The h Bar

2:19 AM

@SillyGoose the ensemble of electrons evolves unitarily into all the different branches. The individual system only sees one branch, giving the illusion of collapse

Silly Goose

3:55 AM

hm but the statement seems to be interpretation dependent?

a separate question: is it accurate to say that $\exp: \mathfrak{su}(2) \oplus \mathfrak{su}(2) \rightarrow SU(4)$ ?

Ryder Rude

4:12 AM

@SillyGoose FlatterMan's comment subscribes to the ensemble interpretation of quantum mechanics, but he doesnt mention it

Silly Goose

okay that makes sense

naturallyInconsistent

Yeah, when it comes to interpretations, quite many people just are not trained to make statements carefully. They often just make interpretation-dependent claims and then make them sound universal

Ryder Rude

@SillyGoose i dont think the pure state of the electron alone evolves unitarily tho. U still have to consider the combined wavefunction of electron+environment for unitary evolution. But this is different from FlatterMan's remark. He is just saying that it doesn't make sense to talk about the wavefunction of an individual system. That is the ensemble interpretation

Or perhaps the electron's state does evolve unitarily aftr we integrate out the environment's state. Is this correct @naturallyInconsistent

naturallyInconsistent

The answer to that question is itself interpretation dependent.

Most people are working in interpretations whereby individual systems are still described by a wavefunction. Flattermann's rather extremist interpretation has its own nice parts, but it is superbly difficult to make judgements. He just doesn't have a critical take of his own interpretation, so it is futile trying to know how much he actually understands of his own interpretation.

naturallyInconsistent

4:31 AM

@RyderRude oh wait. If you integrate out stuff, the resulting evolution is not unitary.

Ryder Rude

I too find it a bit annoying that FlatterMan keeps spamming his interpretation. In this case, ensemble interpretation had nothing to do with SillyGoose's answer

@naturallyInconsistent i thought maybe a time-dependent Hamiltonian can describe can describe unitary evolution of just the electron's pure state, if we ignore the environment

naturallyInconsistent

It will not. The evolution would turn into some Lindbald master equation or some other crazy shit similar to that.

Ryder Rude

Yes. I think u r right. The time dependent Hamiltonian description wud only be possible if the combined environment + electron state can factorize, which is not possible for a general entanglement

So if there is interaction with environement, only the combined wavefunction admits a Hamiltonian evolution description

But i read that, in some special cases, time-dependent Hamiltonians help us in ignoring the environment

Like how they can model non-conservation of energy when energy is lost to environment

naturallyInconsistent

It is much more stereotypical for theorists to be the ones with impenetrable egos. It is weird to find experimentalists who have that kind of egos.

@RyderRude yes, but we lose a simple description of the time evolution operator, and it becomes not clear what is the gain we get back. I suppose not needing to deal with Lindbald is already the gain. Don't want to go bald.

Silly Goose

5:02 AM

i also explicitly state that i use textbook quantum assumptions, which includes copenhagen int :P

so it is strange to respond with a statement that depends entirely on a different interpretation

naturallyInconsistent

Flattermann claims that his is the copenhagen int

Silly Goose

hm i guess maybe i should've explicitly stated that i am idealizing the electron to be representable by a pure state after it is emitted or something

all in all i still think the qualitative description actually presented in the lecture is unnecessarily convoluted for the point that was trying to be made :P

naturallyInconsistent

There is no way to prevent people from making noise when you want answers to specific questions. What you can do to help you improve the signal to noise ratio is to write your question in an interpretation-independent manner, or specifically specify which interpretation you want the answers to conform to. Then at least we can flag the noise for what it is.

Silly Goose

i also preface my answers with "in textbook quantum" or something along those lines :P to signal that my answer is copenhagen dependent

naturallyInconsistent

And copenhagen is right out. There are too many incompatible variants that people all claim their version is it.

Silly Goose

5:11 AM

or more broadly textbook dependent

naturallyInconsistent

@SillyGoose under no circumstance should you expect this to work. MWI is slowly winning ground.

