The h Bar

0celo7

12:00 AM

RPG

You were the bottom I assume :P

ACuriousMind

@0celo7 So? Well-made exams don't expect you to memorize stupid matrices :P

@0celo7 I was, in fact, the (game) master.

0celo7

@ACuriousMind I can't even remember the fucking SE

Bernard Meurer

@ACuriousMind Ah, sweet! Where do you come from again? I had it in my mind you were from Heidelberg

ACuriousMind

@BernardMeurer Nah, I'm originally from Wuppertal

Bernard Meurer

Being the game master is da best

That's near Essen and stuff right?

0celo7

12:02 AM

I have literally no clue what's gong on anymore

what is the Hamiltonian for a spin thingie

ACuriousMind

@BernardMeurer Yeah, that's the general region - although I must be quick to point out that Wuppertal is not technically part of the Ruhrpott.

Bernard Meurer

@ACuriousMind Is it Westphalia?

ACuriousMind

@BernardMeurer No, it belongs to a region called Bergisches Land,

0celo7

German names for regions are terrible.

@ACuriousMind Really?

The problem has to do with spin along the $z$ direction so I need to know the spin operator in that direction

Bernard Meurer

@ACuriousMind Ah, I see, learn something new everyday :)

@DanielSank <3

DanielSank

12:08 AM

@0celo7 Just memorize them. It's worth the neurons.

@BernardMeurer <3

Bernard Meurer

Oh, nice kernel update

DanielSank

@BernardMeurer wut?

Bernard Meurer

kernel.org

DanielSank

@ACuriousMind I agree that exams shouldn't test that sort of memorization.

Bernard Meurer

New 4.7 revision came out few days ago

but I just noticed it now

DanielSank

12:11 AM

@BernardMeurer Wow, they actually took "kernel.org".

Not "linuxkernel.org".

ACuriousMind

@0celo7 Contrary to what the name might suggest, it's named so because it once was ruled by the "Count of Berg", not because it's hilly :P

Bernard Meurer

@DanielSank Awesome right?

DanielSank

-_-

Bernard Meurer

?

kernel 5.0 is getting close, time really goes by quickly

DanielSank

I just discovered something hilarious about Russian and Google translate

Observe the following translations:

бля   = f-----g
бляд  = whore
блядь = damn

It's like build-your-own swear word!

As you add letters to the root, you get different swears. It's beautiful.

They're not even really related. It's like someone sat down and planned this so that when you're mad you can change which swear you want on the fly.

Bernard Meurer

12:20 AM

Lol

That is quite awesome

ACuriousMind

Why are "whore" and "damn" acceptable to type out but "f-----g" isn't? It's not as if it's unclear what the latter stands for :P

Bernard Meurer

Now why were you translating google swears I don't know

@ACuriousMind He's american, they're weird

DanielSank

^ That

@ACuriousMind My morals are all screwed up by the mix of country's Puritan past with modern Hollywood fueled debauchery.

@BernardMeurer I'm brushing up my Russian, and I've always been a scholar of swears.

ACuriousMind

@DanielSank I...see.

Bernard Meurer

@DanielSank You can't swear in portuguese

It's a glorious language for that

DanielSank

12:25 AM

@BernardMeurer empurrão

Bernard Meurer

That doesn't mean anything

That's when you push someone on the street

DanielSank

@BernardMeurer hahahaha ok, well I asked Google translate for "jerk", so there we go.

I was wondering why it looked like what a drunk Spanish speaker would say for "push".

ACuriousMind

lol, it translated the verb "to jerk" instead of the insult "Jerk!" then

DanielSank

Yep.

Wow, yeah, I just translated a longer sentence and it really looks like drunk Spanish.

Bernard Meurer

Lol, bloody google

ACuriousMind

12:29 AM

Translating English swears into German is kinda hard. I often feel that most English swears and insults exist at a level of offensiveness where the German words are either much more harmless or much worse.

Bernard Meurer

Geht zum Teufel

I know that

ACuriousMind

@BernardMeurer That's...rather old-fashioned. Unless you're an angry grandpa most people will rather laugh at that :P

Bernard Meurer

@ACuriousMind I am an angry Grandfather

DanielSank

^ It's true.

Bernard Meurer

That's the best definition I ever heard of myself lol

ACuriousMind

12:31 AM

Shit, the potatos have already procreated?

