The h Bar

Emilio Pisanty

09:18

From Sabine Hossenfelder's Facebook page, facebook.com/590904573/posts/10160251774619574

We are looking to expand the team of our "Talk to a Scientist" service. We offer one-on-one conversations with scientists in which we answer questions about physics. Our typical team members are postdocs who want to top up their salary and get some hands-on experience in science communication. PhD is a requirement. If you are interested, please send your CV/resume to [email protected] with the subject "application for consultation service"

Please do NOT send me a facebook message -- I do not use the facebook messenger.

bolbteppa

09:55

I think that's a good idea, people who spend years on nonsense theories with no outlet for it need someone to evaluate it

Slereah

Andreas Cap pronounces "Lie theory" as "leaf theory"

bolbteppa

Leaf algebras are the easier version of Lie algebras

Slereah

If I can understand Cap's thick Austrian accent correctly, the dimension of the space of torsion-free conformal compatible connection is related to the group extension???

As in the Weyl connection is defined up to a 1-form $\phi$ and the second order of the conformal group is $\mathbb{R}^n$ for the SCTs

If true I guess that would explain why the orthogonal group has a trivial extension

Also I guess that sort of makes sense since the $G^2$ space encodes the symmetric linear connections?

Actually does this mean that the SCT does not impact on the frame but it does change the rep of the non metricity of the connection?

bolbteppa

11:10

> Update: I’ve heard that Lubos himself shutdown the blog, unwilling to agree to follow rules Google was now enforcing.

The Lisi post is gone :'(

Slereah

11:24

F

ACuriousMind

11:42

There seems to be a lot of buzz around the W boson mass measurement but no actually accessible paper?

"To obtain a copy of the paper, please contact [email protected]." are they really trying to tell me that they have a paper on one of the potentially most interesting measurements in a while and they didn't upload it to arXiv?

Slereah

@ACuriousMind Publishing too soon may expose them to ridicule if they're wrong

ACuriousMind

I don't think that makes sense - if you're worried about being wrong you don't make official press releases about the thing that might be wrong, right?

Slereah

You gotta manage the media exposure with the reputation loss

Physics Meta

12:42

0

Q: Cite documents rather than linking only

We should tell people to cite from documentations when they use link. I always look at old posts to learn something new but some answerers give some link but they are dead now (e.g. this). His first link seems to be broken. Even currently there's more than 1000 answers which contains link and tha...

support

fqq

13:34

@ACuriousMind it really seems to be the case

but it's open access on Science

science.org/doi/10.1126/science.abk1781

ACuriousMind

14:37

@fqq ah, thanks!

Nihar Karve

15:44

Hi everyone, I just thought I might let you know: I got an admissions offer from Cambridge University! Hopefully I'll be joining in Autumn.

7

Big thanks to everyone in this chat room - I definitely wouldn't be as interested in physics as I am now without the fascinating conversations that happen here

John Rennie

16:11

@NiharKarve BOOOOOOOOOOOOOOOOOM! :-)

More Anonymous

@NiharKarve Damn!!! Well done!!!

John Rennie

Which college?

Nihar Karve

@JohnRennie Caius!

(for everyone else - that's Gonville and Caius College, the same one where Stephen Hawking was a Fellow)

John Rennie

My niece is currently at Girton. I was at Peterhouse, though that was a loooooooong time ago :-)

user250478

16:39

You are friendly, I read your reply/response to my request in this chat in past days but I need time to elaborate what I wanted to say (in public) to the team of moderators about a post (I don't have internet at home, only in a library). Meanwhile I wish the best week for you and the chat users. Many thanks @ACuriousMind

fqq

17:09

@ACuriousMind that might also explain why it's not on the arxiv, maybe Science has an embargo period?

idk, I've never published there

imbAF

18:22

In Cohen-Tannoudji's book $\lange \vec r| \psi \rangle$ is defined as the component of the ket $|\psi \rangle$ in the position representation. At the same time $\lange \vec r| \psi \rangle=\psi(\vec r)$ is considered later on as the corresponding wave function to the ket $|\psi \rangle$. So is the component of the ket and it's mathematical representation, the same thing?

Semiclassical

20:10

I’m not sure how literally to take the phrase “the component of the ket in the position representation.” It supposes that $|x\rangle $ is a well-defined state vector, which is not really true

It’s probably better to take $\langle x|\psi\rangle$ more like notation, in that you can use it to determine how to compute stuff and use resolutions of identity

imbAF

I see

one more thing

Semiclassical

To put it differently: it’s not hard to make sense of stuff like $\langle \phi_n|\psi$: it’s a component of $|\psi\rangle$ in some orthonormal basis

By contrast, the sense in which $|x\rangle$ is a “basis vector” is rather more delicate

imbAF

yes

I know

it doesn't belong to the wave functions that are square integrate

Semiclassical

Right

You can legitimate it mathematically with rigged Hilbert spaces apparently, but most QM books will not touch that

imbAF

are these equivalent $\mathcal{R}^{-1}_{\vec u}(d\alpha)$, $\mathcal{R}_{-\vec u}(d\alpha)$ and $\mathcal{R}_{\vec u}(-d\alpha)$

?