Silly Goose

well im in no position to learn about quantum interpretations :P

i haven't even learned quantum sufficiently well without the business of interpretations :P

naturallyInconsistent

I think that kind of position should be as widely pooh-poohed upon as people who are seeking for "unbiased sources of news". How is anybody supposed to learn stuff without the interpretations also coming in?

Silly Goose

Well i suppose i don't know where generally interpretations come into play in the theory

my impression was that interpretations mostly come into play when talking about "measurement"

naturallyInconsistent

It is like this: Lorentz and Poincaré had already arrived at the correct Lorentz transformations and had half of the interpretation of SR before Einstein came along. They had, precisely because of that, never actually managed to learn SR. What Einstein did, is mostly interpretations.

Silly Goose

5:20 AM

but a counter example is the ensemble interpretation

PM 2Ring

6:22 AM

@RyderRude It's even more irritating because FlatterMann mostly posts his stuff in comments, not answers, so he's bypassing the voting mechanism. OTOH, we can flag comments...

naturallyInconsistent

@PM2Ring There was a recent meta question precisely about that

Would have been nice if we could have specific tags for the flagging of comments

It is not just FlatterMann. There is also John Doty

Silly Goose

i thought it was just me annoyed by flattermann lol

can i differentiate the exponential map like in step 4? I am a bit skeptical...but at the same time i would expect to be able to differentiate the exponential mapping :P

$H(\theta_1)$ is a Hamiltonian that is parameterized by parameters each corresponding to a GGMM

I guess to be exact I am asking if $\partial_\theta (\exp(-iH(\theta)\tau)$, where $\tau$ is a constant, is $-i\tau*\partial_\theta(H(\theta)) \exp (-iH(\theta)\tau)$

PM 2Ring

6:46 AM

@naturallyInconsistent Yeah, I noticed that, but I think the names got edited out. There are a few other members who've been around for years who persistently post borderline stuff in comments.

naturallyInconsistent

@SillyGoose Are these Hamiltonians constant for each specific value of the parameter? Their time independence will be needed.

Silly Goose

Ah yeah the hamiltonians have no time dependence

PM 2Ring

We can flag such comments as "It's no longer needed", which incorporates the old "Too chatty" flag reason. Or you can use a custom flag with a message like "This comment should be an answer". Another option is:

naturallyInconsistent

Then it should be fine

PM 2Ring

@FlatterMann Please stop posting answers in comments. — PM 2Ring 2 hours ago

Silly Goose

6:49 AM

excellent >:D

ty

where would i formally see why elementary multivariable calculus also seems to work on these functions of operators?

naturallyInconsistent

@PM2Ring This particular thing confuses me. Sometimes I post a short answer and get told it should be just a comment. Other times I post the same in a comment and got told to post it as an answer

@SillyGoose That it works is a non-trivial thing that requires a lot of Lie algebra analysis to prove

Silly Goose

so would this not even be seen in a first course in lie theory?

oh wait is lie analysis its own subject

naturallyInconsistent

But physicists generally tend to just throw random shit into the series definition of the exponential, and profit

in this case, matrices

Silly Goose

lol

Slereah

Exponential boolean

Silly Goose

6:51 AM

i suppose i will justt treat this as multivar calc and check that i havent done anything illegal later XD

naturallyInconsistent

intro to Lie group Lie algebra tends to be immediately bogged down in proving these things.

Slereah

$$e^{\top} = \top \vee (\top \wedge \top) \vee (\top \wedge \top \wedge \top) \vee \ldots$$

Silly Goose

oh hm maybe my choice of book does not go that route...

i am going to do brian c hall's lie theory book next semester

naturallyInconsistent

Anyway, in 1st year at uni I was rather proud to have proved that the binomial theorem could be used to write 2-variable Taylor's expansion, and was feeling weird that my prof was weirdly amused. I only made the realisation that it was nothing a year or so later, because the exponential form makes it trivial to see it for all cases.