Bernard Meurer

@ACuriousMind Yes, they have inherited my latino fire

DanielSank

@BernardMeurer definition?

Bernard Meurer

@DanielSank ~~definition~~ description

DanielSank

@ACuriousMind Potatoes? What's going on in here?

ACuriousMind

@DanielSank Have you somehow missed/forgotten Bernhard's plan to conquer the world with face-clutching linux potatos?

Or...have they already got you? :O :O :O

(I thought that the potatoes would be his children, so if he's a grandfather, they must have created further offspring)

DanielSank

12:39 AM

@ACuriousMind ah

It all makes sense now.

But since @BernardMeurer hasn't even written potatOS yet, I think we're ok.

Bernard Meurer

PotatOS is a project for the future

Some day I'll have the equipment to solder ARM SoCs

that day PotatOS will begin

DanielSank

@ACuriousMind Using BCH twice, I get $\exp(z^* a) \exp(z a^\dagger) \exp(-|z|^2) = \exp(z a^\dagger) \exp(z^* a)$.

ACuriousMind

@DanielSank That appears to be correct

DanielSank

@ACuriousMind Very good then.

Perhaps "complete the square" isn't the right hint.

More like: write down what you want to get, write down what you have, and then add the factors you need to make them equal.

I guess that's pretty close to "complete the square".

user228700

12:56 AM

Hello :) Whenever anybody has the time, can you please tell me what you think this statement means: "One of the limitations of the first law is that it doesn't tell us about the conditions under which heat is transferred."

ACuriousMind

@DanielSank Well, it made sense to me when I typed it, but looking at it now I'm not sure it's a helpful description either

@KaumudiHarikumar Without more context, I find it difficult to tell what that sentence wants to say, but I'd guess it's saying that while the first law involves "heat" in most of its statements it doesn't tell you whether that heat is transferred by heat conduction, by thermal radiation, or some other process.

0celo7

@ACuriousMind I bet the count was named after his hilly land :P

America has some pretty great names for places

Rocky Mountains

Great Plains

Little Rock

ACuriousMind

1:14 AM

@0celo7 No, the count's seat simply was a castle called Berge. How it came by its name is not known to me.

DanielSank

Too bad @IceLord is gone.

0celo7

Why?

@ACuriousMind Long shot, but do you know about the supported hyperplane thm for convex functions?

DanielSank

Could explain how to solve his problem.

ACuriousMind

@0celo7 never heard of it

0celo7

@ACuriousMind have you heard of Jensen's inequality?

ACuriousMind

1:17 AM

yes

Proving the measure-theoretic version was an exercise in my functional analysis course, iirc

0celo7

@ACuriousMind lucky

user218912

@DanielSank I'm back.

user218912

sorry I didn't make much progress on the problem, I had class all day.

user218912

I'll spend the next few hours working on it.

user228700

@ACuriousMind Riight, okay.

DanielSank

1:45 AM

@IceLord dude

BCH says $\exp(A+B) = \exp(A) \exp(B) \exp(-(1/2) [A,B])$, right?

So we rewrite this as $\exp([A,B]/2) \exp(A+B) = \exp(A) \exp(B)$.

The first factor on the left side is just some number because [A,B] is a number (i.e. not an operator).

Now just write the same thing with $A$ and $B$ swapped:

Danu

Now derive BCH ;)

DanielSank

$\exp([B,A]/2) \exp(A+B) = \exp(B) \exp(A)$

@IceLord can you get it from there?

@Danu I don't recall it being that hard.

Danu

I guess it depends on how rigorous one wants to be

@DanielSank By the way you should be careful with this---of course it is still an operator, it's just central (i.e. commutes with everything), as a multiple of the identity

DanielSank

@Danu -_-

Language is hard.

Speaking of which, I owe duolingo my daily Russian exercises...

Danu

I mean... you personally will not have any problems doing this

But I don't think it's right to teach it to someone like that

DanielSank

1:56 AM

@Danu I'm surprised to hear that.

In school "this is a number" was a pretty clear way to convey the meaning.

Danu

You said it is not an operator, which is not correct. The fact that it acts by multiplication by a number is what you're really saying.

It's not a huge deal obviously

But I don't know at what level IceLord is---it might be confusing.

DanielSank

@Danu I see your point.

Thanks.

