let me teach you some slang in that case: "bogan" is our version of the US "redneck" ;)

imbAF

Is with "gauge theory" meant a field theory which is gauge invariant?

Slereah

@qwerty I'm aware

I am a man of the world

qwerty

@Slereah haha okay okay level up, what's an "eshay" :p

Slereah

Don't know that one I fear

@imbAF It depends a little on the context, but generally it's a theory where 1) the Lagrangian has a symmetry 2) It's a Bad Symmetry because it prevents you from solving the equation of motion

imbAF

but why is called gauge ?

Slereah

20:49

@imbAF For completely historical reasons

imbAF

so nothing to do with gauge transformations and invariance ?

Slereah

The original gauge theory was a theory in which the length scale was the gauge degree of freedom, so length scale ~ a gauge (in the measuring device sense)

Ah well that is another thing

If your symmetry is Bad but not Too Bad, in fact it gives rise to a group structure

Which is where the group structure comes from

imbAF

Please do not confuse me even further lmao

But I do want to understand one thing

ACuriousMind

@imbAF A (not the historical) motivating example is the 4-potential of electrodynamics: You can use the gauge transformations to "set the zero" of $A^\mu(x)$ to whatever you want at each point in space, like someone gauging an instrument calibrates the zero of the instrument's display

(metrologists don't get on my case about the difference between gauging and calibrating :P)

imbAF

Ok

Slereah

20:52

Gauge (instrument)

In science and engineering, a dimensional gauge or simply gauge is a device used to make measurements or to display certain dimensional information. A wide variety of tools exist which serve such functions, ranging from simple pieces of material against which sizes can be measured to complex pieces of machinery. Dimensional properties include thickness, gap in space, diameter of materials. == Basic types == All gauges can be divided into four main types, independent of their actual use. Analogue instrument meter with analogue display ("needles"). Until the later decades the most common basic...

many such gauges

qwerty

@Slereah huh I would have thought similar to imbAF

@Slereah which was the OG gauge theory you're referring to?

Slereah

@qwerty That would be Weyl's theory of gravity

imbAF

In my lecture, the following transformation: $\psi \rightarrow \psi'=\psie^{i\alpha}$ and $\psi \rightarrow \psi'=\psie^{i\alpha(x)}$ are classified as global and local gauge transformations. How does one determine whether a transformation is a gauge one or not? Can it be determined immediately or you need to calculate something ?

Tobias Fünke

28

Q: Why are gauge theories called so?

Why are gauge theories called so? I guessed it was because gauge also means to estimate, so when one is trying to find the gauge theory for such and such interactions one has to estimate what might be the best gauge group for that interaction. Does this make sense?

soft-question terminology gauge-theory history

ACuriousMind

@qwerty Whether all "gauge theories" have "gauge transformations" depends a lot on the exact definition you use. The "problem" with solving the equations of motion that Slereah alludes to does not always imply that there are gauge transformations, see e.g. the end of this answer of mine

qwerty

20:56

@Slereah did the example of the EM four-potential not predate that or does it not count?

ACuriousMind

@imbAF You need (one of) the proper definitions of "gauge symmetry" for that. Most elementary approaches to the topic do not attempt such a rigorous classification. There is no brief answer to your question.

Slereah

@qwerty EM four potential was... I think it was done shortly after?

Same era anyway I think

ACuriousMind

the simple heuristic that's usually used is that a transformation that depends on the spacetime point but is not a transformation of spacetime that affects the argument of the fields (like the Lorentz transformations) is gauge

imbAF

@ACuriousMind ok :(

Slereah

@ACuriousMind The elementary approach is "A gauge theory is what EM is"

sometimes you do a fancy EM where the gauge is a matrix

imbAF

20:59

@ACuriousMind if it doesn't affect spacetime coordinates, what does it affect ?

ACuriousMind

@imbAF the value of the fields, like in your example of $\psi(x)\mapsto \mathrm{e}^{\mathrm{i}\alpha(x)}\psi(x)$

imbAF

and with value you mean the operators?