@Slereah no factorials?

Slereah

@naturallyInconsistent What's 2 in boolean algebra

$1 \oplus 1 = 1$

also I don't think it's really a division algebra

PM 2Ring

6:59 AM

@naturallyInconsistent Answers (even short ones) should never be posted in comments because it bypasses the voting system, and tends to encourage comment threads to turn into long discussions.

Silly Goose

Hm, now if $\Lambda$ is a constant matrix, then is $\partial_\theta (e^{-i \theta \Lambda}) = -i \Lambda e^{-i \theta \Lambda}$?

PM 2Ring

OTOH, it can be useful to post info that helps the OP to give their question more focus, or supplements their research. But try to frame that stuff in the form of a question, eg "Have you considered X, as mentioned in some_article?"

naturallyInconsistent

@SillyGoose yes

Silly Goose

ah okay ty

naturallyInconsistent

@PM2Ring comment threads should turn into long discussions... hehehehe

Silly Goose

7:04 AM

man this 4 g of leaves has lasted for 5 brews

how nice

naturallyInconsistent

@Slereah typically the exponential form would be helpful if the series is recognisably there, or if every term quadratic and above are identically zero. Without either, it is weird to insist on using that symbolism

Slereah

Also it's just the identity map

Oh wait I guess $e^{\bot} = \top$

It just maps everything to true

PM 2Ring

The mods tend to disagree with that opinion. ;) Personally, I don't mind the discussions, but I prefer them to happen in chat. Many people hate it when comment threads get moved to chat, but I quite like it.

Slereah

A chat conversation is ephemeral though

John Rennie

@naturallyInconsistent Flatterman's comments are an excellent illustration of why it's not a good idea to answer in comments. Remember that you aren't just answering the OP. You're posting an answer that will be read by generations of physicists to come. A good answer contains all the details those future readers could want, and a pithy comment rarely achieves this.

Also a partial answer in a comment tends to discourage people from making the effort to write a full answer.

PM 2Ring

7:17 AM

@Slereah Comments are even more ephemeral. OTOH, chatrooms get frozen for inactivity.

Slereah

I mean the comments stay under the question

same as an actual answer

Silly Goose

7:30 AM

lol what a nifty formula

hopefully it simplifies greatly :D

naturallyInconsistent

@JohnRennie quite often I'll come back to questions and finding that partial answers are expanded into a full answer. I think that is probably ok

Physics Meta

8:25 AM

0

Q: I can't ask questions even if there are only 2 questions with -ve points, why?

I have been actively asking and answering but now it seems like just because of some negative votes I am unable to ask questions, and downvotes don't even leave the comments.

discussion

john

10:01 AM

does anyone knows tools to easy draw Penrose diagrams?

Slereah

10:27 AM

There's a latex package for it

Amit

11:09 AM

What does qmechanic mean when he mentions "second notion of tensors" here

#3 in his answer

Slereah

11:36 AM

It's the fake tangent bundle

Amit

And does it have more indices for some reason?

Silly Goose

anyone planning to watch the new oppenheimer movie :0

9

Slereah

It has the same number of indices

But now you have two tangent bundles

So twice the indices

Amit

11:58 AM

Okay 👌🏻

john

12:22 PM

@Slereah how is it called?

Silly Goose

do yall think this is a reasonable definition of a group representation

Slereah

@john the one I used was feynmf

I don't know if it is the best but that's what I used

@SillyGoose Seems pretty standard

john

12:38 PM

@Slereah i thought that was for feynman diagrams

i've seen someone using tikz, but it's a real pain

Slereah

Oh penrose diagrams

Hm, I don't know if there's one for Penrose

john

ok thank you anyway

i'll try asking also on tex.stackexchange

Slereah

you can find some examples of code on the tex SE

john

yeah i was hoping in some packages like feynmf or tikz-feynman

user4539917

Perhaps if Penrose teamed up with Wolfram they could fast track a package.