Danu

I mostly care because it was disappointing to discover, after my bachelor's, that I wasn't taught the basic mathematics of e.g. quantum mechanics all that properly :(

user218912

@DanielSank sorry my wifi is kind of broken.

user218912

@DanielSank yes but once I get it in the final form what should I do?

0celo7

2:31 AM

@IceLord are you doing homework

user218912

yes

user218912

I have analysis homework

0celo7

can you prove $|ab|=|a||b|$ yet

user218912

no

user218912

well

0celo7

2:33 AM

Seriously?

user218912

I never tried

user218912

I can probably prove it by taking all the different cases and showing its true

Bernard Meurer

@0celo7 Can I try?

user218912

does that count as a proof?

0celo7

@BernardMeurer yes

@IceLord there's a name for that technique

user218912

2:35 AM

I know

user218912

proof by exhaustion.

user218912

we proved the sum version of that absolute value formula for $a$ and $b$ in class.

0celo7

Yes, but the German name is better

user218912

which is?

user218912

wait let me figure it out

0celo7

2:36 AM

@BernardMeurer listen to Diamonds Dancing by Drake & Future, Fantano really likes it.

Bernard Meurer

So, it is trivial to see that for any given $|a|,|b|$ the results of their multiplication will be the same independent on the sings of $a$ and $b$

0celo7

Is it trivial?

Bernard Meurer

Yep

because $-a * b$ has the same absolute value as $a*b$

and vice-versa

0celo7

@BernardMeurer Proof?

You have to use the definition of the absolute value.

user218912

yes I know how to prove it then.

user218912

2:39 AM

same as proving the sum version.

0celo7

good

Bernard Meurer

@0celo7 Ah

Because the abs of $a$ is it's magnitude

i.e. has no regards for it's sign

0celo7

What is the definition of absolute value?

Bernard Meurer

@0celo7 It's distance from origin on the real line?

0celo7

Distance?

Bernard Meurer

2:40 AM

I don't know dude

I'm just trying

0celo7

@BernardMeurer I'm being pedantic because this is a proof where pedantry is easy.

DanielSank

@IceLord What do you mean "final form"?

0celo7

@BernardMeurer $|a|=a$ if $a\ge 0$, $-a$ if $a<0$.

user218912

@DanielSank like when I find $N$

DanielSank

@IceLord I thought that was the whole point?

Isn't the entire point to find N?

user218912

2:41 AM

yes

user218912

wait

Bernard Meurer

@0celo7 Okay, that makes sense

user218912

really dumb question

user218912

what is

0celo7

@BernardMeurer I need two variables, what should I use

user218912

2:42 AM

$\langle 0 | 0 \rangle =$ ?

user218912

1?

0celo7

Are the states normalized?

user218912

yes

0celo7

Then yes

Bernard Meurer

@0celo7 $b$ and $\psi$

user218912

2:43 AM

@0celo7 what if they weren't?

0celo7

bad, $\psi$ is already the wave function

@IceLord then it would be $\langle 0|0\rangle $ :P

user218912

xD

user218912

just making sure.

Bernard Meurer

@0celo7 $t$ then

user218912

@DanielSank say I change the order of the exponentials so the annihilation operator acts on the ket and the creation operator on the bra with another exponential there in the middle for the $e^{\frac{z^2}{2}}$ term.

user218912

2:45 AM

can I solve for N then?

DanielSank

Hey @IceLord I'm kinda wondering something. You said:

24 hours ago, by IceLord

I did once read shankar front to back

user218912

yep

DanielSank

Do you think maybe you'd retain the info better if you go through quantum again, maybe more slowly?

user218912

yes

user218912

that's what I'm doing right now

DanielSank

2:46 AM

Ah.

user218912

hopefully done by december xP

user218912

I read shankar on the bus everyday.

user218912

and look up stuff in it from QFT that I don't understand.

DanielSank

Ok but you have to solve lots of problems to actually learn stuff.

user218912

so far it's working

user218912

2:47 AM

I know

user218912

@DanielSank what is $\langle 0 | e^{za^\dagger} =$?

DanielSank

First of all, I'm sure that's in Shankar ;)

Second, what have you tried?

user218912

for what?

DanielSank

@IceLord To answer your own question.

