The h Bar

2:14 AM

In the QHO in regular quantum mechanics the creation-annihilation operators connect the energy eigenstates but they don't (obviously) affect occupation number like those used in QFT. Does there exist operators in QFT that do this? i.e. $$a^\dagger(k)|0\rangle=|k\rangle\in \mathcal H,\quad A^\dagger(z)|k\rangle=|k+z\rangle\in\mathcal H,$$ something like this? Essentially an operator that gives a particle more or less energy

fqq

2:54 AM

not sure I understand what you mean, would someghint like $a^\dagger(k+z)a(k)|k\rangle = a^\dagger(k+z)|0\rangle = |k+z\rangle$ work for you?

Nihar Karve

@vzn Is there some kind of classical no-deleting theorem analogue that you know about?

I'm not sure if you're into QC, but you might want to take a look at this

vzn

4:44 AM

@NiharKarve hi, the question reminds me of the thermodynamic heat death of the universe which states entropy is always increasing and that all the stars will eventually burn out. and another theory that black holes will eventually swallow all other matter. anyway think the no-cloning thm is one of the deep principles of QM but speaking freely here, think it has questionable elements. was looking for anyone else questioning it, but nothing seemed to turn up on a cursory search.

RewCie

@JMac 🤦

user487507

5:14 AM

Sup

Nihar Karve

@user131873 Hello

123

6:42 AM

Hi GooD MorNinG...

Yo

ACuriousMind

8:44 AM

@NiharKarve The quantum "no-X theorems" are overhyped and usually don't say anything interesting in their standard formulations, see physics.stackexchange.com/q/296678/50583 :P

skullpatrol

8:58 AM

Games you can play with area:

Nihar Karve

@ACuriousMind I am of the view that they're only really there for the conservation of quantum information, although I don't claim to be an expert

@skullpatrol do you know Zimbax? You might find his videos entertaining

skullpatrol

Nope, thanks I'll check it out.

Nihar Karve

9:14 AM

In general, do your universities help get you a subscription to scientific journals?

Prateek Mourya

11:00 AM

I have a confusion if a rod lying on horizontal which is always perpendicular to one end of rod dow the com remains at rest if yes then how to prove it

ACuriousMind

@PrateekMourya I have no idea what sort of situation "if a rod lying on horizontal which is always perpendicular to one end of rod dow" is supposed to describe

is the rod lying on a table (why "horizontal"?)? what does it mean for that to be perpendicular "to one end of rod"? Things can't be perpendicular to points, only to vectors/lines/directions. What's a "dow"?

skullpatrol

A long cylindrical piece of wood?

ACuriousMind

@skullpatrol That's a dowel, no?

skullpatrol

Yes.

@ACuriousMind do they call that an noun adjunct?

ACuriousMind

11:19 AM

@skullpatrol Why would they?

An adjunct noun/noun adjunct is something entirely different

skullpatrol

ok

ACuriousMind

This is a nominalized adjective, if you really want a fancy word for it

skullpatrol

Yup, that's fancy enough for me, thanks.

satan 29

11:39 AM

DISCLAIMER: PROMOTION IS NOT INTENTED. PLEASE REFRAIN FROM ANSWERING:

1

Q: Proving the fundamental theorem of calculus.

I have tried to prove the fundamental theorem of calculus: Assuming: f,g,F are continuous,differentiable in the appropriate range(s) By definition, $$ \int _{x=a}^{x=b}g(x)dx=\lim_{n \to \infty} \sum_{r=0}^{n}\dfrac{b-a}{n}g(a+ r\dfrac{b-a}{n})----------->E-1$$ Now consider the integral: $I= \int...

how to i get "E-1" to the extreme right side? I want to remove the -------->

Prateek Mourya

I have a confusion whether we take amplitude or principal argument in euler form of complex number

@ACuriousMind suppose a rod is lying on a horizontal table or in x-y plane and a force coplanar is is such that it is always perpendicular to one end of the rod

Nihar Karve

@satan29 you can enclose it in \tag{}

satan 29

@PrateekMourya go on

@NiharKarve thanks!

MY POST CAN BE IGNORED NOW.

skullpatrol

Thanks for the disclaimer :-)

Charlie

12:03 PM

@fqq Yeah I guess that would make sense, thank you

Maybe this is obvious, but if the quantum field operators obtained during second quantisation satisfy an "operator version" of the classical field equations, i.e. we can literally just replace the classical field solutions with the operators in the equations of motion: $$(\partial^2+m^2)\phi(x,t)=0\quad\rightarrow\quad (\partial^2+m^2)\hat \phi(x,t)=0,$$ does this mean that the classical gauge freedom of the $A_\mu$ 4-vector in electrodynamics gains exactly the same freedom as a qft operator?

ACuriousMind

@Charlie Yes and no :P

just wait until you get to QED is probably the best

Charlie

alright then :p

ACuriousMind

the theory you get when you try to do naive canonical quantization of a gauge theory is inconsistent

i.e. the quantization procedure you know now would do exactly what you say but that's not the correct quantization procedure

Charlie

12:19 PM

ah, I did wonder why p&s refer to the quantisation of EM as messy

or complicated, something like that

Slereah

12:37 PM

You can do QED with a "naive" quantization of the gauge and it mostly works

The Gupta Bleuler formalism

Basically doing a quantum version of the Lorenz gauge

ACuriousMind

Yeah, but even in Gupta-Bleuler you have to be careful how you implement the gauge

if you just implement the full gauge $\partial_\mu A^\mu = 0$ you just kill all the photons, that's why there's this "positive part" $(\partial_\mu A^\mu)^+ = 0$ you end up enforcing

Slereah

I say let them die

3

Charlie

12:58 PM

Could I just have a sanity check, two objects of the same mass travelling upwards from the surface of the Earth at different velocities (no acceleration) require exactly the same amount of energy to get $x$ distance from the surface of the Earth, they only differ in the power input required to get them there

Prateek Mourya

1:08 PM

Why h bar not lambda?