Slereah

12:45 PM

there aren't that many commonly used penrose diagrams anyway

You can just reuse existing code

Silly Goose

man representations are really a confluence of the first few concepts of group theory in such an interesting way

Silly Goose

1:07 PM

hm let's consider the poincaré group. so is it accurate to say that we want to distill this "symmetry" that is the principle of relativity. a natural way of distilling symmetries into their essence is to construct an abstract group. distilling this "symmetry" into a group allows us to talk in a consistent way with the math formalism of quantum theory. namely, a linear representation affords us an inner product preserving way to act on a hilbert space

hm i guess im wondering: is group theory about symmetries qua symmetries. that is, sure a reflection on a 2D square does something, but we want to abstract from this. we can characterize an abstract reflection by considering its structure among other abstract elements?

Slereah

1:23 PM

Group theory isn't about symmetries

It's about transformations

But yes it is the abstraction of the group's action, not what it does

All you know is what kind of transformation the composition of two such transformation does, not what it does

Silly Goose

what is your difference between transformations and symmetries

Slereah

A symmetry is a transformation that will leave something unchanged

you can have spaces which are not symmetric wrt the Poincaré group but that still admit a representation of it

Silly Goose

hm what do you mean not symmetric wrt P group? I am thinking that any representation of the P group yields transformations from the object into itself, so the space itself post-poincare transformation is still the original space

Slereah

1:46 PM

What does it mean for two spaces to be the same 🤔

Silly Goose

hm I suppose I just mean the representation of the poincare group is an automorphism on whatever vector space you represent it over

Slereah

I mean every map will lead to an automorphism

A symmetry is usually with respect to some structure

ie the Poincaré group thing is for isometries

Respecting the metric

Silly Goose

oh so you're saying perhaps energy is not conserved by the poincare group

Slereah

I don't know why you even bring up energy here

I'm talking purely math here

Silly Goose

well the metric is something physical as well but it is an example of something left invariant so i was trying to think of something that was not left invariant wrt poincare transformations

well the metric is a mathematical object but in the smae way that energy is a component of a four vector is a mathematical object

Slereah

1:53 PM

You can have various things which are left invariant by some symmetries but not others

The metric is left invariant under Poincaré transformation on Minkowski space, but if you define a vector field on it, it is not left invariant

On the other hand, a volume element has a larger symmetry group

Silly Goose

Okay i see

Slereah

For instance that's why the Hamiltonian breaks Lorentz invariance

you have to define a vector field (the direction of the time coordinate) for it

And your symmetry group drops down to SO(3)

Mr. Feynman

3:50 PM

Given a vector bundle and a connection on it, the connection $1$-form is a matrix of "ordinary" (real valued) differential forms. Intuitively, why for connections on $G$-principal bundles is it $\mathfrak{g}$-valued?

Slereah

I mean the lie algebra is also matrices

at least it is usually represented as such

Mr. Feynman

But why in the transition from vector bundle to principal bundle the connection gets this requirement?

Maybe I should rephrase it. What happens when we pass from a vector bundle to a principal bundle that makes us require that the connection one form is not real valued but lie algebra valued?

Slereah

4:11 PM

The connection is on a vector bundle

Like the actual derivative

Well

It is on the associated bundle

Which is like the product of the vector bundle by the gauge group and then quotiented somewhat

That's what you actually act on

nickbros123

4:26 PM

I remember @Slereah or @Amit talking me about the motivation for matrix multiplication and how it is based upon linear combination or something; jus so u know I'm in that part of Hoffman kunze :D

Slereah

4:41 PM

the case of gravity is a bit more magical than most since the associated vector bundle is the tangent bundle, but for the rest we use the usual ones

Mr. Feynman

@Slereah mhhh from what I know you can define a connection without an associated bundle, which requires an additional structure, namely a left action of the structure group on a manifold (e.g. a representation acting on a vector space)

Slereah

Yeah you can just define a connection from the tangent bundle of the bundle in question