ACuriousMind

my "doesn't affect the argument of the fields" means that on the r.h.s. there is $\psi(x)$ and not some $\psi(\Lambda x)$ or $\psi(x')$ or whatever

@imbAF not necessarily, this notion also makes sense for classical fields

imbAF

ah ok

I had qft in mind

One more thing, so I have a better understanding

things like: scalar field, complex scalar field, spin 1/2 field etc. are all field theories,right? QED, which is what I am doing atm, is a gauge field theory right?

Slereah

yes

qwerty

21:02

@ACuriousMind this answer you wrote is very interesting and has bumped gauge stuff back up my to-study list. thanks for linking

imbAF

@Slereah Isn't it odd, that saying gauge field theory, one gives insight as to under what transformation the field theory is invariant, while in the case of i.e complex scalar field you don't provide such information.

Maybe it's how the nomenclature is

but I find this inconsistency in naming the theories odd

ACuriousMind

I mean neither of those terms fully define the theory

they're all "field theories" and you can stack a bunch of modifiers like "complex scalar" or "gauge" or "with polynomial interaction" or whatever onto that

imbAF

what do you mean? which terms?

ACuriousMind

saying something is an "X field theory" is not meant to imply that X somehow tells you everything about it, just that it's a field theory with property X

imbAF

@ACuriousMind For the moment I will go with this

@ACuriousMind Ok

ACuriousMind

21:06

@qwerty if you found that interesting I highly recommend at least the first chapters of Quantization of Gauge Systems

Signor Feynman

@qwerty it sounds like bogolon

ACuriousMind

it might stop being interesting somewhere around the main theorem of homological perturbation theory in the middle :P

Signor Feynman

If they publish another edition, you deserve to be mentioned in the preface :P

And that will also solve the last name problem

Slereah

I think Henneaux can get by without the @ACuriousMind boost

qwerty

@ACuriousMind yeap you recommended it to me before! I did attempt to read it last year a bit but I think you summarised in the answer some important/fundamental points

ACuriousMind

21:11

I am nothing if not consistent in my shilling for QoGS :P

Silly Goose

@ACuriousMind well i don't think that the paper can possibly be correct. the annihilation operator $\xi_{-q}$ does not annihilate the vacuum $\Psi_0$

Or basically one of the annihilation operators does annihilate the vacuum and the other does not, so the definitions of the vacuum state and the creation/annihilation operators are inconsistent.

ACuriousMind

@SillyGoose maybe, but you didn't give us the definition of any of the symbols so I don't know how you expect anyone here to answer this unless they recognize the paper

Silly Goose

well the authors are just diagonalizing $$v_q =v_{2q}v_{1q}v_{2q} \to \exp[\beta j_2\cos(q)]\begin{pmatrix}
A_q & C_q \\
C_q & B_q
\end{pmatrix} \tag{18a}$$

which is in the basis $\{\Phi_{-q,q}, \Phi_\text{vac}\}$

where $\Phi_{-q,q} := \eta_{-q}^\dagger \eta_q^\dagger \Phi_\text{vac}$

where $\eta_q$ is a set of fermionic c/a operators

ACuriousMind

...and you expected someone to guess that?

Silly Goose

well i didn't think the context was necessary

i thought it was clear that the c/a operators were fermionic based off the text and that that was the essential context

ACuriousMind

21:18

I don't really see how anyone is supposed to judge whether the eigenvectors eq. (3.30) in your screenshot are correct or not without knowing the eigenvectors of what they are supposed to be

Silly Goose

but my question was not really about that. it is more about the fact that the supposed vacuum state $\Psi_0$ is not actually a vacuum state of the fermionic c/a operators

ACuriousMind

but that's not what you asked :P

Silly Goose

separately, i think the authors also made a mistake in writing those eigenstates though...

so i just solved it myself and used my own notation. but i am wondering if there was a way to make sense out of what the authors meant to write

oh wait actually their eigenstates are correct

so i guess their c/a operators are just wrong

okay well i can state the question more precisely now i think

given states $$\Psi_\text{vac}:=\cos(\phi_q)\Phi_{-q,q} + \sin(\phi_q)\Phi_{\text{vac}}, \\
\Psi_{-q,q} := -\sin(\phi_q)\Phi_{-q,q} + \cos(\phi_q) \Phi_\text{vac} \tag{19c}$$

Signor Feynman

@SillyGoose Are you working with bogolons?