ACuriousMind

1:23 PM

@Charlie by "no acceleration" you mean that they propel themselves with a variable acceleration that exactly cancels the gravitational acceleration at the current height?

@PrateekMourya what?

if you form complete sentences the probably of someone getting what you're talking about usually increases :P

RewCie

1:38 PM

@PrateekMourya It's half life is infinite

@Slereah 🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣

@Charlie I remember reading this thing.... I forgot what they call it now....🤔

@skullpatrol Hello bro... Do you know a good book on Systems Programming?

Charlie

@ACuriousMind Yes my bad the situation I was actually thinking of was in a small enough vertical region that $F_g$ is constant

ACuriousMind

@Charlie in that case yes, they both expend $F_gh$ energy to reach $h$.

RewCie

Just In: NASA Writes Back To A 9-year-old Who Asked For A Job To Save Earth From Aliens; Read Here Via: RepublicWorld

Charlie

Ok nice ty

any judgement on my lack of classical mechanics skills will be met with swift legal action

ACuriousMind

@Charlie Nice, I look forward to being awarded fair compensation for having to deal with that lack

Charlie

1:51 PM

D:

ACuriousMind

@RewCie And once again, a years-old story that's certainly not "just in". Cf. e.g. washingtonpost.com/news/speaking-of-science/wp/2017/08/06/… from Aug 2017. Are you trying to only post things that aren't news?

RewCie

Okay... Nvm...

🧐

@ACuriousMind Have you ever been to Switzerland?

ACuriousMind

Yes; I've visited CERN.

Charlie

Did they let you ride inside the LHC like a waterslide

I've heard that's what they do during the down-season

RewCie

Woooow! I heard that Switzerlandish people speak German. So, German classes are compulsary for ZTH

Nihar Karve

1:58 PM

@Charlie heck, you can even get accelerated to 0.5c for a lark

RewCie

@Charlie Wowowowowowowowhat????

If I get LHC for a day, I'll Make a videogame with it.

ACuriousMind

@Charlie nah, they gave us jetpacks and we had a little race inside the collider ring :P

Charlie

:O

Nihar Karve

@ACuriousMind so that's what high energy physicists are talking about when they say "jets"

RewCie

Just In: A cool game trending on Twitter

The correct pronunciation of data is data.

Nihar Karve

2:06 PM

@ACuriousMind when does the Ehresmann connection beat parallel transport in physical applications?

RewCie

/ˈdeɪtə/

ACuriousMind

@NiharKarve yeah, you didn't really buy that ridiculous pretense about particle streams looking like water jets, did you? :P

Nihar Karve

ₙₒ

ACuriousMind

@NiharKarve I'm not sure what you mean - an Ehresmann connection defines a notion of parallel transport via the "horizontal lift"

RewCie

$_{yes}$

Nihar Karve

2:09 PM

not parallel transport, you know, the most common form of it

I actually can't for the life of me remember its name right now

there are three common ways to get around the vectors at different points being part of different tangent spaces: the Ehresmann connection, covariant derivative and ...?

Charlie

the connection one form?

I was actually going to ask a while ago how unique the correspondance between connection 1-forms and Ehresmann connections is

ACuriousMind

if you're asking if the idea of a connection as a horizontal/vertical split as in the Ehresmann formalism is ever physically more relevant than the idea of a connection form, I don't think so

RewCie

Hard

Nihar Karve

@Charlie yeah, thanks

@ACuriousMind oh ok. I liked its formulation so much though :(

ACuriousMind

but an Ehresmann connection is equivalent to having a connection form or a notion of parallel transport, so it's a philosophical difference, not a technical one

RewCie

2:14 PM

Are MO and SO same?

Nihar Karve

er, no?

if you mean MathOverflow and StackOverflow, that is

RewCie

So, I'll have to create duplicate rooms on both sites?

yes

MathOverflow and StackOverflow

Charlie

so the choice is one-to-one in the sense that each one uniquely determines the other?

ACuriousMind

I think the Ehresmann viewpoint is much more geometric and better to visualize, but physically we're not really motivated by the aesthetic of the geometry, but by the idea of having local gauge fields, i.e. we get the bundle by a clutching construction, not by starting with the bundle as the central idea.

so physically the "geometry" of the bundle is more or less accidental, so there's little gain in emphasizing any geometric intuition a mathematician might be interested in

RewCie

We all are mathematicians inside

ACuriousMind

2:17 PM

@Charlie yes

Charlie

I see

ACuriousMind

the connection form is essentially the projection on the vertical part

so you go from the Ehresmann assignment of horizontal/vertical split to a connection form by projecting onto the vertical part at each point, and you go the other way by defining the horizontal space at each point to be the kernel of the connection form

RewCie

How many of you all are using Windows/Linux OS? I've a very interesting trick to show ya here!

Okay, nvm....

b4n

Charlie

As in we define the connection 1-form given the Ehresmann connection as the vector valued 1-form whose image is the vertical subspace?

Nihar Karve

Is it just me or are there way fewer QM/QFT/GR questions around these days?

than, say, a couple of months ago

Charlie

2:26 PM

possibly because peoople aren't cramming for finals :P