Bernard Meurer

whathaveyoutried.com

DanielSank

2:52 AM

^ Can we make that required reading?

user218912

I give up

user218912

gonna sleep

user218912

gnight

0celo7

@DanielSank Would you happen to know the equation for an electron in a uniform constant magnetic field?

pointing along the $z$ axis

nevermind, I got it right

Sir Cumference

3:17 AM

TIL the meter was redefined in terms of the speed of light in 1983, and so the speed of light is now precisely 299,792,458 m/s. No decimals.

0celo7

Correct.

For a nerd, you keep having a lot of TILs about basic facts :/

@DanielSank My QM prof recommended amazon.com/Quantum-Processes-Information-Benjamin-Schumacher/dp/… as a "light" intro to quantum info

Sir Cumference

@0celo7 ._.

0celo7

@SirCumference TIL: Let $Q\subset\Bbb R^n$ be countable. Then $\Bbb R^n-Q$ is path connected.

Sir Cumference

Guess I just haven't come across that info before

No teacher ever mentioned it

@0celo7 Oh screw you XD

DanielSank

@0celo7 You sure like to walk the line between "nice" and "not nice".

0celo7

4:54 AM

Man Addicted To Eating Bricks, Mud and Gravel

►

user116211

@0celo7 WTH is that; well it sounds better than a man who eats knives....

0celo7

@MAFIA36790 He's your countryman...

I don't know what is up with that

user116211

Incredible India

Secret

even more interestingly, why for so many years of eating non nutritious stuff he seemed to be doing fine

I mean, brick is as non nutrious as it can get

0celo7

lol

user116211

5:01 AM

@Secret There is a man here who lives on sand; another one who lived on knives, blades (doctors have found acid-treated rusted knives in his stomach; he is safe now); some eat grasses (mind you, human is not cow) and much more which still didn't not come to the light....

0celo7

@MAFIA36790 ...India is weird.

user116211

hmmm.

Secret

Pica (disorder)

Pica (/ˈpaɪkə/ PY-kə) is characterized by an appetite for substances that are largely non-nutritive, such as ice (pagophagia); hair (trichophagia); paper (papyrophagia); drywall or paint; metal (metallophagia); stones (lithophagia) or earth (geophagia); glass (hyalophagia); or feces (coprophagia). According to DSM-IV (Diagnostic and Statistical Manual of Mental Disorders, 4th Edition) criteria, for these actions to be considered pica, they must persist for more than one month at an age where eating such objects is considered developmentally inappropriate, not part of culturally sanctioned practice...

user116211

@0celo7 I wouldn't generalise this for the whole country; there are many things which make India proud.

user116211

BTW, did anyone here know there is a man who loved his sister and his dogs so much that even after their death, he kept the corpses in his room for more than one year thinking they would come back to life.... after his father's death, he also kept his body in the room; he even celebrated their birthday also..... creepy

0celo7

5:06 AM

...where was he from?

user116211

@0celo7 Robinson Street, Calcutta.

0celo7

Lol, so India?

user116211

I guess so.

ffahim

@0celo7 a and b are non zero vectors now proof a=mb only and only if a and b are parallel............

user116211

Okay, India is weird...

0celo7

5:13 AM

@ffahim I gave you a full proof, stop bugging me please.

ffahim

pls its an different problem@0celo7

0celo7

that's literally the definition of parallel

user116211

@ffahim Prove that they are collinear; and also don't ping somebody unnecessarily.

0celo7

@MAFIA36790 I need a letter for a discrete metric...what's a good letter?

ffahim

hey @MAFIA36790 whats colinear?

0celo7

5:14 AM

google

user116211

@0celo7 maybe $d\;?$

user116211

Lanczos in his book uses that many times.

0celo7

@MAFIA36790 I'm using $d$ already...

user116211

5:26 AM

@0celo7 Then use $d^{\prime}$

0celo7

I'm using $\mu$.

user116211

good.

user116211

@0celo7, Bourbaki in his treatise sometimes used $\eth$ also.

