The h Bar

12:04 AM

Ping me if you have some ideas, just curious.

9 hours later…

8:59 AM

@schn Perhaps elasticity? If you compress something soft in one direction, it elongates in the others.

2 hours later…

11:09 AM

Elongate? Was there an Elon Musk scandal?

satan 29

11:19 AM

bruh

11:32 AM

@Slereah when isn't there one :P

Nihar Karve

11:42 AM

Is $H^k(M, \mathbb{C})=\bigoplus_{p+q=k}H_{\bar{\partial}}^{p, q}(M)$ only true if $M$ is (compact) Kähler

Are those Sobolev spaces?

@NiharKarve yes, you need Kähler otherwise something harmonic in $\mathrm{d}$ need not be harmonic w.r.t $\partial$

@Slereah no, DeRham and Dolbeault cohomologies

ah yes

Nihar Karve

@ACuriousMind right, thanks

I had a brain fade and thought it was for general almost complex manifolds, so I was a little spooked when they mentioned that off-hand in a review

@Slereah you won't catch me unironically doing analysis :P

12:22 PM

@ACuriousMind thanks, elasticity works :)

I'm no fan of analysis but sometimes you gotta

user 85795

12:46 PM

Did anyone hear about the 27 tweets per second recorded on Twitter when # Daft Punk announced their break up?

RIP techno

perhaps now, the market will shift towards heavy metal making a reassurance in the post pandemic music world

techno, in my opinion, is a bit decadent for today's world

4 hours later…

Charlie

5:00 PM

Could someone just clarify a point for me, I was under the impression that including gauge fields in string theory meant defining the gauge field on spacetime (into which the string is embedded) but the Wiki page for "worldsheet" seems to imply that the gauge fields are defined on the worldsheet itself?

Q: How to promote “hidden gems”

@Charlie No, it's saying that the background (including spacetime gauge fields) is encoded in the 2d sCFT on the worldsheet, not that you define some gauge field on the worldsheet

These are the two dual and equivalent ontologies in string theory - either you believe there really is a target space and that induces the sCFT on the worldsheet, or you believe we start with the sCFT and the target space is "derived" from that, see also this answer of mine

Charlie

Ok I think I see a bit more, ty

quite a lot to digest in string theory all at once, including cft

Physics Meta

It often happens to me, and sometime I see really great answers in which the answerer gets little reward. In the discussion section OP explains what they don’t understand and the answerer replies, and this goes back and forth and is never resolved and everything fizzles out. No upvotes, nothing. ...

vzn

@user85795 ah, the days when tens of thousands of ppl could comingle tightly packed in outdoor/ indoor concerts. yeah almost a bit decadent in todays world... my personal delight was daft punk making the tron soundtrack :) ... cmon wheres the sequel? :|

4 hours later…

user 85795

9:07 PM

\o @RyanUnger

10:05 PM

This is about the Born-Oppenheimer approximation, where the wave function can be broken into an electronic part and a nuclear part. How does the second equality follow mathematically? It seems like it's a 1D integral.

If $u(r)=g(r)f(r)$, then $\int u(r) dr=\int g(r)f(r) dr \neq \int g(r)dr \int f(r) dr$, right?

I see.

It is a volume integral.

The electronic part only depends on the radial distance supposedly.

Why is the integral over the electronic space wavefunctions 0?

Franck–Condon principle

The Franck–Condon principle is a rule in spectroscopy and quantum chemistry that explains the intensity of vibronic transitions. Vibronic transitions are the simultaneous changes in electronic and vibrational energy levels of a molecule due to the absorption or emission of a photon of the appropriate energy. The principle states that during an electronic transition, a change from one vibrational energy level to another will be more likely to happen if the two vibrational wave functions overlap more significantly. == Overview == The Franck–Condon principle has a well-established semiclassical...

In the section on Quantum mechanical formulation.

Second equation involving $P$.

10:29 PM

@schn because they're not the same state?

that integral is just the inner product between the $e$ and the $e'$ state

So they are orthogonal? But why isn't the spin integral 0 for example?

I think there's an assumption here that $e$ and $e'$ are not the same state - i.e. the electron actually jumps into a higher/lower state - but you're not assuming anything about its spin

I see. So if it was normalized/eigenfunction to $H$ for example (very handwavy here), that integral would just be 1?

I'm not sure what you mean

the integral is one if the two states are the same state, i.e. $s'$ and $s$ are the same spin

it's zero if they are orthogonal spin states (up and down in the same direction), it's something else if they are something else (e.g. one is a definite spin in x-direction and the other a definite spin in another direction)