It's a serious question

ACuriousMind

::snickers::

Silly Goose

21:23

@SignorFeynman the $\eta$ operator is supposed to be a bogoliubov transform of the $\xi$ operators, but i think they are not quite

although im not sure if that is what a bogolon is

Signor Feynman

@SillyGoose The quasiparticles associated with those oeprators are called bogolons

Silly Goose

oh okay then yes

but the linear transformation $\xi \to \eta$ defined by the authors is linear but clearly not unitary

so either bogoliubov transformations meant something slightly different back then or they made an oopsies

here is the supposed bogoliubov transformation

all operators are fermionic c/a operators

ACuriousMind

@SignorFeynman those öprators

Silly Goose

the angle $\phi_q$ has the property $\sin \phi_{-q} = -\sin \phi_q$ (i think)

Signor Feynman

@ACuriousMind You may have noticed I'm not that careful when I type here :P

ACuriousMind

21:27

öprator is just fun to say

not as funny as bogolon, but still

Signor Feynman

$$(\xi_k, \xi^\dagger_{-k})$$ is the vector you are transforming @SillyGoose

Silly Goose

oh wiat oops i should have written $\eta \to \xi$

Signor Feynman

Why specifically this combination? You get it writing down the Hamiltonian, normal ordering and writing it in matrix form

Then diagonalize the matrix and get bogolons

Silly Goose

right that is why this paper is using this transformation

but i apply 3.33 to the proposed vacuum state and it does not annihilate it

$\eta_q$ and $\eta_{-q}$ should share the same vacuum right?

Signor Feynman

If I'm not too stunned by junctions, yes

The vacuum is annihilated by all $\eta$

imbAF

21:33

Can anyone explain me the one in red?

that's just a statement and there is nothing to back it off

Silly Goose

okay i think i see the error

ACuriousMind

@imbAF It's entirely explained by the rest of the paragraph before it. What do you think "no longer shares the symmetries of $L$" means?

Silly Goose

hm wait actually no i am still confused

the transformation seems fine and actually is unitary (orthogonal) when acting on $(\eta_q, \eta^\dagger_{-q})$

imbAF

@ACuriousMind I don't understand what you are saying

ACuriousMind

well, I don't understand what you're confused by - one of us has to start explaining themselves :P

imbAF

21:36

First of all, it comes as a surprise to me that the same way we have invariance of lagrangian, we have also and the same for a state

Signor Feynman

@SillyGoose That's what it acts on

Silly Goose

right but we have $\Psi_\text{vac} := \cos\phi_q \Phi_\text{vac} + \sin\phi_q\Phi_{-qq}$

imbAF

@ACuriousMind you say that it is explained by the paragraph above. I fail to see how an explanation is given

Signor Feynman

@imbAF Have you tried thinking it in terms of the Hamiltonian formalism? Use the definition of symmetry and apply it to an eigenstate; what can you conclude?

ACuriousMind

@imbAF Just like in classical mechanics where the symmetries of the Lagrangian applied to a solution to the equations of motion yield a solution to the equations of motion, the symmetries of the Lagrangian (and hence the Hamiltonian) in QM applied to an energy eigenstate yield an energy eigenstate of the same energy (this is not hard to show).

Silly Goose

21:38

And, $\xi_q \Psi_\text{vac} = (\cos\phi_q \eta_q + \sin\phi_q \eta_{-q}^\dagger)(\cos\phi_q \Phi_\text{vac} + \sin\phi_q\Phi_{-qq}) = \cos\phi_q\sin\phi_q \Phi_{-q} + \cos\phi_q\sin\phi_q\Phi_{-q}$

Signor Feynman

What does degeneracy entail/not entail?

ACuriousMind

This is a QFT text that expects you to be aware of this basic fact about mechanics

Silly Goose

which means that something has clearly gone wrong as the vacuum is not annihilated

Signor Feynman

In case you're waiting for me, I'm sorry but I'm too tired to be helpful now :(

Silly Goose

this suggests that the typo is $\sin\phi_q \mapsto - \sin \phi_q$ in $\xi_q$

is $$\begin{pmatrix} \cos\phi & -\sin\phi \\ -\sin\phi & \cos\phi \end{pmatrix}$$ a usual bogoliubov transformation?

or is $$\begin{pmatrix} \cos\phi & \sin\phi \\ -\sin\phi & \cos\phi \end{pmatrix}$$? The latter is what is used in the text and leads to incorrect result. The former is what I ended up going with and resolves the vacuum annihilation.