Secret

6:14 AM

Note to self: Break derivatives of H bounded->Break Lebniz integral rule->$$0=\frac{d}{dt}I(f+t\varphi)\rvert_{t=0}=\int_{\omega}\frac{\partial}{\partial t}H(x,f,p)\varphi (x)dx\rvert_{t=0}$$ becomes $$0=\frac{d}{dt}I(f+t\varphi)\rvert_{t=0}=\frac{d}{dt}\int_{\omega}H(x,f,p)\varphi (x)dx\rvert_{t=0}$$ -> Cannot integrate by parts -> Euler Lagrange equation will not pop up on extremising the functional H. -> Discuss with 0celo7 later about this and check understanding with him what is learnt from the

typo: extremising the functional I, not H

(In the event of confusion, explain that the above thought process came from after reading through the proofs, thus it is a follow up thought process after the reading)

2physics

6:30 AM

there is almost nothing on the internet about his death.. this is how most of the scientists/physicists leave us.. sadly

user116211

@2physics :(

user228700

@MAFIA36790 There's a man who eats knives? :o

user116211

@KaumudiHarikumar You can google it.

Xasel

matching guitar rigs is so much fun

user228700

Hey, can you guys help me with something real quick? It's thermo, yet again. Sorry :P

Secret

6:42 AM

ask

user228700

Okay, so I'm trying to understand and solve problems related to PV diagrams.

Secret

ok

user228700

And I'm having some trouble with the following question but I'm unable to find out where I'm going wrong in my reasoning...

user228700

user228700

So looking at the path ABC, it's clear that the volume at A and C are the same and so, technically, the work done should be zero but it's not...because work done is a path function? Correct?

user228700

6:47 AM

..?

Secret

Work done is path dependent. In this system you need to worry about both the pressure and the volume.

Even though after a cycle the volume has not change, work is still being done because the pressure changed

user228700

Riight. Okay. Also, looking at the process AB, I'm inclined to say that it's isothermal because PV at A(6) is the same as PV at B(6).

user228700

But apparently, that's wrong :/ Doesn't it make sense that AB is an isothermal process?

Secret

If this is an ideal gas in a closed system, then yes it is true by PV=nRT that the temperature is the same for A and B. Howeever, to show it is isothermal, one need to show the T along the path AB is indeed constant, which I might need to check my notes to revise myself on how to do it...

John Rennie

@KaumudiHarikumar: the work done is $\int PdV$.

user228700

6:52 AM

Yep, it is indeed an ideal gas in a closed system!

user228700

@JohnRennie Yes, so it's the area under the curve, which is clearly not zero. Got it.

John Rennie

If the pressure is constant the the integral becomes $P \int dV $, so it is just a function of volume so in this case zero net work would be done between A and C.

However $P$ is not a constant so the integral looks like $\int P(V) dV$

Xasel

if thermodyanmics be a troll on the moyntain of physic then @JohnRennie is our life-saving Gandalf

user228700

Yes, OK, that was a stupid mistake. I apologize :P But what about the process AB? :-(

user228700

@Xasel OMG yes! :D

John Rennie

6:54 AM

:-)
Actually thermodynamics is one of my worse subjects. That and electrodynamics. But I can answer simple questions.

Secret

The work doen by AB is nonzero as evidences by the area under the curvie AB

John Rennie

@KaumudiHarikumar what about AB?

user228700

Yes, it's non-zero but isn't the process isothermal in nature?

Xasel

thing that seem simple to you wizards may not be simplr for us muggles:P

John Rennie

@KaumudiHarikumar if it's isothermal then Boyle's law applies

user228700

6:55 AM

@Xasel Are u having a HP/LOTR/The hobbit series marathon?

user228700

@JohnRennie Yes, and PV at A(6) is indeed equal to PV at B(6)!

John Rennie

Which means PV must be constant. Does it look to you as if this is a path along which PV is constant?

Wouldn't a line of constant PV be a hyperbola?

Xasel

Yoy want to have more yhere:Name of the Wind/Got/Brandon sanderson/shkespear etc.. too @KaumudiHarikumar

John Rennie

@Xasel: Have you read The Wise Man's Fear by Patrick Rothfuss?

user228700

@JohnRennie Oh, I didn't think about it like that...

Xasel

6:58 AM

@JohnRennie:yes

user228700

I looked at PV at A and PV at B and concluded that it must be an isothermal process...

Secret

No that's not enough, you need to ensure PV along the whole path be constant

John Rennie

@Xasel I read The Name of the Wind a while ago, and I've been meaning to read The Wise Man's Fear but it's a huge book - 1,100 pages! Is it worth the effort?

user228700

Yes, OK, I realize that now. Thanks very much! I'll be on my way now :-)

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