I assume they are

Semiclassical

20:17

Are these rotations with respect to the u-axis?

imbAF

yes

Semiclassical

Then yeah

imbAF

I understand the 2nd and the 3rd

if I try to visualize them

but not the first

Semiclassical

For the first and last I think algebra is easiest. To make it easier to type on mobile I’ll write “rotation by da with respect to u” as R(u,da)

imbAF

I know

what I am saying is that I don't feel they represent the same thing

i'll give a simple explanation

If you have a power-tool and you put a screw in the wall, and it rotates i counter clockwise, the screw is implanted in the wall

Semiclassical

20:21

My simple explanation would be that, if I draw a circle in front of my face, then another person looking at me will draw it the opposite way

imbAF

if you rotate clockwise, you can remove the screw from the wall correct ?

Semiclassical

Sure

imbAF

That implies

inverse rotation

in other words the first and 3rd eq, mean the same thing

but if you are in the -z axis, and you make counterclockwise rotatition

you are implanting the screw in the wall

when in fact you need a clockwise rotation in the -z axis to pull it out of the wall

Semiclassical

Just to state the algebra: R(u,da) * R(u,-da) = R(u,da-da)=R(u,0) is the identity rotation, ie, no rotation

imbAF

ys

yes*

Semiclassical

20:24

So the two rotations are inverses etc

imbAF

yes same axis direction but opposite angle values

but opposite axis direction and same value of the angle

is not equivalent

unless I am imagining it wrong

Semiclassical

Maybe imagine watching your scenario in a mirror

You’d now presumably see the screw going clockwise as it gets implanted

hmm

imbAF

no

look

Semiclassical

I’m not sure I have that right either, yeah

imbAF

if you imagine the -z +z axis and you rotate $d\alpha$

since it's positive, it implies counterclockwise rotation

Semiclassical

20:28

Watching the screw in a mirror will definitely make the screw appear left-handed, tho

imbAF

regardless whether you consider the -z or +z axis

the screw is being implanted in the wall

I got it

xD the mistake I was doing was so small yet so important

Semiclassical

Yeah,, rotations be like that

imbAF

If you want to do a counterclockwise rotation in the -z axis

you need, your pov needs to be in the direction of the axis

Semiclassical

Right

imbAF

meaning in the -z direction

I was doing it, but my POV was from the +z axis xD

Semiclassical

20:31

Which the mirror is problematic for

imbAF

Yeah

All 3 are equivalent

@Semiclassical One more thing

Semiclassical

It’s like the business of choosing whether to mirror your video feed in Zoom

imbAF

yeah exactly

it's a nice training of your imagination

anyway, can you have a look at this :

0

Q: Geometrical Rotation and Rotation operator relation and ambiguity

In Cohen-Tannoudji's it is said (vol.1 page 694): $|\psi \rangle$ is the state of the system before the rotation. $|\psi ' \rangle$ is the state of the system after the rotation. $|\psi ' \rangle= R |\psi \rangle$ in the bracket notation. Geometrically: $\psi '(\vec r)=\psi (\mathcal{R}^{-1} \v...

angular-momentum reference-frames rotation geometry convection

Semiclassical

20:46

Had to relocate. I don’t have a lot to say, but I think the last point is just this. Suppose $\vec{r}=(x,y,z)=R\vec{r}’$ where $\vec{r}’=(x’,y’,z’).$ Then $\vec{r}’=R^{-1}\vec{r}$, so $\psi’(\vec{r})=\psi(\vec{r}’)=\psi(x’,y’,z’)$

So $\psi’(\vec{r})$ amounts to the original wave function, except it’s evaluated at $\vec{r}’$

I take the “components” here to just mean $x’,y’,z’$, ie, the components of the $\vec{r}’$ vector

imbAF

Yes I am aware

but question 2 is what I am mostly interested

Semiclassical

So in other words, why is the rotation operator unitary?

imbAF

no

because

at this point in the book, we still haven't proved that it's unitary

Semiclassical

Hmm

imbAF

simply, instead of using $R^\dagger$ we use the geometrical rotatation

totally different thing

one is a geometrical entity and the other is an abstract entity acting on the hilbert space

and that is done, as if it's granted for some reason

and explanation ain't necessery

Semiclassical

20:54

Yeah, I feel like phrasing this in terms of bra-ket notation isn’t that useful here

In terms of wavefunctions in position space it should presumably be : $R \psi(\vec{r})=\psi’(\vec{r})=\psi(\mathcal{R}^{-1}\vec{r}).$

imbAF

one second

R is the operator or the geom. rotation

because you didn't use calligraphic notation

Semiclassical

operator. Here I was matching your notation in the post

imbAF

But I assume it must be the geom. rotation, written with a caligraphic

Operator acting on a wave function? and not a ket?

Semiclassical

Yes. But it’s understood that $R\psi(\vec{r})=\langle \vec{r}|R|\psi\rangle$

Which is the convention I know

ACuriousMind

see, that's why you shouldn't post questions here directly after you post them - I already saw the question on the main site and just wrote an answer to it :P

Semiclassical

21:08

Lol

I mean, I’m typing on my phone so low bandwidth conversation

imbAF

@ACuriousMind +2 xD

imbAF

21:21

@ACuriousMind " unitary representation map ρ" I never was aware of this

but this is the link between the inverse and the complex conjugate

Transcript for