ACuriousMind

21:42

the text said that $\phi_q$ is only defined modulo $\pi/2$, so in particular only modulo $\pi$, and $\sin(\phi_q) = -\sin(\phi_q + \pi)$, are you sure there's an actual problem and not just you and the text picking different conventions for how to measure $\phi_q$?

imbAF

@ACuriousMind by "symmetries of the L" you mean transformations for which L is invariant ?

ACuriousMind

@imbAF what else would a symmetry be?

imbAF

Better to double check, since the language used is getting more and more convoluted

Signor Feynman

I haven't read, but it might help you quack @SillyGoose

Silly Goose

@ACuriousMind hm but wouldn't this mean that the sign of $\sin$ is not well-defined?

@SignorFeynman isn't orthogonal a special case of unitary?

ACuriousMind

21:45

@SillyGoose I haven't read the entire text to work out why it wouldn't be problem; I'm just saying the explicit mention of the "modulo $\pi/2$" thing and all your problems just being related to signs and sin/cos mismatches is suspicious

Signor Feynman

(I haven't read, yes it's a special case)

ACuriousMind

because multiples of $\pi/2$ are exactly what turns sins into coss or introduces minus signs

Signor Feynman

Okay, time for some Japanese and then I may rest. I hate Josephson junctions

imbAF

@ACuriousMind that's the point "no longer shares the symmetries of $L$" is just a statement, that wasn't proven at any point. That's the whole point of me highlighting it

Signor Feynman

"phenomenological quantization" is something that will haunt my nightmares

qwerty

21:46

sometimes I feel like ACM in hbar is doing the physics version of en.m.wikipedia.org/wiki/Simultaneous_exhibition in chess

Signor Feynman

@imbAF As I said, have you considered this from the pov of the Hamiltonian formalism of QM?

ACuriousMind

@imbAF It follows from the statements about degeneracy if you think a little bit about it.

Signor Feynman

If you understand it there, just use that the Hamiltonian and the Lagrangian share the same symmetries

I gave you a hint before

And now I can't change my username for a Month T_T

@qwerty I have thought about this ahahahhaha

ACuriousMind

@SignorFeynman what's next, Monsieur Feynmanne?

Signor Feynman

I think that the trend of changing the name was just a German exclusive :P

Limited edition was the expression I was looking for

ACuriousMind

21:54

yo, we got the rare 1st edition German Feinmann

qwerty

lol xD

Signor Feynman

Next time you will is up to Poincaré

I haven't forgotten aboute Feynmate @qwerty

Tobias Fünke

lol

qwerty

what about 'Feynman Sensei' or 'Feynman-san'

Signor Feynman

Dec 23, 2024 at 19:28, by Herr Feinmann

@naturallyInconsistent ファインマンさん could be problematic

qwerty

22:00

I can't read any hiragana or katakana D: you will have to translate

Signor Feynman

Feynman-san

qwerty

why problematic?

Signor Feynman

Imagine if someone new has to write my username

@qwerty I was about to say "it's already good that you recognize they are not kanji", then I remembered past discussions and that you know more hanzi than I can imagine

Silly Goose

@ACuriousMind hm but how can this work at all? then $\sin(\phi_q) = -\sin(\phi_q) = 0$ generically, right?

(i am asking because i am trying to see if what you are saying actually is the case)

Signor Feynman

I still wonder how difficult Chinese must be especially on the first learning stage. One has to write hanzi from teh get-go

qwerty

22:04

@SignorFeynman I do not, I dropped out of weekend Chinese school when I was 6 :p

Signor Feynman

It was a twisted way to say that I remember you have good reasons to know some hanzi :P

ACuriousMind

@SillyGoose I agree, the naive interpretation of the "modulo $\pi/2$" statement does not make sense. But it's suspicious :P

the vague idea I have is that they have an implied range for $\phi_q$ like $[0,\pi/2)$ but you're working in $[\pi,3\pi/2)$ or something

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