A principal bundle connection is just a special case of this

it just needs to obey a few more properties

Amit

@nickbros123 Wasn't me. Lin algebra is beautiful. I am a bit dismayed at how determinants aren't studied deeply enough in a standard course

Mr. Feynman

Because they are a mess

The world is a mess

Amit

Lol, all the more reason to understand them better, this mess is secretly the suffix of every integral more or less

Not line integrals I guess

@Mr.Feynman Mess and energy

Mr. Feynman

5:21 PM

@Amit lol

@Amit or even better of any differential form

One may define differential forms by means of determinants without appealing to the exterior algebra construction. Although you are really using it because the definition of the determinant is in fact that :P

@Amit line integrals too can be expressed in terms of differential forms, so yes :P

Amit

I thought the form of a line integral isn't really a determinant or at least a degenerate one kinda

Mr. Feynman

1×1 matrices are numbers :P

Amit

5:37 PM

Yeah, I see what you mean :]

Mr. Feynman

I have a question for you, people. Recently I have been discussing with my fellows about understanding vs being able to explain.

I think that if one really understands something, then they must be able to explain it well - not necessarily in detail but to be able to convey the idea clearly - to anybody☆.

Let me clarify the meaning of "anybody☆" with an example as a lower bound on the recipient's knowledge is somehow needed, I would say that a physicist who really understands something what he's doing should be able to explain (even simplifying) virtually to any physics student of the first years. If that sounds too much restrictive, then let's drop the "first year" requirement and stick with students of the year corresponding to that course. What do you think about it?

I hear too often that a "Professor is very good in that field but not good at explaining". Admitting that it is true that such Professor is not good at explaining and that's not a student's rant, I would say that is very messed up. How can you understand a concept deeply and not be able to make it understandable for other people that have the right tools to grasp it?

So the final question would be: do you think that the ability to explain and the understanding of something are independent?

WaveInPlace

5:57 PM

They're definitely linked. I've found my own ability to explain something improves with my understanding.

But they are still separate skills. There are some things that I understand intuitively (physical skills in particular), and don't have a more conventional grasp on. Those things I can do well, but have a lot of trouble explaining.

Charlie

Whether or not you're able to explain an idea that you yourself understand could be related to how you learn it yourself. A very popular method of learning is to try to explain an idea back to yourself in increasingly clear terms but I don't doubt there are other ways to do so that might lead to you not developing that skill.

Also, being able to communicate verbally is a skill in and of itself, a person could be able to explain something clearly in their head but struggle to actually get the words out. Possibly being another explanation for "good scientist bad teacher".

Mr. Feynman

6:18 PM

@WaveInPlace Regarding this second part, I would consider this intuitive understanding as yet incomplete and thus not to the deeper form of understanding I'm talking about

Amit

Also, an explanation is usually a dialogue. Unless it's a lecture to a large group.

So being able to figure out how to explain it to a specific person you're talking to, while you're talking is a big part of being a good explainer

Mr. Feynman

@Charlie The only instance in which that makes sense to me is when we restrict to the class itself. Talking to a crowd is not something we're all good at, that's fine. I'm also considering one to one discussions to avoid this other problem

I don't really believe "good scientist bad teacher" is really a thing, unless this person is not really committing to teaching

@Amit oh, you were faster :P

@Amit yes, one also has to be able to adapt to the context

WaveInPlace

@Mr.Feynman, an intuitive understanding isn't necessarily incomplete. Half our brain can't use language, after all.

Link for wild claim: youtube.com/watch?v=wfYbgdo8e-8&pp

Mr. Feynman

@WaveInPlace I think a complete understanding is both intuitive and technical. None should be missing

Amit

6:37 PM

I'm sure some of you happened to hear two people discussing, one trying to explain something to the other, and it's clear the explainer really understands but not managing to put the idea across, and then if you interject and just say a few words that "bridge" the communication in the right way, it often clarifies everything quite quickly

It's being able to understand exactly in what way someone doesn't understand lol

Simulate the faulty process fully before you can correct it :) kinda

Mr. Feynman

@Amit well, it can happen you can't really explain it to a specific person for whatever reason. If it happens systematically for a lot of people though...

Amit

Sure, the first thing is knowing what you understand well enough to explain, and what not... it's not fun for the ego but it's better saying sometimes that you don't know... or as professors usually put it "i'll get back to you on that, good question" :)

Feynman may have said the final word about understanding "What I cannot create, I do not understand"

Shikhar Chamoli

7:25 PM

Hello everyone, Please someone clear this for me. If I have a slit with two holes and I pass light, how will I know that diffraction will occur or interference?? @Amit @Mr.Feynman

Charlie

@Mr.Feynman It's still a factor in 1-to-1 conversation, but I suppose if you remove the interaction at all and consider someone a good explainer even if they're only good at explaining when they're alone (or think they are alone)

Then that probably removes the issue

WaveInPlace

@ShikharChamoli, the Huygens-Fresnel principle is a nice visual representation.
https://en.wikipedia.org/wiki/Huygens%E2%80%93Fresnel_principle

There will definitely be useful questions on the main site that help too. This is a pretty common question.

Shikhar Chamoli

@WaveInPlace ok, will see.

Mr. Feynman

7:51 PM

@Charlie I think you'd need an observer to define a good explainer :P

Amit

@Mr.Feynman The "Yoda" level for explaining is again Feynman for me probably..

Not because of his style

But because he could just try to work out in real time something while explaining it both to himself and to another, and if he failed he would just admit it

Mr. Feynman

Everything I say somehow comes from his legacy

Amit

I recall reading a story that something like that happened in one of his "secret" classes lol

Mr. Feynman

Except when I speak of Physics, I somehow get causally disconnected

geocalc33

$$t \partial_{tt}\varphi_t(x)=-x\partial_x \varphi_t(x)$$

and

$$it \partial_{tt}\varphi_t(x)=x\partial_x \varphi_t(x)$$

These are related by Wick rotation correct?

Amit

8:01 PM

Cf. Example

@Mr.Feynman yeah, I guess it just is ultimately individual

I guess they weren't secret, that was another time he had to conceal the fact he is giving lectures

geocalc33

actually they seem to be related by $t \mapsto -it$ as opposed to Wick rotation which would be $t \mapsto it.$

Mr. Feynman

The Wick rotation I know is the former but I think it's just a matter of convention and metric signature

Slereah

You can rotate time by some arbitrary amount

You can just do $t \to e^{i\alpha} t$

Mr. Feynman

Hey, Slereah

I think I solved my problem with connections of some hours ago

It was a misunderstanding apparently

I though the components (one forms) of the connection form (matrix of one forms) had to be instead lie algebra valued but the whole connection form has to

So we are just saying that matrix lies in a Lie algebra

geocalc33

8:25 PM

Okay so they are indeed related by Wick rotation. I guess that makes sense because the first is essentially a backwards heat equation and the second is essentially the free schrodinger equation. (since this is all one dimensional the double time derivative isn't a problem and time and space can be used interchangeably).

geocalc33

8:38 PM

Then, I'm curious as to why you get solutions to the backwards heat equation, that also form a Cauchy foliation of $\zeta^{1,1} \simeq \Bbb M^{1,1}$ where $\Bbb M^{1,1}$ is the Minkowski plane in Dirac coordinates.

Mr. Feynman

@geocalc33 Wait, how is that the free SE $i\partial_t\phi=-\frac{1}{2m}\partial_x^2\phi$?

geocalc33

well not exactly, due to the pre-multipliers of $t$ and $x$ resp.

I just fail to understand conceptually why a Cauchy foliation of $\zeta^{1,1}$ where it's related to $\Bbb M^{1,1}$ by the diffeomorphism $f(x,y)=(e^x,e^y)$ should satisfy a backwards heat equation. Doesn't make intuitive sense to me. I mean, a Cauchy foliation of $\Bbb M^{1,1}$ i.e. $t/x$ (partition by rectangular hyperbolas) does not satisfy a heat equation...

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